Class 12

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 5 Electrochemistry Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry

1. Choose the most correct option.

Question i.
Two solutions have the ratio of their concentrations 0.4 and ratio of their conductivities 0.216. The ratio of their molar conductivities will be
(a) 0.54
(b) 11.574
(c) 0.0864
(d) 1.852
Answer:
(a) 0.54

Question ii.
On diluting the solution of an electrolyte,
(a) both ∧ and κ increase
(b) both ∧ and κ decrease
(c) ∧ increases and κ decreases
(d) ∧ decreases and κ increases
Answer:
(c) ∧ increases and κ decreases

Question iii.
1 S m² mol^-1 is equal to
(a) 10^-4 S m² mol^-1
(b) 10⁴ Ω^-1 cm² mol^-1
(c) 10^-2 S cm² mol^-1
(d) 10² Ω^-1 cm² mol^-1
Answer:
(b) 10⁴Ω^-1 cm² mol^-1

Question iv.
The standard potential of the cell in which the following reaction occurs
H₂₊ (g, 1 atm) + Cu²⁺ (1 M) → 2H (1 M) + Cu_(s), (\(E_{\mathrm{Cu}}^{0}\) = 0.34 V) is
(a) – 0.34 V
(b) 0.34 V
(c) 0.17 V
(d) -0.17 V
Answer:
(b) 0.34 V

Question v.
For the cell, Pb_(s)|Pb²⁺ (1 M)|| Ag⁺ (1 M)|Ag_(s), if concentration of an ion in the anode compartment is increased by a factor of 10, the emf of the cell will
(a) increase by 10 V
(b) increase by 0.0296 V
(c) decrease by 10 V
(d) decrease by 0.0296 V
Answer:
(d) decrease by 0.0296 V

Question vi.
Consider the half reactions with standard potentials

The strongest oxidising and reducing agents respectively are
(a) Ag and Fe²⁺
(b) Ag⁺ and Fe
(c) Pb²⁺ and I^–
(d) I₂ and Fe²⁺
Answer:
(b) Ag⁺ and Fe

Question vii.
For the reaction
Ni_(s) + Cu²⁺ (1 M) → Ni²⁺ (1 M) + Cu_(s), \(E_{\text {cell }}^{0}\) = 0.57 V. Hence ΔG⁰ of the reaction is
(a) 110 kJ
(b) -110 kJ
(c) 55 kJ
(d) -55 kJ
Answer:
(b) -110 kJ

Question viii.
Which of the following is not correct ?
(a) Gibbs energy is an extensive property
(b) Electrode potential or cell potential is an intensive property.
(c) Electrical work = -ΔG
(d) If half reaction is multiplied by a numerical factor, the corresponding E⁰ value is also multiplied by the same factor.
Answer:
(d) If half reaction is multiplied by a numerical factor, the corresponding E⁰ value is also multiplied by the same factor.

Question ix.
The oxidation reaction that takes place in lead storage battery during discharge is

Answer:
(c) \(\mathrm{Pb}_{(\mathrm{s})}+\mathrm{SO}_{4(\mathrm{aq})}{ }^{2-} \longrightarrow \mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{e}^{-}\)

Question x.
Which of the following expressions represent molar conductivity of Al₂(SO₄)₃ ?

Answer:
(b) \(2 \lambda_{\mathrm{Al}^{3+}}^{0}+3 \lambda_{\mathrm{SO}_{4}^{2-}}^{0}\)

2. Answer the following in one or two sentences.

Question i.
What is a cell constant ?
Answer:
(A) Cell constant of a conductivity cell is defined as the ratio of the distance between the electrodes divided by the area of cross section of the electrodes.

In SI units it is expressed as m-1.

Question ii.
Write the relationship between conductivity and molar conductivity and hence unit of molar conductivity.
Answer:
If k is conductivity and ∧_m is molar conductivity then, ∧_m = \(\frac{\kappa \times 1000}{C}\)
Unit of molar conductivity is, Ω^-1 cm² mol^-1 or S cm² mol^-1.

Question iii.
Write the electrode reactions during electrolysis of molten KCl.
Answer:

Question iv.
Write any two functions of salt bridge.
Answer:
The functions of a salt bridge are :

• It maintains the electrical contact between the two electrode solutions of the half cells.
• It prevents the mixing of electrode solutions.
• It maintains the electrical neutrality in both the solutions of two half cells by a flow of ions.
• It eliminates the liquid junction potential.

Question v.
What is standard cell potential for the reaction
3Ni_(s) + 2Al³⁺ (1M) → 3NI²⁺ (1M) + 2Al_(s)
if \(\boldsymbol{E}_{\mathrm{Ni}}^{0}\) = – 0.25 V and \(\boldsymbol{E}_{\mathrm{Al}}^{0}\) = -1.66V?
Solution :
Given : E⁰_Ni²⁺/Ni = -0.25 V
E⁰_Al³⁺/Al = – 1.66 V; E⁰_cell = ?
Since Ni is oxidised and Al³⁺ is reduced,
\(E_{\text {cell }}^{0}=E_{\mathrm{Al}^{3+} / \mathrm{Al}}^{0}-E_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{0}\)
= – 1.66 – (-0.25)
= – 1.41 V
Ans. \(E_{\text {cell }}^{0}\) = -1.41 V
[Note : Since \(E_{\text {cell }}^{0}\) is negative, the given reaction is not possible but reverse reaction is possible.]

Question vi.
Write Nerst equation. What part of it represents the correction factor for nonstandard state conditions ?
Answer:
(1) Nernst equation for cell potential is,

(2) The part of equation namely,

represents the correction factor for nonstandard state conditions.

Question vii.
Under what conditions the cell potential is called standard cell potential ?
Answer:
In the standard cell, the active masses of the substances taking part in the electrochemical reaction have unit value, i.e., 1 M solution or ions and 1 atm gas.

Question viii.
Formulate a cell from the following electrode reactions :
\(\mathbf{A u}_{(\mathrm{aq})}^{3+}+\mathbf{3 e}^{-} \longrightarrow \mathbf{A} \mathbf{u}_{(\mathrm{s})}\)
\(\mathbf{M g}_{(\mathbf{s})} \longrightarrow \mathbf{M g}_{(\mathrm{aq})}^{2+}+\mathbf{2 e}^{-}\)
Answer:
An electrochemical cell from above electrode reactions is,

Question ix.
How many electrons would have a total charge of 1 coulomb ?
Answer:
Given : 1 Faraday = charge on 1 mol of electrons
= 6.022 × 10²³ electrons and 1 Faraday = 96500 C
∵ 96500 C = 6.022 × 10²³ electrons 6 022 × 10²³
∴ 1 C ≡ \(\frac{6.022 \times 10^{23}}{96500}\) = 6.24 × 10¹⁸ electrons
Ans. Number of electrons = 6.24 × 10¹⁸

Question x.
What is the significance of the single vertical line and double vertical line in the formulation galvanic cell.
Answer:
(i) Consider representation of Daniell cell,

Single vertical line represents separation of two phases, solid Zn_(s) and solution of ions.
(ii) Double vertical lines represent a salt bridge.

3. Answer the following in brief

Question i.
Explain the effect of dilution of solution on conductivity ?
Answer:

• The conductance of a solution is due to the presence of ions in the solution. More the ions, higher is the conductance of the solution.
• Conductivity or the specific conductance is the conductance of unit volume (1 cm³) of the electrolytic solution.
• The conductivity of the electrolytic solution always decreases with the decrease in the concentration of the electrolyte or the increase in dilution of the solution.
• On dilution, the concentration of the solution decreases, hence the number of (current carrying) ions per unit volume decreases. Therefore the conductivity of the solution decreases, with the decrease concentration or increase in dilution. (It is to be noted here that, molar conductivity increases with dilution.)

Question ii.
What is a salt bridge ?
Answer:
A salt bridge is a U-shaped glass tube containing a saturated solution of a strong electrolyte, like KCl, NH₄NO₃, Na₂SO₄ in a solidified agar-agar gel. A hot saturated solution of these electrolytes in 5% agar solution is filled in the U-shaped tube and allowed it to cool and solidify forming a gel.

Fig. 5.9 : Salt bridge
It is used to connect two half cells or electrodes forming a galvanic or voltaic cell.

Question iii.
Write electrode reactions for the electrolysis of aqueous NaCl.
Answer:
Reactions in electrolytic cell :
(i) Reduction half reaction at cathode : There are Na⁺ and H⁺ions but since H⁺ are more reducible than Na⁺, they undergo reduction liberating hydrogen and Na⁺ are left in the solution.
2H₂O_(l) + 2e^– → H_2(g) + 2OH^–_(aq) (reduction) E⁰ = -0.83 V

(ii) Oxidation half reaction at anode : At anode there are Cl^– and OH^–. But Cl^– ions are preferably oxidised due to less decomposition potential.

Net cell reaction : Since two electrons are gained at cathode and two electrons are released at anode for each redox step, the electrical neutrality is maintained. Hence we can write,

Since Na⁺ and OH^– are left in the solution, they form NaOH_(aq).

Question iv.
How many moles of electrons are passed when 0.8 ampere current is passed for 1 hour through molten CaCl₂ ?
Answer:
Given : I = 0.8 A; t = 1 × 60 × 60 = 3600 s
Number of moles of electrons = ?
Q = I × t
= 0.8 × 3600
= 2880 C
1 Faraday = 1 mol electrons
1 Faraday = 96500 C
∵ 96500 C = 1 mol electrons
∴ 2880 C ≡ \(\frac{2880}{96500}\)
= 0.02984 mol electrons
Ans. Number of moles of electrons = 0.02984

Question v.
Construct a galvanic cell from the electrodes Co³⁺|Co and Mn²⁺|Mn. \(\boldsymbol{E}_{\mathrm{Co}}^{0}\) = 1.82 V,
\(\boldsymbol{E}_{\mathrm{Mn}}^{0}\) = – 1.18V. Calculate \(\boldsymbol{E}_{\text {cell }}^{0}\).
Answer:

Question vi.
Using the relationsip between ∆G⁰ of cell reaction and the standard potential associated with it, how will you show that the electrical potential is an intensive property ?
Answer:
(1) For an electrochemical cell involving n number of electrons in the overall cell reaction,
ΔG⁰ = -nF\(E_{\text {cell }}^{0}\)
where ΔG⁰ is standard Gibbs energy change and \(E_{\text {cell }}^{0}\) is a standard cell potential.
(2) ∴ \(E_{\mathrm{cell}}^{0}=\frac{-\Delta G^{0}}{n F}\)
Since ΔG⁰ changes according to number of moles of electrons involved in the cell reaction, the ratio, ΔG⁰/nF remains constant.
(3) Therefore \(E_{\text {cell }}^{0}\) is independent of the amount of substance and it represents the intensive property.

Question vii.
Derive the relationship between standard cell potential and equilibrium constant of cell reaction.
Answer:
For any galvanic cell, the overall cell reaction at equilibrium can be represented as,
Reactants ⇌ Products.
[For example for Daniell cell,
\(\mathrm{Zn}_{(s)}+\mathrm{Cu}_{(\mathrm{aq})}^{2+} \rightleftharpoons \mathrm{Zn}_{(\mathrm{aq})}^{2+}+\mathrm{Cu}_{(\mathrm{s})}\) ]
The equilibrium constant, K for the reversible reaction will be, \(K=\frac{[\text { Products }]}{[\text { Reactants }]}\)

The equilibrium constant is related to the stan-dard free energy change Δ G⁰, as follows,
ΔG⁰ = -RTlnK
If \(E_{\text {cell }}^{0}\) is the standard cell potential (or emf) of the galvanic cell, then ΔG⁰ = -nFE⁰_cell
By comparing above equations,

Question viii.
It is impossible to measure the potential of a single electrode. Comment.
Answer:
(1)

Fig 5.12(a) : Measurement of single electrode potential

Fig 5.12(b) : Measurement of cell potential
According to Nemst theory, electrode potential is the potential difference between the metal and ionic layer around it at equilibrium, i.e. the potential across the electric double layer.

(2) For measuring the single electrode potential, one part of the double layer, that is metallic layer can be connected to the potentiometer but not the ionic layer. Hence, single electrode potential can’t be measured experimentally.

(3) When an electrochemical cell is developed by combining two half cells or electrodes, they can be connected to the potentiometer and the potential difference or cell potential can be measured.
E_cell = E₂ – E₁
where E₁ and E₂ are reduction potentials of two electrodes.

Question ix.
Why do the cell potential of lead accumulators decrease when it generates electricity ? How the cell potential can be increased ?
Answer:
Working of a lead accumulator :
(1) Discharging : When the electric current is withdrawn from lead accumulator, the following reactions take place :

(2) Net cell reaction :
(i) Thus, the overall cell reaction during discharging is

OR
Pb_(s) + PbO_2(s) + 2H₂SO_4(aq) → 2PbSO_4(s) + 2H₂O_(l)
The cell potential or emf of the cell depends upon the concentration of sulphuric acid. During the operation, the acid is consumed and its concentration decreases and specific gravity decreases from 1.28 to 1.17. As a result, the emf of the cell decreases. The emf of a fully charged cell is about 2.0 V.

(ii) Recharging of the cell : When the discharged battery is connected to external electric source and a higher external potential is applied the cell reaction gets reversed generating H₂SO₄.
Reduction at the -ve electrode or cathode :
PbSO4_(s) + 2e^– → Pb_(s) + \(\mathrm{SO}_{4(\mathrm{aq})}^{2-}\)
Oxidation at the + ve electrode or anode :

The emf of the accumulator depends only on the concentration of H₂SO₄.

Question x.
Write the electrode reactions and net cell reaction in NICAD battery.
Answer:
Reactions in the cell :
(i) Oxidation at cadmium anode :
Cd_(s) + 2OH^–_(aq) → Cd(OH)_2(s) + 2e^–
(ii) Reduction at NiO_2(s) cathode :
NiO_2(s) + 2H₂O_(l) + 2e^– → Ni(OH)_2(s) + 2OH^–_(aq)
The overall cell reaction is the combination of above two reactions.
Cd_(s) + NiO_2(s) + 2H₂O_(l) → Cd(OH)_2(s) + Ni(OH)_2(s)

4. Answer the following :

Question i.
What is Kohrausch law of independent migration of ions? How is it useful in obtaining molar conductivity at zero concentration of a weak electrolyte ? Explain with an example.
Answer:
(A) Statement of Kohlrausch’s law : This states that at infinite dilution of the solution, each ion of an electrolyte migrates independently of its co-ions and contributes independently to the total molar conductivity of the electrolyte, irrespective of the nature of other ions present in the solution.

(B) Explanation : Both the ions, cation and anion of the electrolyte make a definite contribution to the molar conductivity of the electrolyte at infinite dilution or zero concentration (∧₀).
If \(\lambda_{+}^{0}\) and \(\lambda_{-}^{0}\) are the molar ionic conductivities of cation and anion respectively at infinite dilution, then
∧₀ = \(\lambda_{+}^{0}\) + \(\lambda_{-}^{0}\).
This is known as Kohlrausch’s law of independent migration of ions.
For an electrolyte, B_xA_y giving x number of cations and y number of anions,
∧₀ = x\(\lambda_{+}^{0}\) + y\(\lambda_{-}^{0}\).

(C) Applications of Kohlrausch’s law :
(1) With this law, the molar conductivity of a strong electrolyte at zero concentration can be determined. For example,
\(\wedge_{0(\mathrm{KCl})}=\lambda_{\mathrm{K}^{+}}^{0}-\lambda_{\mathrm{Cl}^{-}}^{0}\)
(2) ∧₀ values of weak electrolyte with those of strong electrolytes can be obtained. For example,

Molar conductivity of a weak electrolyte at infinite dilution or zero concentration cannot be measured experimentally.
Consider the molar conductivity (∧₀) of a weak acid, CH₃COOH at zero concentration. By Kohlrausch s law,

where λ⁰CH₃COO^– and λ⁰_H⁺ are the molar ionic conductivities of CH₃COO^– and H⁺ ions respectively.
If ∧_0CH3COONa, ∧_0HCl and ∧_0NaCl are molar conductivities of CH₃COONa, HCl and NaCl respectively at zero concentration, then by
Kohlrausch’s law,

Hence, from ∧₀ values of strong electrolytes, ∧₀ of a weak electrolyte CH₃COOH, at infinite dilution can be calculated.

Question ii.
Explain electrolysis of molten NaCl.
Answer:
(1) Construction of an electrolytic cell : It consists of a vessel containing molten (fused) NaCl. Two graphite (carbon) inert electrodes are dipped in it, and connected to an external source of direct electric current (battery). The electrode connected to a negative terminal of the battery is a cathode and that connected to a positive terminal is an anode.

(2) Working of the cell :
(A) In the external circuit, the electrons flow through the wires from anode to cathode of the cell.
(B) The fused NaCl dissociates to form cations (Na⁺) and anions (Cl^–).
\(\mathrm{NaCl}_{\text {(fused) }} \longrightarrow \mathrm{Na}_{(\mathrm{l})}^{+}+\mathrm{Cl}_{(\mathrm{l})}^{-}\)
Na⁺ migrate towards cathode and Cl^– migrate towards anode.

Fig. 5.7 : Electrolysis of fused sodium chloride

(C) Reactions in electrolytic cell :
(i) Reduction half reaction at cathode : The Na⁺ ions get reduced by accepting electrons from a cathode supplied by a battery and form metallic sodium.
\(\mathrm{Na}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Na}_{(\mathrm{s})} \text { (reduction) }\)

(ii) Oxidation half reaction at anode : The Cl^– ions get oxidised by giving up electrons to the anode forming neutral Cl atoms in the primary process, and these Cl atoms combine forming Cl₂ gas in the secondary process.

The released electrons in the anodic oxidation half reaction return to battery through the metallic wires.

Net cell reaction : In order to maintain the electrical neutrality, the number of electrons gained at cathode must be equal to the number of electrons released at anode. Hence the reduction half reaction is multiplied by 2 and both reactions, oxidation half reaction and reduction half reaction are added to obtain a net cell reaction.

Results of electrolysis :

• A molten silvery white Na is formed at cathode which floats on the surface of molten NaCl.
• A pale green Cl₂ gas is liberated at anode.

Question iii.
What current strength in amperes will be required to produce 2.4g of Cu from CuSO₄ solution in 1 hour ? Molar mass of Cu = 63.5 g mol^-1.
Answer:
Given : W_Cu = 2.4 g; t = 1 hr = 1 × 60 × 60 s
M_Cu = 63.5 g mol^-1; I = ?

Ans. Current strength = I = 2.026 A

Question iv.
Equilibrium constant of the reaction,
2Cu⁺_(aq) → Cu²⁺_(aq) + Cu_(s)
is 1.2 × 10⁶. What is the standard potential of the cell in which the reaction takes place ?
Answer:
For the cell reaction, n = 1

Question v.
Calculate emf of the cell
Zn_(s)|Zn²⁺ (0.2M)||H⁺(1.6M)|H₂(g, 1.8 atm)|Pt at 25°C.
Answer:
Given : Zn_(s)|Zn²⁺(0.2M)||H⁺(1.6M)|H₂(g, 1.8 atm)|Pt


= 0.763 – 0.0296 × (- 0.8521)
= 0.763 + 0.02522
= 0.7882
Ans. \(E_{\text {cell }}^{0}\) = 0.7882 V

Question vi.
Calculate emf of the following cell at 25°C.
Zn_(s)| Zn²⁺(0.08M)||Cr³⁺(0.1M)|Cr
E⁰_Zn = – 0.76 V, E⁰_Cr = – 0.74 V
Answer:

Question vii.
What is a cell constant ? What are its units? How is it determined experimentally?
Answer:
(A) Cell constant of a conductivity cell is defined as the ratio of the distance between the electrodes divided by the area of cross section of the electrodes.

In SI units it is expected as m^-1.

The resistance of an electrolytic solution is measured by using a conductivity cell and Wheatstone

Fig. 5.6 : Measurement of conductance
The measurement of molar conductivity of a solution involves two steps as follows :
Step I : Determination of cell constant of the conductivity cell :
KCl solution (0.01 M) whose conductivity is accurately known (κ = 0.00141 Ω^-1 cm^-1) is taken in a beaker and the conductivity cell is dipped. The two electrodes of the cell are connected to one arm while the variable known resistance (R) is placed in another arm of Wheatstone bridge.

A current detector D’ which is a head phone or a magic eye is used. J is the sliding jockey (contact) that slides on the arm AB which is a wire of uniform cross section. A source of A.C. power (alternating power) is used to avoid electrolysis of the solution.

By sliding the jockey on wire AB, a balance point (null point) is obtained at C. Let AC and BC be the lengths of wire.

If R_solution is the resistance of KCl solution and R_x is the known resistance then by Wheatstone’s bridge principle,
\(\frac{R_{\text {solution }}}{\mathrm{BC}}=\frac{R_{x}}{\mathrm{AC}}\)
∴ \(R_{\text {solution }}=\mathrm{BC} \times \frac{R_{x}}{\mathrm{AC}}\)
Then the cell constant ‘ b ’ of the conductivity cell is obtained by, b = κ_Kcl × R_solution.

Step II : Determination of conductivity of the given solution :
KCl solution is replaced by the given electrolytic solution and its resistance (R_s) is measured by Wheatstone bridge method by similar manner by obtaining a null point at D.
The conductivity (κ) of the given solution is,
κ = \(\frac{\text { cell constant }}{R_{\mathrm{s}}}=\frac{b}{R_{\mathrm{s}}}\)

Step III: Calculation of molar conductivity :
The molar conductivity (∧_m) is given by,

Since the concentration of the solution is known, ∧_m can be calculated.

Question viii.
How will you calculate the moles of electrons passed and mass of the substance produced during electrolysis of a salt solution using reaction stoichiometry.
Answer:
Calculation of moles of electrons passed : The charge carried by one mole of electrons is referred to as one faraday (F). If total charge passed is Q C, then moles of electrons passed = \(\frac{Q(\mathrm{C})}{F\left(\mathrm{C} / \mathrm{mol} \mathrm{e}^{-}\right)}\)

Calculation of mass of product : Mass, W of product formed is given by,
W = moles of product × molar mass of product (M)
= \(\frac{Q}{96500}\) × mole ratio × M
= \(\frac{I \times t}{96500}\) × mole ratio × M 96500
When two electrolytic cells containing different electrolytes are connected in series so that same quantity of electricity is passed through them, then the masses W₁ and W₂ of products produced are given by,

Question ix.
Write the electrode reactions when lead storage cell generates electricity. What are the anode and cathode and the electrode reactions during its recharging?
Answer:
Recharging of the cell : When the discharged battery is connected to external electric source and a higher external potential is applied the cell reaction gets reversed generating H₂SO₄.
Reduction at the – ve electrode or cathode :
\(\mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}^{(s)}+\mathrm{SO}_{4(\mathrm{aq})}^{2-}\)
Oxidation at the + ve electrode or anode :

The emf of the accumulator depends only on the concentration of H₂SO₄.

Question x.
What are anode and cathode of H₂-O₂ fuel cell ? Name the electrolyte used in it. Write electrode reactions and net cell reaction taking place in the fuel cell.
Answer:
Construction :
(i) In fuel cell the anode and cathode are porous electrodes with suitable catalyst like finely divided platinum.

(iii) H₂ is continuously bubbled through anode while O, gas is bubbled through cathode.

Working (cell reactions) :
(i) Oxidation at anode : At anode, hydrogen gas is oxidised to H₂O.
2H_2(g) + 4OH^–_(aq) → 4H₂O_(l) + 4e^– (oxidation half reaction)
(ii) Reduction at cathode : The electrons released at anode travel to cathode through external circuit and reduce oxygen gas to OH^–.
O_2(g) + 2H₂O_(l) + 4e^– → 4OH^–_(aq) (reduction half reaction)

(iii) Net cell reaction: Addition of both the above reactions at anode and cathode gives a net cell reaction.
2H_2(g) + O_2(g) → 2H₂O_(l) (overall cell reaction)

Question xi.
What are anode and cathode for Leclanche’ dry cell ? Write electrode reactions and overall cell reaction when it generates electricity.
Answer:
A dry cell has zinc vessel as anode and graphite rod as cathode and moist paste of ZnCl₂, MnO₂, NH₄Cl as electrolytes.
At anode :
Zn_(s) → \(\mathrm{Zn}_{(\mathrm{aq})}^{2+}\) + 2e^– (Oxidation half reaction)
At graphite (c) cathode :
\(2 \mathrm{NH}_{4(\mathrm{e})}^{+}\) + 2e^– → 2NH_3(aq) + H_2(g) (Reduction half reaction)
2MnO_2(s) + H₂ → Mn₂O_3(s) + H₂O_(l)
There is a side reaction inside the cell, between Zn²⁺ ions and aqueous NH₃.
\(\mathrm{Zn}_{(\mathrm{aq})}^{2+}+4 \mathrm{NH}_{3(\mathrm{aq})} \longrightarrow\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}\right]_{(\mathrm{aq})}^{2+}\)

Question xii.
Identify oxidising agents and arrange them in order of increasing strength under standard state conditions. The standard potentials are given in parenthesis.
Al(- 1.66 V), Cl₂ (1.36 V), Cd²⁺ (-0.4 V), Fe (-0.44 V), I₂ (0.54 V), Br^– (1.09 V).
Answer:
The oxidising agents are I₂, Br^– and Cl₂. The increasing strength is

(Note : Actually Br₂ acts as an oxidising agent but not Br^–.)

Question xiii.
Which of the following species are reducing agents? Arrange them in order of increasing strength under standard state conditions. The standard potentials are given in parenthesis.
K (-2.93V), Br₂(1.09V), Mg(-2.36V), Co³⁺(1.61V), Ti²⁺(-0.37V), Ag⁺(0.8V), Ni (-0.23V).
Answer:
Lower the standard reduction potential, higher is reducing power. The reducing agents are Ni, Mg and K. Their increasing strength is,

(Note : Cations don’t act as reducing agent since they are already in oxidised state.)

Question xiv.
Predict whether the following
reactions would occur spontaneously
under standard state conditions.
a. Ca_(s) + Cd²⁺_(aq) → Ca²⁺_(aq) + Cd_(s)
b. 2 Br-_(s) + Sn²⁺(aq) → Br_2(l) + Sn_(s)
c. 2Ag_(s) + Ni²⁺_(aq) → 2 Ag⁺_(aq) + Ni_(s)
(use information of Table 5.1)
Answer:

12th Chemistry Digest Chapter 5 Electrochemistry Intext Questions and Answers

Question 1.
How does electrical resistance depend on the dimensions of an electronic (metallic) conductor?
Answer:
The electrical resistance of an electronic conductor is linearly proportional to its length (l) and inversely proportional to its cross section area a.

Fig. 5.3 : Electronic conductor
Thus, R ∝ l; R ∝ \(\frac{1}{a}\)
∴ R ∝ \(\frac{l}{a}\) or R = ρ × \(\frac{l}{a}\)
where the proportionality constant p is called specific resistance. IUPAC recommends the term resistivity for specific resistance.

Question 2.
What are the units of resistivity ?
Answer:
For an electronic conductor of length l, and cross section area a, the resistance R is represented as
R = ρ × \(\frac{l}{a}\)
where ρ is the resistivity of the conductor.
∴ ρ = R × \(\frac{a}{l}\)
If l = 1 m, a = 1 m², ρ = R

Hence, resistivity is the resistance of a conductor of volume of 1 m³.
(In C.G.S. units, the units of ρ are ohm cm. Hence, ρ is the resistance of a conductor of unit volume or 1 cm³.)

Question 3.
Define resistivity. What are its units ?
Answer:
Resistivity (or specific resistance) : It is the resistance of a conductor that is 1 m in length and 1 m² in cross section area in SI units. (In C.G.S. units, it is the resistance of a conductor that is 1 cm in length and 1 cm² in cross section area.) Hence, the resistivity is the resistance of a conductor of unit volume. (In case of electrolytic solution, ρ is the resistivity i.e., resistance of a solution of unit volume.)
It has SI units, ohm m and C.G.S. units, ohm cm.

Question 4.
Why is alternating current used in the measurement of conductivity of the solution ?
Answer:
If direct current (D.C.) by battery is used, there will be electrolysis and the concentration of the solution is changed. Hence alternating current (A.C.) with high frequency is used.

Try this… (Textbook page No. 93)

Question 1.
What must be the concentration of a solution of silver nitrate to have the molar conductivity of 121.4 Ω^-1 cm² mol^-1 and the conductivity of 2.428 × 10^-3 Ω^-1 cm^-1 at 25 °C ?
Answer:

∴ Concentration of a Solution = 0.02 M

Try this… (Textbook page No. 96)

Question 1.
Obtain the expression for dissociation constant in terms of ∧_c and ∧₀ using Ostwald’s dilution law.
Answer:
Consider a solution of a weak electrolyte, BA having concentration C mol dm^-3. If α is the degree of dissociation, then by Ostwald’s theory of weak electrolytes,

If K is the dissociation constant of the weak electrolyte, then by Ostwald’s dilution law,
K = \(\frac{\alpha^{2} C}{(1-\alpha)}\)
If ∧_m is the molar conductivity of the electrolyte BA at the concentration C and ∧₀ is the molar conductivity at zero concentration or infinite dilution, then

Hence by measuring ∧_m at the concentration C and knowing ∧₀, the dissociation constant can be calculated.
If \(\lambda_{+}^{0}\) and \(\lambda_{-}^{0}\) are the ionic conductivities, then by Kohlrauseh’s law, ∧₀ = \(\lambda_{+}^{0}\) + \(\lambda_{-}^{0}\).

Learn this as well…

Question 1.
How is the cell constant of a conductivity cell determined?
Answer:
The cell constant of a given conductivity cell is obtained by measuring the resistance (R) (or the conductance) of a standard solution whose conductivity (f_c) is accurately known by using Wheatstone’s bridge (discussed in Q. 37). For this purpose, KCl solution of accurately known conductivity is used.
\(\kappa_{\mathrm{KCl}}=\frac{1}{R_{\mathrm{KCl}}} \times \frac{l}{a}\) where \(\frac{l}{a}\) is a cell constant, represented by b.
∴ \(\kappa_{\mathrm{KCl}}=\frac{b}{R_{\mathrm{KCl}}}\)
or b = κ_KCl × R_KCl
For example, the conductivity of 0.01 M KCl is 0.00141 Ω^-1 cm^-1 (S cm^-1). Hence by measuring R KCl the cell constant b can be obtained.

Try this… (Textbook page No. 95)

Question 1.
Calculate ∧₀ (CH₂ClCOOH) if ∧₀ values for HCl, KCl and CH₂ClCOOK are respectively, 4.261, 1.499 and 1.132 Ω^-1 m² mol^-1.
Solution :

Adding equations (i) and (ii) and subtracting equation (iii) we get equation (I).

Can you tell ? (Textbook page No. 103)

Question 1.
You have learnt Daniel cell in XIth standard. Write notations for anode and cathode. Write the cell formula.
Answer:
Daniel cell is represented as,

Try this… (Textbook page No. 104)

Question 1.
Write electrode reactions and overall cell reaction for Daniel cell you learnt in standard XI.
Answer:
Reactions for Daniell cell:

Question 1.
Describe different types of reversible electrodes with examples. (1 mark for each type)
Answer:
A reversible electrochemical cell or a galvanic cell consists of two reversible half cells or electrodes. There are four types of reversible electrodes according to their compositions.
(1) Metal-metal ion electrode : This electrode is set up by dipping a metal in a solution containing its own ions, e.g. Zn rod dipped into ZnSO₄ solution containing Zn⁺⁺ ions of concentration C.
It is represented as,
\(\mathrm{Zn}^{2+}{ }_{(\mathrm{aq})} \mid \mathrm{Zn}_{(\mathrm{s})}\)
The reduction reaction at the electrode is,
Zn⁺⁺_(aq) + 2e^– → Zn_(s)

(2) Metal-sparingly soluble salt electrode : This electrode consists of a metal coated with one of its sparingly soluble salts and immersed in a solution containing an electrolyte having a common anion as that of the salt. For example, silver electrode coated with sparingly soluble AgCl dipped in KCl solution with common anion Cl^–. This electrode is represented as,
Cl^–_(aq) | AgCl_(s) | Ag_(s)
The reduction reaction is,
AgCl_(s) + e → Ag_(s) + Cl^–_(aq)

(3) Gas electrode : This is developed by bubbling pure and dry gas around a platinised platinum foil dipped in the solution containing ions (of the gas) reversible with respect to the gas bubbled.
The gas is adsorbed on the surface of platinum foil and establishes an equilibrium with its ions in the solution. Pt electrode provides electrical contact and also acts as a catalyst.
Some of the gas electrodes are represented as follows :
(i) Hydrogen gas electrode :
H⁺_(aq) | H₂(g, P_H2) | Pt
Reduction reaction : H⁺_(aq) + e^– → \(\frac {1}{2}\)H_2(g)
(ii) Chlorine gas electrode :
Cl^–_(aq) | Cl₂(g, PCl₂) | Pt
Reduction reaction : \(\frac {1}{2}\)Cl_2(g) + e- → Cl^–_(aq)

(4) Redox electrode (Oxidation reduction electrode) : This electrode consists of a platinum wire dipped in a solution containing the ions of the same metal (or a substance) in two different oxidation states, like Fe²⁺ – Fe³⁺, Sn²⁺ – Sn⁴⁺, Mn⁺⁺ – MnO^–₄, etc.
A platinum electrode which provides an electrical contact and acts as catalyst aquires an equilibrium between two ions in the solution, due to their tendency to undergo a change from one oxidation state to another. The electrodes are represented as,
Fe²⁺_(aq), Fe³⁺_(aq) | Pt
Reduction reaction : Fe³⁺_(aq) + e^– → Fe²⁺_(aq)
SnCl_2(aq), SnCl_4(aq) | Pt
Reduction reaction : Sn⁴⁺_(aq) + 2e^– →Sn²⁺_(aq)

Use your brain power! (Textbook page No. 98)

Question 1.
Distinguish between electrolytic and galvanic cells.
Answer:
Electrolytic cell:

1. This device is used to bring about a non-spontaneous chemical reaction by passing an electric current.
2. It is used to bring about a chemical reaction generally for the dissociation (electrolysis) of compounds.
3. In this cell, electrical energy is converted into chemical energy.
4. In this cell, the cathode is negative and the anode is positive.
5. Electrolytic cells are irreversible.
6. Oxidation takes place at the positive electrode and reduction at the negative electrode.
7. The electrons are supplied by the external source and enter through cathode and come out through anode.
8. It is used for electroplating, electrorefining, etc.

Electrochemical cell (Galvanic cell or Voltaic cell):

1. This device is used to produce electrical energy by a spontaneous chemical reaction.
2. It is used to generate electricity.
3. In this cell, chemical energy is converted into electrical energy.
4. In this cell, the cathode is positive and the anode is negative.
5. Electrochernical cells are reversible.
6. Oxidation takes place at the negative electrode and reduction at the positive electrode.
7. The electrons move from anode to cathode in the external circuit.
8. It is used as a source of electric current.

Try this… (Textbook page No. 107)

Question 1.
Write expressions to calculate equilibrium constant from
i. Concentration data
ii. Thermochemical data
iii. Electrochemical data
Answer:
(i) Consider following a reversible cell reaction.
aA + bB ⇌ cC + dD
If [A], [B], [C] and [D] represent concentrations of reactants and products then the equilibrium constant K is,
K = \(\frac{[\mathrm{C}]^{c} \times[\mathrm{D}]^{d}}{[\mathrm{~A}]^{a} \times[\mathrm{B}]^{b}}\)
(ii) If ΔG⁰ is the standard Gibbs free energy change at temperature T then,
ΔG⁰ = – RTlnK = – 2.303 RTlog₁₀K
(iii) From electrochemical data,
if \(E_{\text {cell }}^{0}\) is the standard cell potential and K is the equilibrium constant for the cell reaction at a temperature T, then,
\(E_{\text {cell }}^{0}=\frac{0.0592}{n} \log _{10} K\)

Learn this as well…

Question 1.
The construction and working of the calomel electrode.
Answer:
(1) Since standard hydrogen electrode (SHE) is not convenient for experimental use, a secondary reference electrode like calomel electrode is used.
(2) Construction : It consists of a glass vessel with side arm B for dipping in a desired solution of another electrode like, ZnSO_4(aq) for an electric contact. The vessel is filled with mercury, a paste of Hg and Hg₂Cl₂ (calomel) and saturated KCl solution.

Fig. 5.15 : Determination of standard electrode potential using calomel electrode
(3) The potential developed depends upon the concentration of Cl^– or KCl solution. When saturated KCl solution is used, its reduction potential is 0.242 V.
(4) Consider following cell :
Zn_(s) | ZnSO_4(aq) || KCl_(aq) | Hg₂Cl_2(s) | Hg
OR Zn_(s) | ZnSO_4(aq) || Calomel electrode
Reduction reaction for calomel electrode :
Hg₂Cl_2(s) + 2e^– → 2Hg_(l) + 2Cl^–_(aq)
Hence potential of calomel electrode depends on the concentration of Cl^– or KCl solution.

Can you tell ? (Textbook page No. 114)

Question 1.
In what ways are fuel cells and galvanic cells similar and in what ways are they different ?
Answer:
Similarity between fuel cells and galvanic cells :

• In both the cells, there is oxidation at anode and j reduction at cathode.
• The cell potential is developed due to net redox reactions.
• Both are galvanic cells.

Difference in fuel cells and galvanic cells :

• Fuel cells involve electrodes with large surface area while galvanic cells involve electrodes with j compact surface area.
• Fuel cells involve gaseous materials on a large scale while galvanic cells involve gaseous materials at a definite pressures along with electrolytes or there may not be gases.
• In fuel cells, the cell potential is developed due to exothermic combustion reactions while in galvanic cell, cell potential is developed due to normal redox reactions.
• In fuel cells gaseous electrode materials are continuously supplied from outside while in galvanic cells electrode materials have constant concentration or may change due to reactions.

Use your brain power (Textbook page No. 114)

Question 1.
Indentify the strongest and the weakest oxidizing agents from the electrochemical series.
Answer:
From the electrochemical series,
(a) The strongest oxidising agent is fluorine since it has the highest standard reduction potential (\(E_{\mathrm{F}_{2} / \mathrm{F}^{-}}^{0}\) = + 2.87 V).
(b) The weakest oxidising agent (or the strongest reducing agent) is lithium since it has the lowest standard reduction potential, (\(E_{\mathrm{Li}^{+} / \mathrm{Li}}^{0}\) = -3.045 V).

Use your brain power (Textbook page No. 115)

Question 1.
Identify the strongest and the weakest reducing agents from the electrochemical series.
Answer:
(a) From the electrochemical series, the strongest reducing agent is lithium since it has the lowest standard reduction potential (\(E_{\mathrm{Li}^{+} / \mathrm{Li}}^{0}\) = -3.045 V).
(b) The weakest reducing agent is fluorine since it has the highest standard reduction potential,
(\(E_{\mathrm{F}_{2} / \mathrm{F}^{-}}^{0}\) = +2.87 V).

Question 2.
From E° values given in Table 5.1, predict whether Sn can reduce I₂ or Ni²⁺.
Answer:
From electrochemical series,

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 9 Control and Co-ordination Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 9 Control and Co-ordination

1. Multiple choice questions

Question 1.
The nervous system of mammals uses both electrical and chemical means to send signals via neurons. Which part of the neuron receives impulse?
(a) Axon
(b) Dendron
(c) Nodes of Ranvier
(d) Neurilemma
Answer:
(b) Dendron

Question 2.
……………. is a neurotransmitter.
(a) ADH
(b) Acetyl CoA
(c) Acetyl choline
(d) Inositol
Answer:
(c) Acetyl choline

Question 3.
The supporting cells that produce myelin sheath in the PNS are …………….
(a) Oligodendrocytes
(b) Satellite cells
(c) Astrocytes
(d) Schwann cells
Answer:
(d) Schwann cells

Question 4.
A collection of neuron cell bodies located outside the CNS is called …………….
(a) tract
(b) nucleus
(c) nerve
(d) ganglion
Answer:
(d) ganglion

Question 5.
Receptors for protein hormones are located …………….
(a) in cytoplasm
(b) on cell surface
(c) in nucleus
(d) on Golgi complex
Answer:
(b) on cell surface

Question 6.
If parathyroid gland of man Eire removed, the specific result will be …………….
(a) onset of aging
(b) disturbance of Ca⁺⁺
(c) onset of myxoedema
(d) elevation of blood pressure
Answer:
(b) disturbance of Ca⁺⁺

Question 7.
Hormone thyroxine, adrenaline and non¬adrenaline are formed from ……………
(a) Glycine
(b) Arginine
(c) Ornithine
(d) Tyrosine
Answer:
(d) Tyrosine

Question 8.
Pheromones are chemical messengers produced by animals and released outside the body. The odour of these substance affects …………….
(a) skin colour
(b) excretion
(c) digestion
(d) behaviour
Answer:
(d) behaviour

Question 9.
Which one of the following is a set of discrete endocrine gland?
(a) Salivary glands, thyroid, adrenal, ovary
(b) Adrenal, testis, ovary, liver
(c) Pituitary, thyroid, adrenal, thymus
(d) Pituitary, pancreas, adrenal, thymus
Answer:
(c) Pituitary, thyroid, adrenal, thymus

Question 10.
After ovulation, Graafian follicle changes into …………….
(a) corpus luteum
(b) corpus albicans
(c) corpus spongiosum
(d) corpus callosum
Answer:
(a) corpus luteum

Question 11.
Which one of the following pairs correctly matches a hormone with a disease resulting from its deficiency?
(a) Parathyroid hormone – Diabetes insipidus
(b) Luteinising hormone – Diabetes mellitus
(c) Insulin – Hyperglycaemia
(d) Thyroxine – Tetany
Answer:
(c) Insulin – Hyperglycaemia

Question 12.
……………. is in direct contact of brain in humans.
(a) Cranium
(b) Dura mater
(c) Arachnoid
(d) Pia mater
Answer:
(d) Pia mater

2. Very very short answer questions.

Question 1.
What is the function of red nucleus?
Answer:
Red nucleus plays an important role in controlling posture and muscle tone, modifying some motor activities and motor coordination.

Question 2.
What is the importance of corpora quadrigemina?
Answer:
Corpora quadrigemina consists of 4 solid rounded structures, viz. superior and inferior colliculi. Superior colliculi control visual reflexes while inferior colliculi control auditory reflexes.

Question 3.
What does the cerebellum of brain control?
Answer:
Cerebellum of brain is an important centre which maintains equilibrium of body, posture, balancing orientation, moderation of voluntary movements and maintenance of muscle tone.

Question 4.
Name the three ear ossicles.
Answer:
Malleus [hammer], incus [anvil] and stapes [stirrup].

Question 5.
Name the anti abortion hormone.
Answer:
Progesterone.

Question 6.
Name an organ which acts as temporary endocrine gland.
Answer:
Placenta. Corpus luteum in ovary.

Question 7.
Name the type of hormones which bind to the DNA and alter the gene expression.
Answer:
Steroid hormones.

Question 8.
What is the cause of abnormal elongation of long bones of arms and legs and of lower jaw.
Answer:
Hypersecretion of growth hormones in adults causes abnormal elongation of long bones of arms and legs and of lower jaw i.e. acromegaly.

Question 9.
Name the hormone secreted by the pineal gland.
Answer:
Melatonin.

Question 10.
Which endocrine gland plays important, role in improving immunity?
Answer:
The endocrine gland, thymus plays an important role in improving immunity.

3. Match the organism with the type of nervous system found in them.

Column A	Column B
(1) Neurons	(a) Earthworm
(2) Ladder type	(b) Hydra
(3) Ganglion	(c) Flatworm
(4) Nerve net	(d) Human

Answer:

Column A	Column B
(1) Neurons	(d) Human
(2) Ladder type	(c) Flatworm
(3) Ganglion	(a) Earthworm
(4) Nerve net	(b) Hydra

4. Very short answer questions.

Question 1.
Describe the endocrine role of islets of Langerhans.
OR
Islets of Langerhans.
Answer:
Endocrine cells of pancreas form groups of cells called Islets of Langerhans. There are four kinds of cells in islets of Langerhans which secrete hormones.

1. Alpha (α) cells : They are 20% and secrete glucagon. Glucagon is a hyperglycemic hormone. It stimulates liver for glucogenolysis and increases the blood glucose level.
2. Beta (β) cells : They are 70% and secrete insulin. Insulin is a hypoglycemic hormone. It stimulates liver and muscles for glycogenesis. This lowers blood glucose level.
3. Delta (δ) cells : They are 5% and secrete somatostatin. Somatostatin inhibits the secretion of glucagon and insulin. It also decreases the gastric secretions, motility and absorption in digestive tract. In general it is a growth inhibiting factor.
4. PP cells or F cells : They form 5%. They secrete pancreatic polypeptide (PP) which inhibits the release of pancreatic juice.

Question 2.
Mention the function of testosterone?
Answer:
Testosterone is a steroid sex hormone secreted by testes and cortex of adrenal glands. It controls the secondary sexual characters in males.

Question 3.
Give symptoms of the disease caused by hyposecretion of ADH.
Answer:
Polydipsia, i.e. frequent thirst and polyuria, i.e. frequent urination are the symptoms of the disease caused by hyposecretion of ADH.

5. Short answer questions

Question 1.
Rakesh got hurt on his head when he fell down from his motorbike. Which inner membranes must have protected his brain? What other roles do they have to play
Answer:

1. When Rakesh fell down from his motorbike, the inner membranes that protected his brain were meninges, viz. dura mater, arachnoid membrane and pia mater. Morevover, CSF must have also acted as a shock absorber.
2. Dura mater : It is the outer tough membrane protective in function.
3. Arachnoid membrane : It is the middle web-like membrane which communicates with fluids of upper sub dural space and lower sub arachnoid space.
4. Pia mater : It is the innermost highly vascularised nutritive membrane in close contact with brain and spinal cord.

Question 2.
Injury to medulla oblongata may prove fatal.
OR
Injury to medulla oblongata causes sudden death. Explain.
Answer:

1. Medulla oblongata is the region of the brain that controls all the involuntary activities.
2. Vital activities such as heartbeats, respiration, vasomotor activities, peristalsis, etc. are under the control of medulla oblongata.
3. When medulla oblongata is injured, all these vital functions are instantly stopped.
4. Therefore, injury to medulla oblongata causes sudden death.

Question 3.
Distinguish between the sympathetic and parasympathetic nervous system on the basis of the effect they have on:
Heartbeat andUrinary Bladder.
Answer:

Sympathetic Nervous System	Parasympathetic	Nervous System
(1) Heartbeat	Increases	Decreases
(2) Urinary bladder	Relaxes and stores urine	Contracts causing micturition

Question 4.
While holding a tea cup Mr. Kothari’s hands rattle. Which disorder he may be suffering from and what is the reason for this?
Answer:

1. This condition is due to Parkinson’s disease.
2. It is due to degeneration of dopamine- producing neurons in the CNS.
3. 80% of the patients develop this condition along with stiffness, difficulty in walking, balance and coordination.

Question 5.
List the properties of the nerve fibres.
Answer:

1. Excitability / irritability
2. Conductivity
3. Stimulus
4. Summation
5. All or none
6. Refractory period
7. Synaptic delay
8. Synaptic fatigue
9. Velocity.

Question 6.
How does tongue detect the sensation of taste?
Answer:

1. The surface of tongue is with gustatoreceptors.
2. These receptors are sensitive to the chemicals [sweet, salt, sour, bitter and umami (savory)] present in the food.
3. The receptor cells get stimulated, generate the impulse which is given to the sensory neuron.

Question 7.
State the site of production and function of Secretin, Gastrin and Cholecystokinin.
Answer:

Hormone	Site of production	Functions
1. Secretin	Duodenal mucosa	Stimulates secretion of pancreatic juice and bile from pancreas and liver respectively.
2. Gastrin	Gastric mucosa	Stimulates gastric glands to secrete gastric juice.
3. Cholecystokinin	Duodenal mucosa	Stimulates pancreas and gall bladder to release pancreatic enzymes and bile respectively.

Question 8.
An adult patient suffers from low heart rate, low metabolic rate and low body temperature. He also lacks alertness, intelligence and initiative. What can be this disease? What can be its cause and cure ?
Answer:

1. The above symptoms indicate that the person is suffering from Myxoedema.
2. Myxoedema is condition caused due to hypothyroidism.
3. Hypothyroidism causes deficiency of thyroid hormones like T₃ and T₄ (thyroxine). This results in low BMR.
4. This condition can be cured by giving injections of thyroxine or tablets containing hormone preparation.

Question 9.
Where is the pituitary gland located? Enlist the hormones secreted by anterior pituitary.
Answer:
The pituitary gland is attached to hypothalamus on the ventral surface of brain. It is lodged in a bony depression called sella turcica of sphenoid bone.
For names of hormones:

1. GH : [Growth Hormone/STH : Somatotropic Hormone]
2. TSH/TTH – [Thyroid Stimulating Hormone/ Thyrotropic Hormone]
3. ACTH – [adrenocorticotropic hormone]
4. PRL – [prolactin]

Gonadotropins-

1. FSH [follicle stimulating hormone]
2. LH/ICSH – [leutinizing hormone/ interstitial cells stimulating hormone]

Question 10.
Explain how the adrenal medulla and sympathetic nervous system function as a closely integrated system.
Answer:

1. Adrenal medulla originates from embryonic neuro – ectoderm.
2. It consists of rounded group of large granular cells called chromaffin cells. They are modified post-ganglionic cells of sympathetic nervous system which have lost normal processes and acquired glandular function.
3. These cells are connected with pre-ganglionic fibres of sympathetic nervous system.
4. Hence adrenal medulla is an extension of sympathetic nervous system.
5. Thus adrenal medulla and sympathetic nervous system functions as a closely integrated system.

Question 11.
Name the secretion of alpha, beta and delta cells of islets of Langerhans. Explain their role.
Answer:

Pancreatic islet cells	Secretion	Functions
1. Alpha cells	Glucagon	Stimulates glycogenolysis in the liver
2. Beta cells	Insulin	Stimulates glycogenesis in the liver and muscles
3. Delta cells	Somatostatin	Inhibits the secretion of glucagon and insulin. It also decreases the gastric secretions, motility and absorption in digestive tract.

Question 12.
Which are the two types of goitre? What are their causes?
Answer:
(1) Goitre is the enlargement of thyroid gland. It is easily visible at the base of neck when a person is suffering from it.

(2) Goitre is of two types.

1. Simple goitre : It is also called endemic goitre. This is due to iodine deficiency in the food. This causes iodine deficit in blood. In an attempt to take more iodine from blood, the blood supply to the gland increases. This results in swelling on the thyroid.
2. Exophthalmic goitre : It is also called toxic goitre. This is due to hyperactive thyroid gland. This can happen if there is overstimulation of thyroid due to excess of ACTH. This disorder is also called Grave’s disease or hyperthyroidism.

Question 13.
Name the ovarian hormone and give their functions.
Answer:

Hormone	Functions
Oestrogen	It is responsible for secondary sexual characters in female.
Progesterone	Essential for thickening of uterine endometrium, thus preparing the uterus for implantation of fertilized ovum. It is responsible for development of mammary glands during pregnancy. It inhibits uterine contractions during pregnancy.
Relaxin	It relaxes the cervix of the pregnant female and ligaments of pelvic girdle during parturition.
Inhibin	It inhibits the FSH and GnRH production.

6. Answer the following.

Complete the table.

Location	Cell type	Function
PNS	————-	Produce myelin sheath.
PNS	Satellite cells	————-
————	Oligodendrocytes	Form myelin sheath around central axon.
CNS	————	Pathogens are destroyed by phagocytosis. (Phagocytose)
CNS	————	Form the epithelial lining of brain cavities and central canal.

Answer:

Location	Cell type	Function
PNS	Schwann cells	Produce myelin sheath.
PNS	Satellite cells	Supply nutrients to surrounding neurons, protect and cushion nearby neurons.
CNS	Oligodendrocytes	Form myelin sheath around central axon.
CNS	Microglia	Pathogens are destroyed by phagocytosis. (Phagocytose)
CNS	Ependyma	Form the epithelial lining of brain cavities and central canal.

7. Long answer questions.

Question 1.
Explain the process of conduction of nerve impulses up to development of action potential.
Answer:

1. The origin and maintenance of resting potential depends on the original state of no stimulation.
2. Any stimulus or disturbance to the membrane will make the membrane permeable to Na⁺ ions. This causes rapid influx of Na⁺ ions.
3. The voltage gated Na⁺/K⁺ channels are unique. They can change the potential difference of the membrane as per the stimulus received and also the gates operate separately and are self closing.
4. During resting potential, both gates are closed and resting potential is maintained.
5. However during depolarization, the Na⁺ channels open but not the K⁺ channels. This causes Na⁺ to rush into the axon and bring about a depolarisation. This condition is called action potential.
6. Extra cellular fluid (ECF) becomes electronegative with respect to the inner membrane which becomes electropositive.

Question 2.
Draw the neat labelled diagrams.
a. Human ear.
Answer:

b. Sectional view of human eye.
Answer:

c. Draw the neat labelled diagram of sagittal section or L.S. of human brain
Answer:

d. Draw the neat labelled diagram of Multipolar Neuron.
Answer:

Question 3.
Answer the questions after observing the diagram given below.

a. What do the synaptic vesicles contain?
Answer:
Synaptic vesicles contain a neurotransmitter – acetyl choline.

b. What process is used to release the neurotransmitter ?
Answer:
Exocytosis.

c. What should be the reason for the next impulse to be conducted?
Answer:
Removal of neurotransmitter by the action of acetyl cholinesterase.

d. Will the impulse be carried by post synaptic membrane even if one pre-synaptic neuron is there?
Answer:
As far as impulse is transmitted by pre-synaptic neuron, it will be received by post-synaptic neuron.

e. Can you name the channel responsible for their transmission?
Answer:
Ca⁺⁺ channel

Question 4.
Explain the Reflex Pathway with the help of a neat labelled diagram.
OR
With the help of a neat and labelled diagram, describe reflex arc.
Answer:

I. Reflex action : Reflex action Is defined as a quick, automatic involuntary and often unconscious action brought about when the receptors are stimulated by external or internal stimuli.

II. Reflex arc : Reflex actions are controlled by CNS. Reflex arc is the structural or functional unit of reflex action. Simple reflex arc is formed of the following five components.
(1) Receptor organ : The sensory part that receives the stimulus is called receptor organ. It can be any sense organ that receives the stimulus and converts it into the impulse, e.g. skin, eye, ear, tongue, nasal epithelium, etc.

(2) Sensory neuron or afferent neuron:
Sensory part carrying impulse from receptor organ to CNS is called sensory neuron. Its cyton is located in dorsal root ganglion. Its dendron is long and connected to receptor while the axon enters in the grey matter of spinal cord to form a synapse.

(3) Association, adjustor or intermediate neuron : It is present in the grey matter of spinal cord. Receiving impulse from sensory neuron, interpreting it and generating motor impulse are done by association neuron.
(4) Motor neuron (effector) : The cyton of motor neuron is present in the ventral horn of grey matter and axon travels through ventral root. It conducts motor impulse from spinal cord to effector organ.

(5) Effector organ : Effector organ is a specialized part of the body which is excited by receiving the motor impulse. It gives proper response to the stimulus, e.g. muscles or glands. The path of reflex action is followed by the unidirectional impulse. It originates in the receptor organ and ends in effector organ through CNS.

Question 5.
Krishna was going to school and on the way he saw a major bus accident. His heartbeat increased and hands and feet become cold. Name the part of the nervous system that had a role to play in this reaction.
Answer:

1. The symptoms observed in Krishna were due to sympathetic nervous system. Emergency conditions trigger sympathetic nervous system to stimulate adrenal medulla.
2. The cells of adrenal medulla secrete catecholamines like adrenaline and nor¬adrenaline.
3. These hormones have direct effect on the pacemaker of the heart which causes increase in the heart rate and other associated symptoms.
4. This is a typical fright reaction caused by intervention of sympathetic nervous system.

Question 6.
What will be the effect of thyroid gland atrophy on the human body?
Answer:

1. Atrophy means degeneration. Atrophy of thyroid gland will result in deficient secretion of thyroid hormones leading to hypothyroidism. Deficiency of thyroid hormones [T₃ and T₄] and thyrocalcitonin will cause following effects on the body.
2. Decrease in BMR i.e. basal metabolic rate, decrease in the blood pressure, heart beat, body temperature, etc.
3. Occurrence of myxoedema in which there is abnormal deposition of fats under the skin giving puffy appearance in adults.
4. Irregularities in menstrual cycle in case of female patients.
5. Hair become brittle and fall.
6. Calcium metabolism also disturbs due to lack of thyrocalcitonin.

Question 7.
Write the names of hormones and the glands secreting them for the regulation of following functions
(a) Growth of thyroid and secretion of thyroxine.
Answer:
TSH by adenohypophysis.

(b) Helps in relaxing pubic ligaments to facilitate easy birth of young ones.
Answer:
Relaxin by degenerating corpus luteum of the ovary.

(c) Stimulate intestinal glands to secrete intestinal juice.
Answer:
Secretin by duodenal mucosa.

(d) Controls calcium level in the blood.
Answer:
Calcitonin [hypocalcemic hormone] by thyroid and parathormone [ hypercalcemic hormone] by parathyroid glands.

(e) Controls tubular absorption of water in kidneys.
Answer:
ADH by hypothalamus.

(f) Urinary elimination of water.
Answer:
Atrial natriuretic factor by atria of heart.

(g) Sodium and potassium ion metabolism.
Answer:
Aldosterone by adrenal cortex.

(h) Basal Metabolic rate.
Answer:
T₃ and T₄ by thyroid gland.

(I) Uterine contraction.
Answer:
Oxytocin by hypothalamus.

(j) Heartbeat and blood pressure.
Answer:
Adrenaline, non-adrenaline [stimulation] and acetylcholine [inhibition] by adrenal medulla.

(k) Secretion of growth hormone.
Answer:
GHRF by hypothalamus.

(l) Maturation of Graafian follicle.
Answer:
FSH by anterior pituitary.

Question 8.
Explain the role of hypothalamus and pituitary as a coordinated unit in maintaining homeostasis.
Answer:

1. Homeostasis is maintenance of constant internal environment of the body.
2. When certain hormones from any endocrine glands are secreted in excess quantity, the : inhibiting factors from hypothalamus, automatically exert negative feedback and stop the production of stimulating hormones from pituitary.
3. Similarly, if any hormone is in deficit, then j the concerned gland is given message through releasing factor. This way the hormone production remains in a balanced state or homeostasis.
4. E.g. If thyroxine from thyroid gland is secreted in excess, the secretion of TSH from pituitary is stopped by stopping the production of TRF from hypothalamus.
5. Though most of the endocrine glands are under the influence of pituitary gland, it is in turn controlled by hypothalamus.
6. Hypothalamus secretes releasing factors and inhibiting factors and hence regulate the secretions of pituitary (hypophysis).
7. There is negative feedback mechanism in controlling the secretions of the endocrine glands.
8. Hypothalamus forms the hypothalamo- hypophyseal axis through which transportation of neurohormones take place.

Following are the releasing and inhibiting factors produced by hypothalamus:

1. Somatotropin/GHRF : It stimulates release of growth hormone.
2. Somatostatin/GHRIF : It inhibits the release of growth hormone.
3. Adrenocorticotropin Releasing Hormone / CRF : It stimulates the release of ACTH by the anterior pituitary gland.
4. Thyrotropin Releasing Hormone /TRF : It stimulates the release of TSH by anterior pituitary gland.
5. Gonadotropin Releasing Hormone (GnRH) : It stimulates pituitary to secrete gonadotropins.
6. Prolactin Inhibiting Hormone (Prolactostatin) : It inhibits prolactin released by anterior pituitary gland.
7. Gastrin Releasing Peptide (GRP).
8. Gastric Inhibitory Polypeptide (GIP).

Question 9.
What is adenohypophysis ? Name the hormones secreted by it.
Answer:

1. Adenohypophysis is the large anterior lobe of pituitary gland.
2. It is derived from embryonic ectoderm in the form of Rathke’s pouch which is a small outgrowth from the roof of embryonic stomodaeum.
3. It is made up of epitheloid secretory cells.

It secretes following hormones:

1. GH : [Growth Hormone/STH : Somatotropic Hormone]
2. TSH/TTH – [Thyroid Stimulating Hormone/ Thyrotropic Hormone]
3. ACTH – [adrenocorticotropic hormone]
4. PRL – [prolactin]

Gonadotropins-

1. FSH [follicle stimulating hormone]
2. LH/ICSH – [leutinizing hormone/ interstitial cells stimulating hormone]

Question 10.
Describe, in brief, an account of disorders of adrenal gland.
Answer:
(1) Disorders of adrenal cortical secretions are caused due to hyposecretion and hypersecretion of adrenal corcoid hormones.

(2) Hyposecretion of corticosteroids causes Addison’s disease.

(3) The symptoms of Addison’s disease are low blood sugar, low body temperature, feeble heart action, low BR acidosis, low Na⁺ and K⁺ concentration in plasma, excessive loss of Na⁺ and water in urine, impaired kidney functioning and kidney failure, etc. it leads to weight loss, general weakness, nausea, vomiting and diarrhoea.

(4) Hypersecretion of corticoids causes Cushing’s disease.

(5) The symptoms of Cushing’s disease are high blood sugar level, glucosuria, alkalosis, enhancement of total quantity of electrolytes in extracellular fluid, polydipsia, increased BR muscle paralysis, obesity, wasting of limb muscles, etc.

Question 11.
Explain action of steroid hormones and proteinous hormones.
OR
Explain the mode of action of steroid hormones.
Answer:
The hormones always act on their target organs or tissues to induce their effects. The target tissues have specific binding sites or receptor sites which contain hormone receptors.
I. Steroid hormones:

1. The steroid hormones are lipid soluble and can easily cross the lipoproteinous plasma membrane.
2. The hormone receptors for steroid hormones are present in cytoplasm or in nucleus.
3. Hormone-receptor complex formed in cytoplasm enters the nucleus and regulate the gene expression or chromosome function.
4. In some cases the receptors are present inside the nucleus where hormone receptor complex is formed.
5. These complexes interact with the genome to evoke biochemical changes that result in physiological and developmental functions.

II. Protein hormones:

1. The hormone receptors for protein hormones are present on the cell membrane (i.e. membrane bound receptors).
2. When the hormone binds to its receptor, it forms hormone-receptor complex. Each receptor is specific to a specific hormone.
3. The hormones which interact with membrane bound receptors normally do not enter the target cell but generate second messengers. Such as cyclic AMP Ca⁺⁺ or IP (Inositol triphosphate), etc.
4. This leads to certain biochemical changes : in the target tissue.
5. Thus, the tissue metabolism and consequently the physiological functions are regulated by hormones.

Question 12.
Describe in brief an account of disorders of the thyroid.
OR
What are the functional disorders of thyroid gland? Describe in brief.
Answer:
Disorders of thyroid gland are of three types, viz. hypothyroidism, hyperthyroidism and simple goitre.
(1) Hypothyroidism : Hypothyroidism is deficient secretion of thyroxine. This hyposecretion causes two types of disorders, viz. cretinism in children and myoxedema in adults.
(i) Cretinism : Hyposecretion of thyroxine in childhood causes cretinism. The symptoms of cretinism are retardation of physical and mental growth.

(ii) Myxoedema : Deficiency of thyroxine in adults causes this disorder. It is also referred to as Gull’s disease. Symptoms are thickening and puffiness of the skin and subcutaneous tissue particularly of face and extremities. Patients with low BMR. It also causes mental dullness, loss of memory, slow action.

(2) Hyperthyroidism : Excessive secretion of thyroxine causes exophthalmic goitre or Grave’s disease. There is slight enlargement of thyroid gland. It increases BMR, heart rate, pulse rate and BE Reduction in body weight due to rapid oxidation, nervousness, irritability. Peculiar symptom is exophthalmos, i.e. bulging of eyeballs with staring look and less blinking. This is caused by deposition of fats behind the eye balls in eye sockets. There is muscular weakness and loss of weight.

(3) Simple goitre (Iodine deficiency goitre) : Simple goitre occurs due to deficiency of iodine in diet or drinking water. Simple goitre causes enlargement of thyroid gland. Thyroid gland in an attempt to get more iodine from the blood, swells due to increased blood supply. Prevention of goitre can be done by administering iodized table salt. It is also called endemic goitre as it is common in hilly areas.

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 4 Molecular Basis of Inheritance Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 4 Molecular Basis of Inheritance

1. Multiple Choice Questions

Question 1.
Griffith worked on ………………..
(a) Bacteriophage
(b) Drosophila
(c) Frog eggs
(d) Streptococci
Answer:
(d) Streptococci

Question 2.
The molecular knives of DNA are ………………..
(a) Ligases
(b) Polymerases
(c) Endonucleases
(d) Transcriptase
Answer:
(c) Endonucleases

Question 3.
Translation occurs in the ………………..
(a) Nucleus
(b) Cytoplasm
(c) Nucleolus
(d) Lysosomes
Answer:
(b) Cytoplasm

Question 4.
The enzyme required for transcription is ………………..
(a) DNA polymerase
(b) RNApolymerase
(c) Restriction enzyme
(d) RNase
Answer:
(b) RNA polymerase

Question 5.
Transcription is the transfer of genetic information from ………………..
(a) DNA to RNA
(b) t-RNA to m-RNA
(c) DNA to m-RNA
(d) m-RNA to t-RNA
Answer:
(a) DNA to RNA

Question 6.
Which of the following is NOT part of protein synthesis?
(a) Replication
(b) Translation
(c) Transcription
(d) All of these
Answer:
(a) Replication

Question 7.
In the RNA molecule, which nitrogen base is found in place of thymine?
(a) Guanine
(b) Cytosine
(c) Thymine
(d) Uracil
Answer:
(d) Uracil

Question 8.
How many codons are needed to specify three amino acids?
(a) 3
(b) 6
(c) 9
(d) 12
Answer:
(a) 3

Question 9.
Which out of the following is NOT an example of inducible operon?
(a) Lactose operon
(b) Histidine operon
(c) Arabinose operon
(d) Tryptophan operon
Answer:
(d) Tryptophan operon

Question 10.
Place the following event of translation in the correct sequence ………………..
i. Binding of met-t-RNA to the start codon.
ii. Covalent bonding between two amino acids.
iii. Binding of second t-RNA.
iv. Joining of small and large ribosome subunits.
(a) iii, iv, i, ii
(b) i, iv, iii, ii
(c) iv, iii, ii, i
(d) ii, iii, iv, i
Answer:
(b) i, iv, iii, ii

2. Very Short Answer Questions

Question 1.
What is the function of an RNA primer during protein synthesis?
Answer:
During DNA replication, RNA primer provides 3’ OH to which DNA polymerase enzyme can add nucleotides to synthesize new strand using parental strand of DNA as template.
[Note : RNA primer has no direct role in protein synthesis.]

Question 2.
Why is the genetic code considered as commaless?
Answer:
The triplet codon are arranged one after the other on m-RNA molecule without any gap or space and therefore genetic code is considered as commaless.

Question 3
Genome
Answer:
Genome is the total genetic constitution of an organism or a complete copy of genetic information (DNA) or one complete set of chromosomes (monoploid or haploid) of an organism.

Question 4.
Which enzyme does remove supercoils from replicating DNA?
Answer:
Super-helix relaxing enzyme (Topoisomerase) removes supercoils from replicating DNA.

Question 5.
Why are Okazaki fragments formed on lagging strand only?
Answer:
Okazaki fragments are formed only on lagging template as only short stretch of lagging template becomes available for replication at one time.

Question 6.
When does DNA replication take place?
Answer:
In eukaryotes DNA-replication takes place during S-phase of interphase of cell cycle and in prokaryotes. DNA replicates prior to cell division.

Question 7.
Define term Codogen and Codon
Answer:
Codogen is a triplet of nucleotides present on the DNA which specifies one particular amino acid.
Codon is a triplet of nucleotides present on the m-RNA which specifies one particular amino acid.

Question 8.
What is degeneracy of genetic code?
Answer:
Genetic code is degenerate as 61 codons code for 20 amino acids, that is two or more codons can specify the same amino acid. E.g. Cysteine has two codons, while isoleucine has three codons.

Question 9.
Which are the nucleosomal ‘core’ histones?
Answer:
Two molecules each of histone proteins, viz. H₂A. H₂B, H₃ and H₄ are the nucleosomal ‘core’ histones.

3. Short Answer Questions

Question 1.
DNA packaging in eukaryotic cell.
Answer:

1. In eukaryotic cells, DNA (2.2 metres) is condensed, coiled and supercoiled to be packaged efficiently in the nucleus (10^-16 m).
2. DNA is associated with histone and non-histone proteins.
3. Histones are a set of positively charged, basic proteins, rich in basic amino acid residues lysine and arginine.
4. Nucleosome consists of nucleosome core (two molecules of each of histone proteins viz. H₂A, H₂B, H₃ and H₄ forming histone octamer) and negatively charged DNA (146 bps) that wraps around the histone octamer by 1 3/4 turns.
5. H₁ protein binds the DNA thread where it enters and leaves the nucleosome.
6. Adjacent nucleosomes are linked with linker DNA (varies in length from 8 to 114 bp, average length of linker DNA is about 54 bp).
7. Each nucleosome contains 200 bp of DNA.
8. Packaging involves formation of – Beads on string (10 nm diameter), Solenoid fibre (looks like coiled telephone wire, 30 nm diameter/300Å), Chromatin fibre and Chromosome.
9. Non-Histone Chromosomal Proteins (NHC) contribute to the packaging of chromatin at a higher level.

Question 2.
Enlist the characteristics of genetic code.
Answer:
The characteristics of genetic code are

1. Genetic code is triplet, commaless and non-overlapping.
2. It is degenerate and non-ambiguous.
3. It is universal
4. It has polarity.

Question 3.
Applications of DNA fingerprinting.
Answer:
Applications of DNA fingerprinting are as follows:

1. In forensic science to solve rape and murder cases.
2. Finds out the biological father or mother or both, of the child, in case of disputed parentage.
3. Used in pedigree analysis in cats, dogs, horses and humans.

Question 4.
Explain the role of lactose in ‘Lac Operon’.
Answer:

1. A small amount of beta-galactoside permease enzyme is present in cell even when Lac operon is switched off and it allows a few molecules of lactose to enter into the cell.
2. Lactose binds to repressor and inactivates it.
3. Repressor – lactose complex cannot bind with the operator gene, which is then turned on.
4. RNA polymerase transcribes all the structural genes to produce lac m-RNA which is then translated to produce all enzymes.
5. Thus, lactose acts as an inducer.
6. When the inducer level falls, the operator is blocked again by repressor and structural genes are repressed again. This is negative feedback.

4. Short Answer Questions

Question 1.
Human genome project.
Answer:
1. Human Genome Project (HGP) was initiated in 1990 under the International administration of the Human Genome Organization (HUGO) and it was completed r in 2003.

2. The main aims:

• To sequence 3 billion base pairs of DNA in human genome and to map an estimated 33000 genes.
• To store the information collected from the project in databases.
• To develop tools and techniques for analysis of the data.
• Transfer of the related technologies to the private sectors, such as industries.
• Taking care of the legal, ethical and social issues which may arise from project.
• To sequence the genomes of several other organisms such as bacteria e.g. E.coli, Caenorhabditis elegans, Saccharomyces cerevisiae, Drosophil, rice, Arabidopsis), Mus musculus, etc.

3. Significance:

1. HGP has a major impact in the fields like Medicine, Biotechnology, Bioinformatics and the Life sciences.
2. More understanding of functions of genes, proteins and human evolution.

Question 2.
Describe the structure of operon.
Answer:

1. An operon is a unit of gene expression and regulation.
2. It includes the structural genes and their control elements. Control elements are promoters and operators.
3. The structural genes code for proteins, r-RNA and t-RNA that are necessary for all the cells.
4. Promoters are signal sequences in DNA. They start the RNA synthesis. They also act as sites where the RNA polymerases are bound during transcription.
5. Operators are present between the promoters and structural genes.
6. There is repressor protein that binds to the operator region of the operon.
7. There are regulatory genes which are responsible for the formation of repressors which interact with operators.

Question 3.
In the figure below A, B and C are three types of

Answer:
Answer: A, B and C are A : m-RNA, B : r-RNA, C : t-RNA

Question 4.
Identify the labelled structures on the following diagram of translation.

Part A is the ………………
Part B is the ………………
Part C is the ………………
Answer:
Part A is the anti-codon.
Part B is the amino acid.
Part C is the larger subunit of ribosome.

Question 5.
Match the entries in Column I with those of Column II and choose the correct answer.

Column I	Column II
A. Alkali treatment	i. Separation of DNA fragments on gel slab
B. Southern blotting	ii. Splits DNA fragments into single strands
C. Electrophoresis	iii. DNA transferred to nitrocellulose sheet
D. PCR	iv. X-ray photography
E. Autoradiography	v. Produce fragments different sizes
F. DNA treated with REN	vi. DNA amplification

Answer:

Column I	Column II
A. Alkali treatment	ii. Splits DNA fragments into single strands
B. Southern blotting	iii. DNA transferred to nitrocellulose sheet
C. Electrophoresis	i. Separation of DNA fragments on gel slab
D. PCR	vi. DNA amplification
E. Autoradiography	iv. X-ray photography
F. DNA treated with REN	v. Produce fragments different sizes

5. Long Answer Questions

Question 1.
Explain the process of DNA replication.
Answer:

DNA replication is semi-conservative replication. It involves following steps:
Activation of Nucleotides:

1. Nucleotides (dAMP dGMR dCMP and dTMP) present in the nucleoplasm, are activated by ATP in presence of an enzyme phosphorylase.
2. This phosphorylation results in the formation of deoxyribonucleotide triphosphates i.e. dATE dGTR dCTP and dTTE

Point of Origin or Initiation point:

1. Replication begins at specific point ‘O- Origin and terminates at point ‘T’.
2. At the point ‘O’, enzyme endonuclease nicks (breaks the sugar-phosphate backbone or the phosphodiester bond) one of the strands of DNA, temporarily.

Unwinding of DNA molecule:

1. Enzyme DNA helices breaks weak hydrogen bonds in the vicinity of ‘O’.
2. The strands of DNA separate and unwind. This unwinding is bidirectional.
3. SSBP (Single strand binding proteins) remains attached to both the separated strands and prevent them from recoiling (rejoining).

Replicating fork:

1. Y-shape replication fork is formed due to unwinding and separation of two strands.
2. The unwinding of strands results in strain which is released by super-helix relaxing enzyme.

Synthesis of new strands:

1. Each separated strand acts as a template for the synthesis of new complementary strand.
2. A small RNA primer (synthesized by activity of enzyme RNA primase) get attached to the 3′ end of template strand and attracts complementary nucleotides from surrounding nucleoplasm.
3. These nucleotides bind to the complementary nucleotides on the template strand by hydrogen bonds (i.e. A = T or T = A; G = C or C = G, CEG).
4. The phosphodiester bonds are formed between nucleotides of new strand to form a polynucleotide strand.
5. The enzyme DNA polymerase catalyses synthesis of new complementary strand always in 5′ – 3′ direction.

Leading and Lagging strand:

1. The template strand with free 3′ is called the leading template.
2. The template strand with free 5′ end is called the lagging template.
3. The replication always starts at C-3 end of template strand and proceeds towards C-5 end.
4. New strands are always formed in 5′ → 3′ direction.
5. The new strand which develops continuously towards replicating fork is called the leading strand.
6. The new strand which develops discontinuously away from the replicating fork is called the lagging strand.
7. Maturation of Okazaki fragments : The lagging strand is synthesized in the form of small Okazaki fragments which are joined by enzyme DNA ligase.
8. Later RNA primers are removed by the combined action of RNase H, an enzyme that degrades the RNA strand of RNA-DNA hybrids, and polymerase I.
9. Gaps formed are filled by complementary DNA sequence with the help of DNA polymerase-I in prokaryotes and DNA polymerase-a in eukaryotes.
10. Finally, DNA gyrase (topoisomerase) enzyme forms double helix to form daughter DNA molecules.

Formation of two daughter DNA molecules:

1. In each daughter DNA molecule, one strand is parental and the other one is newly synthesized.
2. Thus, 50% part (i.e. one strand of the helix) is contributed by mother DNA. Hence, it is described as semiconservative replication.

Question 2.
Describe the process of transcription in protein synthesis.
Answer:

Transcription involves three stages, viz. Initiation, Elongation and Termination.
(1) Initiation:

1. RNA polymerase binds to promoter site.
2. It then moves along the DNA and causes local unwinding of DNA duplex into two strands in the region of the gene.
3. Only antisense strand functions as template.

(2) Elongation:

• The complementary ribonucleoside tri-phosphates get attached to exposed bases of DNA template chain.
• As transcription proceeds, the hybrid DNA-RNA molecule dissociates and makes m-RNA molecule free.
• As the m-RNA grows, the transcribed region of DNA molecule becomes spirally coiled and regains double helical form.

(3) Termination:
When RNA polymerase reaches the terminator site on the DNA, both enzyme and newly formed m-RNA (primary transcript) gets released.

Question 3.
Describe the process of translation in protein synthesis.
Answer:

Translation involves the following steps:
1. Activation of amino acids and formation of charged t-RNA (t-RNA – amino acid complex):
i. In the presence of an enzyme amino acyl t-RNA synthetase, the amino acid is activated and then attached to the specific t-RNA molecule at 3’ end to form charged t-RNA (t-RNA – amino acid complex).

ii. ATP is essential for the reaction.

2. Initiation of Polypeptide chain:

• Small subunit of ribosome binds to the m-RNA at 5’ end.
• Start codon is positioned properly at P-site.
• Initiator t-RNA, (carrying amino acid methionine in eukaryotes or formyl methionine in prokaryotes) binds with initiation codon (AUG) of m-RNA, by its anticodon (UAC) through hydrogen bonds.
• The large subunit of ribosome joins with the smaller subunit in the presence of Mg⁺⁺.
• Thus, initiator charged t-RNA occupies the P-site and A – site is vacant.

3. Elongations of polypeptide chain:
Addition of amino acid occurs in 3 Step cycle-
i. Codon recognition.
Anticodon of second (and subsequent) amino acyl t-RNA molecule recognizes and binds with codon at A-site by hydrogen bonds.

ii. Peptide bond formation.

1. Ribozyme catalyzes the peptide bond formation between amino acids on the initiator t-RNA at P-site and t-RNA at A-site.
2. It takes less than 0.1 second for formation of peptide bond.
3. Initiator t-RNA at ‘P’ site is then released from E-site.

iii. Translocation.

1. Translocation is the process in which sequence of codons on m-RNA is decoded and accordingly amino acids are added in specific sequence to form a polypeptide on ribosomes.
2. Due to this A’-site becomes vacant to receive next charged t-RNA molecule.
3. The events like arrival of t-RNA – amino acid complex, formation of peptide bond, ribosomal translocation and release of previous t-RNA, are repeated.
4. As ribosome move over the m-RNA, all the codons on m-RNA are exposed oiie by one for translation.

4. Termination and release of polypeptide:
When stop codon (UAA, UAG, UGA) gets exposed at the A-site, the release factor binds to the stop codon, thereby terminating the translation process
The polypeptide gets released in the cytoplasm.
Two subunits of ribosome dissociate and last t-RNA and m-RNA are released in the cytoplasm.
m-RNA gets denatured by nucleases immediately.

Question 4.
Describe Lac ‘Operon’.
Answer:

Lac operon consists of the following components:
(1) Regulator gene:

• Regulator gene precedes the promoter gene.
• It may not be present immediately adjacent to operator gene.
• Regulator gene codes for a repressor protein which binds with operator gene and represses (stops) its action.

(2) Promoter gene:

• It precedes the operator gene.
• It is present adjacent to operator gene.
• RNA Polymerase enzyme binds at promoter site.
• Promoter gene base sequence determines which strand of DNA acts a template.

(3) Operator gene:

• It precedes the structural genes.
• When operator gene is turned on by an inducer, the structural genes get transcribed to form m-RNA.

(4) Structural gene:

• There are 3 structural genes in the sequence lac-Z, lac-Y and lac-A.
• Enzymes produced are β-galactosidase, β-galactoside permease and transacetylase respectively.
  Inducer Allolactose acts as an inducer. It inactivates the repressor by binding with it.

Question 5.
Justify the statements. If the answer is false, change the underlined word(s) to make the statement true.
(i) The DNA molecule is double stranded and the RNA molecule is single stranded.
Answer:

1. DNA as the genetic material has to be chemically and structurally stable.
2. It should be able to generate its replica.
3. Sugar-phosphate backbone and complementary base pairing between the two strands, give stability to DNA.
4. Both the strands of DNA act as template for synthesis of their complementary strands. This allows accurate replication of DNA.
5. Single stranded RNA can be folded to form complex structures and perform specific functions such as synthesis of proteins.

(ii) The process of translation occurs at the ribosome.
Answer:

1. Translation is the process in which sequence of codons of m-RNA is decoded and accordingly amino acids are added in specific sequence to form a polypeptide on ribosomes.
2. Ribosome has one binding site for m-RNA. It orients m-RNA molecule in such a way that all the codons are properly read.
3. Ribosome has three binding sites for t-RNA : P-site (peptidyl t-RNA-site), A-site (aminoacyl t-RNA-site) and E-site (exit site).
4. t-RNAs place the required amino acids in correct sequence and translate the coded message of RNA.
5. In eukaryotes, a groove which is present between two subunits of ribosomes, protects the polypeptide chain from the action of cellular enzymes and also protects m-RNA from the action of nucleases.
6. Thus ribosome plays an important role in translation.

(iii) The job of m-RNA is to pick up amino acids and transport them to the ribosomes.
Answer:
The job of t-RNA is to pick up amino acids and transport them to ribosomes. t-RNA is an adapter molecule. It reads the codons of m-RNA and also simultaneously transfer specific amino acid to m-RNA Ribosome complex. It binds with amino acid at its 3′ end.

(iv) Transcription must occur before translation may occur.
Answer:
In prokaryotes, translation can start before transcription is complete, as both these processes occur in the same compartment, i.e. cytoplasm. But in eukaryotes, transcription and processing of hnRNA occurs in nucleus. hnRNA then comes out of the nucleus through nuclear pores and then it is translated at ribosomes in the cytoplasm.

Question 6.
Guess
(i) the possible locations of DNA on the collected evidence from a crime scene and
(ii) the possible sources of DNA.

Evidence	Possible location of DNA on the evidence	Sources of DNA
e.g. Eyeglasses	e.g. Earpieces	e.g. Sweat, Skin
Bottle, Can, Glass	Sides, mouthpiece	—————-
————–	Handle	Sweat, skin, blood
Used cigarette	Cigarette butt	—————–
Bite mark	—————–	Saliva
————-	Surface area	Hair, semen, sweat, urine

Answer:

Evidence	Possible location of DNA on the evidence	Sources of DNA
e.g. Eyeglasses	e.g. Earpieces	e.g. Sweat, Skin
Bottle, Can, Glass	Sides, mouthpiece	Saliva
Door	Handle	Sweat, skin, blood
Used cigarette	Cigarette butt	Saliva
Bite mark	Teeth impression	Saliva
Clothes	Surface area	Hair, semen, sweat, urine

 

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 13 Amines Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 13 Amines

Question 1.
What are amines?
Answer:
Amines : The alkyl or aryl derivatives of ammonia in which one, two or all the three hydrogen atoms attached to nitrogen are replaced by same or different alkyl or aryl groups are called amines. OR Amines are nitrogen-containing organic compounds having basic character.

Example : methyl amine : CH₃ – NH₂

Question 2.
Classify the following amines as primary, secondary and tertiary.
Answer:

Question 3.
Mention the functional group in :
(1) Primary amine
(2) Secondary amine
(3) Tertiary amine.
Answer:
(1) A primary amine has a functional group – NH₂ (amino group).
Example : ethylamine, C₂H₅ – NH₂
(2) A secondary amine has a functional group – NH – (imino group).
Example :
(3) A tertiary amine has a functional group (tertiary nitrogen atom)

Example :

Question 4.
Write common and IUPAC names of following compounds :
Answer:
(A) Primary amines :


(B) Secondary amine :

(C) Tertiary Aimines :
.

Question 5.
Give the structures of the following :
Answer:

Question 6.
Give the IUPAC names of the following amines :
Answer:

Question 7.
Write the IUPAC names of the following amines :
Answer:

Question 8.
Give the structures and IUPAC names of the following amines :
Answer:

Question 9.
Classify the following amines as primary, secondary and tertiary and write the IUPAC names.
Answer:

Question 10.
Write the structures and classify the following amines as primary, secondary, tertiary amines.
Answer:

Question 11.
Write the common and IUPAC name of a tertiary amine in which one methyl, one ethyl and one w-propyl group is attached to nitrogen.
Answer:

Question 12.
How will you prepare ethanamine from ethyl iodide?
Answer:
When ethyl iodide is heated with excess of alcoholic ammonia, under pressure at 373 K ethanamine is obtained as a major product.

Question 13.
How is a nitroalkane converted to a primary amine?
OR
What is the action of LiAlH4/ether on (i) 1-Nitropropane (ii) 2-MethyI-l-nitropropane?
Answer:
When a nitroalkane is refluxed with tin (or iron) and concentrated HCl it gives corresponding primary amine.

For example, (1) nitromethane on reduction by refluxing with Sn and concentrated HCl gives methylamine.

(2) 1-Nitropropane on reduction with Sn and concentrated HCl gives propan-1-amine.

(3) Niirobenzcnc on reducion with tin and concentrated HCI or by using H₂/Pd in ethanol gives anilinc.

(4) When nitropropane is reduced in the presence of LiAlH₄ in ether, n-propyl amine is obtained.

(5) When 2-methyl-1-nitropropane is reduced in the presence of LiAlH₄ in ether, 2-methyl propan-1-amine is obtained.

Question 14.
How will you prepare aniline from nitrobenzene?
OR
How is aniline prepared from nitro compounds?
Answer:
Nitrobenzene is reduced to aniline by passing hydrogen gas in the presence of finely divided nickel, palladium or platinum.

Question 15.
Identify the compounds A and B in the following reactions

Answer:

Question 16.
How will you obtain a primary amine from an alkyl cyanide (nitrile)?
OR
Write a short note on Mendius reduction.
Answer:
Alkyl cyanides (nitriles) on reduction by sodium and ethyl alcohol form corresponding primary amines. This reaction is called Mendius reduction.

For example; propionitrile on reduction by sodium and ethanol gives n-propyl amine (Propan-1-amine).

Methyl cyanide or acetonitrile on reduction by sodium and ethanol gives ethanaminc.

Question 17.
How will you prepare ethylamine from acetonitrile?
OR
How is ethanamine prepared from methyl cyanide?
OR
What is the action of a mixture of sodium and alcohol on acetonitrile?
Answer:
Methyl cyanide or acetonitrile on reduction by sodium and ethyl alcohol forms ethanamine. The reaction is called Mendius reduction.

Question 18.
How will convert phenyl acetonitrile to β-phenylethylamine?
Answer:
When phenyl acetonitrile is reduced in the presence of sodium and ethanol, β-phenyl ethylamine is obtained.

Question 19.
How will you obtain primary amine from an acid amide?
Answer:
Acid amides on reduction with lithium aluminium hydride or sodium, ethanol form corresponding primary amines.

For example : Acetamide on reduction with lithium aluminium hydride or sodium, ethanol gives ethylamines.

Question 20.
Explain Hoffmann degradation of amides.
Write a note on Hoffmann bromamide degradation.
Answer:
The conversion of amides into amines in the presence of bromine and alkali is known as Hoffmann degradation of amides. An important characteristic of this reaction is that an amine with one carbon less than those in the amide is formed. Thus, decreasing the length of carbon chain. This reaction is an example of molecular rearrangement and involves the migration of an alkyl or aryl group from the carbonyl carbon to the adjacent nitrogen atom. For example,

(1) When propanamide is treated with bromine and aqueous or alcoholic sodium hydroxide, ethanamine is obtained which has one carbon atom less.

(2) When benzamide is treated with bromine and aqueous or alcoholic sodium hydroxide, aniline is obtained.

Question 21.
How will you obtain methyl amine from acetamide?
Answer:
When acetamide is treated with bromine and aq or alcoholic solution of KOH, methyl amine is obtained, which has one cabon atom less.

Question 22.
How will you convert the following?

(1) Ethyl bromide to ethylamine.
Answer:

(2) Propionitrile to n-propyl amine.
Answer:

(3) Acetonitrile to ethylamine.
Answer:

(4) Phenyl acetonitrile to β-phenylethyl amine.
Answer:

(5) Acetamide to ethylamine.
Answer:

(6) Nitropropane to propan-l-amine.
Answer:

(7) Nitrobenzene to Aniline.
Answer:

(8) Benzamide to aniline.
Answer:

Question 23.
How will you prepare propan-l-amine from (1) butane nitrile (2) 1-nitropropane (3) propanamide (4) butanamide?
Answer:
(1) From butane nitrile :
When butane nitrile is reduced by sodium and ethanol, it gives propan-l-amine.

(2) From 1-nitropropane :
When 1-nitropropane is reduced in the presence of tin and cone, hydrochloric acid, propan-l-amine is obtained.

(3) From propanamide :
When propanamide is reduced in the presence of lithium aluminium hydride, propan-l-amine is obtained.

(4) From butanamide :
When butanamide is treated with bromine and aq. KOH, propan-l-amine is obtained.

Question 24.
Write a reaction to, convert acetic acid into methyl amine.
Answer:

Question 25.
Primary and secondary amines have boiling points higher than the tertiary amines. Explain why?
Answer:
(1) The N – H bond in amines is polar in nature because of electronegativities of nitrogen (3.0) and hydrogen (2.1) are different.
(2) Due to the polar nature of N – H bond, primary and secondary have strong intermolecular hydrogen bonding. Tertiary amines do not have intermolecular hydrogen bonding as there is no hydrogen atom on nitrogen of tertiary amine. Thus, intermolecular forces of attraction are strongest in primary and secondary amines and weakest in to tertiary amines. Hence, primary and secondary amines have boiling points higher than the tertiary amines.

Question 26.
Amines have boiling points higher than the hydrocarbon but lower than the alcohols of comparable masses. Explain, why?
Answer:
Amines are polar than alkanes but less polar than alcohols. Primary and secondary amines form intermolecular hydrogen bonds. This hydrogen bonding leads to an associated structure. The association is more in primary amines than that in secondary amines as there are two hydrogen atoms attached to the nitrogen atom. However, tertiary amines do not form intermolecular hydrogen bonds because they do not contain any hydrogen atoms attached to the nitrogen atom. Hence, amines have higher boiling points than the hydrocarbons but lower boiling points than the alcohols of comparable masses.

Compound	Molar mass	Boiling points (K)
nC2H5CH(CH3)2	72	300
nC4H9NH2	73	350.8
nC4H9OH	74	391

Question 27.
Arrange the following compounds in the decreasing order of their solubility in water.
(a) Ethyl amine, diethyl amine and triethyl amine.
Answer:
Diethyl amine > triethyl amine > ethyl amine
(The reason that ethyl group has greater +1 effect than methyl group)

(b) Ethyl amine, n-propyl amine and n-butyl amine.
Answer: n-butyl amine < n-propyl amine < ethyl amine

(c) n-Butane, n -butyl alcohol and n-butyl amine
Answer:
n-butyl alcohol < n-butyl amine < n-butane

Question 28.
Arrange the following compounds in the decreasing order of their boiling points.
(a) Ethane, ethyl amine and ethyl alcohol.
Answer:
Ethyl alcohol < ethyl amine < ethane

(b) Ethyl amine, n-propyl amine and n-butyl amine.
Answer:
n-butyl amine < n-propyl amine < ethyl amine

(c) n-propyl amine, ethyl methyl amine and trimethyl amine.
Answer:
n-propyl amine < ethyl methyl amine < trimethyl amine.

(d) Ethyl alcohol, dimethyl amine and ethyl amine.
Answer:
Ethyl alcohol < ethyl amine < dimethyl amine.

Question 29.
Explain the basic nature of amines with a suitable examples.
OR
Explain why amines are basic.

Question 38.
Tertiary amine (R₃N) or 3° amine is weaker base than secondary amine R₂NH or 2° amine. Explain.
Answer:


The increase in basic strength from 1° amine to 2° amine is explained on the basis of increased stabilization of conjugate acids by +1 effect of the increased number of the alkyl group. However, decreased basic strength of 3° implies that the conjugate acid of 3° amine is less stabilized and is weak base though the +1 effect of three alkyl groups in R₃NH is large.

R₂NH is best stabilized by solvation while the stabilization by solvation is very poor in R₃NH. Hence (R₃N) or tertiary amine or 3° amine is weaker base than secondary amine (R₂NH) or 2° amine.

Question 30.
Primary or aliphatic amine is a stronger base than ammonia. Explain.
Answer:
(1) The alkyl group in primary amines has +I effect i.e. (electron releasing).

The alkyl group tends to increase the electron density on the nitrogen atom. As a result, amines can donate the lone pair of electrons on nitrogen more easily than ammonia.

(2) The amine being a base, can donate a pair of electrons to an acid. The alkyl group with +I effect will disperse the positive charge on the cation more than ammonia.

Due to +I effect of alkyl group cation formed by primary amine is more stable compared to cation formed from ammonia. Also it is seen that observed increasing basic strength from ammonia to primary amine is explained on the basis of increased stabilization of conjugate acids by +I effect for the presence of alkyl (R) groups. Hence, primary or aliphatic amine is a stronger base than ammonia.

Question 31.
Aniline is less basic than ammonia. Explain.
Answer:
The less basic character of aniline can be explained on the basis of resonance shown by aniline.

Due to resonance, the nitrogen atom of amino group in aniline acquires a positive charge, hence, lone pair of electrons is less available for protonation as compared to that of ammonia. Aniline is resonance stabilized by five resonance structures. On the other hand, aniline in aqueous medium, accepts a proton does not have lone pair of electrons on nitrogen to produce a very low concentration of anilium ion and anilium ion shows only two resonance structures and therefore less stabilized than anline.

Thus, aniline is more stable than anilium ion. Hence aniline accepts proton less readily or less basic in nature than ammonia.

Question 32.
Explain the order of basicity in ammonia and aliphatic amines.
Answer:
Since nitrogen atom in ammonia molecule has a lone pair of electrons, it is a Lewis base.
Greater the availability of an electron pair, more is the basic character.

Since alkyl group (R -) is an electron releasing group with (+I) inductive effect, alkyl amines act as a stronger base than ammonia.

The decreasing order of basicity is –

The availability of a lone pair of electrons on a nitrogen atom in amines is influenced by steric factor due to crowding of alkyl groups which affects solvation along with inductive effect of alkyl groups.

Due to high energy of solvation of \(\mathrm{NH}_{4}^{+}\) ions, they acquire higher stability in aqueous solutions.

The presence of alkyl groups in secondary and tertiary amines, due to steric hindrance decrease the solvation energy.

This effect is more in tertiary amines making the tertiary ammonium ions (R₃NH⁺) unstable as compared to secondary ammonium ion (R₂N⁺H₂).
Hence the cumulative effect on the order of basicity of amines is, secondary amine > primary amine > tertiary amine > ammonia (NH₃).

Question 33.
Arrange the following amines in the decreasing order of their basic nature.
(a) Aniline, propan-l-amine and N-methylethanamine.
Answer:
N-methylethanamine < propan-l-amine < aniline

(b) Benzene-1, 4-diamine, ammonia and 4-aminobenzoic acid.
Answer:
Ammonia < benzene-1, 4-diamine < 4-aminobenzoic acid

(c) N-Methylaniline, phenylmethylamine and N-phenylaniline.
Answer:
N-Methylaniline < N-phenylaniline < phenylmethylamine

Question 34.
Arrange the following amines in the increasing order of their pKb values.
(a) Aniline, N-methylaniline and cyclohexalamine.
Answer:

(b) Phenylmethylamine, 2-aminotoluene and 2-fluoroaniline.
Answer:

(c) Aniline, 4-methoxyaniline and 4-nitroaniline.
Answer:

Question 35.
Arrange the following compounds in the decreasing order of their basic nature in the gaseous phase.
Ammonia, N-methylhexanamine, propan-1-amine and N, N-dimethylethanolamine.
Answer:
Propan-1-amine < N-methylethanamine < N,N-dimethylmethanamine < ammonia

Question 36.
Explain laboratory test for amines.
Answer:
(1) All amines are basic compounds. Aqueous solution of water soluble amines turns red litmus blue.

(2) When water insoluble amine is dissolved in aqueous HCl, forms water soluble substituted ammonium chloride, further a substituted ammonium chloride on reaction with excess aqueous NaOH regenerates the original insoluble amine.

(3) Diazotization reaction/ Orange dye test: In a sample of aromatic primary amine, 1-2 mL of cone. HCl is added. The aqueous solution of NaNO₂ is added with cooling. This solution is transfered to a test tube containing solution of β naphthol in NaOH. Formation of orange dye indicates presence of aromatic primary amino group. (It may be noted that temperature of all the solutions and reaction mixtures is maintained near 0 °C throughout the reaction).

Question 37.
Explain Hofmann’s exhaustive alkylation.
OR
Explain Hofmann’s exhaustive methylation of amines.
Answer:
Hofmann’s Exhaustive alkylation : When a primary amine is heated with excess of primary alkyl halide it gives a mixture of secondary amine, tertiary amine along with tetraalkylammonium halide

If excess of alkyl halide is used, tetraalkyl ammonium halide is obtained as major product. The reaction is known as exhaustive alkylation of amines.

Hofmann’s Exhaustive Methylation : The process of converting a primary, secondary or tertiary amine into quaternary ammonium halide by heating them with excess of methyl iodide, is called exhaustive methylation or Hoffmann’s exhaustive methylation.

Thus when methyl amine is heated with excess of methyl iodide it forms dimethylamine (secondary amine), then trimethylamine (a tertiary amine) and finally of quaternary ammonium iodide. The reaction is carried out in the presence of mild base NaHCO₃, to neutralize the large quantity of HI formed.

Question 38.
Predict the products of exhaustive methylation of following compounds.
(1) Ethylamine.
Answer:
A primary amine, ethylamine (CH₃ – CH₂ – NH₂) on exhaustive methylation, i.e., on heating with excess methyl iodide, forms secondary amine, tertiary amine and finally a quaternary ammonium salt, ethyl-trimethyl ammonium iodide.

(2) Benzylamine.
Answer:
Benzylamine C6H5CH2NH2 on exhaustive methylation i.e., on heating with excess methyl iodide forms benzylmethyl amine, benzyldimethyl ammonium chloride and finally benzyltrimethyl ammonium iodide.

Question 39.
Explain Hofmann elimination.
OR
Write a note on Hoffmann elimination.
Answer:
When tetra alkyl ammonium halide is heated with moist silver hydroxide, a quaternary ammonium hydroxide is obtained. Quaternary ammonium hydroxides are deliquescent crystalline solids and are basic in nature. Quaternary ammonium hydroxides on strong heating undergo ^-elimination to give tertiaryamine, alkenes and water, the reaction is called Hofmann elimination. The major product is least substituted alkene.

Question 40.
Write the bond line formula of the alkene which is obtained as major product from the following amines, on heating with excess of methyl iodide followed by strong heating with moist silver oxide.

Answer:

Answer:

Answer:

Question 41.
Compound X with a molecular formula C₅H₁₃N did not react with nitrous acid, but reacted with one mole of CH3I to form a salt. What is the structure of X?
Answer:
The structure of compound X is ethyl-N-methylethanamine since compound X is tertiary amine. It reacts with one mole of CH₃I to give a quaternary ammonium salt.

Question 42.
What is the action of acetyl chloride on :
(1) ethyl amine (ethanamine)
(2) diethyl amine (N-Ethylethanamine)
(3) triethyl amine?
OR
Write a short note on acylation of amines.
Answer:
The reaction of amines with acetyl chloride is called acetylation of amines.

(1) Acetyl chloride on reaction with ethylamine forms monoacetyl derivative, N-ethylacetamide (or N-acetyl ethylamine).

(2) Diethyl amine on reaction with acetyl chloride forms N-acetyl dimethylamine.

(3) Triethyl amine, being a tertiary amine does not have H atom attached to nitrogen of amine, hence it does not react with acetyl chloride.

Question 43.
What is the action of acetic anhydride on aniline?
Answer:
Aniline on reaction with acetic anhydride forms N-phenyl acetamide.

Question 44.
What is the action of benzoyl chloride on ethanamine?
Answer:
When benzoyl chloride is treated with ethanamine, N-ethyl benzamide is obtained.

Question 45.
What is the action of nitrous acid on ethylamine?
Answer:
Ethyl amine on reaction with nitrous acid in cold forms aliphatic diazonium salt, (unstable intermediate), which decomposes immediately by reaction with solvent water to produce ethyl alcohol and nitrogen gas.

Question 46.
What is the action of nitrous acid on aniline?
Answer:
Aniline reacts with nitrous acid in cold to form diazonium salt which has reasonable solubility at 273 K

Question 47.
How is benzenediazon|um chloride prepared?
Answer:
Benzenediazonium chloride is prepared by the action of nitrous acid on aniline at 273-278 K. Nitrous acid being unstable, is prepared in situ by the reaction between sodium nitrite and dilute hydrochloric acid.

Question 48.
Write resonance stabilized structures of aryl diazonium salt.
Answer:

Question 49.
Write a note on Sandmeyer’s reaction.
OR
How is aryl chloride or aryl bromide or aryl cyanide prepared from diazonium salt?
Answer:
[Replacement by Cl^–, Br^– and -CN : Sandmeyer reaction.] Freshly prepared aromatic diazonium salt on reaction with cuprous chloride gives aryl chloride, on reaction with cuprous bromide gives aryl bromide and on reaction with cuprous cyanide give aryl cyanide. The reaction in which copper (I) salts are used to replace nitrogen in diazonium salt is called Sandmeyer reaction.

Question 50.
How is aryl chloride or aryl bromide prepared by Gattermann reactions?
Answer:
The aryl chloride or bromides can also be prepared by Gattermann reactions in which diazonium salt reacts with

Question 51.
How is aryl iodide obtained from diazonium salt?
Answer:
When diazonium salt is warmed with potassium iodide, aryl iodide is obtained.

Question 52.
Explain the reduction of arene diazonium salt?
OR
How is arene obtained from arene diazonium salt?
OR
What is the action of benzene diazonium chloride on ethanol?
Answer:
Arene diazonium salt on treatment with mild reducing agents like phosphinic acid (hypophosphoric acid) or ethanol, arene is obtained.

Question 53.
How is phenol obtained from arene diazonium salt?
Answer:
When arene diazonium salt is slowly added to a large volume of boiling dilute sulphuric acid, phenol is obtained,

Question 54.
How is aryl fluoride obtained from diazonium salt?
Answer:
When fluoroboric acid is treated with the solution of diazonium salt, a precipitate of diazonium fluoroborate is obtained, which is filtered and dried. When dry diazonium fluoroborate is heated, it decomposes to give aryl fluoride. This reaction is called Balz-Schiemann reaction.

Question 55.
How is nitrobenzene obtained from benzene diazonium fluoroborate?
Answer:
When benzene diazonium fluoroborate is heated with aqueous solution of sodium nitrite in the presence of copper powder, nitrobenzene is obtained.

Benzene diazonium fluorobate can be obtained by reaction of benzene diazonium chloride with HBF₄.

Question 56.
What is meant by a coupling reaction? Explain with suitable examples.
OR
What is the action of benzene diazonium chloride on (a) phenol in alkaline medium (b) aniline?
OR
Write a note on the coupling reaction.
Answer:
Diazonium salts react with certain aromatic compounds having an electron-rich group (e.g.-OH, – NH₂, etc.) to form azo compounds. This reaction is an electrophilic substitution and is called coupling reaction. Azo compounds are brightly coloured and are used as dyes and indicators. Coupling reaction is an electrophilic substitution reaction. Benzene diazonium chloride reacts with alkaline solution of phenol to give p-hydroxy azo benzene (orange dye).

Benzene diazonium chloride reacts with aniline in mild alkaline medium to give p-aminobenzene (yellow dye).

Question 57.
What is the action of p-toluene sulphonyl chloride on ethyl amine and diethyl amine?
Answer:
(1) When ethyl amine is treated with p-toluene sulphonyl chloride, N-ethyl p-toluene sulphonamide is obtained.

(2) When diethyl amine is treated with p-toluene suiphonyl chloride. N.N-dicthyl p-toluene suiphonyl amide is formed.

Question 58.
How will you distinguish between :
(1) Ethylamine, diethyl amine and triethyl amine by using (i) nitrous acid (ii) Hinsberg’s reagent.
(2) Diethyl amine and triethyl amine by using acetic anhydride.
Answer:

Question 59.
Give a chemical test to distinguish between following pairs of compounds.
(i) Ethylamine and diethyl amine :
Answer:
Ethylamine (C₂H₅NH₂) is a primary amine while diethyl amine ( (C₂H₅)₂NH) is a secondary amine. So the two can be distinguished by the following test.

(ii) Ethyl amine and aniline :
Answer:
Ethylamine is an aliphatic amine, while aniline is an aromatic amine. So the two can be distinguished by the following test :

(iii) Aniline and benzyl amine :
Answer:

(iv) Aniline and N-ethylaniline :
Answer:

Question 60.
Compound ‘X’ with a molecular formula C₄H₁₁N did not react with Hinsberg’s reagent, but reacted with one mole of CH₃I to form a salt. What is the structure of ‘X’?
Answer:
The structure of compound ‘X’ is :

Since the compound ‘X’ does not react with NaN02 and HC1 i.e. nitrous acid (HO – N = O), it must be a tertiary amine.

The tertiary amine reacts with one mole of CH3I to give a quaternary ammonium salt.

Question 74.

p-(dimethylamino) azobenzene is yellow dye which was formerly used as a colouring agent in margarine. Write the structures of the reactants used in the preparation of this dye.
Answer:

Question 61.
Convert 3-Methyl aniline into 3-nitrotoluene.
Answer:

Question 62.
How will you bring about following conversions?
(1) N.Methyl aniline into N-methyl benzanilide.
Answer:

(2) 1.4-Dichlorobutane Into hexane-1,6-diamlne.
Answer:

(3) Benzene into 3-bromo aniline.
Answer:

(4) Chlorobenzene into 4-chioroanilinc.
Answer:

(5) 11enaniide into toluene.
Answer:

Question 63.
What is the action of aqueous bromine on aniline?
Answer:
Action of aqueous bromine on aniline : When aniline is treated with bromine water at room temperature, a white precipitate of 2, 4, 6-tri bromoaniline is obtained.

Question 64.
Explain the action of cone, nitric acid (nitrating mixture) on aniline.
Answer:
When aniline is warmed with a mixture of cone, nitric acid and cone, sulphuric acid (a nitrating mixture), a mixture of ortho, meta and para nitroaniline is obtained.

Question 65.
What is the action of acetic anhydride on aniline?
Answer:
When aniline is heated with acetic anhydride, an acetanilide is obtained.

Question 66.
How will you convert aniline to p-nltroanhline? (major product)
Answer:

Question 67.
What is the action of cone, sulphuric acid on aniline?
Answer:
Aniline on treatment with cold sulphuric acid forms anilium hydrogen sulphate which on heating with sulphuric acid at 453 K-475 K gives sulphanilic acid, (p-aminobenzene sulphonic acid) as major product.
Sulphanilic acid exists as a salt; called dipolar ion or zwitter ion. It is produced by the reaction between an acidic group and a basic group present in the same molecule.

Question 68.
How will you convert the following?
(1) Ethylamine to ethyl alcohol.
Answer:

(2) N-Methyl aniline to N-Nitroso-N-methyl aniline.
Answer:

(3) Diethylamine to N-nitrosodiethylamine
Answer:

(4) Triethylamine to triethyl ammonium nitrite.
Answer:

(5) Ethyl amine to N-ethylacetamide.
Answer:

(6) Diethyl amine to N-acetyl diethylamine.
Answer:

(7) Aniline to acetanilide.
Answer:

(8) Aniline to N-ethyl henzamide.
Answer:

(9) Ethylamine to ethyl isocyanide.
Answer:

(10) Aniline to phenyl isocyanide.
Answer:

(11) Aniline to 2,4,6-tribromoaniline.
Answer:

Question 69.
Give a plausible explanation for each of the following statements :
(1) Ethylamine is soluble in water whereas aniline is not.
Answer:
Ethylamine is soluble in water due to intermolecular hydrogen bonding resulting in the formation of C₂H₅NH₃ ion. Whereas in anline the hydrogen bonding with water is negligible due to the phenyl group (C₆H₅) is bulky and has -I effect. Therefore, aniline is nearly insoluble in water.

(2) Butan-1-ol is more soluble in water than butani-amine.
Answer:
Rutan- l-al is more soluble in watcr duc to intermoiccular hydrogen bonding. In alcohols, hydrogen bonding is through oxygen atoms. WIereas hutani-amine is less soluble in water due to the larger hydrocarbon part is hydrophobic in nature. Hence, butan-l-ol is more soluble in water than butani-amine.

(3) Butan-1-amlne has higher boiling point than N-ethylethanamine.
Answer:
Due to the presence of two H-atoms on N-atom in butait- I -amine, they undergo extensive intermolecular H-bonding while in N-cthylethanamine due to the presence of one-H atom on the N-atom, they undergo least intermolecular H-bonding. Hence, butan- l-amine has higher boiling point than-N-ethyl ethanamine.

(4) AnIline Is less basic than ethyl afine.
Answer:
Aniline (K_b4-2 x 10^-10) is less basic than ethyl amine (K_b5.1 x 10^-4). This is because -I effect of phenyl group in aniline as compared to + 1 effect of ethyl group in ethyl amine.

Due to resonance, the lone pair of electrons on the nitrogen atom gets delocalized over the benzene ring and thus less available for protonation. On the other hand, in ethyl anine, delocalization of the lone pair of electrons on the nitrogen atom by resonance is not possible. Further more, the electron density on the nitrogen atom is increased by +1 effect of the ethyl group. Hence, aniline is less basic than ethyl amine.

(5) pK_b value of diethyl amine is less than that of ethyl amine.
Answer:
The basic strength of amines is expressed in terms of pK_b values. Smaller is the value of pK_b more basic is the amine. The pK_b value of ethyl amine is 3.29 and that of diethyl amine is 3.00. Therefore, diethyl amine is more basic than ethyl amine.

(6) Aniline cannot be prepared by Gabriel phthalimide synthesis.
Answer:
In Gabriel-phthalimide synthesis of aniline, potassium phthalimide requires the treatment with chlorobenzene or bromobenzene. Since aryl halides do not undergo nucleophilic substitution reaction. Therefore, chlorobenzene or bromobenzene does not react with potassium phthalimide to give N-phenylphthalimide and hence aniline cannot be prepared by Gabriel phthalimide synthesis.

(7) Gabriel phthalimide synthesis is preferred for the preparation of aliphatic primary amines.
Answer:
In aromatic amines, the lone pair of electrons on the N-atom is delocalized over the benzene ring. As a result electron density on the nitrogen atom decreases. Whereas in aliphatic primary amines, due to +1 effect of alkyl group, electron density on nitrogen atom increases. As the pKh value of aliphatic amines is more than that of aromatic amines, aromatic amines are less basic than primary aliphatic amines. Hence, Gabriel phthalimide synthesis is preferred for the preparation of aliphatic amines.

(8) Arere diazonium salts are relatively more stable than alkyl diazonium salts.
Answer:
Arene diazonium salts are stable due to the dispersal of the positive charge over the benzene ring as shown below :

Alkane diazonium salts are unstable due to their tendency to eliminate a stable molecule of nitrogen to form carbocation. Aromatic diazonium salts have much lower tendency to remove nitrogen than aliphatic diazonium salts. Hence, arene diazonium salts are relatively more stable than alkyl diazonium salts.

(9) Tertiary amines cannot be acylated.
Answer:
Tertiary amines do not react with acetic anhydride or acetyl chloride i.e. they can be acylated because they do not contain a H-atom on the N-atom.

(10) Besides the ortho and para derivatives, considerable amount of meta derivatives is also formed during nitration of aniline.
OR
Although amino group is o- and p-directing in electrophilic substitution reactions, aniline on nitration gives substantial amount of m-nitroaniline.
Answer:
In aromatic amines, -NH₂ is an electron releasing or activating group. It activates the ortho and para positions in the benzene ring towards electrophilic substitution. When aniline is treated with nitrating mixture (cone. HNO₃+ cone. H₂SO₄), a mixture of ortho and para nitroaniline is obtained. However, a substantial amount of m-nitroaniline is also formed. Aniline being a base gets protonated in acidic medium to form anilium cation, which deactivates the ring and the substitution takes place at the meta position.

Question 70.
How will you convert :
(1) Aniline into benzyl alcohol.
Answer:

(2) Aniline into 4-bromoaniline.
Answer:

(3) Aniline into 1,3,5-tribromo benzene.
Answer:

(4) Aniline into 2,4,6-tribromo fluoro benzene.
Answer:

Question 71.
How will you convert :
(1) Propanoic acid into ethanoic acid.
Answer:

(2) Propanoic acid into ethanol
Answer:

(3) Ethanamine into propan-l-amine.
Answer:

(4) Propan-l-amine into ethanamine.
Answer:

(5) Propanoic acid into ethanamine.
Answer:

(6) Ethanamine into propanoic acid.
Answer:

(7) Benzene to aniline.
Answer:

(8) Aniline to Benzene.
Answer:

(9) Aniline into benzoic acid.
Answer:

(10) Benzoic acid into aniline.
Answer:

(11) Aniline into benzamide.
Answer:

(12) 3-Nitrotoluene into 3-methyl aniline.
Answer:

(13) 3-Methyl aniline into 3-Nitrotoluene.
Answer:

Question 72.
An organic compound ‘A’ having molecular formula C₂H₆O evolves hydrogen gas on treatment with sodium metal and on treatment with red phosphorous and iodine gives compound ‘B’. The compound ‘B’ on treatment with alcoholic KCN and on subsequent reduction gives compound ‘C’. The compound ‘C’ on treatment with nitrous acid evolves nitrogen gas. Write the balanced chemical equations for all the reactions involved and identify the compounds ‘A’, ‘B’ and ‘C;.
Answer:
A = C₂H₅OH ethanol
B = C₂H₅I ethyl iodide
C = C₂H₅CH₂NH₂ n-propyl amine
Compound C₂H₆O = C₂H₅OH

Question 73
Identify B, C and D write complete reactions :

Answer:

Question 74.
Identify the compounds B, C and D in the following series of reactions and rewrite the complete equations :

Answer:

Question 75.
Identify the compounds ‘A’ and ‘B’ in the following equation :

Answer:

Question 76.
Answer in one sentence :

(1) Arrange the following compounds in decreasing order of basic strength in their aqueous solutions. NH₃, C₂H₅NH₂, (CH₃)₂NH, (CH₃)₃N
Answer:
The decreasing order of basic strength is – (C₂H₅)₂NH > (C₂H₅)₃N > (C₂H₅)₂NH > NH₃
(The reason that ethyl group has greater +1 effect than methyl group).

(2) Arrange the following compounds in an increasing order of their solubility in water.

Answer:
The solubility increases in order in which molecular mass decreases.

(3) What is Hinsberg’s reagent?
Answer:
Benzenesulphonyl chloride (C₆H₅SO₂Cl) is known as Hinsberg’s reagent.

(4) Name the reaction in which a primary amine is formed from amide.
Answer:
Hoffmann bromamide degradation.

(5) NH₃ is a Lewis base.
Answer:
Since nitrogen in ammonia molecule has a lone pair of electrons, it is a Lewis base.

(6) How many primary amines are possible for the compound C₃H₉N?
Answer:
For the compound C₃H₉N, two primary amines are possible.

(7) State the hybridization of the nitrogen atom in amines.
Answer:
The hybridization of nitrogen atom in amines is sp³.

(8) Arrange the following compounds in an increasing order of basic strength. Aniline, p-nitroaniline, p-toluidine.
Answer:
p-nitroaniline < aniline < p-toluidine.

(9) Which of the two is more basic and why? CH₃NH₂ or NH₃
Answer:
Due to +1 effect of -CH₃ group, electron density on N-atom increases, hence methyl amine is a stronger base than ammonia.

(10) Which of the two is more basic and why? p-toluidine or aniline.
Answer:
p-toluidine is more basic due to the presence of -CH₃ group at para position. Due to +1 effect of -CH₃ group, electron density on nitrogen increases, hence the tendency to donate pair of electrons increases.

Multiple Choice Questions

Question 77.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. Which of the following is an amine?
(a) C₂H₅N(COCH₃)₂
(b) (C₂H₅)₂N – N = 0
(c) (C₂H₅)₃N
(d) All of these
Answer:
(d) All of these

2. N-methyl-N-ethyl-n-propyl amine is
(a) a primary amine
(b) a secondary amine
(c) a tertiary amine
(d) an alkyl nitrile
Answer:
(c) a tertiary amine

3. Which of the following is a tertiary amine?

Answer:
(d)

4. Tertiary butyl amine is a
(a) primary amine
(b) secondary amine
(c) tertiary amine
(d) quaternary ammonium salt
Answer:
(a) primary amine

5. The IUPAC name of

(a) ethyl propanamine
(b) ethyl butylamine
(c) 2-pentanamine
(d) 3-hexanamine
Answer:
(d) 3-hexanamine

6. The IUPAC name of ethyl dimethyl amine is ……………..
(a) 2-amino propane
(b) N,N-dimethyl ethanolamine
(c) ethyl methanamine
(d) propanamine
Answer:
(b) N,N-dimethyl ethanolamine

7. Isopropyl amine and trimethyl amine are ……………..
(a) acidic in nature
(b) electrophilic compounds
(c) structural isomers
(d) optically active compounds
Answer:
(c) structural isomers

8. N, N-dimethylethanolamine is ……………

Answer:
(b)

9. IUPAC name of diethylmethyl amine is ………………
(a) methyl amino propane
(b) N-Ethyl-N-methylhexanamine
(c) methyl diethanamine
(d) amino pentane
Answer:
(b) N-Ethyl-N-methylhexanamine

10. Ethyl bromide reacts with excess of alcoholic ammonia, the major product is …………..
(a) ethyl amine
(b) diethylamine
(c) triethylamine
(d) tetraethyl ammonium bromide
Answer:
(a) ethyl amine

11. Isopropylamine is obtained by the reduction of
(a) acetoxime
(b) acetaldoxime
(c) formaldoxime
(d) aldoxime
Answer:
(a) acetoxime

12. Which of the following compounds can be converted into amines in the presence of Na and alcohol?
(a) Alkyl nitriles
(b) Aldoxime
(c) Ketoxime
(d) All of these
Answer:
(d) All of these

13. Chloroethane when boiled with excess of aqueous-alcoholic ammonia gives hydrochloric acid and
(a) triethyl amine
(b) trimethyl amine
(c) diethyl amine
(d) ethyl amine
Answer:
(d) ethyl amine

14. How many hydrogen atoms are required for the reduction of 1-nitropropane to n-propyl amine?
(a) Four
(b) Three
(c) Six
(d) Two
Answer:
(c) Six

15. A secondary alkyl halide is heated with excess of ammonia, the major product obtained is
(a) primary amine
(b) secondary amine
(c) tertiary amine
(d) quaternary ammonium salt
Answer:
(a) primary amine

16. The true statement about ethylamine is
(a) it is weaker base than ammonia
(b) it is stronger base than diethyl amine
(c) it is stronger base than triethyl amine
(d) it is stronger base than alkali
Answer:
(c) it is stronger base than triethyl amine

17. The reaction which is given only by primary amines is
(a) acetylation
(b) alkylation
(c) reaction with HNO₂
(d) carbyl amine test
Answer:
(d) carbyl amine test

18. The amine which reacts with NaNO₂ and dil. HCl to give yellow oily compound is
(a) ethylamine
(b) isopropylamine
(c) sec-butylamine
(d) dimethylamine
Answer:
(d) dimethylamine

19. The name of the compound ‘C’ in the following series of reactions, is
(a) propan-l-ol
(b) propan-2-ol
(c) butan-l-ol
(d) butan-2-ol
Answer:
(b) propan-2-ol

20. Triethylamine when treated with nitrous acid gives
(a) an alcohol
(b) a nitrosamine
(c) a monoacetyl derivative
(d) a soluble nitrite salt
Answer:
(d) a soluble nitrite salt

21. Ammes are basic in nature because
(a) of the nitrogen atom contain or lone pair of electrons
(b) they give H+ ions in aqueous medium
(c) they form quaternary ammonium salts when heated with acids
(d) both (a) and (c)
Answer:
(a) of the nitrogen atom contain or lone pair of electrons

22. An aqueous solution of primary amine contains

Answer:
(d)

23. The basic nature of amines in an aqueous solution is in the order of
(a) tert. > sec. > pri.
(b) sec. > pri. > tert.
(b) pri. > sec. > tert.
(d) pri. > tert. > sec.
Answer:
(b) pri. > sec. > tert.

24. In trimethyl ammonium ion, the number of sigma bonds attached to nitrogen are
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

25. The number of coordinate bond/bonds in a trialkyl ammonium ion is
(a) one
(b) two
(c) three
(d) four
Answer:
(a) one

26. The number of electrons in the valence shell of nitrogen in methyl amine is
(a) 5
(b) 3
(c) 8
(d) 7
Answer:
(c) 8

27. Ethanamine reacts with excess of acetyl chloride to form
(a) C₂H₅NHCOCH₃
(b) C₂H₅N(CH₃)₂
(c) C₂H₅N(COCH₃)₂
(d) C₂H₅N+H₃Cl
Answer:
(c) C₂H₅N(COCH₃)₂

28. The compound used for acylation of amine is
(a) (CH₃CO)₂O
(b) CH₃COOH
(c) CH₃COCl
(d) both (a) and (c)
Answer:
(d) both (a) and (c)

29. Dimethyl amine reacts with acetyl chloride to give
(a) N-acetyl methyl amine
(b) N-acetyl ethyl amine
(c) N-acetyl dimethyl amine
(d) N-acetyl diethyl amine
Answer:
(c) N-acetyl dimethyl amine

30. Identify ‘A’ in the following reaction :

Answer:
(c)

31. n-propyl alcohol is obtained when HNO₂ is treated with

Answer:
(c)

32. A mixture of CH₃NH₂, (CH₃)₂NH, (CH₃)₃N can be distinguished by using
(a) HCI
(b) HNO₂
(c) HNO₃
(d) H₂SO₄
Answer:
(b) HNO₂

33. In the acetylation reaction the H-atom of an amine is replaced by
(a) a carbonyl group
(b) an alkyl group
(c) an acetyl group
(d) an imino group
Answer:
(c) an acetyl group

34. Amines are basic in nature
(a) as they have a fishy odour
(b) as they form quaternary ammonium salts with alkyl halides
(c) due to the presence of an unshared pair of electrons on the nitrogen atom
(d) all of these
Answer:
(c) due to the presence of an unshared pair of electrons on the nitrogen atom

35. The correct order of increasing basic strength is
(a) NH₃ < CH₃NH₂ < (CH₃)₂NH
(b) CH₃NH₂ < (CH₃)₂NH < NH₃
(c) CH₃NH₂ < NH₃ < (CH₃)₂NH
(d) (CH₃)₂NH < NH₃ < CH₃NH₂
Answer:
(a) NH₃ < CH₃NH₂ < (CH₃)₂NH

36. Which of the following is the strongest base?

Answer:
(d)

37. Identify the weakest base amongst the following :
(a) p-methoxyaniline
(b) o-toluidine
(c) benzene-1, 4-diamine
(d) 4-aminobenzoic acid
Answer:
(d) 4-aminobenzoic acid

38. Amine that cannot be prepared by Gabriel phthalimide synthesis is
(a) aniline
(b) benzyl amine
(c) methyl amine
(d) iso-butyl amine
Answer:
(a) aniline

39. Which of the following exist as Zwitter ion?
(a) Salicylic acid
(b) Suphanilic acid
(c) p-Aminophenol
(d) p-Amino acetophenone
Answer:
(b) Suphanilic acid

40. Reduction of benzene diazonium chloride with Zn/HCl gives
(a) phenyl hydrazine
(b) hydrazine hydrate
(c) aniline
(d) ozo benzene
Answer:
(c) aniline

41. When primary amine reacts with CHCl₃ in alcoholic KOH, the product is
(a) aldehyde
(b) alcohol
(c) cyanide
(d) an isocyanide
Answer:
(d) an isocyanide

42. Which of the following amines cannot be prepared by Gabriel phthalimide synthesis?
(a) sec-Propylamine
(b) tert-Butylamine
(c) 2-Phenylethylamine
(d) N-Methyl benzyl amine
Answer:
(d) N-Methyl benzyl amine

43. Which of the following compounds has highest boiling point?
(a) Ethane
(b) Ethanoic acid
(c) Ethanol
(d) Ethanamine
Answer:
(b) Ethanoic acid

44. Identify the statement about the basic nature of amines.
(a) Alkylamines are weaker bases than ammonia.
(b) Arylamines are stronger bases than alkylamines.
(c) Secondary aliphatic amines are stronger bases than primary aliphatic amines.
(d) Tertiary aliphatic amines are weaker bases than arylamines.
Answer:
(c) Secondary aliphatic amines are stronger bases than primary aliphatic amines.

45. The compounds ‘A’, ‘B’ and ‘C’ react with methyl iodide to give finally quaternary ammonium iodides. Only ‘C’ gives carbylamines test while only ‘A’ forms yellow oily compound on reaction with nitrous acid. The compounds ‘A’, ‘B’ and ‘C’ are respectively.
(a) butan-1-amine, N-ethylethanamine and
N, N-dimethylethanamine.
(b) N-ethylethanamine, N, N-dimethylethanamine and butan-1 – amine.
(c) N, N-dimethylethanamine, N-ethylethanamine and butan-1-amine.
(d) N-ethylethanamine, butan-1-amine and N-ethylethanamine.
Answer:
(b) N-ethylethanamine, N, N-dimethylethanamine and butan-1 – amine.

46. Which of the following amines is most basic in nature?
(a) 2, 4-Dichloroaniline
(b) 2, 4-Dimethylaniline
(c) 2, 4-Dinitroaniline
(d) 2, 4-Dibromoaniline
Answer:
(b) 2, 4-Dimethylaniline

47. How many moles of methyl iodide are required to convert ethylamine, diethylamine and triethylamine into quaternary ammonium salt, respectively?
(a) 1, 2 and 3
(b) 2, 3 and 1
(c) 3, 2 and 1
(d) 3, 1 and 2
Answer:
(c) 3, 2 and 1

48. Which of the following amines does not undergo acetylation?
(a) t-Butylamine
(b) Ethylamine
(c) Diethylamine
(d) Triethylamine
Answer:
(d) Triethylamine

49. n-Propylamine can be prepared by catalytic reduction of
(a) n-propyl cyanide
(b) propionaldoxime
(c) acetoxime
(d) nitroethane
Answer:
(b) propionaldoxime

50. Identify the compound ‘B’ in the following series of reactions :

Answer:
(c)

51. Chloropicrin is used as
(a) antiseptic
(b) antibiotic
(c) insecticide
(d) anaesthetic
Answer:
(c) insecticide

52. Identify the compound B in the following series of reactions.
(a) n-propyl chloride
(b) propanamine
(c) n-propyl alcohol
(d) Isopropyl alcohol
Answer:
(c) n-propyl alcohol

53. Which of the following amines yields foul smelling product with haloform and alcoholic KOH?
(a) Ethyl amine
(b) Diethyl amine
(c) Triethyl amine
(d) Ethyl methyl amine
Answer:
(a) Ethyl amine

54. Which of the following compounds is NOT prepared by the action of alcoholic NH₃ on alkyl halide?
(a) CH₃NH₂
(b) CH₃-CH₂-NH₂
(c) CH₃ – CH₂ – CH₂ – NH₂
(d) (CH₃)₃CNH₂
Answer:
(d) (CH₃)₃CNH₂

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 8 Respiration and Circulation Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 8 Respiration and Circulation

1. Multiple choice questions

Question 1.
The muscular structure that separates the thoracic and abdominal cavity is …………………..
(a) pleura
(b) diaphragm
(c) trachea
(d) epithelium
Answer:
(b) diaphragm

Question 2.
What is the minimum number of plasma membrane that oxygen has to diffuse across to pass from air in the alveolus to haemoglobin inside a R.B.C.?
(a) two
(b) three
(c) four
(d) five
Answer:
(a) two

Question 3.
…………………. is a sound producing organ.
(a) Larynx
(b) Pharynx
(c) Tonsils
(d) Trachea
Answer:
(a) Larynx

Question 4.
The maximum volume of gas that is inhaled during breathing in addition to T.V. is …………………..
(a) residual volume
(b) IRV
(c) GRV.
(d) vital capacity
Answer:
(b) IRV

Question 5.
………………….. muscles contract when the external intercostals muscles contract.
(a) Internal abdominal
(b) Jaw
(c) Muscles in bronchial walls
(d) Diaphragm
Answer:
(d) Diaphragm

Question 6.
Movement of cytoplasm in unicellular organisms is called …………………..
(a) diffusion
(b) cyclosis
(c) circulation
(d) thrombosis
Answer:
(b) cyclosis

Question 7.
Which of the following animals do not have closed circulation?
(a) Earthworm
(b) Rabbit
(c) Butterfly
(d) Shark
Answer:
(c) Butterfly

Question 8.
Diapedesis is performed by …………………..
(a) erythrocytes
(b) thrombocytes
(c) adipocytes
(d) leucocytes
Answer:
(d) leucocytes

Question 9.
Pacemaker of heart is …………………..
(a) SA node
(b) AV node
(c) His bundle
(d) Purkinje fibers
Answer:
(a) SA node

Question 10.
Which of the following is without nucleus?
(a) Red blood corpuscle
(b) Neutrophil
(c) Basophil
(d) Lymphocyte
Answer:
(a) Red blood corpuscle

Question 11.
Cockroach shows which kind of circulatory system?
(a) Open
(b) Closed
(c) Lymphatic
(d) Double
Answer:
(a) Open

Question 12.
Diapedesis can be seen in …………………..
(a) RBC
(b) WBC
(c) Platelet
(d) neuron
Answer:
(b) WBC

Question 13.
Opening of inferior vena cava is guarded by …………………..
(a) bicuspid valve
(b) tricuspid valve
(c) Eustachian valve
(d) Thebesian valve
Answer:
(c) Eustachian valve

Question 14.
…………………. wave in ECG represent atrial depolarization.
(a) P
(b) QRS complex
(c) Q
(d) T
Answer:
(a) P

Question 15.
The fluid seen in the intercellular spaces in Human is …………………..
(a) blood
(b) lymph
(c) interstitial fluid
(d) water
Answer:
(b) lymph

2. Match the columns

Question 1.
Respiratory surface Organism

Respiratory surface	Organism
(1) Plasma membrane	(a) Insect
(2) Lungs	(b) Salamander
(3) External gills	(c) Bird
(4) Internal gills	(d) Amoeba
(5) Trachea	(e) Fish

Answer:

Respiratory surface	Organism
(1) Plasma membrane	(d) Amoeba
(2) Lungs	(c) Bird
(3) External gills	(b) Salamander
(4) Internal gills	(e) Fish
(5) Trachea	(a) Insect

3. Very Short Answer Questions

Question 1.
Why does trachea have ‘C’-shaped rings of cartilage?
Answer:
Trachea is supported by ‘C’-shaped rings of J cartilage which prevent it from collapsing and always keep it open.

Question 2.
Why is respiration in insect called direct respiration?
Answer:
Respiration in insect is called direct because tracheal tubes exchange O₂ and CO₂ directly with the haemocoel which then exchange them with tissues.

Question 3.
Why is gas exchange very rapid at alveolar level?
OR
Why does gas exchange in the alveolar region very rapid?
Answer:
Gas exchange is very rapid at alveolar level because numerous alveoli (about 700 millions) in the lungs provide large surface area for gaseous exchange.

Question 4.
Name the organ which prevents the entry of food into the trachea while eating.
Answer:
Epiglottis prevents the entry of food into trachea while eating.

4. Short Answer Questions

Question 1.
Why is it advantageous to breathe through the nose than through the mouth?
Answer:
Breathing through nose is better than breathing through the mouth because of the following reasons:

1. The nostrils are smaller than the mouth so air exhaled through the nose creates a backflow of air into the lungs.
2. As we exhale more slowly through the nose than we do through the mouth, the lungs have more time to extract oxygen from the air that we have already taken in.
3. The hairs inside nostrils filter any dust particles and microbes in the air and it only lets the clean air pass through.
4. The air gets warm and humidified in nostrils as it passes into our bodies.
5. Moreover breathing through the mouth can dry the oral cavity and lead to bad breath, gum disease and tooth decay.

Question 2.
Identity the incorrect statement and correct it.
(a) A respiratory surface area should have a. large surface area.
(b) A respiratory surface area should be kept dry.
(c) A respiratory surface area should be thin, may be 1 mm or less.
Answer:
Statement (a) and statement (c) are correct whereas statement (b) is incorrect. A respiratory surface area should be kept moist, is the correct statement.

Question 3.
Given below are the characteristics of some modified respiratory movement. Identify them.
a. Spasmodic contraction of muscles of expiration and forceful expulsion of air through nose and mouth.
Answer:
Sneezing

b. An inspiration followed by many short convulsive expiration accompanied by facial expression.
Answer:
Laughing, Crying.

Question 4.
Blood plasma.
Answer:

1. Plasma is a straw coloured, slightly alkaline viscous fluid part of the blood, having 90-92% water and 8-10% soluble proteins.
2. Serum albumin, serum globulin, heparin, fibrinogen and prothrombin are the plasma proteins which form 7% of the plasma.
3. Glucose, amino acids, fatty acids and glycerol are the nutrients dissolved in plasma.
4. Nitrogenous wastes (urea, uric acid, . ammonia and creatinine) and respiratory gases (oxygen and carbon dioxide) is present in plasma.
5. Enzymes and hormones too are transported Ada plasma.
6. Inorganic minerals are also present in plasma such as bicarbonates, chlorides, phosphates and sulphates of sodium, potassium, calcium and magnesium.

Question 5.
Blood clotting/Coagulation of blood.
OR
Explain blood clotting in short.
Answer:

1. The process of converting the liquid blood into a semisolid form is called blood clotting or coagulation.
2. The process of clotting may be initiated by contact of blood with any foreign surface (intrinsic process) or with damaged tissue (extrinsic process).
3. Intrinsic and extrinsic processes involve interaction of various substances called clotting factors by a step wise or cascade mechanism.
4. There are in all twelve clotting factors numbered as I to XII (factor VI is not in active use).
5. Interaction of these factors in a cascade manner leads to formation of enzyme, Thromboplastin which helps in the formation of enzyme prothrombinase.
6. Prothrombinase inactivates heparin and also converts inactive prothrombin into active thrombin.
7. Thrombin converts soluble blood protein- fibrinogen into insoluble fibrin. Fibrin forms a mesh in which platelets and other blood cells are trapped to form the clot.
8. These reactions occur in 2 to 8 minutes. Therefore, clotting time is said to be 2 to 8 minutes.

Question 6.
Describe pericardium.
Answer:

1. Pericardium is the double layered peritoneum that encloses the heart. It consists of two layers, viz. fibrous pericardium and serous pericardium.
2. Fibrous pericardium is the outer layer having tough, inelastic fibrous connective tissue whereas serous pericardium is the v inner double layered membrane. It has in turn an outer parietal layer and inner visceral layer.
3. Parietal layer of serous pericardium lies on the inner side of fibrous pericardium.
4. Visceral layer also known as epicardium adheres to heart and thus forms outer covering over the heart.
5. There is a pericardial fluid in the pericardial space which is present in between the parietal and visceral layers of serous pericardium.

Question 7.
Describe valves in the human heart.
Answer:
Human heart has following main valves:

1. Tricuspid valve : Tricuspid valve is present between the right atrium and right ventricle. It has three cusps or flaps. It prevents the backflow of blood into right atrium.
2. Bicuspid valve : Bicuspid valve, also called mitral valve is present between the left atrium and left ventricle. It has two flaps. It prevents the backflow of blood in left atrium. Both tricuspid and bicuspid valves are attached to papillary muscles with tendinous chords or chordate tendinae to prevent valves from turning back into atria at the time of systole.
3. Semilunar valve : These are present at the opening of pulmonary artery and systemic aorta. They prevent the back flow of blood when ventricles undergo systole.
4. Thebesian valve : Thebesian valve is present at the opening of coronary sinus.
5. Eustachian valve : Eustachian valve is present at the opening of inferior vena cava.

Question 8.
What is the role of papillary muscles and chordae tendinae in human heart?
Answer:

1. Papillary muscles are large and well- developed muscular ridges present along the inner surface of the ventricles.
2. Bicuspid and tricuspid valves are attached to papillary muscles of ventricles by chordae tendinae.
3. Chordae tendinae are inelastic fibres present in the lumen of ventricles.
4. The chordae tendinae prevent the valves from turning back into the atria during the contraction of ventricles and regulate the opening and closing of bicuspid and tricuspid valves.

Question 9.
Explain in brief the factors affecting blood pressure.
Answer:

1. Cardiac output : Normal cardiac output is 5 lit/min. Increase in cardiac output increases systolic pressure.
2. Peripheral resistance : Peripheral resistance depends upon the diameter of blood vessels. Decrease in diameter of arterioles and capillaries under the effect of vasopressin cause increase in peripheral resistance and thereby increase in blood pressure.
3. Blood volume : Loss of blood in accidents decreases blood volume and thus cause decrease in blood pressure.
4. Viscosity of blood : Blood pressure is directly proportional to viscosity of blood.
5. Age : Blood pressure increases with age due to increase in inelasticity of blood vessels.
6. Venous return : Amount of blood brought to the heart via the veins per unit time is called the venous return and it is directly proportional to blood pressure.
7. Length and diameter of blood vessels : Blood pressure is directly proportional to the total length of the blood vessel. Blood pressure can also be affected by vasoconstriction or vasodilation.
8. Gender : Females have slightly lower BP than males of her age before menopause. However, the risk of high B. P increases in the females after menopause sets in.

5. Give Scientific Reason

Question 1.
Closed circulation is more efficient than open circulation.
Answer:

1. Closed circulation considerably enhances the speed, precision and efficiency of circulation.
2. The blood flows more rapidly, it takes less time to circulate through the closed system and return to the heart.
3. This fastens the supply and removed of materials to and from the tissues by the blood as compared to open circulation.
4. In open circulation, there are no blood vessels such as arteries or veins, to pump the blood.
5. Therefore, the blood pressure is very low.
6. Organisms with an open circulatory system typically have a relatively high volume of hemolymph and low blood pressure. Closed circulation is thus more efficient than open circulation.

Question 2.
Human heart is called as myogenic and autorhythmic?
Answer:

1. The heart shows auto rhythmicity because the impulse for its rhythmic movement develops inside the heart. Such heart is called myogenic.
2. Some of the cardiac muscle fibres become auto rhythmic (self-excitable) and start generating impulse during development.
3. These autorhythmic fibres perform two important function, viz. acting as a pacemaker and setting the rhythm for heart.
4. They also form conducting system for conduction of nerve impulses throughout the heart muscles.

Question 3.
In human heart, the blood flows only in one direction.
Answer:

1. In veins there are valves, which prevent the back flow of the blood.
2. In arteries, blood flows with unidirectional pressure.
3. Hence the circulation takes place only in one direction.

Question 4.
Arteries are thicker than veins.
Answer:

1. Arteries have relatively thick walls to enable them to withstand the high pressure of blood ejected from the heart.
2. Arteries expand when the pressure increases as the heart pushes blood out but then recoil (shrink) Wn the pressure decreases when the heart relaxes between heartbeats.
3. This expansion and recoiling occurs to maintain a smooth blood flow.
4. Veins, on the other hand, have thinner walls and larger lumen veins have no need for thick walls as then need not have to withstand high pressure like arteries.
5. Moreover, as veins transport relatively low pressure blood, they are commonly equipped with valves to promote the unidirectional flow of blood towards the heart.

Question 5.
Left ventricle is thick than all other chambers of heart.
OR
Left ventricle has thicker wall than the right ventricle.
Answer:

1. Left ventricle pumps oxygenated blood to all parts of the body. Therefore, there is greater pressure from the blood in left ventricle.
2. Right ventricle sends deoxygenated blood to lungs for oxygenation. This does not put more pressure and lungs are in vicinity of the heart.
3. Due to these functional differences between the two ventricles, left ventricle has thicker wall than that of the right ventricle.

6. Distinguish Between

Question 1.
Open circulation and Closed circulation
Answer:

Open circulation	Closed circulation
1. In open circulation, blood flows through large open spaces and channels called lacunae and haemocoels among the tissues.	1. In closed circulation, blood flows through a network of blood vessels all over the body.
2. Tissues are in direct contact with the blood.	2. Blood does not come in direct contact with tissue.
3. Blood flows with low pressure and usually does not contain any respiratory pigment like haemoglobin.	3. Blood flows with high pressure and contains respiratory pigment like haemoglobin.
4. Exchange of material takes place directly between blood and cells or tissues of the body.	4. Exchange of material takes place between blood and body tissues through an intermediate fluid called lymph.
5. Volume of blood flowing through a tissue cannot be controlled as blood flows out in open space.	5. Volume of blood can be regulated by the contraction and relaxation of the smooth muscles of the blood vessels.
6. Open circulatory system is found in arthropods and some molluscs.	6. Closed circulatory system is found in annelids, echinoderms and all vertebrates.

Question 2.
Arteries and veins.
Answer:

Arteries	Veins
1. The blood vessels that arise from the heart and carry blood away from heart are called arteries.	1. The blood vessels that bring blood to the heart are called veins.
2. Arteries are thick walled blood vessels, situated in deep layers in the body.	2. Veins cure thin walled blood vessels, situated superficially in the body.
3. Arteries do not have valves.	3. Veins have valves.
4. Tunica adventitia, the outermost layer of arteries is thick and elastic.	4. Tunica externa, the outermost layer of veins is thin.
5. Tunica media is very thick and contain elastic fibres.	5. Tunica media is thin layer and contain involuntary muscle fibres.
6. The lumen of arteries is small.	6. The lumen of the veins is very spacious.
7. With the exception of pulmonary arteries, all other arteries carry oxygenated blood.	7. With the exception of pulmonary veins, all other veins carry deoxygenated blood.
8. Blood in the arteries show high blood pressure.	8. Blood in the veins show lesser blood pressure.

Question 3.
Blood and Lymph.
Answer:

Blood	Lymph
1. Contains blood plasma with proteins and all three types of blood cells namely RBCs, WBCs and blood platelets.	1. Contains blood plasma without blood proteins, RBCs and platelets and contains lymphocytes.
2. Red in colour due to presence of RBCs.	2. Light yellow in colour and does not contain RBCs.
3. Carries oxygen in the body.	3. Does not carry oxygen.
4. The flow of blood in blood vessels is fast.	4. The flow of lymph in lymph capillaries is slow.
5. Lymphocytes are present.	5. Lymphocytes are present, more in number than those present in the blood.

Question 4.
Blood capillary and Lymph capillary.
Answer:

Blood capillary	Lymph capillary
1. Reddish, easy to observe.	1. Colourless, difficult to observe.
2. Joined to arterioles at one end and to venules at another end.	2. Blind (closed at the tip).
3. Narrower than lymph capillaries.	3. Wider than blood capillaries.
4. Wall consists of normal endothelium and basement membrane.	4. Wall consists of thin endothelium and poorly developed basement membrane.
5. Contains red blood.	5. Contains colourless lymph.
6. Have relatively high pressure.	6. Have relatively low pressure.

Question 5.
Intrinsic and Extrinsic process of clotting.
Answer:

Intrinsic process	Extrinsic process
1. The intrinsic pathway requires only clotting factors found within the blood itself – in particular, clotting factor XII (Hageman factor) from the platelets.	1. The extrinsic pathway is initiated by factors external to the blood, in the tissues adjacent to damaged blood vessel – in particular, it is initiated by clotting factor III, thromoboplastin from the damaged tissues.
2. It is a longer, multistep process and it takes a little longer for the blood to clot by this mechanism.	2. It involves fewer chemical reaction steps and produce a clot a little more quickly than the intrinsic pathway.

7. Long Answer Questions

Question 1.
Smita was working in a garage with the doors closed and automobiles engine running. After some time she felt breathless and fainted. What would be the reason? How can she be treated
OR
While working with the car engine in a closed garage, John suddenly felt dizzy and fainted what is the possible reason?
Answer:

1. As Smita and John were working with the car engine running in a closed garage, they must be suffering from carbon monoxide poisoning.
2. Carbon monoxide (CO) is a highly toxic gas produced when fuels burn incompletely from automobile engines.
3. Because of strong affinity of haemoglobin with carbon monoxide, it readily combines with carbon monoxide to from a stable compound, carboxyhaemoglobin. Thus, less haemoglobin is available for oxygen transport depriving the cells of oxygen.
4. Exposure to carbon monoxide can usually leads to throbbing headache, drowsiness, breathlessness and often person gets fainted. In extreme cases carbon monoxide poisoning usually leads to unconsciousness, convulsions, cardiovascular failure, coma and eventually death.

The breathless persons can be treated by following method:

1. Oxygen treatment : The best way to treat carbon monoxide poisoning is to breathe in pure oxygen (high-dose oxygen treatment)
2. Oxygen chamber : Doctor may temporarily place her in a pressurized oxygen chamber (also known as a hyperbaric oxygen chamber)

Question 2.
Shreyas went to a garden on a wintry morning. When he came back, he found it difficult to breath and stated wheezing. What could be the possible condition and how can he be treated?
Answer:
(1) It indicates that Shreyas might be suffering from allergic reactions. He may have come in contact with allergens such as pollen, dust, pet dander or other environmental substances on his way in the garden. Or Shreyas may be already a patient of Asthma and his symptoms may have aggrevated due to wintry climate.

(2) If a person is allergic to a substance, such as pollen, his immune system reacts to the substance as if it was foreign and harmful, and tries to destroy it.

(3) The body reacts to these allergens by making and releasing substances known as IgE antibodies. These IgE antibodies attach to most cells in the body which release histamine. Histamine is the main substance responsible for pollen allergy symptoms such as difficulty in breathing, wheezing, sneezing, itchy throat, etc.

(4) Treatment : There are several drugs to treat the allergic reactions:

• Antihistamines such as cetirizine or diphenhydramine.
• Decongestants, such as pseudoephedrine or oxymetazoline.
• Medications that combine an antihistamine and decongestant such as Actifed and Claritin-D.

Question 3.
Why can you feel a pulse when you keep a finger on the wrist or neck but not when you keep them on a vein?
Answer:
(1) When the heart contracts, it creates pressure that pushes blood out of heart. This pressure acts like a wave. This “wave” of pressure is the pulse you feel. But this pressure is not constant.

(2) When the heart pumps the blood out of it at the time of systole, there is maximum pressure in the arteries. This pressure weakens considerably when it reaches capillaries, and so the veins which are away from the heart are under less pressure. Due to low pressure veins have valves to prevent backflow of blood.

(3) The pressure in the arteries can be felt every time the heart beats, especially in arteries which come to surface of the body like that of the wrist and neck but not in veins.

(4) The pressure in veins is always weaker than in arteries, resulting in a weaker pulse to the point that it is undetectable by touch
alone.

(5) Owing to this, when we keep finger on the arteries of wrist or neck, we feel a pulse but not when we keep it on a vein.

Question 4.
A man’s pulse rate is 68 and cardiac output is 5500 cm3. Find the stroke volume.
Answer:
Cardiac output is the volume of blood pumped out per min for a normal adult human being it is calculated as follows:
Cardiac output = Heart rate × Stroke volume
Given : Cardiac output = 5500 cm³
Pulse rate = Heart rate = 68
By using these values stroke volume of is calculated as follows:
∴ Cardiac output = Heart rate × Stroke volume
∴ Stroke volume = Cardiac output/Heart rate
= 5500/68
= Approx. 80. ∴ Stroke volume is 80 ml.

Question 5.
Which blood vessel leaving from the heart will have the maximum content of oxygen and why?
Answer:

1. The Aorta leaving the heart from left ventricle carry the maximum content of oxygen.
2. Deoxygenated blood becomes oxygenated in the pulmonary capillaries surrounding the alveoli of lungs. The oxygenated blood from lungs is collected by the four pulmonary veins.
3. These pulmonary veins carry that oxygenated blood to left atrium of heart. During atrial systole that blood is carried to left ventricle.
4. Left ventricle then pumps that oxygenated blood to Aorta during ventricular systole. Therefore, aorta has the maximum content of oxygen.

Question 6.
If the duration of the atrial ‘systole is 0.1 second and that of complete diastole is 0.4 second, then how does one cardiac cycle complete in 0.8 second?
Answer:

1. The time duration required to complete one cardiac cycle is 0.8 second.
2. Cardiac cycle is divided into three important phases, viz, atrial systole, ventricular systole and joint diastole.
3. Atrial systole in normal condition lasts for 0.1 second, ventricular systole follows atrial systole and lasts for 0.3 second whereas joint diastole or complete diastole lasts for about 0.4 second.
4. In this way one cardiac cycle is completed in 0.8 second.

Question 7.
How is blood kept moving in the large veins of the legs?
Answer:
1. When heart undergoes systole, it pushes the blood with pressure in aorta. This pressure moves the entire circulation of the blood throughout the body. Aorta gives rise to dorsal aorta after supplying to upper parts of body. Then it divides into two arteries which enter two legs. The blood is forced to move in the legs due to blood pressure and also aided by gravity.

2. In addition, the muscles in legs help transport blood back to our heart. As the muscles of our body contract and relax to move our limbs, they squeeze the blood in veins and the blood is then pushed towards the heart.

3. The veins in legs also have valves to keep this process going and prevent blood from flowing back down towards the feet.

4. In this way blood is kept moving in the large veins of the legs.

Question 8.
Describe histological structure of artery, vein and capillary.

Answer:
Histological structure of artery and vein.

1. Artery is a thick walled blood vessel that carries oxygenated blood. (Exception is pulmonary artery which carries deoxygenated blood from heart to lungs for oxygenation.)
2. All the arteries arise from heart and carry blood away from the heart.
3. Each artery is made up of three layers, viz. tunica externa, tunica media and tunica interna.
4. Tunica externa or adventitia is the thickest layer of all. It is the outermost coat made up of connective tissue with elastic and collagen fibres.
5. Tunica media is the middle coat made up of smooth muscle fibres and elastic fibres. It withstands high blood pressure during ventricular systole. It is also thick.
6. Tunica interna or intima is the innermost coat made of endothelium and elastic layer.

Histology of Capillaries:

1. Capillaries are the smallest and thinnest blood vessels. Capillaries are formed by the division and re-division of the arterioles.
2. The wall of the capillary is made up of endothelium or squamous epithelium.
3. The capillary wall is permeable to water and dissolved substances.
4. Exchange of respiratory gases, nutrients, excretory products, etc. takes place through the capillary wall.
5. Capillaries unite to form venules.

Question 9.
What is blood pressure? How is it measured? Explain factors affecting blood pressure.
Answer:
1. Blood pressure:

1. The pressure exerted by blood on the wall of the blood vessels is called blood pressure. Pressure exerted by blood on the wall of arterial wall is arterial blood pressure. Blood pressure is described in two terms viz. systolic blood pressure and diastolic blood pressure.
2. Systolic blood pressure is the pressure exerted on arterial wall during ventricular contraction (systole). For a normal healthy adult the average value is 120 mmHg.
3. Diastolic blood pressure is the pressure on arterial wall during ventricular relaxation (diastole). For a normal healthy adult it is 80 mmHg.
4. B. E = SP/DP = 120/80 mmHg. Blood pressure is normally written as 120/80 mmHg. Difference between systolic and diastolic pressure is called pulse pressure normally, it is 40 mmHg.

2. Measurement of blood pressure:

1. Blood pressure is measured with the help of an instrument called sphygmomanometer.
2. The instrument consists of inflatable rubber bag cuff covered by a cotton cloth. It is connected with the help of tubes to a mercury manometer on one side and a rubber bulb on the other side.
3. During measurement, the person is asked to lie in a sleeping position. The instrument is placed at the level of heart and the cuff is tightly wrapped around upper arm.
4. The cuff is inflated till the brachial artery is blocked due to external pressure. Then pressure in the cuff is slowly lowered till the first pulsatile sound is produced. At this moment, pressure indicated in manometer is systolic pressure. Sounds heard during this measurement of blood pressure are called as Korotkoff sounds.
5. Pressure in the cuff is further lowered till any pulsatile sound cannot be heard due to smooth blood flow. At this moment, pressure indicated in manometer is diastolic pressure an optimal blood pressure (normal) level reads 120/80 mmHg.

3. Factors affecting blood pressure:

1. Cardiac output : Normal cardiac output is 5 lit/min. Increase in cardiac output increases systolic pressure.
2. Peripheral resistance : Peripheral resistance depends upon the diameter of blood vessels. Decrease in diameter of arterioles and capillaries under the effect of vasopressin cause increase in peripheral resistance and thereby increase in blood pressure.
3. Blood volume : Loss of blood in accidents decreases blood volume and thus cause decrease in blood pressure.
4. Viscosity of blood : Blood pressure is directly proportional to viscosity of blood.
5. Age : Blood pressure increases with age due to increase in inelasticity of blood vessels.
6. Venous return : Amount of blood brought to the heart via the veins per unit time is called the venous return and it is directly proportional to blood pressure.
7. Length and diameter of blood vessels : Blood pressure is directly proportional to the total length of the blood vessel. Blood pressure can also be affected by vasoconstriction or vasodilation.
8. Gender : Females have slightly lower BP than males of her age before menopause. However, the risk of high B. P increases in the females after menopause sets in.

Question 10.
Describe human blood and give its functions.
Answer:
Blood Composition:

1. Blood is a red coloured fluid connective tissue derived from embryonic mesoderm.
2. It has two components – the fluid plasma (55%) and the formed elements i.e. blood cells (44%).
3. Plasma is a straw coloured, slightly alkaline and viscous fluid having 90% water and 10% solutes such as proteins, nutrients, nitrogenous wastes, salts, hormones, etc.
4. Blood corpuscles are of three types, viz. erythrocytes (RBCs), white blood corpuscles (WBCs) and thrombocytes (platelets).
   

(5) Red blood corpuscles or Erythrocytes:

1. Erythrocytes or red blood corpuscles. They are circular, biconcave, enucleated cells.
2. The RBC size : 7 pm in diameter and 2.5 pm in thickness.
3. The RBC count : 5.1 to 5.8 million RBCs/ cu mm of blood in an adult male and 4.3 to 5.2 million/cu mm in an adult female.
4. The average life span of RBC : 120 days.
5. RBCs are formed by the process of erythropoiesis. In foetus, RBC formation takes place in liver and spleen whereas in adults it occurs in red bone marrow.
6. The old and worn out RBCs are destroyed in liver and spleen.
7. Polycythemia is an increase in number of RBCs while erythrocytopenia is decrease in their (RBCs) number.

Functions of RBCs:

1. Transport of oxygen from lungs to tissues and carbon dioxide from tissues to lungs with the help of haemoglobin.
2. Maintenance of blood pH as haemoglobin acts as a buffer.
3. Maintenance of the viscosity of blood.

(6) White blood corpuscles / Leucocytes:

1. Leucocytes or White Blood Corpuscles (WBCs) are colourless, nucleated, amoeboid and phagocytic cells.

2. Their size ranges between 8 to 15 pm. Total WBC count is 5000 to 9000 WBCs/cu mm of blood. The average life span of a WBC is about 3 to 4 days.

3. They are formed by leucopoiesis in red bone marrow, spleen, lymph nodes, tonsils, thymus and Payer’s patches, whereas the dead WBCs are destroyed by phagocytosis in blood, liver and lymph nodes.

4. Leucocytes are mainly divided into two types, viz., granulocytes and agranulocytes.

5. Granulocytes : Granulocytes are cells with granular cytoplasm and lobed nucleus. Based on their staining properties and shape of nucleus, they are of three types, viz. neutrophils, eosinophils and basophils.

(I) Neutrophils:

1. In neutrophils, the cytoplasmic granules take up neutral stains.
2. Their nucleus is three to five lobed.
3. It may undergo changes in structure hence they are called polymorphonuclear leucocytes or polymorphs.
4. Neutrophils are about 70% of total WBCs.
5. They are phagocytic in function and engulf microorganisms.

(II) Eosinophils or acidophils:

1. Cytoplasmic granules of eosinophils take up acidic dyes such as eosin. They have bilobed nucleus.
2. Eosinophils are about 3% of total WBCs.
3. They are non-phagocytic in nature.
4. Their number increases (i.e. eosinophilia) during allergic conditions.
5. They have antihistamine property.

(III) Basophils:

1. The cytoplasmic granules of basophils take up basic stains such as methylene blue.
2. They have twisted nucleus.
3. In size, they are smallest and constitute about 0.5% of total WBCs.
4. They too are non-phagocytic.
5. Their function is to release heparin which acts as an anticoagulant and histamine that is involved in inflammatory and allergic reaction.

6. Agranulocytes : There are two types of agranulocytes, viz. monocytes and lymphocytes. Agranulocytes do not show cytoplasmic granules and their nucleus is not lobed. They are of two types, viz. lymphocytes and monocytes.
(I) Lymphocytes:

1. Agranulocytes with a large round nucleus are called lymphocyte.
2. They are about 30% of total WBCs.
3. Agranulocytes are responsible for immune response of the body by producing antibodies.

(II) Monocytes:

1. Largest of all WBCs having large kidney shaped nucleus are monocytes. They are about 5% of total WBCs.
2. They are phagocytic in function.
3. They can differentiate into macrophages for engulfing microorganisms and removing cell debris. Hence they are also called scavengers.
4. At the site of infections they are seen in more enlarged form.

(7) Thrombocytes/Platelets:

1. Thrombocytes or platelets are non- nucleated, round and biconvex blood corpuscles.
2. They are smallest corpuscles measuring about 2.5 to 5 mm in diameter with a count of about 2.5 lakhs/cu mm of blood.
3. Their life span is about 5 to 10 days.
4. Thrombocytes are formed from megakaryocytes of bone marrow. They break from these cells as fragments during the process of thrombopoiesis.
5. Thrombocytosis is the increase in platelet count while thrombocytopenia is decrease in platelet count.
6. Thrombocytes possess thromboplastin which helps in clotting of blood.
7. Therefore, at the site of injury platelets aggregate and form a platelet plug. Here they release thromboplastin due to which further blood clotting reactions take place.

(8) Functions of blood:

1. Transport of oxygen and carbon dioxide
2. Transport of food
3. Transport of waste product
4. Transport of hormones
5. Maintenance of pH
6. Water balance
7. Transport of heat
8. Defence against infection
9. Temperature regulation
10. Blood clotting/coagulation
11. Helps in healing

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Solutions Chapter 8 Company Accounts – Issue of Shares Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Book Keeping & Accountancy Solutions Chapter 8 Company Accounts – Issue of Shares

1. Objective Questions:

A. Select the appropriate answer from the alternative given below and rewrite the sentence.

Question 1.
The balance of Share Forfeiture A/c is transferred to ______________ Account after re-issue of these share.
(a) Reserve Capital
(b) Capital Reserve
(c) Profit & Loss
(d) Share Capital
Answer:
(b) Capital Reserve

Question 2.
Premium received on issue of shares is shown to ______________
(a) Liability side of Balance Sheet
(b) Asset side of Balance Sheet
(c) Profit & Loss A/c debit side
(d) Profit & Loss A/c credit side
Answer:
(a) Liability side of Balance Sheet

Question 3.
Shareholders get ______________ on shares.
(a) interest
(b) commission
(c) rent
(d) dividends
Answer:
(d) dividends

Question 4.
The document inviting to subscribe the shares of a company is ______________
(a) Prospectus
(b) Memorandum of Association
(c) Articles of Association
(d) Share certificate
Answer:
(a) Prospectus

Question 5.
As per SEBI guidelines, minimum amount payable on share application should be ______________ Nominal Value of shares.
(a) 10%
(b) 15%
(c) 2%
(d) 5%
Answer:
(d) 5%

Question 6.
When shares are forfeited the Share Capital Account is ______________
(a) credited
(b) debited
(c) neither debited nor credited
(d) None of the given
Answer:
(b) debited

Question 7.
The liability of shareholder in Joint Stock Company is ______________
(a) joint and several
(b) limited
(c) unlimited
(d) huge
Answer:
(b) limited

Question 8.
The Share Capital which a company is authorized to issue by its Memorandum of Association is ______________
(a) Nominal Capital/Authorised Capital
(b) Issued Capital
(c) Paid-up Capital
(d) Reserve Capital
Answer:
(a) Nominal Capital/Authorised Capital

Question 9.
The unpaid amount on allotment and calls may be transferred to ______________ Account.
(a) Calls-in-Advance
(b) Calls
(c) Calls-in-Arrears
(d) Allotment
Answer:
(c) Calls-in-Arrears

Question 10.
There must be provision in ______________ for forfeiture of shares.
(a) Articles of Association
(b) Memorandum of Association
(c) Prospectus
(d) Balance Sheet
Answer:
(a) Articles of Association

B. Give one word/term/phrase for each of the following statements.

Question 1.
Amount called up on shares by the company but not received.
Answer:
Calls-in-Arrears

Question 2.
Issue of share at its face value.
Answer:
Issue at par

Question 3.
The person who purchases the shares of a company.
Answer:
Shareholder

Question 4.
The form of business organisation where a huge amount of capital can be raised.
Answer:
Joint-stock company

Question 5.
The capital is subscribed by the public.
Answer:
Subscribed capital

Question 6.
The shares having preferential rights at the time of winding up of the company.
Answer:
Preference shares

Question 7.
The shares on which dividend is not fixed.
Answer:
Equity shares

Question 8.
The part of subscribed capital is not called up by the company.
Answer:
Uncalled capital

C. State true or false with reasons.

Question 1.
Directors can forfeit the shares for any reason.
Answer:
This statement is False.
After paying money on share application, When share applicant fails to pay the call money or premium on shares in spite of repeated reminders and warnings directors/company can forfeit the shares.

Question 2.
Once the application money is received, directors can immediately proceed with the allotment of shares.
Answer:
This statement is False.
Directors can proceed for allotment of shares only after receiving the minimum subscription amount of the issued amount by cheque or other instrument complying with all legal requirements.

Question 3.
Joint-stock company forms of business organisations came into existence after the industrial revolution.
Answer:
This statement is True.
As the volume and scale of trade and industry expanded, especially after the industrial revolution, a very large unit of the commercial organisation requiring large capital and greater managerial skill, called Joint-stock company came into existence.

Question 4.
Equity shareholders get a guaranteed rate of dividend every year.
Answer:
This statement is False.
One of the features of equity shares is the rate of dividend payable on equity shares keeps on changing from one year to another. So, there is no question of guaranteed dividend every year for equity shareholders.

Question 5.
The face value of shares and market value of shares is always the same.
Answer:
This statement is False.
Face value of shares means the issue price of shares while the market value of shares means the trading price of shares at the stock exchange. The face value of shares remains the same and fixed. However, market price changes as per the performance of the company. Hence face value and market value of shares is not the same.

Question 6.
Sweat shares are issued to the public.
Answer:
This statement is False.
Sweat shares are issued by a company to its directors or employees at a discount or for consideration other than cash. Sweat shares are not issued to the public.

D. State whether you agree or disagree with the following statements.

Question 1.
In the case of Pro-rata allotment the excess application money received must be refunded.
Answer:
Disagree

Question 2.
Calls-in-Advance account is shown on the asset side of the Balance Sheet.
Answer:
Disagree

Question 3.
The Authorised Capital is also known as Nominal Capital.
Answer:
Agree

Question 4.
Paid-up capital can be more than Called-up Capital.
Answer:
Disagree

Question 5.
The joint-stock company can raise a huge amount of capital.
Answer:
Agree

Question 6.
When shares are Forfeited Shares Capital Account is credited.
Answer:
Disagree

Question 7.
Directors can re-issue forfeited shares.
Answer:
Agree

Question 8.
When the issued price of a share is ₹ 12 and face value is ₹ 10, the share is said to be issued at a premium.
Answer:
Agree

Question 9.
A public limited company can issue its share without issuing its prospectus.
Answer:
Disagree

Question 10.
Shares can be issued for consideration other than cash.
Answer:
Agree

E. Answer in one sentence only.

Question 1.
What are Preference Shares?
Answer:
Preference Shares are a type of share which enjoys priority or preference over equity share for the repayment of dividends at a predetermined fixed rate and for the repayment of capital.

Question 2.
What is Registered Capital?
Answer:
Registered Capital or Authorised Capital means the maximum limit up to which a company is authorized to raise share capital.

Question 3.
What is Reserve Capital?
Answer:
Reserve Capital is that part of the subscribed capital which is reserved to be called up only at the time of winding up or liquidation of the company.

Question 4.
What is Over Subscription of Shares?
Answer:
When a company received more applications of shares than those actually offered or issued to the public, known as Over Subscription of Shares.

Question 5.
Which account is debited when share first call money is received?
Answer:
The bank account will be debited when share first call money is received.

Question 6.
When are shares allotted on a pro-rata basis?
Answer:
Shares are said to be allotted on a pro-rata basis when the applications are received for more shares than the number of shares issued and shares are allotted in the proportion to the number of shares applied for.

Question 7.
What is Forfeiture of Shares?
Answer:
When a shareholder fails to pay the call money or premium on the shares in spite of repeated reminders and warnings, the company forfeits the shares of such defaulters known as forfeiture of shares.

Question 8.
What is Calls-in-Arrears?
Answer:
Non-payment of allotment or call money by the applicants in spite of repeated reminders are called Calls-in-Arrears.

Question 9.
What do you mean by Shares Issued at Premium?
Answer:
When shareholders are supposed to pay a price higher than the face value of the shares, their shares are said to be issued at a premium.

Question 10.
What is Paid-up Capital?
Answer:
Part of the called-up capital which is actually paid by the shareholders is called Paid-up Share Capital.

F. Complete the following sentences.

Question 1.
When the face value of the share is ₹ 100 and the issued price is ₹ 120, then it is said that the shares are issued at ______________
Answer:
premium

Question 2.
______________ Capital is the capital which a company is authorized to issue by its Memorandum of Association.
Answer:
Authorized

Question 3.
The difference between Called-up Capital and Paid-up Capital is known as ______________
Answer:
Calls-in-Arrears

Question 4.
______________ shareholders get fixed rate of dividend.
Answer:
Preference

Question 5.
______________ shareholders are the real owners of the company.
Answer:
Equity

Question 6.
______________ form of business organisation in which capital is raised through the issue of shares.
Answer:
Joint-stock company

Question 7.
______________ Capital is the part of Issued capital which is subscribed by the public.
Answer:
Subscribed

Question 8.
The part of Authorised Capital which is not issued to the public is known as ______________ Capital.
Answer:
Unissued

G. Calculate the following.

Question 1.
One shareholder holding 500 equity shares paid share application money @ ₹ 3, Allotment money @ ₹ 4 per share and failed to pay a final call of ₹ 3 per share his share was forfeited calculate the amount of forfeiture.
Solution:
Amount of forfeiture = Amount received by the company (In case of non-payment of ‘calls’)
Here, shareholders paid ₹ 3 per share on application and ₹ 4 per share on the allotment on 500 shares.
So, total amount received by company = 500 × ₹ 3 + 500 × ₹ 4
= 1,500 + 2,000
= ₹ 3,500
∴ Amount of share forfeiture = ₹ 3,500.

Question 2.
10,000 equity shares of ₹ 10 each issued at a 10% premium. Calculate the total amount of share premium.
Solution:
Equity shares = 10,000
Face value = ₹ 10 per share
Premium @ 10% = 10,000 × 10 × \(\frac{10}{100}\) = ₹ 10,000
So, premium 10,000 shares of ₹ 10 each at 10% = ₹ 10,000

Question 3.
The company received excess applications for 5000 shares @ ₹ 4 per share. The application of 1000 shares was rejected and a pro-rata allotment was made. Calculate the amount of application money adjusted with allotment.
Solution:
Excess application money received for 5000 shares @ ₹ 4 per share = ₹ 20,000
Less: Application of 1000 shares rejected and money refunded = ₹ 4,000
Excess money received to be adjusted with allotment = ₹ 16,000

Question 4.
80,000 equity shares of ₹ 10 each issued and fully subscribed and called up at 20% premium. Calculate the amount of Equity Share capital.
Answer:
Equity Share capital = No. of equity shares × face value of each share
= 80,000 × ₹ 10
= ₹ 8,00,000
Note: Equity Share capital has no concern with premium or discount amount.

Question 5.
Directors issued 20,000 equity shares of ₹ 100 each at par. These were fully subscribed and called up. All money was received except one shareholder holding 100 equity shares failed to pay a final call of ₹ 20 per share. Calculate the amount of Paid-up capital of the company.
Solution:
Fully subscribed and called-up amount = 20,000 equity shares × ₹ 100 each share
= ₹ 20,00,000
But one share holder failed to pay final call of ₹ 20 per share of 100 equity shares means
Non-payment of shares = 100 equity shares × ₹ 20 per share = ₹ 2,000
∴ Total Paid-up capital amount = ₹ 20,00,000 – ₹ 2,000 = ₹ 19,98,000

Question 6.
The company sends a regret letter for 100 shares and an Allotment letter to 25,000 shareholders. Application money per ₹ 20 per share. Calculate the amount of application money that the company is refunding.
Solution:
The company sends a Regret letter for 100 shares for ₹ 20 per share application money received i.e. only that much amount the company will refund.
Amount of refund = No. of shares × Value of per share
= 100 × ₹ 20
= ₹ 2,000

Practical Problems

Question 1.
Vijay Ltd. was registered with an authorized capital of ₹ 15,00,000 divided into 1,50,000 equity shares of ₹ 10 each.
The company issued 1,00,000 equity shares of ₹ 10 each at a premium of ₹ 2 per share. The company received applications for 80,000 equity shares and was allotted the shares.
The company received application money ₹ 3 per share, allotment money ₹ 4 per share
(Including premium) and first, call money ₹ 3 per share.
The Directors have not made the final call of ₹ 2 per share. All money was received except one shareholder holding 500 shares did not pay the first call.
Show Authorised Capital, Issued Capital, Subscribed Capital, Called-up Capital,
Paid-up Capital, Calls in Arrears, and Share Premium amount in the company balance sheet.
Solution:
In the books of Vijay Ltd.
Balance Sheet as on ______________

Working Notes:
1. Bank balance at the end = Amount received on application + Amount received on allotment + Amount received on 1st call + Premium amount received
= 80,000 × 3 + 80,000 × 2 × 79,500 × 3 + 80,000 × 2
= 2,40,000 + 1,60,000 + 2,38,500 + 1,60,000
= ₹ 7,98,500

2. Directors have not made the final call of ₹ 2 per share means total called-up amount = ₹ 10 – ₹ 2 = ₹ 8

3. Calls-in-Arrears on 500 shares at ₹ 3 = ₹ 1,500 of the first call

4. Share premium on 80,000 shares @ ₹ 2 received at allotment stage i.e. share premium amount = 80,000 x ₹ 2 = ₹ 1,60,000

Question 2.
Anand Company Limited issued 1,00,000 preference shares of ₹ 10 each payable as-
On Application ₹ 4
On Allotment ₹ 3
On First call ₹ 2
On Second & Final call?
The company received applications for all these shares and received all money.
Pass Journal Entries in the books of Anand Company Ltd.
Solution:
Journal Entries in the books of Anand Company Ltd.

Question 3.
Rohini Company Limited issued 25,000 equity shares of ₹ 100 each payable as follows:
On Application ₹ 20
On Allotment ₹ 30
On First call ₹ 20
On the Second & Final call ₹ 30
The application was received for 22,000 equity shares and allotment of shares was made to them. All money was received by the company.
Pass Journal Entries in the books of Rohini Co. Ltd.
Solution:
Journal Entries in the books of Rohini Company Limited

Question 4.
Deepak Manufacturing Co. Ltd. issued a prospectus inviting applications for 1,00,000 equity shares of ₹ 10 each payable as follows :
₹ 2 on Application
₹ 4 on Allotment
₹ 2 on the First call
₹ 2 on Final call
The application was received for 1,20,000 equity shares. The Directors decided to reject excess applications and refunded application money on that. The company received all money.
Pass Journal Entries in the books of a company.
Solution:
Journal Entries in the books of Deepak Manufacturing Co.Ltd

Question 5.
Sucheta Company Limited issued ₹ 20,00,000 new capital divided into ₹ 100 equity shares at a premium of ₹ 20 per share payable as ₹ 10 on Application, ₹ 40 on Allotment and ₹ 10 premium ₹ 50 on Final call and ₹ 10 premium.
The issue was oversubscribed to the extent of 26,000 equity shares. The applicants on 2,000 shares were sent a letter of regret and their application money was refunded.
The remaining applicants were allotted shares on a Pro-rata basis. All the money due on Allotment and Final call was only received.
Make necessary Journal Entries in the books of Sucheta Company Ltd.
Answer:
Solution:
Journal Entries in the books of Sucheta Company Limited

Working Note:
Calculation of Application money transferred to Share Allotment:
Application money received (26,000 × 10) = 2,60,000
Less: Application money refunded (2,000 × 10) = 20,000
Less: Application money transferred to Share Capital: (20,000 × 10) = 2,00,000
Excess money received on application transferred to Share Allotment = 40,000
Bifurcation of calls amount:

Question 6.
Suhas Limited issued 10,000 equity shares of ₹ 10 each at a premium of ₹ 2 per share payable ₹ 3 on application, ₹ 5 (including premium) on the allotment, and the balance in two calls of an equal amount. Applications were received for 11,000 equity shares and pro-rata allotment was made for all the applicants. The excess application money was adjusted towards allotment.
Mrs. Shobha who was allowed 200 equity shares failed to pay F/F/C and her shares were forfeited after the final call.
Show Journal Entries in the books of Suhas Ltd. and also show its presence in Balance Sheet.
Solution:
Journal Entries in the books of Suhas Limited

Balance Sheet of Suhas Limited

Working Notes:
1. Excess amount received at the time of application ₹ 3,000 adjusted at allotment stage, so allotment amount received in the bank is ₹ 47,000.

2. Amount called-up per share: ₹ 3 on application, ₹ 5 (including premium) on allotment i.e. ₹ 2 premium + ₹ 3 capital and balance amount ₹ 4 in two calls of the equal amount i.e. ₹ 2 on the first call and ₹ 2 on final call.

3. Mrs. Shobha was not able to pay F/F/C i.e. first and final call means 200 × ₹ 2 first call money = ₹ 400 and 200 × ₹ 2 final call money = ₹ 400.
Mrs. Shobha paid ₹ 6 per share towards capital which the company received and the company has the right to forfeit only paid amount means the company forfeited ₹ 1,200 of Mrs. Shobha.

Question 7.
Subhash Company Limited issues 2000 Equity shares of ₹ 100 each payable as ₹ 30 on application, ₹ 30 on the allotment, ₹ 40 on first and final call.
All the shares were subscribed and duly allotted. The company made all the calls. All cash was duly received except the first and final call on 100 equity shares. These shares were forfeited by the company and were re-issued as fully paid for ₹ 75 per share.
Show the Journal Entries in the books of Subhash Company Ltd.
Solution:
Journal Entries in the books of Subhash Company Limited

Working Notes:
1. Amount forfeited by the company on 100 shares forfeited = 100 × (30 + 30)
= 100 × 60
= ₹ 6,000

2. Calls-in-Arrears = 100 × 40 = ₹ 4,000.

3. Amount received on re-issue of 100 forfeited shares = 100 × 75 = ₹ 7,500.
Balance of ₹ 2,500 (i.e. loss 25 × 100) is transferred to Share Forfeiture A/c.

4. Amount transfer from Share Forfeiture A/c to Capital Reserve is ascertained by preparing Share Forfeiture A/c.

Question 8.
Pass Journal Entries for the forfeiture and re-issue of shares in the following cases:
(A) Asha Ltd. forfeited 100 equity shares of ₹ 20 each fully called-up for non-payment of the first call of ₹ 3 per share and final call of ₹ 5 per share. 80 shares of these were re-issued at ₹ 15 per share as fully paid.
(B) Bhakti Ltd. forfeited 100 equity shares of ₹ 10 each, ₹ 6 called-upon which the shareholder paid application and allotment of ₹ 5 per share. Of these 80 shares were re-issued as fully paid-up for ₹ 16 per share.
(C) Konark Ltd. forfeited 50 shares of ₹ 10 each, ₹ 8 called-up. The shareholder failed to pay the first call of ₹ 3 per share. Later on, 30 shares of these were re-issued at ₹ 7 per share.
Solution:
Journal Entries [For Asha Ltd.]

Working Notes for A:
1. Out of 100 forfeited shares, 80 shares were re-issued accordingly Equity Share Capital A/c is debited and credited.
2. To find the proportionate amount for Forfeiture A/c:
For 100 shares-share forfeiture amount = ₹ 1,200
∴ 80 shares – share forfeiture amount = ₹ 960
Now, out of this ₹ 960 we used ₹ 400 from Share Forfeiture A/c at the time of re-issue of shares.
So, balance of Share Forfeiture A/c = ₹ 960 – ₹ 400 = ₹ 560

Journal Entries [For Bhakti Ltd.]

Working Notes for B:
1. Out of 100 forfeited shares, 80 shares were re-issued accordingly Equity Share Capital A/c is debited for ₹ 600 and credited for ₹ 480.

2. The proportionate amount debited to Forfeiture A/c:
For 100 shares-share forfeiture amount debited = ₹ 500 1 Qn
∴ 80 shares – share forfeiture amount = ₹ \(\frac{80}{100} \times \frac{500}{1}\) = ₹ 400
Now, shares were re-issued at ₹ 6 per share which is a called-up amount.
∴ The proportionate amount for Forfeiture A/c ₹ 400 will be transferred to Capital Reserve A/c.

Journal Entries (For Konark Ltd.)

Working Note for C:
The proportionate amount debited to Forfeiture A/c:
For 50 shares – share forfeiture amount debited is ₹ 250
∴ 30 shares-share forfeiture amount = ₹ \(\frac{30}{50} \times 250\) = ₹ 150
Out of this ₹ 30 used for re-issue of forfeited shares.
∴ Balance of Share Forfeiture A/c = ₹ 150 – ₹ 30 = ₹ 120.

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Solutions Chapter 7 Bills of Exchange Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Book Keeping & Accountancy Solutions Chapter 7 Bills of Exchange

Objective Questions

A. Select the correct option and rewrite the sentence:

Question 1.
The person on whom a bill is drawn is called a ______________
(a) Drawee
(b) Payee
(c) Drawer
(d) Acceptor
Answer:
(a) Drawee

Question 2.
Before acceptance the bill is called a ______________
(a) Order
(b) Request
(c) Draft
(d) Instrument
Answer:
(c) Draft

Question 3.
When the due date of the bill drawn falls due on a public holiday, the payment must be made on the ______________ day.
(a) same
(b) preceding
(c) next
(d) any
Answer:
(b) preceding

Question 4.
The due date of the bill drawn for 2 months on 23rd Nov. 2019 will be ______________
(a) 23rd Jan. 2020
(b) 25th Jan. 2019
(c) 26th Jan. 2019
(d) 25th Jan. 2020
Answer:
(d) 25th Jan. 2020

Question 5.
Noting charges are borne by ______________
(a) Notary Public
(b) Drawee
(c) Drawer
(d) Endorsee
Answer:
(b) Drawee

Question 6.
There are ______________ parties to bill of exchange.
(a) five
(b) four
(c) three
(d) two
Answer:
(c) three

Question 7.
When a bill is drawn for 2 months after date on 3rd Jan. 2020, its due date will be ______________
(a) 3rd Jan. 2020
(b) 3rd Mar. 2020
(c) 5th Mar. 2020
(d) 6th Mar. 2020
Answer:
(d) 6th Mar. 2020

Question 8.
Notary Public is ______________
(a) Govt. Officer
(b) Drawer
(c) Payee
(d) Endorsee
Answer:
(a) Govt. Officer

Question 9.
When Acceptor or Drawee does not pay the amount of bill to the holder on the due date it is known as ______________ the bill.
(a) returning
(b) discounting
(c) honouring
(d) dishonouring
Answer:
(d) dishonouring

Question 10.
The person who accepts the bill treats the bill as ______________
(a) Bills Payable
(b) Promissory Note
(c) Draft
(d) Bills Receivable
Answer:
(a) Bills Payable

B. Write the word/phrase/term, which can substitute each of the following statements:

Question 1.
Three extra days are allowed over and above the term of the bill.
Answer:
Grace days

Question 2.
Fees charged by Notary Public for getting the fact of dishonour noted.
Answer:
Noting Charges

Question 3.
A person who is entitled to receive the amount of bill of exchange.
Answer:
Payee

Question 4.
A person in whose favour a bill endorsed.
Answer:
Endorsee

Question 5.
Officer appointed by the government for noting of dishonour of bill.
Answer:
Notary Public

Question 6.
Cancellation of the bill on maturity in return for a new bill for an extended period of credit.
Answer:
Renewal of Bill

Question 7.
Bill of exchange drawn and accepted without any valuable consideration.
Answer:
Accommodation bill

Question 8.
A person who is in possession of the Bill of Exchange.
Answer:
Holder

Question 9.
Conversion of Bill of Exchange into its present value.
Answer:
Discounting of the bill

Question 10.
The amount is not recoverable from Drawee on account of insolvency.
Answer:
Bad debts

C. State whether the following statements are True or False with reasons:

Question 1.
An Inland bill is one that is drawn in one country and payable in another country.
Answer:
This statement is False.
Inland bill means, a bill drawn, accepted, and made payable within the territory of one and same, country. So, a bill is drawn in one country and payable in another country can’t be an inland bill.

Question 2.
Retirement of the bill means payment of the bill before the due date.
Answer:
This statement is True.
Payment of the bill, by the acceptor of the bill to the holder of the bill before the due date, is known as Retirement of the bill. So retirement of the bill means payment of the bill before the due date.

Question 3.
Drawee can transfer the ownership of the bill.
Answer:
This statement is False.
Drawee is a debtor. He has to pay the amount of the bill to its holder on the due date. Hence he cannot transfer its ownership to other people. The drawer can transfer the ownership of the bill as he is the owner of the bill.

Question 4.
Acceptance of the bill without making any changes in the terms of the bill is called qualified acceptance.
Answer:
This statement is False.
Acceptance of the bill with some changes as regards the terms, amount, place, etc. of a bill is known as qualified acceptance. Acceptance of the bill without making changes as regards the term is called general acceptance.

Question 5.
Discounting is a device to convert the bill into its present value.
Answer:
This statement is True.
When the drawer or holder of the bill approaches the bank to discount the bill, the bank pays the bill amount after deducting a certain amount (which is known as discounting charges). It means conversion of the bill into its present value in cash. So, we can say that discounting is a device to convert the bill into its present value.

Question 6.
A bill of exchange must be presented to the acceptor on the due date.
Answer:
This statement is True.
To get the payment of the bill from the acceptor, the holder of the bill is required to present it to the acceptor on its due date. Acceptor either honours the bill or dishonours the bill.

Question 7.
If a bill is discounted by the holder, no entry is passed in his book when the bill is honoured on the due date.
Answer:
This statement is True.
On discounting the bill the holder gives the possession of the bill to the bank. On the maturity date, the bank has to present the bill to the drawee to collect the payment. When the discounted bill is honoured, the transaction takes place between drawee and bank.

Question 8.
Noting charges are to be borne by the drawer.
Answer:
This statement is False.
Noting charges are to be borne by the drawee only as due to his act of non-payment, the bill is dishonoured and the drawer is not able to get money on its due date.

Question 9.
If a bill is drawn payable ‘on demand’ no grace days are allowed.
Answer:
This statement is True.
‘On demand’ means the amount of the bill is to be paid by drawee immediately on presentation of the bill as no time period is mentioned on it. In demand bill, 3 days grace is not allowed by law.

Question 10.
There are three parties to a promissory note.
Answer:
This statement is False.
There are only two parties to a promissory note, i.e. Drawer or maker of the note and drawee or payee of the note.

D. Find the odd one:

Question 1.
(a) Retaining
(b) Noting
(c) Discounting
(d) Endorsing
Answer:
(b) Noting

Question 2.
(a) Trade bill
(b) Accommodation bill
(c) After date bill
(d) Demand bill
Answer:
(d) Demand bill

Question 3.
(a) Notary public
(b) Drawer
(c) Drawee
(d) Payee
Answer:
(a) Notary public

Question 4.
(a) Discounting charges
(b) Rebate
(c) Bank charges
(d) Noting charges
Answer:
(d) Noting charges

Question 5.
(a) Stamp
(b) Acceptance
(c) Draft
(d) Amount
Answer:
(c) Draft

E. Complete the sentences:

Question 1.
Making payment of bill before the due date of maturity is known as ______________
Answer:
Retirement of Bill

Question 2.
A person whose liabilities are more than his assets and is not in a position to pay off his liabilities is ______________
Answer:
Insolvent person

Question 3.
Amount that cannot be paid by acceptor on account of insolvency is known as ______________
Answer:
Deficiency

Question 4.
A bill of exchange payable after certain period is known as ______________
Answer:
After date bill

Question 5.
A bill which is drawn and accepted with valuable consideration is known as ______________
Answer:
Trade Bill

Question 6.
A person who draws the bill of exchange is known as ______________
Answer:
Drawer

Question 7.
A bill whose due date is calculated from the date of acceptance is known as ______________
Answer:
After sight bill

Question 8.
Recording the fact of dishonour of bill is known as ______________
Answer:
Noting

Question 9.
When drawee accepts the bill payable at a particular place only, it is known as ______________
Answer:
qualified acceptance as to place

Question 10.
Fees charged by the bank for collection of bill on behalf of holder is ______________
Answer:
bank charges

F. Answer in a sentence:

Question 1.
What do you mean by Bill of Exchange?
Answer:
A Bill of Exchange is a written order signed by the drawer, directing a certain person to pay a certain sum of money on-demand or on a certain future date to a certain person or as per his order.

Question 2.
What are Days of Grace?
Answer:
The three extra days allowed to the drawee or the acceptor of a bill for making payment on it are called Days of Grace.

Question 3.
What do you mean by Discounting a Bill of Exchange?
Answer:
Encashment of a bill of exchange with the bank for certain cash which is less than the face value of the bill, before its due date by its drawer or holder is called Discounting of a Bill of Exchange.

Question 4.
What is Noting of the Bill?
Answer:
Noting of a Bill of Exchange is the recording of the facts of its dishonour by a Notary public.

Question 5.
What are Noting Charges?
Answer:
Noting Charges are the fees charged by the Notary public for noting the facts of dishonour on the face of the bill and in his official register.

Question 6.
What is the relationship between drawer and drawee?
Answer:
The relationship between the drawer and the drawee is that of the creditor and debtor.

Question 7.
Who is the Payee of the Bill?
Answer:
The Payee of a Bill is the person to whom the bill is made payable or in whose favour the bill is drawn.

Question 8.
What do you mean by Rebate?
Answer:
Any concession or discount in monetary terms given by the holder of the bill of exchange to the drawee or acceptor, when a bill is retired is called a Rebate.

Question 9.
What is the Legal Due Date?
Answer:
The date which is arrived at after adding three days of grace to the nominal due date is known as Legal Due Date.

Question 10.
What are Bills Payable on Demand?
Answer:
When the amount of bill is payable by a drawee on the presentation of a bill, in which time period is not mentioned and grace days are not allowed is known as Bills Payable on Demand.

G. Do you agree or disagree with the following statements:

Question 1.
A bill of exchange is a conditional order.
Answer:
Disagree

Question 2.
The party which is ordered to pay the amount is known as the payee.
Answer:
Disagree

Question 3.
The person in whose favour the bill is endorsed is known as the endorsee.
Answer:
Agree

Question 4.
Rebate or discount given on retiring a bill is an income to the Drawee.
Answer:
Agree

Question 5.
A bill from the point of view of the debtor is called Bills payable.
Answer:
Agree

Question 6.
In case of bill drawn payable ‘on demand,’ no grace days are allowed.
Answer:
Agree

Question 7.
A bill is required to be accepted by Drawer.
Answer:
Disagree

Question 8.
A bill of exchange need not be dated.
Answer:
Disagree

Question 9.
A bill before acceptance is called Promissory Note.
Answer:
Disagree

Question 10.
Renewal is requested by the drawee to extend the credit period of the bill.
Answer:
Agree

H. Calculations:

Question 1.
Ganesh draws a bill for ₹ 40,260 on 15th Jan. 2020 for 50 days. He discounted the bill with the Bank of India @ 15 % p.a. on the same day. Calculate the amount of discount.
Solution:
Discount = Amount of Bill × \(\frac{\text { Rate }}{100} \times \frac{\text { Unexpired days }}{366}\)
= 40,260 × \(\frac{15}{100} \times \frac{50}{366}\)
= ₹ 825
(Note: 2020 is a Leap year, so the total number of days = 366)

Question 2.
Shefali Traders drew a bill on Maya for ₹ 30,000 on 1st Oct. 2019 payable after 3 months.
Calculate the amount of discount in the following cases:
(i) Shefali Traders discounted the bill on the same day @ 12 % p.a.
(ii) Shefali Traders discounted the bill on 1st Nov. 2019 @ 12 % p.a.
(iii) Shefali Traders discounted the bill on 1st Dec. 2019 @ 12 % p.a.
Solution:
Discount = Amount of Bill × \(\frac{\text { Rate }}{100} \times \frac{\text { Unexpired days }}{365}\)
(i) Discount = 30,000 × \(\frac{12}{100} \times \frac{3}{12}\) = ₹ 900
(ii) Discount = 30,000 × \(\frac{12}{100} \times \frac{2}{12}\) = ₹ 600
(iii) Discount = 30,000 × \(\frac{12}{100} \times \frac{1}{12}\) = ₹ 300

Question 3.
Veena who had accepted Sudha’s bill for ₹ 28,000 was declared bankrupt and only 35 paise in a rupee could be recovered from her estate. Calculate the amount of bad debts.
Solution:
From Veena, only 35 paise in a rupee could be recovered i.e. 65 paise in a rupee is bad debt for Sudha. So 65% of ₹ 28,000 = ₹ 18,200 is the amount of bad debts.

Question 4.
Nitin renewed his acceptance for ₹ 72,000 by paying ₹ 22,000 in cash and accepting a new bill for the balance plus interest @ 18%. p.a. for 4 months. Calculate the amount of the new bill.
Selution:
For Nitin,
Total outstanding = ₹ 72,000
Nitin paid in cash= ₹ 22,000
Remaining dues = ₹ 50,000
Now, on this ₹ 50,000 we have to calculate interest @ 18% for 4 months
I = \(\frac{\mathrm{PRN}}{100}\)
= 50,000 × \(\frac{18}{100} \times \frac{4}{12}\)
= ₹ 3,000
So, amount of new bill = Remaining dues + Interest
= 50,000 + 3,000
= ₹ 53,000

Question 5.
Nisha’s acceptance for ₹ 16,850 sent to the bank for the collection was honoured and bank charges debited were ₹ 125. Find out the amount actually received by Drawer.
Solution:
Bill of ₹ 16,850 sent to the bank for collection and it is honoured and bank charges = ₹ 125
So, actual amount received by drawer = 16,850 – 125 = ₹ 16,725.

Question 6.
A bill of ₹ 16,000 was drawn by Keshav on Gopal on 12th June 2019 for 2 months, what will be the due date, if all of sudden, the legal due date is declared as an emergency holiday?
Solution:
Consider immediate or next working day as the due date in case the legal due date is declared as an emergency holiday.

∴ The legal due date is 16th August 2019 (The next day).

I. Prepare the following specimens:

Question 1.
Prepare a bill of exchange from the following information:
Drawer: Shankar, Vadodara, Gujarat
Drawee: Vinayak, Somwar Peth, Pune
Amount: ₹ 16,000
Period: 3 months
Date of Bill: 6th Sept. 2019
Date of acceptance: 11th Sept. 2019
Solution:

Question 2.
Prepare a bill of exchange from the following information:
Drawer: Dinesh, P. R. Road, Andheri (West)
Drawee: Mahesh, L. B. S. Road, Mulund
Payee: Amit, Thane (West)
Amount: ₹ 9,500
Period of Bill: 4 months after sight
Date of Bill: 26th Nov. 2019
Date of acceptance: 29th Nov. 2019
Solution:

Question 3.
Kantilal, 343/D, Palm Heights, Jogeshwari, drew a bill on 10th Oct. 2019 for ₹ 63,490 for 45 days after the date on Shantilal, B2, Himalaya Towers, Baramati, payable to Priyanka, Satara. The bill was accepted on 13th Oct. 2019 for 60 days.
Prepare a format of bill of exchange from the above details.
Solution:

Question 4.
Prepare a format of bill exchange from the following:
Rahul Sane, 86-D, Raviwar Peth, Nagpur accepted the bill drawn on him by Prithviraj, Icon Heights, Wardha for ₹ 87,000 on 30th July 2019.
The bill was drawn on 26th July 2019 for ₹ 1,00,000 for 90 days after the date.
Solution:

Question 5.
Prepare a format of bill of exchange from the following.
Drawer: Kashmira Shah, Partner M/S Shah, and Shah, 2 – C, Matruchhaya Building, Akola
Drawee: Dhanashree Traders, Bangalore Road, Belgaum (Signed by Jayshree, Partner)
Payee: M/S Janki Traders, Akola
Amount: ₹ 64,500
Period of Bill: 3 months
Date of drawing: 12th Sept. 2019
Date of acceptance: 15th Sept. 2019
Solution:

Question 6.
Prepare a format Bill of Exchange with imaginary Drawer, Drawee, Address, Amount, Dates.
Drawer: Dhanesh Shah, 24/c, Amir Mahal, Borivali, Mumbai
Drawee: Kalpana Shah, 33, Sharadashram, Dadar (West), Mumbai
Amount: ₹ 80,500
Period: 60 days
Date of the bill: 2nd December 2020
Accepted on: 5th December 2020
Solution:

J. Complete the following Table.

Question 1.

Answer:

Question 2.

Answer:

S.No.	Date of Drawing	Date of Acceptance	Tenure	Type	Nominal due Date	Legal due Date
(i)	3rd January, 2020	5th January, 2020	45 days	after date	17th Feb. 2020	20th Feb. 2020
(ii)	9th April, 2019	12th April, 2019	4 months	after sight	12th Aug. 2019	14th Aug. 2019
(iii)	23rd November, 2019	23rd November, 2019	2 months	after date	23rd Jan. 2020	25th Jan. 2020
(iv)	16th August, 2019	20th August, 2019	4 months	after sight	20th Dec. 2019	23rd Dec. 2019
(v)	23rd December, 2018	24th December, 2018	60 days	after date	21st Feb. 2019	24th Feb. 2019

Practical Problems

Question 1.
On 1st Jan., 2020 Hemant sold goods of ₹ 18,500 to Nitin. On the same date Hemant drew a bill of exchange for ₹ 18,500 at 2 months. On the due date the bill was duly honoured.
Give Journal Entries in the books of Hemant and Nitin. Prepare Hamant’s Account in the books of Nitin.
Solution:
In the books of Hemant
Journal Entries

In the books of Nitin
Journal Entries

Question 2.
Neha sold goods to Rohan ₹ 42,000 on 6th Sept. 2019. Neha drew a bill of exchange at 3 months for the amount which was accepted by Rohan. Neha discounted the bill with her bankers at ₹ 41,000. On the due date of the bill Rohan dishonoured the bill and bank paid ₹ 300 as Noting Charges.
Show Journal Entries in the books of Neha and Rohan.
Solution:
In the books of Neha
Journal Entries

In the books of Rohan
Journal Entries

Question 3.
Jyoti owes ₹ 31,000 to Swati for which she draws a bill on Jyoti for 2 months. The bill was duly accepted by Jyoti. Swati sends the bill to bank for collection. Jyoti honoured the bill on the due date and bank charges ₹ 475 as bank charges.
Give Journal Entries in the books of Swati.
Solution:
In the books of Swati
Journal Entries

Question 4.
Pankaj purchased goods of ₹ 20,000 from Omprakash on credit on 15th April, 2019. Omprakash draws After Sight bill for the amount due on Pankaj for 3 months which was accepted by Pankaj on 18th April, 2019. On 20th April, 2019 Omprakash endorsed the bill to his creditor Jagdish in full settlement of his amount ₹ 21,000. On the due date the bill was dishonoured by Pankaj.
Give Journal Entries in the books of Omprakash, Pankaj and Jagdish.
Solution:
In the books of Omprakash
Journal Entries

In the books of Pankaj
Journal Entries

In the books of Jagdish
Journal Entries

Question 5.
Siddhant sold goods to Sudhir of ₹ 43,800 on 18th March, 2019. Siddhant draws a bill on Sudhir on the same day for ₹ 43,800 for 3 months which was duly accepted by Sudhir. Siddhant discounted the bill on the same day at 8% p.a. The bill was dishonoured on the due date and Sudhir requested Siddhant to accept ₹ 13,800 and interest in cash on remaining amount at 12% p. a. Siddhant agreed and for the balance amount accepted a new bill at 2 months. Before the due date of new bill Sudhir retired the bill by paying ₹ 29,700.
Pass necessary Journal Entries in the books of Siddhant.
Solution:
In the books of Siddhant
Journal Entries

Working Notes:
1. March 18, Discount = 43,800 × \(\frac{3}{12} \times \frac{8}{100}\) = ₹ 876

2. March 21, calculation of interest balance amount:
I = \(\frac{\mathrm{PRN}}{100}\)
= 30,000 × \(\frac{12}{100} \times \frac{2}{12}\) (for 2 months on remaining amount ₹ 30,000)
= ₹ 600

3. Before due date bill was retired by Sudhir by paying ₹ 300 less which is considered as discount and as date is not given, here it is not recorded.

Question 6.
Sangeeta accepted a bill for ₹ 18,000 drawn by Geeta at 3 months. Geeta discounted the bill for ₹ 17,400. Before the due date Sangeeta approached Geeta for renewal of the bill. Geeta agreed on the condition that Sangeeta should pay ₹ 6,000 immediately and for the balance she should accept a new bill for 4 months along with interest ₹ 550. The arrangements were carried through. But on the due date of new bill Sangeeta became insolvent and 35 paise in a rupee could be recovered from her estate.
Give Journal Entries in the books of Sangeeta and prepare Sangeeta’s Account in the books of Geeta.
Solution:
In the books of Sangeeta
Journal Entries

Working Notes:
1. It is advisable to write journal entries in the books of Geeta also to get entries in ‘Sangeeta’s Account’ property.
In the books of Geeta
Journal Entries

Question 7.
Priyanka owed Meena ₹ 18,000, Priyanka accepted a bill drawn by Meena for the amount at 4 months. Meena endorsed the same bill to Sagar. Before due date Priyanka approached Meena for renewal of bill. Meena agreed on condition that ₹ 6,000 be paid immediately together with interest on the remaining amount of 8% p.a. for 3 months and Priyanka should accept a new bill for the balance amount. These arrangements were carried through. However, before the due date Priyanka became insolvent and only 50% of the amount could be recovered from her estate.
Give Journal Entries in the books of Meena.
Solution:
In the books of Meena
Journal Entries

Working Note:
Calculation of interest on remaining amount ₹ 12,000 @ 8 % p.a. and for 3 months
I = \(\frac{\mathrm{PRN}}{100}\)
= 12,000 × \(\frac{8}{100} \times \frac{3}{12}\)
= ₹ 240

Question 8.
Seema purchased goods from Roma on credit on 1st August, 2019 for ₹ 37,000. Seema accepts bill for 2 months drawn by Roma for the same amount. On the same day, Roma discounts the bill with the bank for ₹ 36,200 on 3rd August, 2019. On the due date the bill is dishonoured and Noting Charges of ₹ 160 is paid by the bank. Seema pays ₹ 19,000 and Noting Charges in cash immediately. A new bill is drawn by Roma for the balance including interest ₹ 650 for 2 months, which is accepted by Seema. The new bill is retired one month before the due date at a rebate of ₹ 300.
Give Journal Entries in the books of Seema and prepare Seema’s Account in the books of Roma.
Solution:
In the books of Seema
Journal Entries

Question 9.
Uday purchased goods from Shankar on credit for ₹ 35,000 at 10 % trade discount. Uday paid ₹ 1,500 immediately and for the balance accepted a bill for 3 months. Before due date Uday approached Shankar with a request to renew the bill. Shankar agreed but with condition that Uday should accept a new bill for 3 months including interest at 12% p.a.
Give Journal Entries in the books of Shankar.
Solution:
In the books of Shankar
Journal Entries

Working Note:
I = \(\frac{\text { PRN }}{100}\)
= 30,000 × \(\frac{3}{12} \times \frac{12}{100}\)
= ₹ 900

Question 10.
Sagar drawn an after sight bill on 21st Nov., 2019 for ₹ 21,000 at 3 months on Prasad. The bill is discounted by Sagar at 8% p.a. with his bank. On maturity. Prasad finds himself unable to make payment of the bill and requests Sagar to renew it. Sagar accepts the request and draws a new bill at one month for ₹ 21,750 including interest which was duly accepted by Prasad. Sagar deposits the bill into bank for the collection. Prasad honours the bill on the due date and bank charges ₹ 250 as bank charges.
Pass necessary Journal Entries in the books of Sagar and prepare Sagar’s Account in the books of Prasad.
Solution:
In the books of Sagar
Journal Entries

Question 11.
Journalise the following transaction in the books of Abhishek:
(a) Siddhant informs Abhishek that Vineet’s acceptance for ₹ 23,000 endorsed to Siddhant has been dishonoured. Noting Charges amounted to ₹ 430.
(b) Kajal renews her acceptance to Abhishek for ₹ 39,000 by paying ₹ 3,000 in cash and accepting a fresh bill for the balance along with interest at 11.5% p.a. for 3 months.
(c) Radhika retired her acceptance to Abhishek for ₹ 23,000 by paying ₹ 22,250 by cheque.
(d) Abhishek sent a bill of Subodh for ₹ 9,000 to bank for collection. Bank informed that the bill has been dishonoured by Subodh.
Solution:
In the books of Abhishek
Journal Entries

Working Note:
Amount of interest = 36,000 × \(\frac{3}{12} \times \frac{11.5}{100}\) = ₹ 1,035.

Question 12.
Journalise the following transaction in the books of Narendra:
(a) Narendra retires his acceptance to Upendra by paying ₹ 4,000 in cash and endorsing a bill accepted by Ramlal for ₹ 5,000.
(b) Vikram’s acceptance to Narendra ₹ 6,000 retired one month before the due date at rebate of 12% p.a.
(c) Dilip renews his acceptance to Narendra for ₹ 12,000 by paying ₹ 4,000 in cash and accepting a fresh bill for the balance plus interest at 12% p.a. for 3 months.
(d) Bank informed Narendra that, Kartik’s acceptance for ₹ 13,000 to Narendra, discounted with the bank was dishonoured and Noting Charges paid by bank ₹ 140.
Solution:
In the books of Narendra
Journal Entries

Question 13.
Journalise the following transaction in the books of Bharti:
(a) Bank informed that Amit’s acceptance for ₹ 15,750 sent to bank for collection was honoured and bank charges debited were ₹ 150.
(b) Nitin renewed his acceptance for ₹ 22,200 by paying ₹ 2,200 in cash along with interest on balance amount at 10% and accepted a fresh bill for the balance for 3 months.
(c) Dhanshri who had accepted Bharti’s bill for ₹ 17,500 was declared insolvent and only 40% of the amount due could be recovered from her estate.
(d) Discharged our acceptance to Savita for ₹ 9,450 by endorsing Pravin’s acceptance to us ₹ 9,000.
Solution:
In the books of Bharti
Journal Entries

Question 14.
Journalise the following transaction in the books of Sudha:
(a) Endorsed Sonali’s acceptance at 2 months for ₹ 6,000 in favour of Urmila and paid cash ₹ 3,500 in full settlement of her account ₹ 10,000.
(b) Discounted 2 months acceptance of Surya for ₹ 7,800 with bank at 10% p.a.
(c) Bank informed that Anuradha’s acceptance of ₹ 4,800 which was discounted was dishonoured and bank paid Noting Charges ₹ 125.
(d) Pooja honoured her acceptance of ₹ 16,400 which was deposited into bank for collection.
Solution:
In the books of Sudha
Journal Entries

Question 15.
Journalise the following transaction in the books of Mrunal:
(a) Bank informed that Aishwarya’s acceptance of ₹ 24,000 which was discounted had been dishonoured and bank paid Noting Charges ₹ 220. Bill was renewed at the request of Aishwarya for 2 months with interest of ₹ 480.
(b) Received ₹ 4,630 from private estate of Ankur who was declared insolvent against bill accepted by him for ₹ 6,000.
(c) Accepted a bill of ₹ 15,000 at 3 months drawn by Anushka for the amount due to her ₹ 20,000 and balance paid by cheque.
(d) Dishonoured our acceptance to Vivek ₹ 27,000 and Noting Charges paid by Vivek ₹ 700.
Solution:
In the books of Mrunal
Journal Entries

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Solutions Chapter 9 Analysis of Financial Statements Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Book Keeping & Accountancy Solutions Chapter 9 Analysis of Financial Statements

Objective Questions

A. Select the most appropriate alternative from those given below and rewrite the sentences:

Question 1.
Gross Profit Ratio indicates the relationship of gross profit to the ____________
(a) Net cash
(b) Net sales
(c) Net purchases
(d) Gross sales
Answer:
(b) Net sales

Question 2.
Current ratio = \(\frac{……………….}{Current liabilities}\)
(a) Quick assets
(b) Quick liabilities
(c) Current assets
(d) None of these
Answer:
(c) Current assets

Question 3.
Liquid assets = ____________
(a) Current assets + Stock
(b) Current assets – Stock
(c) Current assets – Stock + Prepaid Expenses
(d) None of these
Answer:
(b) Current assets – Stock

Question 4.
Cost of goods sold = ____________
(a) Sales – Gross profit
(b) Sales – Net profit
(c) Sales proceeds
(d) None of these
Answer:
(a) Sales – Gross profit

Question 5.
Net profit ratio is equal to ____________
(a) Operating ratio
(b) Operating net profit ratio
(c) Gross profit ratio
(d) Current ratio
Answer:
(a) Operating ratio

Question 6.
The common size statement requires ____________
(a) common base
(b) journal entries
(c) cashflow
(d) current ratio
Answer:
(a) common base

Question 7.
Bill payable is ____________
(a) long-term loan
(b) current liabilities
(c) liquid assets
(d) net loss
Answer:
(b) current liabilities

Question 8.
Generally current ratio should be ____________
(a) 2 : 1
(b) 1 : 1
(c) 1 : 2
(d) 3 : 1
Answer:
(a) 2 : 1

Question 9.
From financial statement analysis the creditors are specially interested to know ____________
(a) Liquidity
(b) Profits
(c) Sale
(d) Share capital
Answer:
(a) Liquidity

B. Give one word/term/phrase for each of the following statements.

Question 1.
The statement showing the profitability of two different periods.
Answer:
Comparative Income Statement

Question 2.
The ratio measures the relationship between gross profit and net sales.
Answer:
Gross Profit Ratio

Question 3.
Critical evaluation of financial statement to measure profitability.
Answer:
Analysis of Financial Statement

Question 4.
A particular mathematical number showing the relationship between two accounting figures.
Answer:
Ratio

Question 5.
An asset that can be converted into cash immediately.
Answer:
Liquid Asset

Question 6.
The ratio measuring the relationship between net profit and ownership capital employed.
Answer:
ROCE

Question 7.
The statement showing financial position for different periods of the previous year and the current year.
Answer:
Comparative Balance Sheet

Question 8.
Statement showing changes in cash and cash equivalent during a particular period.
Answer:
Cash Flow Statement

Question 9.
Activity related to the acquisition of long-term assets and investment.
Answer:
Financing Activities

Question 10.
The ratio that establishes a relationship between Quick Assets and Current Liabilities.
Answer:
Liquid Ratio

C. State true or false with reasons.

Question 1.
Financial statements include only the Balance Sheet.
Answer:
This statement is False.
Financial statements include Balance Sheet and Profit and Loss A/c. This is because financial statements are prepared by business organisations to find out the efficiency, solvency, profitability, growth, strength, and status of the business. For this, they need information from the balance sheet as well as from Profit and Loss A/c.

Question 2.
Analysis of financial statements is a tool but not a remedy.
Answer:
This statement is True.
Based on analysis of the financial statement one can get an idea of the financial strength and weakness of the business. However, based on this one cannot take decisions about the business on various issues. Hence analysis of financial statements is a tool but not a remedy.

Question 3.
Purchase of fixed assets is operating cash flow.
Answer:
This statement is False.
Purchase of fixed assets is cash flow from investing activities. It is not a day-to-day operations activity like office/selling/distribution finance expenses/activities.

Question 4.
Dividend paid is not a source of funds.
Answer:
This statement is True.
The dividend is always paid on shares issued by a company as an expense. Shares itself is a source of funds. In payment of dividends, cash goes out from the company. It is an outflow of cash and not a source of funds.

Question 5.
Gross profit depends upon net sales.
Answer:
This statement is True.
The gross profit ratio discloses the relation between gross profit and total net sales. The gross profit ratio is an income-based ratio, where gross profit is an income. There is a direct relation between net sales and gross profit. Higher the net sales higher the gross profit.

Question 6.
Payment of cash against the purchase of stock is the use of funds.
Answer:
This statement is True.
Cash payment for the purchase of stock is made from cash balance or/and from bank balance which is a part of the business fund. When stock or materials we purchase we use cash for payment.

Question 7.
Ratio Analysis is useful for inter-firm comparison.
Answer:
This statement is True.
The comparison of the operating performance of a business entity with the other business entities is known as an inter-firm comparison. This ratio analysis assists to know-how and to what extent a business entity is strong or weak as compared to other business entities.

Question 8.
The short-term deposits are considered as cash equivalent.
Answer:
This statement is True.
The short-term deposits are liquid assets. It means deposits are kept for some period (usually less than one year) and they are kept with an intention to get money quickly as and when required.
They are as good as cash and considered as cash equivalent.

Question 9.
Activity ratios and Turnover ratios are the same.
Answer:
This statement is True.
Turnover ratio is an efficiency ratio to check how efficiently a company is using different assets to extract earnings from them.
Activity ratios are financial analysis tools used to measure a business’s ability to convert its assets into cash. From both these definitions, we can say that Activity ratios and Turnover ratios are the same.

Question 10.
The current ratio measures the liquidity of the business.
Answer:
This statement is True.
The current ratio shows the relationship between current assets and current liabilities. If the proportion of current assets is higher than current liabilities, the liquidity position of the business entity is considered good. More liquidity means more short-term solvency. From the above, it is proved that the current ratio measures the liquidity of the business.

Question 11.
Ratio analysis measures profitability efficiency and financial soundness of the business.
Answer:
This statement is True.
With the help of profitability ratios (Gross profit, Net profit, and Operating profit) one can get the idea of profitability efficiency of the firm, and with the help of liquidity ratios (Current ratio and liquid ratio) one can get the idea of solvency or financial soundness of the business.

Question 12.
Usually, the current ratio should be 3 : 1.
Answer:
This statement is False.
Usually, the current ratio should be 2 : 1. It means current assets are double of current liabilities. It shows the short-term solvency of business enterprises.

D. Answer in one sentence only.

Question 1.
Mention two objectives of the comparative statement.
Answer:
Objectives of comparative statements are:

• Compare financial data at two points of time and
• Helps in deriving the meaning and conclusions regarding the changes in financial positions and operating results.

Question 2.
State three examples of cash inflows.
Answer:
Examples of cash inflows are:

• Interest received
• Dividend received
• Sale of asset/investment
• Rent received.

Question 3.
State three examples of cash-out flows.
Answer:
Examples of cash outflows are:

• Interest paid
• Loss on sale of an asset
• Dividend paid
• Repayment of short-term borrowings.

Question 4.
Give the formula of Gross Profit Ratio.
Answer:
Gross profit ratio = \(\frac{\text { Gross profit }}{\text { Net sales }} \times 100\)
Where Gross profit = Net sales – Cost of goods sold.
Cost of goods sold = Opening stock + Purchase – Purchase return + Direct expense – Closing stock
Net sales = Sales – Sales return.

Question 5.
Give the formula of Gross profit.
Answer:
Gross profit = Net sales – Cost of goods sold.
Cost of goods sold = Opening stock + Purchase – Purchase return + Direct expense – Closing stock
Net sales = Sales – Sales return.

Question 6.
Give any three examples of current assets.
Answer:
Cash or cash equivalent short-term lending and advances, expenses paid in advance, taxes paid in advance, etc. are examples of current assets.

Question 7.
Give the formula of the current ratio.
Answer:
Current ratio = \(\frac{\text { Current assets }}{\text { Current liabilities }}\)

Question 8.
Give the formula of quick assets.
Answer:
Quick assets = Current assets – (Stock + Prepaid expense)

Question 9.
State the formula of Cost of goods sold.
Answer:
Cost of goods sold = Opening stock + Purchase – Purchase return + Direct expense – Closing stock

Question 10.
State the formula of Average stock.
Answer:
Average stock = \(\frac{\text { Opentng stock of goods }+\text { Closing stock of goods }}{2}\)

Practical Problems

Question 1.
From the Balance Sheet of Amar Traders as of 31st March 2018 and 31st March 2019 prepare a Comparative Balance Sheet.

Solution:
Comparative Balance Sheet of Amar Traders as of 31st March 2018 and 31st March 2019


Percentage change = \(\frac{\text { Amount of absolute change }}{\text { Amount of previous year }} \times 100\)

Question 2.
From the Balance Sheet of Alpha Limited prepare a Comparative Balance Sheet as of 31st March 2018 and 31st March 2019:
Balance Sheet as of 31st March 2018 and 31st March 2019

Solution:
Comparative Balance Sheet of Alpha Limited as of 31st March 2018 and 31st March 2019

Question 3.
Prepare Comparative Balance Sheet for the year ended 31-3-18 and 31-3-19. Assets & Liabilities as follows:

Solution:
Comparative Balance Sheet as of 31st March 2018 and 31st March 2019

Question 4.
Prepare Comparative Balance Sheet for the year ended 31-3-17 and 31-3-18.

Solution:
Comparative Balance Sheet as of 31st March 2017 and 31st March 2018

Question 5.
Prepare Comparative Income statement of Noha Limited for the year ended 31-3-17 and 31-3-18.

Solution:
Comparative Income Statement of Noha Limited
For the year ended 31st March 2017 and 31st March 2018

Question 6.
Prepare Comparative Income Statement of Sourabh Limited for years ended 31-3-17 and 31-3-18.

Solution:
Comparative Income Statement of Sourabh Limited
For the year ended on 31st March 2017 and 31st March 2018

Question 7.
Following is the Balance Sheet of Sakshi Traders for the year ended 31-3-17 and 31-3-18

Prepare Common Size Balance Sheet for the years 31-03-17 and 31-03-18.
Solution:
Common Size Statement of Balance Sheet of Sakshi Traders as of 31st March 2017 and 31st March 2018

Note: Taking Total borrowed funds and Tota Funds applied as a base (100), Calculation is done.

Question 8.
Prepare Common Size Income Statement for the year ended 31-3-17 and 31-3-18.

Solution:
Common Size Statement for the year ended on 31st March 2017 and 31st March 2018

Note: Taking the amount of sales as base (100) other percentage figures are calculated.

Question 9.
Following is the Balance Sheet of Sakshi Limited. Prepare Cash Flow Statement:

Solution:
Cash Flow Statement
For the year ended 31st March 2017 and 31st March 2018

Question 10.
From the following Balance Sheet of Konal Traders prepare a Cash Flow Statement.

Solution:
Cash Flow Statement
For the year ended on 31st March 2017 and 31st March 2018

Question 11.
A company had following Current Assets and Current Liabilities:
Debtors = ₹ 1,20,000
Creditors = ₹ 60,000
Bills Payable = ₹ 40,000
Stock = ₹ 60,000
Loose Tools = ₹ 20,000
Bank Overdraft = ₹ 20,000
Calculate Current ratio.
Solution:
1. Current Assets = Debtors + Stock + Loose Tools
= 1,20,000 + 60,000 + 20,000
= ₹ 2,00,000

2. Current liabilities = Creditors + Bills payable + Bank overdraft
= 60,000 + 40,000 + 20,000
= ₹ 1,20,000

3. Current ratio = \(\frac{\text { Current assets }}{\text { Current liabilities }}\)
= \(\frac{2,00,000}{1,20,000}\)
= \(\frac{5}{3}\)
= 5 : 3

Question 12.
Current assets of company ₹ 6,00,000 and its Current ratio is 2 : 1. Find Current liabilities.
Solution:
Current ratio = \(\frac{\text { Current assets }}{\text { Current liabilities }}\)
\(\frac{2}{1}=\frac{6,00,000}{\text { Current liabilities }}\)
2 × Current liabilities = 6,00,000 × 1
Current liabilities = \(\frac{6,00,000}{2}\) = ₹ 3,00,000

Question 13.
Current liabilities = ₹ 3,00,000
Working capital = ₹ 8,00,000
Inventory = ₹ 2,00,000
Calculate Quick ratio.
Solution:
Current assets = Current liabilities + Working capital
= 3,00,000 + 8,00,000
= ₹ 11,00,000
Quick assets = Current assets – Inventory
= 11,00,000 – 2,00,000
= ₹ 9,00,000
Quick liability = Current liabilities – Bank O/D = ₹ 3,00,000
Quick ratio = \(\frac{\text { Quick assets }}{\text { Quick liabilities }}\)
= \(\frac{9,00,000}{3,00,000}\)
= \(\frac{3}{1}\)
= 3 : 1

Question 14.
Calculate Gross Profit ratio
Sales = ₹ 2,70,000
Net purchases = ₹ 1,50,000
Sales Ratio = ₹ 20,000
Closing Stock = ₹ 25,000
Operating Stock = ₹ 45,000
Solution:
Net sales = Sales – Sales return
= 2,70,000 – 20,000
= ₹ 2,50,000
Cost of goods sold = Opening stock + Net purchase – Closing stock
= 45,000 + 1,50,000 – 25,000
= ₹ 1,70,000
Gross profit = Net sales – Cost of goods sold
= 2,50,000 – 1,70,000
= ₹ 80,000
Gross Profit ratio = \(\frac{\text { Gross profit }}{\text { Net sales }} \times 100\)
= \(\frac{80,000}{2,50,000} \times 100\)
Gross profit ratio = 32%

Question 15.
Calculate Net Profit ratio from the following:
Sales = ₹ 3,80,000
Cost of goods sold = ₹ 2,60,000
Indirect expense = ₹ 60,000
Solution:
Sales = ₹ 3,80,000
Less: Cost of goods sold = ₹ 2,60,000
Gross profit = ₹ 1,20,000
Less: Indirect expense = ₹ 60,000
Net profit = ₹ 60,000
Net profit ratio = \(\frac{\text { Net profit }}{\text { Sales }} \times 100\)
= \(\frac{60,000}{3,80,000} \times 100\)
= 15.79%

Question 16.
Calculate Operating ratio:
Cost of goods sold = ₹ 3,50,000
Operating expense = ₹ 30,000
Sales = ₹ 5,00,000
Sales return = ₹ 30,000
Solution:
Net sales = Sales – Sales return
= 5,00,000 – 30,000
= ₹ 4,70,000
Operating ratio = \(\frac{\text { Cost of goods sold }+\text { Operating expense }}{\text { Net sales }} \times 100\)
= \(\frac{3,50,000+30,000}{4,70,000} \times 100\)
= \(\frac{3,80,000}{4,70,000} \times 100\)
= 80.85%

Question 17.
Calculate Current ratio.
1. Current assets = ₹ 3,00,000
2. Current liabilities = ₹ 1,00,000
Solution:
Current ratio = \(\frac{\text { Current assets }}{\text { Current liabilities }}\)
= \(\frac{3,00,000}{1,00,000}\)
= \(\frac{3}{1}\)
= 3 : 1

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 7 Elements of Groups 16, 17 and 18 Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18

1. Select appropriate answers for the following.

Question i.
Which of the following has the highest electron gain enthalpy?
A. Fluorine
B. Chlorine
C. Bromine
D. Iodine
Answer:
B. Chlorine

Question ii.
Hydrides of group 16 are weakly acidic. The correct order of acidity is
A. H₂O > H₂S > H₂Se > H₂Te
B. H₂Te > H₂O > H₂S > H₂Se
C. H₂Te > H₂Se > H₂S > H₂O
D. H₂Te > H₂Se > H₂O > H₂S
Answer:
C. H₂Te > H₂Se > H₂S > H₂O

Question iii.
Which of the following element does not show oxidation state of +4 ?
A. O
B. S
C. Se
D. Te
Answer:
A. O

Question iv.
HI acid when heated with conc. H₂SO₄ forms
A. HIO₃
B. KIO₃
C. I₂
D. KI
Answer:
C. I₂

Question v.
Ozone layer is depleted by
A. NO
B. NO₂
C. NO₃
D. N₂O₅
Answer:
A. NO

Question vi.
Which of the following occurs in liquid state at room temperature?
A. HIO₃
B. HBr
C. HCl
D. HF
Answer:
D. HF

Question vii.
In pyrosulfurous acid oxidation state of sulfur is
A. Only +2
B. Only +4
C. +2 and +6
D. Only +6
Answer:
B. Only + 4

Question viii.
Stability of interhalogen compounds follows the order
A. BrF > IBr > ICl > ClF > BrCl
B. IBr > BeF > ICl > ClF > BrCl
C. ClF > ICl > IBr > BrCl > BrF
D. ICl > ClF > BrCl > IBr > BrF
Answer:
C. ClF > ICl > IBr > BrCl > BrF

Question ix.
BrCl reacts with water to form
A. HBr
B. Br₂ + Cl₂
C. HOBr
D. HOBr + HCl
Answer:
D. HOBr + HCl

Question x.
Chlorine reacts with excess of fluorine to form.
A. ClF
B. ClF₃
C. ClF₂
D. Cl₂F₃
Answer:
B. ClF₃

Question xi.
In interhalogen compounds, which of the following halogens is never the central atom.
A. I
B. Cl
C. Br
D. F
Answer:
D. F

Question xii.
Which of the following has one lone pair of electrons?
A. IF₃
B. ICl
C. IF₅
D. ClF₃
Answer:
C. IF₅

Question xiii.
In which of the following pairs, molecules are paired with their correct shapes?
A. [I₃] : bent
B. BrF₅ : trigonal bipyramid
C. ClF₃ : trigonal planar
D. [BrF₄] : square planar
Answer:
A. [I₃] : bent

Question xiv.
Among the known interhalogen compounds, the maximum number of atoms is
A. 3
B. 6
C. 7
D. 8
Answer:
D. 8

2. Answer the following.

Question i.
Write the order of the thermal stability of the hydrides of group 16 elements.
Answer:
The thermal stability of the hydrides of group 16 elements decreases in the order of H₂O > H₂S > H₂Se > H₂Te.

Question ii.
What is the oxidation state of Te in TeO₂?
Answer:
The oxidation state of Te in TeO₂ is + 4.

Question iii.
Name two gases which deplete ozone layer.
Answer:
Nitrogen oxide (NO) released from exhaust systems of car or supersonic jet aeroplanes and chlorofluorocarbons (Freons) used in aerosol sprays and refrigerators deplete ozone layer.

Question iv.
Give two uses of ClO₂
Answer:
(i) ClO₂ is used as a bleaching agent for paper pulp and textiles.
(ii) It is also used in water treatment.

Question v.
What is the action of bromine on magnesium metal?
Answer:
Bromine reacts instantly with magnesium metal to give magnesium bromide.

Question vi.
Write the names of allotropic forms of selenium.
Answer:
Selenium has two allotropic forms as follows :
(i) Red (non-metallic) form
(ii) Grey (metallic) form

Question vii.
What is the oxidation state of S in H₂SO₄.
Answer:
The oxidation state of S in H₂SO₄ is + 6.

Question viii.
The pKa values of HCl is -7.0 and that of HI is -10.0. Which is the stronger acid?
Answer:
For HCl, pK_a = -7.0, hence its dissoClation constant is, K_a = 1 x 10^-7.
For HI pK_a = – 10.0, hence its dissoClation constant is K_a = 1 x 10^-7. Hence HCl dissoClates more than HI.
Therefore HCl is a stronger acid than HI.

Question ix.
Give one example showing reducing property of ozone.
Answer:
Ozone decomposes to liberate nascent oxygen, hence it is a powerful oxidising agent. O_3(g) → O_2(g) + O

For example :
(i) It oxidises lead sulphide (PbS) to lead sulphate (PbSO₄).
pbS_(s) + 4O_3(g) → PbSO_(s) + 4O_2(g)
(ii) Potassium iodide, KI is oxidised to iodine, I₂ in the solution.
2KI_(aq) + H₂O₍₁₎ + O_3(g) → 2KOH_(aq) + I_2(s) + O_2(g)

Question x.
Write the reaction of conc. H₂SO₄ with sugar.
Answer:
Concentrated sulphuric acid when added to sugar, it is dehydrated giving carbon.

The carbon that is left behind is called sugar charcoal and the process is called char.

Question xi.
Give two uses of chlorine.
Answer:
Chlorine is used for :

• for sterilization of drinking water.
• bleaching wood pulp required for the manufacture of paper and rayon, cotton and textiles are also bleached using chlorine.
• in the manufacture of organic compounds like CHCl₃, CCl₄, DDT, dyes and drugs.
• in the extraction of metals like gold and platinum.
• in the manufacture of refrigerant like Freon (i.e., CCl₂F₂).
• in the manufacture of several poisonous gases like mustard gas (Cl-C₂H₄-S-C₂H₄-Cl), phosgene (COCl₂) used in warfare.
• in the manufacture of tear gas (CCl₃NO₂).

Question xii.
Complete the following.
1. ICl₃ + H₂O …….. + …….. + ICl
2. I₂ + KClO₃ ……. + KIO₂
3. BrCl + H₂O ……. + HCl
4. Cl₂ + ClF₃ ……..
5. H₂C = CH₂ + ICl …….
6. XeF₄ + SiO₂ ……. + SiF₄
7. XeF₆ + 6H₂O …….. + HF
8. XeOF₄ + H₂O ……. + HF
Answer:
1. 2ICI₃ + 3H₂O → 5HCl + HlO₃ + ICl
2. I₂ + KCIO₃ → ICl + KIO₃
3. BrCl + H₂O → HOBr + HCl
4. Cl₂ + C1F₃ → 3ClF
5. CH₂ = CH₂ + ICl → CH₂I – CH₂Cl
6. 2XeF₆ + SiO₂ → 2XeOF₄ + SiF₄
7. XeF₆ + 3H₂O → XeO₃ + 6HF
8. XeOF₄ + H₂O→  XeO₂F₂ + 2HF

Question xiii.
Match the following
A – B
XeOF₂ – Xenon trioxydifluoride
XeO₂F₂ – Xenon monooxydifluoride
XeO₃F₂ – Xenon dioxytetrafluoride
XeO₂F₄ – Xenon dioxydifluoride
Answer:
XeOF₂ – Xenon monooxydifluoride
XeO₂F₂ – Xenon dioxydifluoride
XeO₃F₂ – Xenon trioxydifluoride
XeO₂F₄ – Xenon dioxytetrafluoride

Question xiv.
What is the oxidation state of xenon in the following compounds?
XeOF₄, XeO₃, XeF₅, XeF₄, XeF₂.
Answer:

Compound	Oxidation state of Xe
XeOF4	+ 6
XeO3	+ 6
XeF6	+ 6
XeF4	+ 4
XeF2	+ 2

3. Answer the following.

Question i.
The first ionisation enthalpies of S, Cl and Ar are 1000, 1256 and 1520 kJ/mol^-1, respectively. Explain the observed trend.
Answer:
(i) The atomic number increases as, ₁₆S < ₁₇Cl < ₁₈Ar₁.
(ii) Due to decrease in atomic size and increase in effective nuclear charge, Cl binds valence electrons strongly.
(iii) Hence ionisation enthalpy of Cl (1256 kJ mol^-1) is higher than that of S(1000 kJ mol^-1)
(iv) Ar has electronic configuration 3s²3p⁶. Since all electrons are paired and the octet is complete, it has the highest ionisation enthalpy, (1520 kJ mol^-1)

Question ii.
“Acidic character of hydrides of group 16 elements increases from H₂O to H₂Te” Explain.
Answer:
(i) The thermal stability of the hydrides of group 16 elements decreases from H₂O to H₂Te. This is because the bond dissociation enthalpy of the H-E bond decreases down the group.
(ii) Thus, the acidic character increases from H₂O to H₂Te.

Question iii.
How is dioxygen prepared in laboratory from KClO₃?
Answer:
By heating chlorates, nitrates and permanganates.
Potassium chlorate in the presence of manganese dioxide on heating decomposes to form potassium chloride and oxygen.

Question iv.
What happens when
a. Lead sulfide reacts with ozone (O₃).
b. Nitric oxide reacts with ozone.
Answer:
(i) It oxidises lead sulphide (PbS) to lead sulphate (PbSO₄) changing the oxidation state of S from – 2 to +6.
PbS_(s) + 4O_3(g) → PbSO_(s) + 4O_2(g)

(ii) Ozone oxidises nitrogen oxide to nitrogen dioxide.
NO_(g) + O_3(g) → NO_2(g) + O_2(g)

Question v.
Give two chemical reactions to explain oxidizing property of concentrated H₂SO₄.
Answer:
Hot and concentrated H₂SO₄ acts as an oxidising agent, since it gives nascent oxygen on heating.

Question vi.
Discuss the structure of sulfur dioxide.
Answer:
(i) SO₂ molecule has a bent V shaped structure with S-O-S bond angle 119.5° and bond dissoClation enthalpy is 297 kJ mol^-1.
(ii) Sulphur in SO₂ is sp² hybridised forming three hybrid orbitals. Due to lone pair electrons, bond angle is reduced from 120° to 119.5°.
(iii) In SO₂, each oxygen atom is bonded to sulphur by σ and a π bond.
(iv) a bond between S and O are formed by sp²-p overlapping.
(v) One of π bonds is formed by pπ – pπ overlapping while other n bond is formed by pπ – dπ overlap.
(vi) Due to resonance both the bonds are identical having observed bond length 143 pm due to resonance,

Question vii.
Fluorine shows only -1 oxidation state while other halogens show -1, +1, +3, +5 and +7 oxidation states. Explain.
Answer:

• Halogens have outer electronic configuration ns² np⁵.
• Halogens have tendency to gain or share one electron to attain the stable configuration of nearest inert element with configuration ns²np⁶.
• Hence they are monovalent and show oxidation state – 1.
• Since fluorine does not have vacant d-orbital, it shows only one oxidation state of – 1 while all other halogens show variable oxidation states from – 1 to +7.
• These oxidation states are, – 1, +1, + 3, +5 and + 7. Cl and Br also show oxidation states + 4 and + 6 in their oxides and oxyaClds.

Question viii.
What is the action of chlorine on the following
a. Fe
b. Excess of NH₃
Answer:
(a) Chlorine reacts with Fe to give ferric chloride.
2Fe + 3Cl₂ → 2FeCl₃

(b) Chlorine reacts with the excess of ammonia to form ammonium chloride, NH₄Cl and nitrogen.

Question ix.
How is hydrogen chloride prepared from sodium chloride?
Answer:

1. In the laboratory, hydrogen chloride, HCl is prepared by heating a mixture of NaCl and concentrated H₂SO₄.
   
2. Hydrogen chloride gas, is dried by passing it through a dehydrating agent like concentrated H₂SO₄ and then collected by upward displacement of air.

Question x.
Draw structures of XeF₆, XeO₃, XeOF₄, XeF₂.
Answer:

Question xi.
What are interhalogen compounds? Give two examples.
Answer:
Interhalogen compounds : Compounds formed by the combination of atoms of two different halogens are called interhalogen compounds. In an interhalogen compound, of the two halogen atoms, one atom is more electropositive than the other. The interhalogen compound is regarded as the halide of the more electropositive halogen.
For example ClF, BrF₃, ICl

Question xii.
What is the action of hydrochloric acid on the following?
a. NH₃
b. Na₂CO₃
Answer:
a. Hydrochloric acid reacts with ammonia to give white fumes of ammonium chloride.
NH₃ + HCl → NH₄Cl

b. Hydrochloric acid reacts with sodium carbonate to give sodium chloride, water with the liberation of carbon dioxide gas.
Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Question xiii.
Give two uses of HCl.
Answer:
Hydrogen chloride (OR hydrochloric acid) is used :

• in the manufacture of chlorine and ammonium chloride,
• to manufacture glucose from com, starch
• to manufacture dye
• in mediClne and galvanising
• as an important reagent in the laboratory
• to extract glue from bones and for the purification of bone black.
• for dissolving metals, Fe + 2HCl_(aq) → FeCl₂ + H_2(g)

Question xiv.
Write the names and structural formulae of oxoacids of chlorine.
Answer:

Question xv.
What happens when
a. Cl₂ reacts with F₂ in equal volume at 437 K.
b. Br₂ reacts with excess of F₂.
Answer:
(a) Cl₂ reacts with F₂ in equal volumes at 437 K to give chlorine monofluoride ClF.

(b) Br₂ reacts with excess of F₂ to give bromine trifluoride BF₃.

Question xvi.
How are xenon fluorides XeF₂, XeF₄ and XeF₆ obtained ? Give suitable reactions.
Answer:
Xenon fluorides are generally prepared by the direct reaction of xenon and fluorine in different ratios and under appropriate experimental conditions, such as temperature, in the presence of an electric discharge and by a photochemical reaction.
(i) Preparation of XeF₂ :

(ii) Preparation of XeF₄ :

(iii) Preparation of XeF₆ :

Question xvii.
How are XeO₃ and XeOF₄ prepared?
Answer:
Preparation of XeO₃ : Xenon trioxide (XeO₃) is prepared by the hydrolysis of XeF₄ or XeF₆.

• By hydrolysis of XeF₄ :
  3XeF₄ + 6H₂0 → 2Xe + XeO₃ + 12 HF + \(1 \frac{1}{2} \mathrm{O}_{2}\)
• By hydrolysis of XeF₆ :
  XeF₆ + 3H₂O → XeO₃ + 6HF
• Preparation of XeOF₄ :
  Xenon oxytetrafluoride (XeOF₄) is prepared by the partial hydrolysis of XeF₆.
  XeF₆ + H₂O → XeOF₄ + 2HF

Question xviii.
Give two uses of neon and argon.
Answer:
Uses of neon (Ne) :

• Neon is used in the production of neon discharge lamps and signs by filling Ne in glass discharge tubes.
• Neon signs are visible from a long distance and also have high penetrating power in mist or fog.
• A mixture of neon and helium is used in voltage stabilizers and current rectifiers.
• Neon is also used in the production of lasers and fluorescent tubes.

Uses of argon (Ar) :

• Argon is used to fill fluorescent tubes and radio valves.
• It is used to provide inert atmosphere for welding and production of steel.
• It is used along with neon in neon sign lamps to obtain different colours.
• A mixture of 85% Ar and 15% N2 is used in electric bulbs to enhance the life of the filament.

Question xix.
Describe the structure of Ozone. Give two uses of ozone.
Answer:
(A)

• Ozone has molecular formula O₃.
• The lewis dot and dash structures for O₃ are :
  
• Infrared and electron diffraction spectra show that O₃ molecule is angular with 0-0-0 bond angle 117°.
• Both 0-0 bonds are identical having bond length 128 pm which is intermediate between single and double bonds.
• This is explained by considering resonating structures and resonance hybrid.
  

(B) Uses of Ozone :

• Ozone sterilises drinking water by oxidising germs and bacteria present in it.
• It is used as a bleaching agent for ivory, oils, starch, wax and delicate fabrics like silk.
• Ozone is used to purify the air in crowded places like Clnema halls, railways, tunnels, etc.
• In industry, ozone is used in the manufacture of synthetic camphor, potassium permanganate, etc.

Question xx.
Explain the trend in following atomic properties of group 16 elements.
i. Atomic radii
ii. Ionisation enthalpy
iii. Electronegativity.
Answer:
(1) Atomic and ionic radii :

1. As compared to group 15 elements, the atomic and ionic radii of group 16 elements are smaller due to higher nuclear charge.
2. The atomic and ionic radii increase down the group from oxygen to polonium. This is due to the addition of a new shell at each successive elements on moving down the group. The atomic radii increases in the order O < S < Se < Te < Po

(2) Ionisation enthalpy :

• The ionisation enthalpy of group 16 elements has quite high values.
• Ionisation enthalpy decreases down the group from oxygen to polonium. This is due to the increase in atomic volume down the group.
• The first ionisation enthalpy of the lighter elements of group 16 (O, S, Se) have lower values than those of group 15 elements in the corresponding periods. This is due to difference in their electronic configurations.

Group 15 : (valence shell) ns² np_x¹ np_y¹ np_z¹
Group 16 : (valence shell) ns² np_x² np_y¹ np_z¹

Group 15 elements have extra stability of half-filled and more symmetrical orbitals, while group 16 elements acquire extra stability by losing one of paired electrons from npx- orbital forming half-filled p-orbitals.

Hence group 16 elements have lower first ionisation enthalpy than group 15 elements.

(3) Electronegativity :

• The electronegativity values of group 16 elements have higher values than corresponding group 15 elements in the same periods.
• Oxygen is the second most electronegative elements after fluorine. (O = 3.5, F = 4)
• On moving down the group electronegativity decreases from oxygen to polonium.
• On moving down the group atomic size increases, hence nuclear attraction decreases, therefore electro-negativity decreases.

Elements	O	S	Se	Te	Po
Electronegativity	3.5	2.44	2.48	2.01	1.76

4. Answer the following.

Question i.
Distinguish between rhombic sulfur and monoclinic sulfur.
Answer:

Rhombic sulphur	Monoclinic sulphur
1. It is pale yellow.	1. It is bright yellow.
2. Orthorhombic crystals	2. Needle-shaped monoclinic crystals
3. Melting point, 385.8 K	3. Melting point, 393 K
4. Density, 2.069 g/cm3	4. Density: 1.989 g/cm3
5. Insoluble in water, but soluble in CS2	5. Soluble in CS2
6. It is stable below 369 K and transforms to α-sulphur above this temperature.	6. It is stable above 369 K and transforms into β-sulphur below this temperature.
7. It exists as S8 molecules with a structure of a puckered ring.	7. It exists as S8 molecules with a structure of a puckered ring.
8. It is obtained by the evaporation of roll sulphur in CS2	8. It is prepared by melting rhombic sulphur and cooling it till a crust is formed. Two holes are pierced in the crust and the remaining liquid is poured to obtain needle-shaped crystals of monoclinic sulphur (β-sulphur).

Question ii.
Give two reactions showing oxidizing property of concentrated H₂SO₄.
Answer:
Hot and concentrated H₂SO₄ acts as an oxidising agent, since it gives nascent oxygen on heating.

Question iii.
How is SO₂ prepared in the laboratory from sodium sulfite? Give two physical properties of SO₂.
Answer:
(A) Laboratory method (From sulphite) :

• Sodium sulphite on treating with dilute H₂SO₄ forms SO₂.
  Na₂SO₃ + H₂SO_4(aq) → Na₂SO₄ + H₂O₍₁₎ + SO_2(g)
• Sodium sulphite, Na₂SO₃ on reaction with dilute hydrochloric acid solution forms SO₂.
  Na₂SO_3(aq) + 2HCl_(aq) → 2NaCl_9aq0 + H₂O₍₁₎ + SO_2(g)

(B) Physical properties of SO₂

• It is a colourless gas with a pungent smell.
• It is highly soluble in water and forms sulphurous acid, H₂SO₃ SO_2(g) + H₂O₍₁₎ → H₂SO_3(aq)
• It is poisonous in nature.
• At room temperature, it liquefies at 2 atmospheres. It has boiling point 263K.

Question iv.
Describe the manufacturing of H₂SO₄ by contact process.
Answer:
Contact process of the manufacture of sulphuric acid involves following steps :

(1) Preparation of SO₂ : Sulphur or pyrite ores like iron pyrites, FeS₂ on burning in excess of air, form SO₂.

(2) Oxidation of SO₂ to SO₃ : SO₂ is oxidised to SO₃ in the presence of a heterogeneous catalyst V₂O₅ and atmospheric oxygen. This oxidation reaction is reversible.

To avoid the poisoning of a costly catalyst, it is necessary to make SO₂ free from the impurities like dust, moisture, As₂O₃ poison, etc.

The forward reaction is exothermic and favoured by increase in pressure. The reaction is carried out at high pressure (2 bar) and 720 K temperature. The reacting gases, SO₂ and O₂ are taken in the ratio 2:3.

(3) Dissolution of SO₃ : SO₃ obtained from catalytic converter is absorbed in 98%. H₂SO₄ to obtain H₂S₂O₇, oleum or fuming sulphuric acid.


Flow diagram for the manufacture of sulphuric acid

Question 7.1 (Textbook Page No 141)

12th Chemistry Digest Chapter 7 Elements of Groups 16, 17 and 18 Intext Questions and Answers

Question 1.
Elements of group 16 generally show lower values of first ionisation enthalpy compared to the elements of corresponding period of group 15. Why?
Answer:
Group 15 elements have extra stable, half filled p-orbitals with electronic configuration (ns²np³). Therefore more amount of energy is required to remove an electron compared to that of the partially filled orbitals (ns²np⁴) of group 16 elements of the corresponding period.

Question 7.2 (Textbook Page No 141)

Question 1.
The values of first ionisation enthalpy of S and Cl are 1000 and 1256 kJ mol^-1, respectively. Explain the observed trend.
Answer :
The elements S and Cl belong to second period of the periodic table.
Across a period effective nuclear charge increases and atomic size decreases with increase in atomic number. Therefore the energy required for the removal of electron from the valence shell (I.E.) increases in the order S < Cl.

Question 7.4 (Textbook Page No 141)

Question 1.
Fluorine has less negative electron gain affinity than chlorine. Why?
Answer :
The size of fluorine atom is smaller than chlorine atom. As a result, there are strong inter electronic repulsions in the relatively small 2p orbitals of fluorine and therefore, the incoming electron does not experience much attraction. Thus fluorine has less negative electron gain affinity than chlorine.

Try this… (Textbook Page No 140)

Question 1.
Explain the trend in the following properties of group 17 elements.

(1) Atomic size,
(2) Ionisation enthalpy,
(3) Electronegativity,
(4) Electron gain enthalpy.
Answer:
(1) Atomic size :

• Atomic and ionic radii increase down the group as atomic number increases due to the addition of new electronic valence shell to each succeeding element.
• The atomic radii increase in the order F < Cl < Br < 1
• Halogens possess the smallest atomic and ionic radii in their respective periods since the effective nuclear charge experienced by valence electrons in halogen atoms is the highest.

(2) Ionisation enthalpy :

• The ionisation enthalpies of halogens are very high due to their small size and large nuclear attraction.
• The ionisation ethalpies decrease down the group since the atomic size increases.
• The ionisation enthalpy decreases in the order F > Cl > Br > I.
• Among halogens fluorine has the highest ionisation enthalpy due to its smallest size.

Element	F	Cl	Br	I
Ionisation enthalpy kJ/mol	1680	1256	1142	1008

(3) Electronegativity :

• Halogens have the highest values for electronegativity due to their small atomic radii and high effective nuclear charge.
• Each halogen is the most electronegative element of its period.
• Fluorine has the highest electronegativity as compared to any element in the periodic table.
• The electronegativity decreases as,
  F > Cl > Br > I
  4.0 3.2 3.0 2.7 (electronegativity)

(4) Electron gain enthalpy (Δ_egH) :

• The halogens have the highest negative values for electron gain enthalpy.
• Electron gain enthalpies of halogens are negative indicating release of energy.
• Halogens liberate maximum heat by gain of electron as compared to other elements.
• Since halogens have outer valence electronic configuration, ns² np⁵, they have strong tendency to accept an electron to complete an octet and acquire electronic configuration of the nearest inert elements.
• In case of fluorine due to small size of 2 p-orbitals and high electron density, F has less negative electron gain enthalpy than Cl.
  F_(g) + e^– → F^–_(g) Δ_egH = – 333 klmol^-1
  Cl_(g) + e^– → Cl^–_(g) Δ_egH = – 349 kJ mol^-1
• The variation in electron gain enthalpy is in the order of, Cl > F > Br > I.

Question 2.
Oxygen has less negative electron gain enthalpy than sulphur. Why?
Answer:

• Oxygen has a smaller atomic size than sulphur.
• It is more electronegative than sulphur.
• It has a larger electron density.
• Due to high electron density, oxygen does not accept the incoming electron easily and therefore has less electron gain enthalpy than sulphur.

Question 7.3 (Textbook Page No 141)

Question 1.
Why is there a large difference between the melting and boiling points of oxygen and sulphur?
Answer :
Oxygen exists as diatomic molecule (O₂) whereas sulphur exists as polyatomic molecule (S₈). The van der Waals forces of attraction between O₂ molecules are relatively weak owing to their much smaller size. The large van der Waals attractive forces in the S₈ molecules are due to large molecular size. Therefore oxygen has low m.p. and b.p. as compared to sulphur.

Question 7.5 (Textbook Page No 141)

Question 1.
Bond dissoClation enthalpy of F₂ (158.8 kj mol^-1) is lower than that of Cl₂ (242.6 kj mol^-1) Why?
Answer :
Fluorine has small atomic size than chlorine. The lone pairs on each F atom in F₂ molecule are so close together that they strongly repel each other, and make the F – F bond weak. Thus, it requires less amount of energy to break the F – F bond. In Cl₂ molecule the lone pairs on each Cl atom are at a larger distance and the repulsion is less.

Thus Cl – Cl bond is comparatively stronger. Therefore bond dissoClation enthalpy of F₂ is lower than that of Cl₂.

Question 7.6 (Textbook Page No 142)

Question 1.
Noble gases have very low melting and boiling points. Why?
Answer :
Noble gases are monoatomic, the only type of interatomic interactions which exist between them are weak van der Waals forces. Therefore, they can be liquefied at very low temperatures and have very low melting or boiling points.

Can you tell? (Textbook Page No 142)

Question 1.
The first member of the a group usually differs in properties from the rest of the members of the group. Why?
Answer:
The first member of a group usually differs in properties from the rest of the members of the group for the following reasons :

• Its small size
• High electronegativity
• Absence of vacant d-orbitals in its valence shell.

Use your brain power! (Textbook Page No 142)

Question 1.
Oxygen forms only OF₂ with fluorine while sulphur forms SF₆. Explain. Why?
Answer:

• Oxygen combines with the most electronegative element fluorine to form OF₂ and exhibits positive oxidation state (+ 2). Since, oxygen does not have vacant J-orbitals it cannot exhibit higher oxidation states.
• Sulphur has vacant d-orbitals and hence can exhibit + 6 oxidation state to form SF₆.

Question 2.
Which of the following possesses hydrogen bonding? H₂S, H₂O, H₂Se, H₂Te
Answer:

• Oxygen being more electronegative, is capable of forming hydrogen bonding in the compound H₂O.
• The other elements S, Se and Te of Group 16, being less electronegative do not form hydrogen bonds.
• Thus, hydrogen bonding is not present in the other hydrides H₂S, H₂Se and H₂Te.

Question 3.
Show hydrogen bonding in the above molecule with the help of a diagram.
Answer:

Try this….. (Textbook Page No 143)

Question 1.
Complete the following tables :

Answer:

Can you tell? (Textbook Page No 146)

Question 1.
What is allotropy?
Answer:
The property of some elements to exist in two or more different forms in the same physical state is called allotropy.

Question 2.
What is the difference between allotropy and polymorphism?
Answer:

• Allotropy is the existence of an element in more than one physical form. It means that under different conditions of temperature and pressure an element can exist in more than one physical forms.
• Coal, graphite and diamond etc., are different allotropic forms of carbon.
• Polymorphism is the existence of a substance in more than one crystalline form.
• It means that under different conditions of temperature and pressure, a substance can form more than one type of crystal. For example, mercuric iodide exists in the orthorhombic and trigonal form.

Question 7.7 (Textbook Page No 146)

Which form of sulphur shows paramagnetic behaviour?
Answer :
In the vapour state, sulphur partly exists as S₂ molecule, which has two unpaired electrons in the antibonding π^* orbitals like O₂. Hence it exhibits paramagnetism.

Try this….. (Textbook Page No 149)

Question 1.
Why water in a fish pot needs to be changed from time to time?
Answer:
A fish pot is an artificial ecosystem and the fish in it are selective and maintained in a restricted environment.

In a fish pot, the unwanted food and waste generated by the fish mix with the water and remain untreated due to lack of decomposers.

Accumulation of waste material will decrease the levels of dissolved oxygen in the water pot.

Hence, it is necessary to change the water from time to time.

Question 7.8 (Textbook Page No 149)

Dioxygen is paramagnetic in spite of having an even number of electrons. Explain.
Answer :
Dioxygen is a covalently bonded molecule.
The paramagnetic behaviour of O₂ can be explained with the help of molecular orbital theory.
Electronic configuration O₂
KK σ(2s)² σ(2s)² σ^*(2p_z)² π(2p_x)² π(2p_x)² π(2p_y)² π^*(2p_x)¹ π^*(2p_y)¹. Presence of two unpaired electrons in antibonding orbitals explains paramagnetic nature of dioxygen.

Question 7.9 (Textbook Page No 150)

High concentration of ozone can be dangerously explosive. Explain.
Answer :
Thermal stability : Ozone is thermodynamically unstable than oxygen and decomposes into O₂. The decomposition is exothermic and results in the liberation of heat (ΔH is – ve) and an increase in entropy (ΔS is positive). This results in large negative Gibbs energy change (ΔG). Therefore high concentration of ozone can be dangerously explosive. Eq O₃ → O₂ + O

Try this…… (Textbook Page No 151)

(a) Ozone is used as a bleaching agent. Explain.
Answer:

• Ozone due to its oxidising property can act as a bleaching agent. O_3(g) → O_2(g) + O
• It bleaches coloured matter. coloured matter + O → colourless matter
• Ozone bleaches in the absence of moisture, so it is also known as dry bleach.
• Ozone can bleach ivory and delicate fabrics like silk.

(b) Why does ozone act as a powerful oxidising agent?
Answer:
Ozone decomposes to liberate nascent oxygen, hence it is a powerful oxidising agent. O_3(g) → O_2(g) + O
For example :

• It oxidises lead sulphide (PbS) to lead sulphate (PbSO₄).
  pbS_(s) + 4O_3(g) → PbSO_(s) + 4O_2(g)
• Potassium iodide, KI is oxidised to iodine, I₂ in the solution.
  2KI_(aq) + H₂O₍₁₎ + O_3(g) → 2KOH_(aq) + I_2(s) + O_2(g)

Question 7.10 : (Textbook Page No 154)

What is the action of concentrated H₂SO₄ on (a) HBr (b) HI
Answer :
Concentrated sulphuric acid oxidises hydrobromic acid to bromine.

2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O
It oxidises hydroiodic acid to iodine.
2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O

Try this….. (Textbook Page No 156)

Question 1.
Give the reasons for the bleaching action of chlorine.
Answer:

• Chlorine acts as a powerful bleaching agent due to its oxidising nature.
• In moist conditions or in the presence of water it forms unstable hypochlorous acid, HOCl which decomposes giving nascent oxygen which oxidises the vegetable colouring matter of green leaves, flowers, litmus, indigo, etc.
  Cl₂ + H₂O → HCl + HOCl
  HOCl → HCl + [O]
  Vegetable coloured matter + [O] → colourless matter.

Question 2.
Name two gases used in war.
Answer:
Phosgene : COCl₂
Mustard gas: Cl – CH₂ – CH₂ – S – CH₂ – CH₂ – Cl

Use your brain power! (Textbook Page No 157)

Question 1.
Chlorine and fluorine combine to form interhalogen compounds. The halide ion will be of chlorine or fluorine?
Answer:
Among the- two halogens, chlorine is more electropositive than fluorine (Electronegativity values: F = 4.0, Cl = 3.2)

The interhalogen compound is regarded as the halide of the more electropositive halogen. Hence, the interhalogen compound is the fluoride of chlorine, i.e. chlorine monofluoride, CiF.

Question 2.
Why does fluorine combine with other halogens to form maximum number of fluorides?
Answer:
Since fluorine is the most electronegative element and has the smallest atomic radius compared to other halogen fluorine forms maximum number of fluorides.

Use your brain power! (Textbook Page No 158)

Question 1.
What will be the names of the following compounds: ICl, BrF?
Answer:
ICl : Iodine monochloride
BrF : Bromine monofluoride

Question 2.
Which halogen (X) will have maximum number of other halogen (X ) attached?
Answer:
The halogen Iodine (I) will have the maximum number of other halogens attached.

Question 3.
Which halogen has tendency to form more interhalogen compounds?
Answer:
The halogen fluorine (F) has the maximum tendency to form more interhalogen compounds as it has a small size and more electronegativity.

Question 4.
Which will be more reactive?
(a) ClF₃ or ClF,
(b) BrF₅ or BrF
Answer:
ClF₃ is more reactive than ClF, while BrF₅ is more reactive than BrF. Both ClF₃ and BrF₅ are unstable compared to ClF and BrF respectively due to steric hindrance hence are more reactive.

Question 5.
Complete the table :

Formula	Name
ClF	Chlorine monofluoride
ClF3	…………………………………
…………………………………	Chlorine pentachloride
BrF	…………………………………
…………………………………	Bromine pentafluoride
ICl	…………………………………
ICl3	…………………………………

Answer:

Formula	Name
ClF	Chlorine monofluoride
ClF3	Chlorine trifluoride
CIF5	Chlorine pentafluoride
BrF	Bromine monofluoride
BrF5	Bromine pentafluoride
ICl	Iodine monochloride
ICl3	Iodine trichloride

Use your brain power! (Textbook Page No 159)

Question 1.
In the special reaction for ICl, identify the oxidant and the reductant? Denote oxidation states of the species.
Answer:

Potassium chlorate, KClO₃ is the oxidising agent or oxidant and iodine is the reducing agent or reductant.

Use your brain power! (Textbook Page No 162)

Question 1.
What are missing entries?

Formula	Name
XeOF2 …………… XeO3F2 XeO2F4	Xenon monooxyfluoride Xenon dioxydifluoride …………………………………….. ……………………………………..

Answer:

Formula	Name
XeOF2 XeO2F2 XeO3F2 XeO2F4	Xenon monooxydifluoride Xenon dioxydifluoride Xenon trioxydifluoride Xenon dioxytetrafluoride

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Solutions Chapter 6 Dissolution of Partnership Firm Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Book Keeping & Accountancy Solutions Chapter 6 Dissolution of Partnership Firm

1. Objective Questions.

A. Select the most appropriate answer from the alternatives given below and rewrite the sentences.

Question 1.
In case of dissolution assets and liabilities cire transferred to ______________ Account.
(a) Bank Account
(b) Partner’s Capital Account
(c) Realisation Account
(d) Partner’s Current Account
Answer:
(c) Realisation Account

Question 2.
Dissolution expenses are credited to ______________ Account.
(a) Realisation Account
(b) Cash/Bank Account
(c) Partner’s Capital Account
(d) Partner’s Loan Account
Answer:
(b) Cash/Bank Account

Question 3.
Deficiency of insolvent partner will be suffered by solvent partners in their ______________ ratio.
(a) capital ratio
(b) profit sharing ratio
(c) sale ratio
(d) liquidity ratio
Answer:
(b) profit sharing ratio

Question 4.
If any asset is taken over by partner from firm his Capital Account will be ______________
(a) credited
(b) debited
(c) added
(d) divided
Answer:
(b) debited

Question 5.
If any unrecorded liability is paid on dissolution of the firm ______________ account is debited.
(a) Cash/Bank Account
(b) Realisation Account
(c) Partner’s Capital Account
(d) Loan Account
Answer:
(b) Realisation Account

Question 6.
Partnership is completely dissolved when the partners of the firm become ______________
(a) solvent
(b) insolvent
(c) creditor
(d) debtors
Answer:
(b) insolvent

Question 7.
Assets and liabilities are transferred to Realisation Account at their ______________ values.
(a) market
(b) purchase
(c) sale
(d) book
Answer:
(d) book

Question 8.
If the number of partners in a firm falls below two, the firm stands ______________
(a) dissolved
(b) established
(c) realisation
(d) restructured
Answer:
(a) dissolved

Question 9.
Realisation Account is ______________ on realisation of asset.
(a) debited
(b) credited
(c) deducted
(d) closed
Answer:
(b) credited

Question 10.
All activities of partnership firm ceases on ______________ of firm.
(a) dissolution
(b) admission
(c) retirement
(d) death
Answer:
(a) dissolution

B. Write a word/phrase/term which can substitute each of the following statements.

Question 1.
Debit balance of Realisation Account.
Answer:
Realization Loss

Question 2.
Winding up of partnership business.
Answer:
Dissolution of Partnership

Question 3.
An account is opened to find out the profit or loss on sale of assets and settlement of liabilities.
Answer:
Realization A/c

Question 4.
Debit balance of an Insolvent Partner’s Capital Account.
Answer:
Capital Deficiency

Question 5.
The credit balance of the Realisation Account.
Answer:
Realization Profit

Question 6.
Conversion of asset into cash on the dissolution of the firm.
Answer:
Realisation

Question 7.
Liability is likely to arise in the future on the happening of certain events.
Answer:
Contingent Liabilities

Question 8.
Assets that are not recorded in the books of accounts.
Answer:
Unrecorded Assets

Question 9.
The account shows the realization of assets and discharge of liabilities.
Answer:
Realization A/c

Question 10.
Expenses incurred on the dissolution of the firm.
Answer:
Dissolution/Realisation Expenses

C. State whether the following statements are True or False with reasons.

Question 1.
The firm must be dissolved on the retirement of a partner.
Answer:
This statement is False.
On the retirement of a partner, if the partnership agreement allows, then the remaining partner can continue the business activities. It means the firm is not to dissolve.

Question 2.
On dissolution Cash/Bank Account is closed automatically.
Answer:
This statement is True.
As the firm is dissolved, there is no question of any business activities to be carried out further and so Cash/Bank Account is also not necessary. Therefore on dissolution Cash/Bank Account is closed automatically.

Question 3.
On dissolution, Bank overdraft is transferred to Realisation Account.
Answer:
This statement is True.
As a sundry liability of the business, bank overdraft is a liability of a firm and hence, it is transferred to Realisation Account at the time of dissolution and paid a third party Liability.

Question 4.
A solvent partner having a debit balance to his Capital Account does not share the deficiency of insolvent partner Capital Account.
Answer:
This statement is False.
In the partnership, the partner’s liability is unlimited so, a solvent partner having a debit balance to his Capital Account should share the deficiency of the insolvent partner capital account.

Question 5.
At the time of dissolution of the partnership, all assets should be transferred to Realisation Account.
Answer:
This statement is False.
At the time of dissolution of the partnership, the cash account and Bank A/c are not transferred to Realisation A/c. Similarly, if an asset is taken over by a partner or by any creditor then that asset is transferred to the concerned person’s account and not to the Realisation Account.

Question 6.
The debit balance of an insolvent partner’s Capital Account is known as a capital deficiency.
Answer:
This statement is True.
Debit balance of Partners’ Capital Account means the excess of drawings than the capital credit balance. In the case of an insolvent partner, the debit balance of the Capital Account means liabilities which he cannot pay. It means capital deficiency.

Question 7.
At the time of dissolution, a loan from a partner will be transferred to Realisation Account.
Answer:
This statement is False.
At the time of dissolution, a loan from a partner will be paid after the payment of liabilities of third parties to the firm. It is not transferred to Realisation Account. Partner’s Loan A/c is separately opened and paid accordingly.

Question 8.
Dissolution takes place when the relationship among the partners comes to an end.
Answer:
This statement is True.
As per definition, Dissolution means to wind up or to close down, and it is possible only when relations among the partners in a partnership firm come to an end.

Question 9.
The insolvency loss at the time of dissolution of the firm is shared by the solvent partners in their profit sharing ratio.
Answer:
This statement is True.
In the partnership, partners’ liability is unlimited and in case of insolvency loss, legally solvent partners are ultimately liable and are suppose to bear the loss of an insolvent partner in their profit sharing ratio.

Question 10.
Realization loss is not transferred to insolvent partner’s Capital Account.
Answer:
This statement is False.
All partners of the firm are responsible for Loss on realization and hence loss on realization is supposed to be transferred to all Partners’ Capital Account, without any discrimination of solvent or insolvent.

D. Calculate the following:

Question 1.
Vinod, Vijay, and Vishal are partners in a firm sharing profit and losses in the ratio of 3 : 2 : 1. Vishal becomes insolvent and his capital deficiency is ₹ 6000. Distribute the capital deficiency among the solvent partner.
Answer:
Here, capital deficiency of ₹ 6000 is to be distributed among continuing partners in their profit and loss sharing ratio, i.e. 3 : 2
Share of deficiency for Vinod = 6,000 × \(\frac{3}{5}\) = ₹ 3,600
Share of deficiency for Vijay = 6,000 × \(\frac{2}{5}\) = ₹ 2,400
Vinod and Vijay will bear ₹ 3,600 and ₹ 2,400 of Vishal’s capital deficiency.

Question 2.
Creditors ₹ 30,000, Bills Payable ₹ 20,000, and Bank Loan ₹ 10,000. Available Bank balance ₹ 40,000. What will be the amount that creditors will get in case of all partner’s insolvency?
Answer:
Ratio of creditors, Bills payable and Bank Loan = 30,000 : 20,000 : 10,000 i.e., 3 : 2 : 1
Amount received by creditors = \(\frac{3}{3+2+1}\) × 40,000
= \(\frac{3}{6}\) × 40,000
= ₹ 20,000.

Question 3.
Insolvent Partner Capital A/c debit side total is ₹ 10,000 and credit side total is ₹ 6,000. Calculate deficiency.
Answer:
Deficiency of insolvent partner = Debit side total – Credit side total
= 10,000 – 6,000
= ₹ 4,000.

Question 4.
Insolvent Partners Capital A/c debit side is ₹ 15,000 and insolvent partner brought cash ₹ 6,000. Calculate the amount of insolvency loss to be distributed among the solvent partners.
Answer:
₹ 9,000 (15,000 – 6,000) is the amount of insolvency loss to be distributed among the solvent partners.

Question 5.
The realization profit of a firm is ₹ 6,000, partners share profit and loss in the ratio of 3 : 2 : 1. Calculate the amount of realization profit to be credited to Partners’ Capital A/c.
Answer:
Distribution of ₹ 6,000 in 3 : 2 : 1 ratio
6,000 × \(\frac{3}{6}\) = ₹ 3,000, 6,000 × \(\frac{2}{6}\) = ₹ 2,000, 6,000 × \(\frac{1}{6}\) = ₹ 1,000
Amount of realisation profit ₹ 3,000, ₹ 2,000 and ₹ 1,000 is to be credited to Partner’s Capital A/c respectively.

E. Answer in one sentence only.

Question 1.
What is the dissolution of the partnership firm?
Answer:
Dissolution of the partnership firm means complete closure of business activities and stoppage of partnership relations among all the partners.

Question 2.
When is Realisation Account opened?
Answer:
Realisation Account is opened at the time of dissolution of the partnership firm.

Question 3.
Which accounts are not transferred to Realisation Account?
Answer:
Cash/Bank balance, Reserve funds, Profit and Loss A/c balance, Partners’ Loan accounts, etc. are not transferred to Realisation Account.

Question 4.
Who is called an insolvent person?
Answer:
Whose capital A/c shows debit balance and who is not in a position to meet his capital deficiency even from his private property is called an insolvent person.

Question 5.
What is capital deficiency?
Answer:
The debit balance of the insolvent partner’s Capital Account which the insolvent partner cannot pay is called a capital deficiency.

Question 6.
In what proportion is the balance on Realisation Account transferred to Partners Capital/Current Accounts?
Answer:
The balance on the Realisation Account is transferred to Partners Capital/Current Accounts in their profit sharing ratio.

Question 7.
Who should bear the capital deficiency of insolvent partners?
Answer:
The capital deficiency of insolvent partners should be borne by the solvent partners.

Question 8.
Which account is debited on repayment of partner’s loan?
Answer:
Partner’s Loan Account is debited on repayment of partner’s loan.

Question 9.
Which account is debited on payment of dissolution expenses?
Answer:
Realisation Account is debited on payment of dissolution expenses.

F. Complete the table.

Question 1.

Answer:

Practical Problems

(Simple Dissolution)

Question 1.
Ganesh and Kartik are partners sharing profits and losses equally. They decided to dissolve the firm on 31st March 2018. Their Balance Sheet was as under:
Balance Sheet as of 31st March 2018

Assets were realised as under:
Building ₹ 82,000, Debtors ₹ 22,000, Stock ₹ 20,000. Bills Receivable ₹ 3,200 and Ganesh agreed to take over Furniture for ₹ 10,000. Realisation Expenses amounted to ₹ 2,000.
Show Realisation A/c, Partners’ Capital A/c, and Cash A/c.
Solution:
In the books of Ganesh and Kartik


Working Notes:
1. Amount paid to Ganesh and Kartik are ₹ 27,600 and ₹ 77,600 respectively.
2. Loss on Realisation and Reserve fund amounts are equally distributed.
3. Furniture is taken over by Ganesh so his Capital A/c is debited.

Question 2.
Leela, Manda, and Kunda are partners in the firm ‘Janki Stores’ sharing profits and losses in the ratio of 3 : 2 : 1 respectively. On 31st March 2018, they decided to dissolve the firm when their Balance Sheet was as under.
Balance Sheet as of 31st March 2018

Leela agreed to take over the Building at ₹ 1,23,600. Manda took over Goodwill, Stock, and Debtors at book values and agreed to pay Creditors and Bills payable. Motor car and Machinery realized ₹ 1,51,080 and ₹ 31,680 respectively. Investments were taken by Kunda at an agreed value of ₹ 55,440. Realisation expenses amounted to ₹ 6,800.
Pass necessary entries in the books of ‘Janki Stores’.
Solution:
In the books of ‘Janki Stores’
Journal Entries


Working Notes:
In the books of Leela, Manda, and Kunda

Question 3.
Shailesh and Shashank were partners sharing profits and losses in the ratio of 3 : 2. Their Balance Sheet as of 31st March 2019 was as follows:
Balance Sheet as of 31st March 2019

The firm was dissolved on the above date and the assets realised as under:
1. Plant ₹ 8,000, Building ₹ 6,000, Stock ₹ 4,000 and Debtors ₹ 12,000.
2. Shailesh agreed to pay off the Bills Payable.
3. Creditors were paid in full.
4. Dissolution expenses were ₹ 1,400.
Prepare Realisation A/c, Partners’ Current A/c, Partners’ Capital A/c, and Bank A/c.
Solution:
In the books of Shailesh and Shashank

Question 4.
Asha, Usha, and Nisha were partners sharing profits and losses in the ratio of 2 : 2 : 1. The following is the Balance Sheet as of 31st March 2019.
Balance Sheet as of 31st March 2019

On the above date, the partners decided to dissolve the firm.
1. Assets were realised at: Machinery ₹ 90,000, Stock ₹ 36,000, Investment ₹ 42,000 and Debtors ₹ 90,000.
2. Dissolution expenses were ₹ 6,000.
3. Goodwill of the firm realized ₹ 48,000.
Pass Journal Entries to close the books of the firm.
Solution:
In the books of Asha, Usha, and Nisha
Journal Entries


Working Notes:
In the books of Asha, Usha, and Nisha

Question 5.
Seeta and Geeta are partners in the firm sharing profits and losses in the ratio of 4 : 1. They decided to dissolve the partnership on 31st March 2020 on which date their Balance Sheet stood as follows:
Balance Sheet as of 31st March 2020

Additional Information:
1. Plant and Stock took over by Seeta at ₹ 78,000 and ₹ 22,000 respectively.
2. Debtors realised 90% of the book value and Trademark at ₹ 5,000 and Goodwill was realised for ₹ 27,000.
3. Unrecorded assets estimated at ₹ 4,500 were sold for ₹ 1,500.
4. ₹ 1,000 Discounts were allowed by creditors while paying their claim.
5. The Realisation expenses amounted to ₹ 3,500.
You are required to prepare Realisation A/c, Cash A/c, and Partners’ Capital A/c.
Solution:
In the books of Seeta and Geeta


Working Notes:
1. Bank Loan is an external liability of the firm and therefore it is transferred to Realisation A/c.
2. Amount recovered from Debtors = 90% of Gross Debtors = \(\frac {90}{100}\) × 48,000 = ₹ 43,200.
3. Amount paid to creditors = Value of Creditors – Discount given = 35,000 – 1,000 = ₹ 34,000.
4. Sale of unrecorded assets for ₹ 1,500 is recorded on the credit side of Realisation A/c and debit side of Cash A/c.
5. It is presumed that Furniture realised nothing.

Question 6.
Sangeeta, Anita, and Smita were in partnership sharing profits and losses in the ratio 2 : 2 : 1. Their Balance Sheet as of 31st March 2019 was as under:
Balance Sheet as of 31st March 2019

They decided to dissolve the firm as follows:
1. Assets realised as; Land recovered ₹ 1,80,000; Goodwill for ₹ 75,000; Loans and Advance realised ₹ 12,000; 10% of the Debts proved bad.
2. Sangeeta took Plant at book value.
3. Creditors and Bills payable paid at 5% discount.
4. Sandhya’s loan was discharged along with ₹ 6,000 as interest.
5. There was a contingent liability in respect of bills of ₹ 1,00,000 which was under discount. Out of them, a holder of one bill of ₹ 20,000 became insolvent.
Show Realisation Account, Partners’ Capital Account, and Bank Account.
Solution:
In the books of Sangeeta, Anita, and Smita


Working Notes:
1. Amount paid towards Sandhya’s Loan = Loan amount + Interest due on loan
= 1,20,000 + 6,000
= ₹ 1,26,000

2. Amount received from Debtors = Debtors – Bad debts
= 1,25,000 – 10% of 1,25,000
= 1,25,000 – 12,500
= ₹ 1,12,500

3. Amount paid to Creditors = Creditor – 5% discount
= 1,20,000 – 5% on 1,20,000
= 1,20,000 – 6,000
= ₹ 1,14,000

4. Amount paid towards Bills payable = Bills payable – 5% discount
= 20,000 – 5% on 20,000
= 20,000 – 1,000
= ₹ 19,000

5. Bill of ₹ 1,00,000 was discounted with the Bank. On the due date, bank could not recover ₹ 20,000 from one bill holder as he was declared insolvent. Therefore, we are required to settle that contingent liability of ₹ 20,000.

Question 7.
Saiesh, Sumit, and Hemant were in partnership sharing Profits and Losses in the ratio 2 : 2 : 1. They decided to dissolve their partnership firm on 31st March 2019 and their Balance Sheet on that date stood as;
Balance Sheet as of 31st March 2019

It was agreed that;
1. Sailesh to discharge Loan and to take Debtors at book value.
2. Plant realised ₹ 1,35,000.
3. Stock realised ₹ 72,000.
4. Creditors were paid off at a discount of ₹ 45.
Show Realisation Account, Partners’ Capital Account, and Bank Account.
Solution:
In the books of Sailesh, Sumit, and Hemant

(When one partner become Insolvent)

Question 8.
Sitaram, Gangaram, and Rajaram are partners sharing profits and losses in the ratio of 4 : 2 : 3. On 1st April 2019 they agreed to dissolve the partnership, their Balance Sheet was as follows:
Balance Sheet as of 31st March 2019

The assets realised: Building ₹ 46,750; Machinery ₹ 18,550; Furniture ₹ 9,600; Investment ₹ 10,650; Bill Receivable and Debtors ₹ 20,750. All the liabilities were paid off. The cost of realisation was ₹ 800. Rajaram becomes bankrupt and ₹ 1,100 only was recovered from his estate.
Show Realisation Account, Bank Account, and Capital Account of the partners.
Solution:
In the books of Sitaram, Gangaram and Rajaram


Working Notes:
1. ₹ 1,100 is recovered from Rajaram’s estate which is recorded on the credit side of Rajaram’s Capital Account and on the debit side of Bank A/c.

2. Capital deficiency of Rajaram = Debit total of Capital A/c – Credit total of Capital A/c
= 18,000 – 15,900
= ₹ 2,100
The deficit amount of Rajaram A/c ₹ 2,100 is distributed among continuing partners’ in 2 : 1 ratio.

Question 9.
Following is the Balance Sheet of Vaibhav, Sanjay, and Santosh
Balance Sheet as of 31st March 2019

Santosh is declared insolvent so the firm is dissolved and assets realised as follows:
1. Stock and Debtors ₹ 54,000, Goodwill – NIL, Machinery at book value.
2. Creditors allowed a discount of 10%.
3. Santosh could pay only 25 paise in the rupee of the balance due.
4. Profit sharing ratio was 8 : 4 : 3.
5. A contingent liability against the firm ₹ 9,000 is cleared.
Give Ledger Account to close to books of the firm.
Solution:
In the books of Vaibhav, Sanjay, and Santosh


Working Notes:
1. Contingent liability paid, so Realisation A/c is debited and Bank A/c is credited.
2. Santosh could pay only 25 paise in a rupee of the balance due i.e.
Balance due from Santosh (Debit side of Partners Capital A/c) = ₹ 10,560
25% of ₹ 10,560 = ₹ 2,640 (Amount recorded on debit side of Bank A/c)
Capital deficiency of Santosh = 10,560 – 2,640 = ₹ 7,920
₹ 7,920 to be distributed among continuing partner in their profit-loss ratio = 8 : 4 i.e. 2 : 1.
7,920 × \(\frac{2}{3}\) = ₹ 5,280
7,920 × \(\frac{1}{3}\) = ₹ 2,640

(When Two Partners become Insolvent)

Question 10.
Shweta, Nupur, and Sanika are partners sharing profits and losses in the ratio of 3 : 2 : 1. Their Balance Sheet as of 31st March 2019 was as follows:
Balance Sheet as of 31st March 2019

The firm is dissolved on 31st March 2019. Sundry assets realised @ 60% of its book value. Realisation expenses ₹ 2,000 paid by Shweta. Nupur and Sanika both are insolvent.
Nupur’s private estate has got a surplus of ₹ 3,000 and that of Sanika ₹ 8,000.
Show necessary Ledger Accounts to close the books of the firm.
Solution:
In the books of Shweta, Nupur and Sanika

(When All Partners become Insolvent)

Question 11.
Following is the Balance Sheet as of 31st March 2019 of a firm having three partners Priti, Priya, and Prachi.
Balance Sheet as of 31st March 2019

The firm was dissolved due to the insolvency of all the partners. Machinery was sold for ₹ 18,000, while Furniture fetched ₹ 14,000, Stock realized ₹ 35,000. Realisation expenses amounted to ₹ 2,000. Nothing could be recovered from Priya and Prachi, but ₹ 3,400 could be collected from Priti’s private estate.
Close the books of accounts of the firm.
Solution:
In the books of Priti, Priya, and Prachi



Working Notes:
1. Amount paid to loan from sale of machinery = ₹ 18,000
Balance of Loan 30,000 – 18,000 = ₹ 12,000

2. Ratio of Trade creditors and Loan = 50,000 : 12,000
= 50 : 12
= 25 : 6

3. Balance of cash available = 10,000 + 67,000 + 3,400 – 18,000 – 2,000
= 80,400 – 20,000
= ₹ 60,400
Amount paid towards loan = \(\frac{6}{31} \times \frac{60,400}{1}\) = ₹ 11,690
Amount paid to Trade creditors = \(\frac {25}{31}\) × 60,400 = ₹ 48,710
Amount paid towards loan = 18,000 + 11,690 = ₹ 29,690.

Question 12.
Shashwat and Shiv are equal partners. Their Balance Sheet stood as under:
Balance Sheet as of 31st March 2019

Due to weak financial position, all partners were declared bankrupt.
The Assets were realised as follows:
Stock ₹ 3,500, Furniture ₹ 2,000, Debtors ₹ 5,000 and Machinery ₹ 7,000.
The cost of collection and distributing the estate amounted to ₹ 1,500. Shashwat’s private estate is not sufficient even to pay his private debts, whereas in Shiv’s private estate there is a surplus of ₹ 500.
Prepare necessary Ledger Accounts to close the books of the firm.
Solution:
In the books of Shashwat and Shiv


Working Note:
As partners we’re not able to pay their loss amount, a difference of amount is considered as deficiency of partners.