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Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 5 Chemical Bonding Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 5 Chemical Bonding

Question 1.
Explain electronic theory of valence.
Answer:
Electronic theory of valence:

• Electronic theory of valence was proposed by Kossel and Lewis in 1916.
• They gave a logical explanation of valence which was based on the inertness of noble gases (that is, octet rule developed by Lewis).
• According to Lewis, the atom can be pictured in terms of a positively charged ‘kernel’ (the nucleus plus inner electrons) and outer shell that can accommodate a maximum of eight electrons. This octet of electrons represents a stable electronic arrangement.
• Thus, according to this theory, during the formation of a chemical bond, each atom loses, gains or shares outer electrons so that it achieves stable octet.
• The formation of NaCl involves transfer of one electron from sodium (Na) to chlorine (Cl). Na⁺ and Cl^– ions are formed which are held together by chemical bond. The formation of H₂, F₂, Cl₂, HCl, etc., involves sharing of a pair of electrons between the atoms. In both the cases, each atom attains a stable outer octet of electrons.

Question 2.
Give the significance of octet rule. Explain why this rule is not valid for H and Li atoms.
Answer:
i. Significance: Octet rule is found to be very useful:

• in explaining the normal valence of elements
• in the study of the chemical combination of atoms leading to the formation of molecule.

ii. Octet rule is not valid for H and Li atoms. According to octet rule, during the formation of a chemical bond, each atom loses, gains or shares electrons so that it achieves stable octet (eight electrons in the valence shell). However, H and Li atoms tend to have only two electrons in their valence shell similar to that of Helium (1s²), which called duplet. Hence, octet rule is not valid for H and Li atoms.

Question 3.
Define ionic bond.
Answer:
The bond formed by complete transfer of one or more electrons from an electropositive atom to an electronegative atom, leading to formation of ions which are held together by electrostatic attraction is called ionic bond or electrovalent bond.

Question 4.
Explain the formation of ionic bond in sodium chloride (NaCl).
Answer:
Formation of sodium chloride (NaCl):
i. The electronic configurations of sodium and chlorine are:
Na (Z = 11): 1s² 2s² 2p⁶ 3s¹ or (2, 8, 1)
Cl (Z = 17): 1s² 2s² 2p⁶ 3s² 3p⁵ or (2, 8, 7)
ii. Sodium has one electron in its valence shell. It has tendency to lose one electron to acquire the electronic configuration of the nearest inert gas, neon (2, 8).
iii. Chlorine has seven electrons in its valence shell. It has tendency to gain one electron and thereby acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
iv. During the combination of sodium and chlorine atoms, the sodium atom transfers its valence electron to the chlorine atom.
v. Sodium atom changes into Na+ ion while the chlorine atom changes into Cl^– ion. The two ions are held together by strong electrostatic force of attraction.
vi. The formation of ionic bond between Na and Cl can be shown as follows:

Question 5.
Explain the formation of ionic bonds in calcium chloride (CaCl₂).
Answer:
Formation of calcium chloride (CaCl₂):
i. The electronic configurations of calcium and chlorine are:
Na (Z = 11): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² or (2, 8, 8, 2)
Cl (Z = 17): 1s² 2s² 2p⁶ 3s² 3p⁵ or (2, 8, 7)
ii. Calcium has two electrons in its valence shell. It has tendency to lose two electrons to acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
iii. Chlorine has seven electrons in its valence shell. It has tendency to gain one electron and thereby acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
iv. During the combination of calcium and chlorine atoms, the calcium atom transfers its valence electrons to two chlorine atoms.
v. Calcium atom changes into Ca²⁺ ion while the two chlorine atoms change into two Cl^– ions. These ions are held together by strong electrostatic force of attraction.
vi. The formation of ionic bond(s) between Ca and Cl can be shown as follows:

Question 6.
What are ionic solids?
Answer:
Ionic solids are solids which contain cations and anions held together by ionic bonds.
e.g. Sodium chloride (NaCl), Calcium chloride (CaCl₂)

Question 7.
Define lattice enthalpy.
Answer:
Lattice enthalpy of an ionic solid is defined as the energy required to completely separate one mole of solid ionic compound into the gaseous components.
Note: Lattice enthalpy values of some ionic compounds:

Question 8.
Arrange NaCl, CaCl₂ and AlCl₃ in increasing order of lattice enthalpy (positive value). Justify your answer.
Answer:
Compounds having cations with higher charge have large lattice enthalpy (higher positive value) than compounds having cations with lower charge.
Hence, the correct order is NaCl < CaCl₂ < AlCl₃.

Question 9.
Lattice enthalpy of LiF is more than that of NaF. Explain.
Answer:
As the size of the cation decrease, lattice enthalpy increases. Li⁺ ion is smaller than Na⁺ ion. Hence, lattice enthalpy of LiF is more than that of NaF.

Question 10.
Define covalent bond.
Answer:
The attractive force which exists due to the mutual sharing of electrons between the two atoms of similar electronegativity or having small difference in electronegativities is called a covalent bond.

Question 11.
Explain the formation of covalent bond in H₂ molecule.
Answer:
Formation of H₂ molecule:
i. The electronic configuration of H atom is 1s¹.
ii. It needs one more electron to complete its valence shell.
iii. When two hydrogen atoms approach each other at a certain internuclear distance, they share their valence electrons.
iv. The shared pair of electrons belongs equally to both the hydrogen atoms. The two atoms are said to be linked by a single covalent bond and a H₂ molecule is formed.

Question 12.
Explain the formation of covalent bond in Cl₂ molecule.
Answer:
Formation of Cl₂ molecule:
i. The electronic configuration of Cl atom is [Ne] 3s² 3p⁵.
ii. It needs one more electron to complete its valence shell.
iii. When two chlorine atoms approach each other at a certain internuclear distance, they share their valence electrons. In the process, both the atoms attain the valence shell of octet of nearest noble gas, argon.
iv. The shared pair of electrons belongs equally to both the chlorine atoms. The two atoms are said to be linked by a single covalent bond and a Cl₂ molecule is formed.

Question 13.
What are the important features of covalent bond?
Answer:

• Each covalent bond is formed as a result of sharing of electron pair between the two atoms.
• When a covalent bond is formed, each combining atom contributes one electron to the shared pair.
• The combining atoms attain the outer shell noble gas configuration as a result of the sharing of electrons.

Question 14.
Explain the types of covalent bond with suitable examples.
Answer:
The three types of covalent bonds are as follows.
i. Single bond: When two combining atoms share one electron pair, the covalent bond between them is called single bond.
Single bond is observed in number of molecules.
e.g. H₂, Cl₂, water molecule, etc.

ii. Double bond: When two combining atoms share two pairs of electrons, the covalent bond between them is called a double bond, e.g. Double bond is present in C₂H₄ molecule

iii. Triple bond: When two combining atoms share three pairs of electrons, the covalent bond between them is called a triple bond, e.g. Triple bond is present in N₂ molecule.

Note: Formation of covalent bonds in CO₂, CCl₄ and C₂H₂:

Question 15.
Distinguish between ionic bond and covalent bond.
Answer:
Ionic bond:

1. It is formed by the transfer of electrons from one atom to another.
2. It is formed by the transfer of electrons from one atom to another.
3. In this, oppositely charged ions are formed.
4. There are no multiple ionic bonds.
5. This bond usually exists between metal and non-metal atoms.
6. e.g. NaCl, CaCl₂

Covalent bond:

1. It is formed by sharing of electrons.
2. Atoms are held together due to shared pair of electrons.
3. In this, oppositely charged ions are not formed.
4. Covalent bonds may be single or double or triple bonds.
5. This bond usually exists between non-metal atoms.
6. e.g. H₂, Cl₂

Question 16.
What are the steps to write Lewis dot structure?
Answer:
Steps to write Lewis dot structures:

• Add the total number of valence electrons of combining atoms in the molecule.
• In anions, add one electron for each negative charge.
• In cations, subtract one electron from valence electrons for each positive charge.
• Write skeletal structure of the molecule to show the atoms and number of valence electrons forming the single bond between the atoms.
• Add remaining electron pairs to complete the octet of each atom.
• If octet is not complete form multiple bonds between the atoms such that octet of each atom is complete.
• In polyatomic atoms and ions, the least electronegative atom is the central atom.
  e.g. In \(\mathrm{SO}_{4}^{2-}\) ion, ‘S’ is the central atom and in \(\mathrm{NO}_{3}^{-}\), ‘N’ is the central atom.
• After writing the number of electrons as shared pairs forming single bonds, the remaining electron pairs are used either for multiple bonds or they remain as lone pairs.

Question 17.
Write the Lewis structure of nitrite ion, \(\mathrm{NO}_{2}^{-}\).
Answer:
Step I: Count the total number of valence electrons of nitrogen atom, oxygen atom and one electron of additional negative charge.
Valence shell configuration of nitrogen and oxygen are:
N ⇒ (2s² 2p³), O ⇒ (2s² 2p⁴)
The total electrons available are:
5 + (2 × 6) + 1 = 18 electrons
Step II: The skeletal structure of \(\mathrm{NO}_{2}^{-}\) is written as O N O
Step III: Draw a single bond i.e., one shared electron pair between the nitrogen and each oxygen atoms. Then distribute the remaining electrons to achieve noble gas configuration for each atom. This does not complete the octet of nitrogen.

Hence, there is a multiple bond between nitrogen and one of the oxygen atoms (a double bond). The remaining two electrons constitute a lone pair on nitrogen.
Following are Lewis dot structures of \(\mathrm{NO}_{2}^{-}\).

Question 18.
Write the Lewis structure of CO molecule.
Answer:
Step I: Count number of electrons of carbon and oxygen atoms. The valence shell configuration of carbon and oxygen atoms are: 2s² 2p² and 2s² 2p⁴, respectively. The valence electrons available are:
4 + 6=10
Step II: The skeletal structure of CO is written as
C O
Step III: Draw a single bond (One shared electron pair) between C and O and complete the octet on O. The remaining two electrons is a lone pair on C.

The octet on carbon is not complete. Hence, there is a multiple bond between C and O (a triple bond between C and O atom). This satisfies the octet rule for carbon and oxygen atoms.
The Lewis structure of CO molecule is:

Question 19.
Explain the term formal charge.
Answer:
i. Formal charge is the charge assigned to an atom in a molecule, assuming that all electrons are shared equally between atoms, regardless of their relative electronegativities.
ii. While determining the best Lewis structure per molecule, the structure is chosen such that the formal charge is as close to zero as possible.
iii. The structure having the lowest formal charge has the lowest energy.
iv. Formal charge is assigned to an atom based on electron dot structures of the molecule/ion.
v. Formal charge on an atom in a Lewis structure of a polyatomic species can be determined using the following formula:

Question 20.
Explain the calculation of the formal charge on oxygen atoms in case of O₃ (ozone) molecule.
Answer:
i. Lewis dot structure of O₃ (ozone) molecule is:

Three oxygen atoms are present in the O₃ molecule and are labelled as 1, 2 and 3.
ii. Formal charges on oxygen atoms labelled as 1, 2, 3 are calculated as shown below:

iii. On the basis of the formal charge values, O₃ is shown as

[Note: Indicating the charges on the atoms in the Lewis structure helps in keeping track of the valence electrons in the molecule. Formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given species.]

Question 21.
CO₂ can be represented by following three structures:

Calculate the formal charge on each atom in all the three structures of CO₂ molecule. Identify the structure with lowest energy.
Answer: Formal charges on atoms labelled as 1, 2, 3 are calculated as shown below:
Structure (I):

Structure (II):

Structure (III):

While determining the best Lewis structure per molecule, the structure is chosen such that the formal charge is as close to zero as possible. The structure having the lowest formal charge has the lowest energy.

In structure (I), the formal charge on each atom is 0 while in structures (II) and (III) formal charge on carbon is 0 while oxygens have formal charge -1 or +1. Hence, the possible structure with the lowest energy will be structure (I). Thus, formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given species.

Question 22.
Find out the formal charges on S, C and N.
(S = C = N)^– ; (S – C ≡ N)^– ; (S ≡ C – N)^–
Answer:
Step I:
Write Lewis dot diagrams for the structures:

Step II:
Assign formal charges for all the atoms:
F.C. = V.E. – N.E. – 1/2 (B.E.)
Structure A:
Formal charge on S = 6 – 4 – 1/2(4) = 0
Formal charge on C = 4 – 0 – 1/2 (8) = 0
Formal charge on N = 5 – 4 – 1/2 (4) = -1
Structure B:
Formal charge on S = 6 – 6 – 1/2(2) = -1
Formal charge on C = 4 – 0 – 1/2 (8) = 0
Formal charge on N = 5 – 2 – 1/2 (6) = 0
Structure C:
Formal charge on S = 6 – 2 – 1/2(6) = +1
Formal charge on C = 4 – 0 – 1/2(8) = 0
Formal charge on N = 5 – 6 – 1/2(2) = -2

Question 23.
Give the limitations of octet rule:
Answer:
Limitations of octet rule:
i. Octet rule does not explain stability of some molecules.
The octet rule is based on the inert behaviour of noble gases, which have their octet complete i.e., have eight electrons in their valence shell. It is very useful to explain the structures and stability of organic molecules. However, there are many molecules whose existence cannot be explained by the octet theory. The central atoms in these molecules does not have eight electrons in their valence shell, and yet they are stable.
Such molecules can be categorized as having:
a. Incomplete octet
b. Expanded octet
c. Odd electrons

a. Molecules with incomplete octet: e.g. BF₃, BeCl₂, LiCl
In these covalent molecules, the atoms B, Be and Li have less than eight electrons in their valence shell but these molecules are stable.
Li in LiCl has only two electrons, Be in BeCl₂ has four electrons while B in BF₃ has six electrons in the valence shell.

b. Molecules with expanded octet: Some molecules like SF₆, PCl₅, H₂SO₄ have more than eight electrons around the central atom.

c. Odd electron molecules:
Some molecules like NO (nitric oxide) and NO₂ (nitrogen dioxide) do not obey the octet rule. These molecules, have odd number of valence electrons.

ii. The observed shape and geometry of a molecule, cannot be explained, by the octet rule.
iii. Octet rule fails to explain the difference in energies of molecules, though all the covalent bonds are formed in an identical manner, that is, by sharing a pair of electrons. The rule fails to explain the differences in reactivities of different molecules.
Note: Sulphur also forms many compounds in which octet rule is obeyed. For example, in sulphur dichloride, the sulphur atom has eight electrons around it.

Question 24.
State the basic idea on which VSEPR theory was proposed by Sidgwick and Powell.
Answer:
Valence Shell Electron Pair Repulsion (VSEPR) theory is based on the basic idea that the electron pairs on the atoms shown in the Lewis diagram repel each other. In the real molecule, they arrange themselves in such a way that there is minimum repulsion between them.

Question 25.
Give the rules of VSEPR Theory.
Answer:
Rules of VSEPR Theory:
i. Electron pairs arrange themselves in such a way that repulsion between them is minimum.
ii. The molecule acquires minimum energy and maximum stability.
iii. Lone pair of electrons also contribute in determining the shape of the molecule.
iv. Repulsion of other electron pairs by the lone pair (L.P.) stronger than that of bonding pair (B.P.).
Trend for repulsion between electron pair is as follows:
L.P. – L.P. > L.P. – B.P. > B.P. – B.P.
Lone pair-Lone pair repulsion is maximum because this electron pair is under the influence of only one nucleus while the bonded pair is shared between two nuclei.
Thus, the number of lone pair and bonded pair of electrons decide the shape of the molecules. Molecules having no lone pair of electrons have a regular geometry.

Note:
i. Electron pair geometry: The arrangement of electrons around the central atom is called as electron pair geometry. These electron pairs may be shared in a covalent bond or they may be lone pairs.
ii. Geometry of some molecules (having no lone pair of electrons):


iii. Geometry of some molecules (having one or more lone pairs of electrons):

Question 26.
Match the following:

Answer:
i – a,
ii – b,
iii – d,
iv – c

Question 27.
Complete the following table:

Answer:

Question 28.
Explain geometry of NH₃ molecule according to VSEPR theory.
Answer:

• In NH₃ molecule, the central atom nitrogen has five electrons in its valence shell. On bond formation with three hydrogen atoms, there are 8 electrons in the valence shell of nitrogen. Out of these, three pairs are bond pairs (N – H covalent bonds) and one forms lone pair. The expected geometry is tetrahedral and bond angle is 109° 28′.
• There are two types of repulsions between the electron pairs: Lone pair – bond pair and bond pair – bond pair
• The lone pair – bond pair repulsions are stronger and the bonded pairs are pushed inwards. Thus, reducing the bond angle to 107°18′ and shape of the molecule becomes trigonal pyramidal.

Diagram:

Question 29.
Explain geometry of H₂O molecule according to VSEPR theory.
Answer:

• In H₂O molecule, the central atom oxygen has six electrons in its valence shell. On bond formation with two hydrogen atoms, there are 8 electrons in the valence shell of oxygen. Out of these, two pairs are bond pairs and two are lone pairs.
• Due to lone pair – lone pair repulsion, the lone pairs are pushed towards the bond pairs and bond pair – lone pair repulsions become stronger thereby reducing the H-O-H bond angle from 109° 28′ to 104° 35′ and the geometry of the molecule becomes angular (bent).

Diagram:

Question 30.
Write the postulates of Valence Bond Theory.
Answer:
Postulates of Valence Bond Theory:

• A covalent bond is formed when the half-filled valence orbital of one atom overlaps with the half-filled valence orbital of another atom.
• The electrons in the half-filled valence orbitals must have opposite spins.
• During bond formation, the half-filled orbitals overlap and the opposite spins of the electrons get neutralized. The increased electron density decreases the nuclear repulsion and energy is released during overlapping of the orbitals.
• Greater the extent of overlap, stronger is the bond formed. However, complete overlap of orbitals does not take place due to intemuclear repulsions.
• If an atom possesses more than one unpaired-electrons, then it can form more than one bond. So, number of bonds formed will be equal to the number of half-filled orbitals in the valence shell i.e., number of unpaired electrons.
• The distance at which the attractive and repulsive forces balance each other is the equilibrium distance between the nuclei of the bonded atoms. At this distance, the total energy of the two bonded atoms is minimum and stability of the molecule is maximum.
• Electrons which are paired in the valence shell cannot participate in bond formation. However, in an atom if there is one or more vacant orbital present then these electrons can unpair and participate in bond formation provided the energies of the filled and vacant orbitals differ slightly from each other.
• During bond formation, the ‘s’ orbital which is spherical can overlap in any direction. The ‘p’ orbitals can overlap only in the x, y or z directions. Similarly, ‘d’ and ‘f orbitals are oriented in certain directions in space and overlap only in these directions. Thus, the covalent bond is directional in nature.

Note: In order to explain the covalent bonding, Heitler and London developed the valence bond theory on the basis of wave mechanics. This theory was further extended by Pauling and Slater.

Question 31.
Explain the formation of hydrogen molecule with the help of potential energy curve.
OR
Explain the formation of H₂ on the basis of VBT.
Answer:
Formation of H₂ on the basis of VBT:

• Hydrogen atom has electronic configuration 1s¹. It contains one unpaired electron in its valence shell.
• When the two hydrogen atoms containing unpaired electrons with opposite spins are separated by a large distance, they can neither attract nor repel each other (there are no interactions between them). The energy of the system is the sum of the potential energies of the two atoms which is arbitrarily taken as zero.
• When the two atoms approach each other, attractive and repulsive forces begin to operate on them. Experimentally, it has been found that during formation of hydrogen molecule, the magnitude of the newly developed attractive forces contributes more than the newly developed repulsive forces. As a result, the potential energy of the system begins to decrease.
• As the atoms come closer to one another the energy of the system decreases. The overlap of the atomic orbitals increases only up to a certain distance between the two nuclei, where the attractive and repulsive forces balance each other and the system attains minimum energy. At this stage, a stable bond is formed between the two atoms.
• If the distance between the two atoms is decreased further, the repulsive forces exceed the attractive forces and the energy of the system increases and stability decreases.
• When two hydrogen atoms with electrons having parallel spin approach each other, the potential energy of the system increases and bond formation does not take place.

Question 32.
Define:
i. Sigma overlap
ii. pi overlap
Answer:
i. Sigma overlap (σ bond): When two half-filled orbitals of two atoms overlap along the internuclear axis, it is called as sigma overlap or sigma bond.
ii. pi overlap (π bond): When two half-filled orbitals of two atoms overlap side-ways (laterally), it is called as π overlap or π bond.

Question 33.
Explain with example:
i. s-s σ overlap
ii. p-p σ overlap
iii. s-p a overlap
Answer:
i. s-s σ overlap:
a. The overlap between two half-filled s orbitals of two different atoms containing unpaired electrons with opposite spins is called s-s overlap.
e.g. Formation of H₂ molecule by s-s overlap:
Hydrogen atom (Z = 1) has electronic configuration: 1s¹. The 1s¹ orbitals of two hydrogen atoms overlap along the internuclear axis to form a σ bond between the atoms in H₂ molecule.
b. Diagram:

ii. p-p σ overlap:
a. This type of overlap takes place when two p orbitals from different atoms overlap along the internuclear axis.
e.g. Formation of F₂ molecule by p-p overlap:
Fluorine atom (Z = 9) has electronic configuration 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{2}\) \(2 \mathrm{p}_{\mathrm{y}}^{2}\) \(2 \mathrm{p}_{\mathrm{z}}^{1}\).
During the formation of F₂ molecule, half-filled 2p_z orbital of one F atom overlaps with similar half-filled 2p_z orbital containing electron with opposite spin of another F atom axially and a p-p σ bond is formed.
b. Diagram:

iii. s-p σ overlap:
a. In this type of overlap one half filled s orbital of one atom and one half filled p orbital of another orbital overlap along the internuclear axis.
e.g. Formation of HF molecule by s-p overlap:
Hydrogen atom (Z = 1) has electronic configuration: 1s¹ and fluorine atom (Z = 9) has electronic configuration 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{2}\) \(2 \mathrm{p}_{\mathrm{y}}^{2}\) \(2 \mathrm{p}_{\mathrm{z}}^{1}\). During the formation of HF molecule, half-filled 1s orbital of hydrogen atom overlaps coaxially with half-filled 2p_z orbital of fluorine atom with opposite electron spin and an s-p σ bond is formed.
b. Diagram:

Question 34.
Explain the formation of π bond with diagram.
OR
Explain π overlap with diagram.
Answer:

• When two half-filled p orbitals of two atoms overlap side-ways (laterally), it is called π overlap and the bond formed is called π bond.
• π bond is perpendicular to the intemuclear axis.

Diagram:

Question 35.
Identify the type of bond formed:


Answer:
i. σ bond
ii. π bond
iii. σ bond
iv. σ bond

Question 36.
Explain divalency of beryllium, though number of unpaired electrons in a beryllium atom is zero.
Answer:
Beryllium (Z = 4) has electronic configuration 1s² 2s² in its ground state.
When one of the 2s electrons of Be is promoted to the vacant 2p orbital, the electronic configuration of Be in its excited state becomes 1s² 2s¹ \(2 \mathrm{p}_{\mathrm{x}}^{1}\). This is called formation of an excited state and it has two unpaired electrons. Hence, though the number of unpaired electrons in the ground state of Be atom is zero, beryllium shows divalency.

Question 37.
Define the term hybridization.
Answer:
Hybridization is defined as the process of mixing of valence orbitals of same atom and recasting them into equal number of new equivalent orbitals (hybrid orbitals).

Question 38.
Explain in detail the steps involved in hybridization.
Answer:
Steps involved in hybridization:
i. Formation of the excited state:
a. The paired electrons in the ground state are uncoupled and one electron is promoted to the vacant orbital having slightly higher energy.
b. Now, total number of half-filled orbitals is equal to the valency of the element in the stable compound, e.g. In BeF₂, valency of Be is two. In the excited state, one electron from 2s orbital is uncoupled and promoted to 2p orbital.

ii. Mixing and recasting:

• In this step, the two ‘s’ and ‘p’ orbitals having slightly different energies mix with each other.
• Redistribution of electron density and energy takes place and two new orbitals having exactly same shape and energy are formed.
• These new orbitals arrange themselves in space in such a way that there is minimum repulsion and maximum separation between them. e.g. During formation of sp hybrid orbitals as in Be, the two sp hybrid orbitals form an angle of 180° with each other.

Question 39.
List the important conditions required for hybridization.
Answer:
Conditions for hybridization:

• Orbitals belonging to the same atom can participate in hybridization.
• Orbitals having nearly same energy can undergo hybridization.

[Note: 2s and 2p orbitals of the same atom undergo hybridization but 3s and 2p orbitals of the same atom do not.]

Question 40.
Enlist the characteristic features of hybrid orbitals.
Answer:
Characteristic features of hybrid orbitals:

• Number of hybrid orbitals formed is exactly the same as the participating atomic orbitals.
• They have same energy and shape.
• Hybrid orbitals are oriented in space in such a way that there is minimum repulsion and thus are directional in nature.
• The hybrid orbitals are different in shape from the participating atomic orbitals, but they bear the characteristics of the atomic orbitals from which they are derived.
• Each hybrid orbitals can hold two electrons with opposite spins.
• A hybrid orbital has two lobes on the two sides of the nucleus. One lobe is large and the other small.
• Covalent bonds formed by hybrid orbitals are stronger than those formed by pure orbitals, because the hybrid orbital has electron density concentrated on the side with a larger lobe and the other is small allowing greater overlap of the orbitals.

Question 41.
Explain with diagrams:
i. sp³ hybridization
ii. sp² hybridization
iii. sp hybridization
Answer:
i. sp³ hybridization:
In this type, one s and three p orbitals having comparable energy mix and recast to form four sp³ hybrid orbitals, ‘s’ orbital is spherically symmetrical while the p_x, p_y, p_z, orbitals have two lobes and are directed along x, y and z axes, respectively.

The four sp³ hybrid orbitals formed are equivalent in energy and shape. They have one large lobe and one small lobe. They are at an angle of 109° 28′ with each other in space and point towards the comers of a tetrahedron. CH₄, NH₃, H₂O are examples where the orbitals on central atom undergo sp³ hybridization.

Diagram:

ii. sp² hybridization:
This hybridization involves the mixing of one s and two p orbitals to give three sp² hybrid orbitals of same energy and shape. The three orbitals are maximum apart and oriented at an angle of 120° and are in one plane. The third p orbital does not participate in hybridization and remains at right angles to the plane of the sp² hybrid orbitals. BF₃, C₂H₄ molecules are examples of sp² hybridization.
Diagram:

iii. sp hybridization:
In this type, one s and one p orbital undergo mixing and recasting to form two sp hybrid orbitals of same energy and shape. The two hybrid orbitals are placed at an angle of 180°. Other two p orbitals do not participate in hybridization and are at right angles to the hybrid orbitals. For example, BeCl₂ and acetylene molecule (HC ≡ CH).
Diagram:

Question 42.
Explain the formation of an ammonia molecule on the basis of hybridization.
Answer:
Formation of an ammonia (NH₃) molecule on the basis of sp³ hybridization:
i. Ammonia molecule (NH₃) has one nitrogen atom and three hydrogen atoms.
ii. The ground state electronic configuration of nitrogen (Z = 7) is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\) \(2 \mathrm{p}_{\mathrm{y}}^{1}\) \(2 \mathrm{p}_{\mathrm{z}}^{1}\)
Electronic configuration of nitrogen:

iii. The ground state electronic configuration explains the observed valency of nitrogen in NH₃ which is three.
iv. The 2s, 2p_x, 2p_y and 2p_z orbitals of nitrogen atom mix and recast to form four sp³ hybrid orbitals of equivalent energy. These orbitals are tetrahedrally oriented in space. One of the sp³ hybrid orbital contains a lone pair of electrons.
v. Three half-filled sp³ hybrid orbitals of N atom overlap axially with half-filled 1s orbital of three different hydrogen atoms to form three N-H (sp³-s) sigma covalent bonds.
vi. Since, there is one lone pair of electrons in one of the sp³ hybrid orbitals of nitrogen, there is repulsion between lone pair and bonding pair of electrons. As a result, the H-N-H bond angle is reduced from regular tetrahedral angle 109° 28′ to 107° 18′. Geometry of NH₃ molecule is pyramidal or distorted tetrahedral.

Question 43.
Explain the formation of water (H₂O) molecule on the basis of hybridization.
Answer:
Formation of water (H₂O) molecule on the basis of sp³ hybridization:
i. Water molecule (H₂O) has one oxygen atom and two hydrogen atoms.
ii. The ground state electronic configuration of oxygen (Z = 8) is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{2}\) \(2 \mathrm{p}_{\mathrm{y}}^{1}\) \(2 \mathrm{p}_{\mathrm{z}}^{1}\).
Electronic configuration of oxygen:

iii. The ground state electronic configuration explains the observed valency of oxygen in H₂O molecule which is 2.
iv. The 2s, 2p_x, 2p_y and 2p_z orbitals of oxygen atom mix and recast to form four sp³ hybrid orbitals of equivalent energy. These orbitals are tetrahedrally oriented in space. Two of the sp hybrid orbitals contain lone pair of electrons.
v. Two half-filled sp³ hybrid orbitals of O atom overlap axially with half-filled 1s orbitals of two different hydrogen atoms to fonn two O-H (sp³-s) sigma covalent bonds.
vi. Since, there are two lone pairs of electrons in two of the sp³ hybrid orbitals of oxygen, there is repulsion between lone pair and bonding pair of electrons. As a result, the H-O-H bond angle is reduced from regular tetrahedral angle 109°28′ to 104°35′. The geometry of H₂O molecule is angular or V shaped.

Diagram:

Question 44.
Explain the formation of an ethene molecule on the basis of hybridization.
Answer:
Formation of an ethene (ethylene) molecule on the basis of sp² hybridization:
i. Ethene molecule (C₂H₄) has two carbon atoms and four hydrogen atoms.
ii. The ground state electronic configuration of C (Z = 6) is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\) \(2 \mathrm{p}_{\mathrm{y}}^{1}\) \(2 \mathrm{p}_{\mathrm{z}}^{0}\).
Electronic configuration of carbon:

iii. One electron from 2s orbital of each carbon atom is excited to the 2p_z orbital. Then each carbon atom undergoes sp2 hybridization.
iv. One ’s’ orbital and two ‘p’ orbitals on carbon hybridize to form three sp² hybrid orbitals of equal energy and symmetry.
v. Two sp² hybrid orbitals overlap axially two ‘s’ orbitals of hydrogen to form sp²-s σ bond. The unhybridized ‘p’ orbitals on the two carbon atoms overlap laterally to form a π bond. Thus, the C₂H₄ molecule has four sp²-s σ bonds, one sp²-sp² σ bond and one p-p π bond.
vi. Each H-C-H and H-C-C bond angle in ethene molecule is 120°. All the six atoms in ethene (ethylene) molecule are in one plane. Geometry of the molecule at each carbon atom is trigonal planar.

Question 45.
Explain the formation of boron trifluoride on the basis of hybridization.
Answer:
Formation of boron trifluoride on the basis of sp² hybridization:
i. Boron trifluoride (BF₃) has one boron atom and three fluorine atoms.
ii. Observed valency of boron in BF₃ molecule is three and its geometry is trigonal planar. This can be explained on the basis of sp² hybridization.
iii. The ground state electronic configuration of B (Z = 5) is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\) \(2 \mathrm{p}_{\mathrm{y}}^{0}\) \(2 \mathrm{p}_{\mathrm{z}}^{0}\).
Electronic configuration of boron:

iv. One electron from 2s orbital of boron atom is uncoupled and promoted to vacant 2p_y orbital.
v. The three orbitals i.e. 2s, 2p_x of and 2p_y of boron undergoes sp² hybridization to form three sp² hybrid orbitals of equivalent energy, which are oriented along the three comers of an equilateral triangle making an angle of 120°.
vi. Each sp² hybrid orbital of boron atom having unpaired electron overlaps axially with half-filled 2p_z orbital of fluorine atom containing electron with opposite spin to form three B-F sigma bonds by sp²-p type of overlap.
vii. Each F-B-F bond angle in BF₃ molecule is 120°. The geometry of BF₃ molecule is trigonal planar.

Diagram:

Question 46.
Explain the formation of an acetylene molecule on the basis of hybridization.
Answer:
Formation of acetylene (ethyne) molecule on the basis of sp hybridization:
i. Acetylene molecule (C₂H₂) has two carbon atoms and two hydrogen atoms.
ii. The ground state electronic configuration of C (Z = 6) is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\) \(2 \mathrm{p}_{\mathrm{y}}^{1}\) \(2 \mathrm{p}_{\mathrm{z}}^{0}\).
Electronic configuration of carbon:

iii. Each carbon atom undergoes sp hybridization. One s and one p orbitals mix and recast to give two sp hybrid orbitals arranged at 180° to each other.
iv. Out of the two sp hybrid orbitals of carbon atom, one overlaps axially with s orbital of hydrogen while the other sp hybrid orbital overlaps with sp hybrid orbital of other carbon atom to form the sp-sp σ bond. The C H σ bond is formed by sp-s overlap.
v. The remaining two unhybridized p orbitals overlap laterally to form two p-p π bonds. So, there are three bonds between the two carbon atoms: one C-C σ bond (sp-sp) overlap, two C-C π bonds (p-p) overlap. There are two sp-s σ bonds in acetylene (one between each C and H).
vi. Each H-C-C bond angle in ethyne molecule is 180°. All the four atoms in ethyne molecule are in a straight line. The geometry of acetylene molecule is linear.

Diagram:

Question 47.
Explain the formation of BeCl₂.
Answer:
Formation of BeCl₂:
i. BeCl₂ molecule has one Be atom and two chlorine atoms.
ii. Electronic configuration of Be is 1s² 2s² \(2 \mathrm{p}_{\mathrm{z}}^{0}\).
Electronic configuration of beryllium:

iii. The 2s and 2p_z orbitals undergo sp hybridization to form two sp hybrid orbitals oriented at 180° with each other. 2p_z orbitals of two chlorine atoms overlap with the sp hybrid orbitals to form two sp-p σ bonds.
Cl – Be – Cl bond angle is 180°. The geometry of the molecule is linear.

Diagram:

Question 48.
Match the following:

Answer:
i – b
ii – a
iii – d
iv – c

Question 49.
Give the importance of valence bond theory.
Answer:
Valence Bond theory introduced five new concepts in chemical bonding:

• Delocalization of electron over the two nuclei
• Shielding effect of electrons
• Covalent character of bond
• Partial ionic character of a covalent bond
• The concept of resonance and connection between resonance energy and molecular stability

Question 50.
What are the limitations of valence bond theory?
Answer:
Limitations of valence bond theory:

• Valence Bond theory explains only the formation of covalent bond in which a shared pair of electrons comes from two bonding atoms. However, it offers no explanation for the formation of a coordinate covalent bond in which both the electrons are contributed by one of the bonded atoms.
• Oxygen molecule is expected to be diamagnetic according to this theory. The two atoms in oxygen molecule should have completely filled electronic shells which give no unpaired electrons to the molecule making it diamagnetic. However, experimentally the molecule is found to be paramagnetic having two unpaired electrons. Thus, this theory fails to explain paramagnetism of oxygen molecule.
• Valence bond theory does not explain the bonding in electron deficient molecules like B₂H₆ in which the central atom possesses a smaller number of electrons than required for an octet of electron.

Question 51.
What are the two ways in which two atomic orbitals combine to form molecular orbitals (MOs)?
Answer:
Two atomic orbitals can combine in two ways to form molecular orbitals:
i. By addition of their wave functions.
ii. By subtraction of their wave functions.
Addition of the atomic orbtials wave functions results in formation of a molecular orbital which is lower in energy than atomic orbitals and is termed as Bonding Molecular Orbital (BMO). Subtraction of the atomic orbitals results in the formation of a molecular orbital which is higher in energy than the atomic orbitals and is termed as Antibonding Molecular Orbital (AMO).

Question 52.
State True or False. Correct the false statement.
i. According to MO theory, the formation of molecular orbitals from atomic orbitals is expressed in terms of Linear Combination of Atomic Orbitals (LCAO).
ii. An MO contains maximum two electrons with opposite spins.
iii. Interference of electron waves of combining atoms can only be constructive.
iv. In bonding molecular orbital, the large electron density is observed between the nuclei of the bonding atoms than the individual atomic orbitals.
v. In the antibonding molecular orbital, the electron density is nearly zero between the nuclei.
Answer:
i. True
ii. True
iii. False
Interference of electron waves of combining atoms can be constructive or destructive.
iv. True
v. True

Question 53.
What are the conditions required for linear combination of atomic orbitals to form molecular orbitals?
Answer:
The following conditions are required for the linear combination of atomic orbitals (LCAO) to form molecular orbitals:
i. The combining atomic orbitals must have comparable energies.
So, Is orbitals of one atom can combine with 1 s orbital of another atom but not with 2s orbital, because energy of 2s orbital is much higher than that of 1 s orbital.

ii. The combining atomic orbitals must have the same symmetry along the molecular axis. Conventionally, z axis is taken as the internuclear axis. So even if atomic orbitals have same energy but their symmetry is not same they cannot combine. For example, 2s orbital of an atom can combine only with 2p_z orbital of another atom, and not with 2p_x or 2p_y orbital of that atom because the symmetries are not same. p_z is symmetrical along z axis while px is symmetrical along x axis.

iii. The combining atomic orbitals must overlap to the maximum extent. Greater the overlap, greater is the electron density between the nuclei and so stronger is the bond formed.

Question 54.
Explain and draw an energy level diagram obtained by the linear combination of two 1s atomic orbitals.
Answer:
The s-orbitals are spherically symmetrical along x, y and z axis. Two Is atomic orbitals combine to form σ 1s (bonding molecular orbital) and σ*1s (antibonding molecular orbital). Both the a bonding and σ* antibonding orbitals are symmetrical along the bond axis.
Diagram:

[Note: If we consider ‘z’ to be internuclear axis then linear combination of p_z orbitals from two atoms can form σ 2p_z bonding σ*(2p_z) molecular orbitals.]

Question 55.
Explain the formation of π and π* molecular orbitals with the help of a diagram.
Answer:
When the atomic orbitals overlap laterally, a pi (π) molecular orbital is formed.
The px and py orbitals are not symmetrical along the bond axis. They have a positive lobe above the axis and negative lobe below the axis. Hence, linear combination of such orbitals leads to the formation of molecular orbitals with positive and negative lobes above and below the bond axis. These are designed as π bonding and π antibonding orbitals. The electron density in such π orbitals is concentrated above and below the bond axis. The π molecular orbitals has a node between the nuclei.
Diagram:

Question 56.
Write the increasing order of energies of molecular orbitals in various diatomic molecules of second row elements.
Answer:
The increasing order of energies of molecular orbitals for molecules (except O₂ and F₂) is:
σ1s < σ*1s < σ2s < σ*2s < (π2p_x = π2p_y) < σ2p_z < (π*2p_x = π*2p_y) < σ*2p_z
The increasing order of energy of molecular orbitals for diatomic molecules like O₂ and F₂ is:
σ1s < σ*1s < σ2s < σ*2s < σ2p_z < (π2p_x = π2p_y) < (π*2p_x = π*2p_y) < σ*2p_z

Question 57.
Explain briefly the information provided by the electronic configuration of molecules.
Answer:
The electronic configuration of molecules provides the following information:

• Stability of molecules: If the number of electrons in bonding MOs is greater than the number in antibonding MOs the molecule is stable.
• Magnetic nature of molecules: If all MOs in a molecule are completely filled with two electrons each, the molecule is diamagnetic (i.e., repelled) by magnetic field. However, if at least one MO is half-filled with one electron, the molecule is paramagnetic (i. e., attracted by magnetic field).
• Bond order of molecule: The bond order of the molecule can be calculated from the number of electrons in bonding MOs (N_b) and antibonding MOs (N_a).

Question 58.
What are the key ideas of MO theory?
OR
What are the salient features of MO theory?
Answer:
Key ideas of MO Theory:

• MOs in molecules are similar to AOs of atoms. Molecular orbital describes region of space in the molecule representing the probability of an electron.
• MOs are formed by combining AOs of different atoms. The number of MOs formed is equal to the number of AOs combined.
• Atomic orbitals of comparable energies and proper symmetry combine to form molecular orbitals.
• MOs those are lower in energy than the starting AOs are bonding MOs and those higher in energy are antibonding MOs.
• The electrons are filled in MOs beginning with the lowest energy.
• Only two electrons occupy each molecular orbital and they have opposite spins, that is, their spins are paired.
• The bond order of the molecule can be calculated from the number of bonding and antibonding electrons.

Question 59.
Explain the formation of the following molecules on the basis of MOT. Also find the bond order.
i. H₂
ii. Li₂
iii. N₂
iv. O₂
v. F₂
Answer:
i. Hydrogen molecule (H₂):

a. Hydrogen atom (Z = 1) has electronic configuration as 1s¹.
b. Hydrogen atom contains one electron, hence hydrogen molecule which is diatomic contains two electrons.
c. Linear combination of two 1s atomic orbitals gives rise to two molecular orbitals σ1s and σ*1s.
d. The two electrons from the hydrogen atoms occupy the σ1s molecular orbital and σ*1s remains vacant.
e. Thus, electronic configuration of H₂ molecule is σ1s².
f. Since, no unpaired electron is present in hydrogen molecule, it is diamagnetic.
g. There are no electrons in the antibonding molecular orbital (σ*1s).
h. The bond order of H₂ molecule is
Bond order = \(\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}=\frac{2-0}{2}\) = 1
Thus, a single covalent bond is present between two hydrogen atoms.
[Note: The bond length is 74 pm and the bond dissociation energy is 438 kJ mol^-1.]

ii. Lithium molecule (Li₂):

a. Lithium atom (Z = 3) has electronic configuration as 1s² 2s¹.
b. Lithium atom has 3 electrons, hence Li₂ molecule has 6 electrons.
c. Linear combination of four atomic orbitals gives rise to four molecular orbitals namely σ1s, σ*1s, σ2S and σ*2s.
d. The electronic configuration of Li₂ molecule is (σ1s)² (σ*1s)² (σ2s)².
e. Since no unpaired electron is present in lithium molecule, it is diamagnetic.
f. Bond order of Li₂ molecule
\(=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}=\frac{4-2}{2}=1\)
Thus, a single covalent bond is present between two Li atoms. Hence, Li₂ is a stable molecule.

iii. Nitrogen molecule (N₂):

a. Nitrogen atom (Z = 7) has electronic configuration as 1s² 2s² 2p³.
b. Nitrogen atom contains 7 electrons, hence nitrogen molecule contains 14 electrons.
c. Linear combination of atomic orbitals gives rise to different molecular orbitals.
d. The electronic configuration of N₂ molecule is
N₂: (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (π2p_x)² (π2p_y)² (σ2p_z)²
e. Since N₂ molecule does not have unpaired electron, it is diamagnetic.
f. Bond order of N₂ molecule
\(=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}=\frac{10-4}{2}=3\)
Thus, there are three bonds in N₂ molecule.

Question 60.
Study the following tables showing bond enthalpies of single and multiple bonds.

i. Among single bonds, which bond is the strongest?
ii. How is bond enthalpy related to bond strength?
Answer:
i. Among single bonds, O-H bond is the strongest.
ii. Larger the bond enthalpy, stronger is the bond.

Question 61.
Write a short note on bond length.
Answer:
Bond length:

• Bond length is defined as the equilibrium distance between the nuclei of two covalently bonded atoms in a molecule.
• Each atom of the bonded pair contributes to the bond length.
• Bond length depends upon the size of atoms and multiplicity of bonds. It increases with increase in size of atom and decreases with increase in multiplicity of bond.
  e.g. C – C single bond is longer than C ≡ C triple bond.
• Bond lengths are measured by X-ray and electron diffraction techniques.

Question 62.
Cl-Cl covalent bond length is smaller than Br-Br covalent bond length. Explain.
Answer:
Bond length increases with increase in size of atom. Cl atom is smaller than Br atom. Hence, Cl-Cl covalent bond length is smaller than Br-Br covalent bond length.

Question 63.
Arrange the following bonds in decreasing order of bond strength: C-N, C=N, C≡N
Answer:
C≡N > C=N > C-N
Note: Average bond lengths for some single, double and triple bonds:

Question 64.
Write a short note on bond order.
Answer:
i. According to the Lewis theory, bond order is given by the number of bonds between the two atoms in a molecule.
e.g. a. In hydrogen molecule, bond order between hydrogen atoms is one as one electron pair is shared.
b. In oxygen molecule, bond order between oxygen atoms is two as two electron pairs are shared.
c. In acetylene molecule, bond order between two carbon atoms is three as three electron pairs are shared.

ii. The isoelectronic molecules and ions have identical bond orders.
e.g. a. The bond order of F₂ and \(\mathrm{O}_{2}{ }^{2-}\) is one.
b. The bond order of N₂, CO and NO⁺ is 3.

iii. As the bond order increases, the bond enthalpy increases and bond length decreases.
iv. With the help of bond order, the stability of a molecule can be predicted.
[Note: N₂ molecule has bond enthalpy of 946 kJ mol^-1. It is one of the highest for diatomic molecules.]

Question 65.
Explain how polarity (ionic character) is developed in a covalent bond.
Answer:
i. Covalent bonds are formed between two atoms of the same or different elements.
ii. When a covalent bond is formed between atoms of same element such as H-H, F-F, Cl-Cl, etc., the shared pair of electrons is attracted equally to both atoms and is situated midway between two atoms. Such covalent bond is termed as nonpolar covalent bond.
iii. When a covalent bond is formed between two atoms of different elements that have different electronegativities, the shared electron pair does not remain at the centre. The electron pair is pulled towards the more electronegative atom resulting in the separation of charges. This give rise to dipole. The more electronegative atom acquires a partial -ve charge and the other atom gets a partial +ve charge. Such a bond is called as polar covalent bond. The examples of polar molecules include HF, HC1, etc.

Fluorine is more electronegative than hydrogen, therefore, the shared electron pair is more attracted towards fluorine and the atoms acquire partial +ve and -ve charges, respectively.
iv. Polarity of the covalent bond increases as the difference in the electronegativity between the bonded atoms increases. When the difference in electronegativities of combining atom is about 1.7, ionic percentage in the covalent bond is 50%.

Question 66.
Define and explain the term dipole moment.
Answer:
i. Dipole moment (μ) is the product of the magnitude of charge and distance between the centres of +ve and -ve charges.
ii. It is given by, µ = Q × r
where, Q = charge, r = distance of separation.
iii. Unit of dipole moment is Debye (D).
iv. Dipole moment being a vector quantity is represented by a small arrow with the tail on the positive centre and head pointing towards the negative centre.

Note: 1 D = 3.33564 × 10^-30 C m
where C is coulomb and m is meter.

Question 67.
Dipole moment in case of BeF2 is zero. Explain.
Answer:

• Dipole moment is a vector quantity. Therefore, the resultant dipole moment of a molecule is the vector sum of dipole moments of various bonds in the molecule.
• In BeF₂ molecule, Be-F bond is polar and has a bond dipole moment.
• BeF₂ is a linear molecule with two Be-F bonds oriented at 180° (opposite to each other).
• The two bond dipoles are equal in magnitude and act in opposite direction to cancel each other. Therefore, the net dipole moment in case of BeF₂ is zero.

Question 68.
Dipole moment in case of BF₃ is zero. Explain.
Answer:
i. Dipole moment is a vector quantity. Therefore, the resultant dipole moment of a molecule is the vector sum of dipole moments of various bonds in the molecule.
ii. In BF₃ molecule, B-F bond is polar and has a bond dipole moment.
iii. Also, in BF₃, the three B-F bonds are oriented at an angle of 120° to one another.
iv. The resultant of any two bond moments is equal in magnitude and opposite in direction to that of third. Hence, the net sum is zero and the dipole moment of tetra-atomic BF₃ molecule is zero.

Question 69.
Dipole moment of H₂O is higher than that of NH₃. Explain.
Answer:
In both NH₃ and H₂O, the central atom undergoes sp³ hybridization. In both the molecules, the orbital dipole due to the lone pair increases the effect of resultant dipole moment. However, in NH₃, nitrogen has only one lone pair while in H₂O, oxygen has two lone pairs. Hence, dipole moment of H₂O is higher than that of NH₃.

Question 70.
Dipole moment of NF₃ is less than that of NH₃, even though N-F bond is more polar than N-H bond. Explain.
Answer:

• Both NH₃ and NF₃ have pyramidal structure with a lone pair on the N atom. In NF₃, F is more electronegative than N while in NH₃, N is more electronegative.
• In NH₃, the orbital dipole due to the lone pair is in the same direction as the resultant dipole moment of N-H bonds, whereas in NF₃ the orbital dipole is in the direction opposite to the resultant dipole moment of three N-F bonds.
• The orbital dipole because of lone pair decreases the effect of the resultant N-F bond moments, which results in the low dipole moment of NF₃ (0.8 × 10^-30 C m) as compared to NH₃ (4.90 × 10^-30 C m).

Question 71.
CHCl₃ is polar. Explain.
Answer:
In CHCl₃, the dipoles are not equal and do not cancel each other. Hence, CHCl₃ is polar with a non-zero dipole moment.

Note: Dipole moments and geometry of some molecules are given in the following table:

Question 72.
Explain Fajan’s rule with suitable examples.
Answer:
Fajan’s rule:
i. Smaller the size of the cation and larger the size of the anion, greater is the covalent character of the ionic bond. For example, Li⁺ Cl^– is more covalent than Na⁺Cl. Similarly, Li⁺I^– is more covalent than Li⁺Cl^–.

ii. Greater the charge on cation, more is covalent character of the ionic bond. For example, covalent character of AlCl₃, MgCl₂ and NaCl decreases in the following order Al³⁺(Cl^–)₃ > Mg²⁺(Cl^–)₂ > Na⁺ Cl^–

iii. A cation with the outer electronic configuration of the s²p⁶d¹⁰ type possesses greater polarising power compared to the cation having the same size and same charge but having outer electronic configuration of s²p⁶ type.

This is because d electrons of the s²p⁶d¹⁰ shell screen nuclear charge less effectively compared to s and p electrons of s²p⁶ shell. Hence, the effective nuclear charge in a cation having s²p⁶d¹⁰ configuration is greater than that of the one having s²p⁶ configuration. For example: Cu⁺Cl^– is more covalent than Na⁺Cl^–. Here,
(Cu⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰; Na⁺ = 1s² 2s² 2p⁶)

Question 73.
Explain resonance with respect to \(\mathrm{CO}_{3}^{2-}\) ion.
Answer:
i. Three structures written for \(\mathrm{CO}_{3}^{2-}\) as follows:

ii. Each structure differs from the other only in the position of electrons without changing positions of the atoms. None of these individual structures is adequate to explain the properties of \(\mathrm{CO}_{3}^{2-}\).
iii. The actual structure of \(\mathrm{CO}_{3}^{2-}\) is a combination of three Lewis structures and is called as the resonance hybrid.
iv. Energy of the resonance hybrid structure is less than the energy of any single canonical form. Hence, resonance stabilizes certain polyatomic molecules or ions.
v. The average of all resonating structures contributes to overall bonding characteristic features of the molecule or ion.

Question 74.
Explain O₃ molecule is the resonance hybrid.
Answer:
Ozone is a resonance hybrid of structures I and II. The structures I and II are canonical forms while structure III is a resonance hybrid. The energy of structure III is less than that of I and II.

Question 75.
Define resonance energy.
Answer:
Resonance energy is defined as the difference in energy of the most stable contributing structure and the resonating forms.

Question 76.
Write the resonance structures of \(\mathrm{NO}_{3}^{-}\) ion.
Answer:
Resonance structures of \(\mathrm{NO}_{3}^{-}\) :

Question 77.
A student represents the Lewis dot structure of AlCl₃ molecule as shown below:

i. Is the representation correct? Justify your answer.
ii. If the chlorine atoms are replaced by bromine atoms, what will be the number of electrons present in the valence shell of aluminium?
Answer:
i. No, the representation is incorrect. There will be no lone pair of electrons on aluminium.
ii. The number of electrons present in the valence shell of aluminium will be six.

Question 78.
Below is an incomplete Lewis structure for glycine. Complete the following Lewis structure and answer the following questions. (Hint: Add lone pairs and multiple bonds to the structure below to give each atom a formal charge of zero.)

i. How many lone pairs of electrons are present on N-atom in the structure?
ii. How many pi bonds are present in the structure?
iii. How many sigma bonds are present in the structure?
Answer:
The correct Lewis structure is:

i. The number of lone pairs of electrons on N-atom is 1.
ii. The number of pi bonds in the structure is 1.
iii. The number of sigma bonds in the structure is 9.

Question 79.
Consider the following four species and answer the below given questions.
\(\mathrm{O}_{2}^{-}\), O₂ \(\mathrm{O}_{2}^{+}\), \(\mathrm{O}_{2}^{2-}\)
i. What is the bond order of \(\mathrm{O}_{2}^{+}\) ?
ii. Which species is least stable?
Answer:
i. Electronic configuration of \(\mathrm{O}_{2}^{+}\) can be given as:
(σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (σ2p_z)² (π2p_x)² (π2p_y)² (π*2p_x)¹ (π*2p_y)⁰
∴ Bond order = \(\frac {1}{2}\) (10 – 5) = 2.5

ii. Stability of the molecule or species ∝ Bond order
Bond order decreases as: \(\mathrm{O}_{2}^{+}\) > O₂ > \(\mathrm{O}_{2}^{-}\) > \(\mathrm{O}_{2}^{2-}\)
∴ Stability decreases as: \(\mathrm{O}_{2}^{+}\) > O₂ > \(\mathrm{O}_{2}^{-}\) > \(\mathrm{O}_{2}^{2-}\)
Hence, least stable species is \(\mathrm{O}_{2}^{2-}\).

Multiple Choice Questions

1. The CORRECT Lewis structure of CO molecule is:

Answer:

2. Which of the following molecule does NOT obey octet rule?
(A) BF₃
(B) CO₂
(C) H₂O
(D) N₂
Answer:
(A) BF₃

3. In BF₃, bond angle is .
(A) 90°
(B) 109°
(C) 120°
(D) 180°
Answer:
(C) 120°

4. Identify the geometry represented by the following diagram.

(A) Trigonal bipyramidal
(B) T-shape
(C) square planar
(D) square pyramidal
Answer:
(D) square pyramidal

5. The geometry of H₂S is ………….
(A) tetrahedral
(B) angular
(C) linear
(D) trigonal planar
Answer:
(B) angular

6. What will be the shape of molecule whose central atom is associated with 3 bonds and one lone pair?
(A) Trigonal pyramidal
(B) Tetrahedral
(C) Square planar
(D) Triangular planar
Answer:
(A) Trigonal pyramidal

7. Pair of molecules having identical geometry is …………..
(A) BF₃, NH₃
(B) BF₃, AlF₃
(C) BeF₂, H₂O
(D) BCl₃, PCl₃
Answer:
(B) BF₃, AlF₃

8. Which of the following molecule has bent shape?
(A) PCl₃
(B) OF₂
(C) BH₃
(D) BeBr2
Answer:
(B) OF₂

9. Which of the following is INCORRECT?
(A) The strength of the bond depends on the extent of overlap of the atomic orbitals.
(B) The extent of overlap depends on the shape and size of the atomic orbitals.
(C) The energy of the bonded atoms is more than that of the free atoms.
(D) During overlap of atomic orbitals, the electron density increases in between the two nuclei.
Answer:
(C) The energy of the bonded atoms is more than that of the free atoms.

10. In the potential energy curve for hydrogen molecule, the maximum stability is achieved when …………..
(A) potential energy of the system is maximum
(B) potential energy of the system is minimum
(C) force of repulsion become greater than force of attraction
(D) no bond formation takes place
Answer:
(B) potential energy of the system is minimum

11. In acetylene, C-C σ bond is formed by …………. overlap.
(A) sp²-sp²
(B) sp-sp
(C) sp-s
(D) p-p
Answer:
(B) sp-sp

12. The formation of O-H bonds in a water molecule involves …………. overlap.
(A) sp³-s
(B) sp¹-s
(C) sp-p
(D) sp³-p
Answer:
(A) sp³-s

13. The molecular orbital shown in the diagram can be described as ………….

(A) σ
(B) σ*
(C) π*
(D) π
Answer:
(C) π*

14. The bond order of lithium molecule is ………….
(A) one
(B) two
(C) three
(D) four
Answer:
(A) one

15. The bond order in N₂ molecule is …………
(A) 1
(B) 2
(C) 3
(D) 4
Answer:
(C) 3

16. The bond energies of F₂, Cl₂, Br₂ and I₂ are 37, 58, 46, and 36 kcal/mol respectively. The strongest bond is present in …………..
(A) Br₂
(B) I₂
(C) Cl₂
(D) F₂
Answer:
(C) Cl₂

17. The common features among the species CO and NO⁺ are: ……………
(A) isoelectronic species and bond order 3
(B) isoelectronic species and bond order 2
(C) odd electron species and unstable
(D) odd electron species and bond order 1
Answer:
(A) isoelectronic species and bond order 3

18. Which of the following is CORRECT for H₂O ?

Answer:
(C)

19. Each of the following molecules has a non-zero dipole moment EXCEPT:
(A) NF₃
(B) BF₃
(C) SO₂
(D) LiH
Answer:
(B) BF₃

20. Which of the following compounds is non-polar?
(A) HCl
(B) CH₂Cl₂
(C) CHCl₃
(D) CCl₄
Answer:
(D) CCl₄

21. The dipole moment of …………..
(A) NF₃ is higher than that of NH₂
(B) BF₃ is higher than that of NH₃
(C) H₂S is higher than that of H₂O
(D) HCl is higher than that of HBr
Answer:
(D) HCl is higher than that of HBr

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 4 Structure of Atom Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 4 Structure of Atom

Question 1.
Complete the information about the properties of subatomic particles in the following table:

Answer:

Question 2.
What are three important subatomic particles of an atom?
Answer:
Electron, proton and neutron are the three important subatomic particles of an atom.

Question 3.
Write a short note on discovery of electron.
Answer:

• In the year 1897, J J. Thomson studied the properties of cathode rays through a cathode ray tube experiment and found that the cathode rays are a stream of very small, negatively charged particles.
• These particles are 1837 times lighter than a hydrogen atom and are present in all atoms.
• These particles were later named as electrons.

Question 4.
Draw labelled diagram of Rutherford’s α-particle scattering experiment.
Answer:

Question 5.
Write a short note on discovery of proton.
Answer:

• After the discovery of nucleus in an atom, Rutherford found that the fast moving α-particles transmuted nitrogen into oxygen with simultaneous liberation of hydrogen.
  \({ }_{7}^{14} \mathrm{~N}+{ }_{2}^{4} \alpha \longrightarrow{ }_{8}^{17} \mathrm{O}+{ }_{1}^{1} \mathrm{H}\)
• He further showed that other elements could also be transmuted similarly and hydrogen was always emitted in the process.
• Based on these observations, he proposed that the hydrogen nucleus must be contained inside nuclei of all the elements. Hence, he renamed hydrogen nucleus as proton.

Question 6.
Write a short note on discovery of neutron.
Answer:

• In 1920, Ernest Rutherford proposed the existence of an electrically neutral and massive particle in the nucleus of an atom in order to account for the disparity in atomic number and atomic mass of an element.
• In 1932, James Chadwick measured the velocity of protons ejected from paraffin by an unidentified radiation from beryllium (Be).
• From that he determined the mass of the particles of this unidentified neutral radiation, which was found to be almost same as that of the mass of a proton.
• He named this neutral particle as ‘neutron’, which was earlier predicted by Rutherford.

Question 7.
State true or false. Correct the false statement.
i. An electron is 1837 times lighter than a proton.
ii. In Rutherford’s experiment of scattering of α-particles by thin gold foil, most of the α-particles
bounced back.
iii. Cathode rays are a stream of very small, positively charged particles.
Answer:
i. True
ii. False,
In Rutherford’s experiment of scattering of α-particles by thin gold foil, very few α-particles bounced back.
iii. False,
Cathode rays are a stream of very small, negatively charged particles.

Question 8.
Explain the term: Atomic number
Answer:

• Atomic number is defined as the number of protons present in the nucleus of an atom of a particular element.
• Atomic number is represented by Z.
• An atom is electrically neutral. Hence, the number of protons equals to the number of electrons. In other words, the atomic number of an atom is equal to the number of electrons.

∴ Atomic Number (Z) = Number of protons = Number of electrons

Question 9.
Give reason: The approximate atomic mass in Daltons is numerically equal to the number of nucleons in the atom.
Answer:

• The electrons possess negligible mass. They do not contribute much to the mass of an atom.
• Therefore, the entire mass of an atom is supposed to be present in the nucleus which consists of protons and neutrons, which are collectively called as nucleons.

Hence, approximate atomic mass in Daltons is numerically equal to the number of nucleons in the atom.

Question 10.
Explain: Atomic mass number.
Answer:

• The sum of the total number of protons and neutrons present in the nucleus of an atom is called the atomic mass number of that atom.
• Atomic mass number is represented by A
• Mass number (A) = Number of protons (Z) + Number of neutrons (N) = Total number of nucleons
  ∴ A = Z + N OR N = A – Z

Question 11.
How is an atom of an element ‘X’ having atomic number ‘Z’ and atomic mass number ‘A’ represented?
Answer:
An atom of an element ‘X’ having atomic number ‘Z’ and atomic mass number ‘A’ is represented as: \({ }_{Z}^{A} X\)

Question 12.
What is a nuclide?
Answer:
The atom or nucleus having a unique composition as specified by \({ }_{Z}^{A} X\) is called as a nuclide.

Question 13.
If an element ‘X’ has 6 protons and 8 neutrons, then write its representation.
Answer:
The representation of the given element is \({ }_{6}^{14} \mathrm{X}\).

Question 14.
Three elements Q, R and T have mass number 40. Their atoms contain 22, 21 and 20 neutrons respectively. Represent their atomic composition with appropriate symbol.
Answer:
Mass number (A) = Number of protons (Z) + Number of neutrons (N) .
A = Z + N
∴ Z = A – N
For the given three elements, A = 40. Values of their atomic numbers Z, are calculated from the given values of the number of neutrons, N, using the above formula.
For Q: Z = A – N = 40 – 22 = 18
For R: Z = A – N = 40 – 21 = 19
For T: Z = A – N = 40 – 20 = 20
Now, atomic composition of an element (X) is represented as \({ }_{Z}^{A} X\).
The atomic compositions of the three elements are written as follows:
\({ }_{18}^{40} \mathrm{Q}\), \({ }_{19}^{40} \mathrm{R}\), \({ }_{20}^{40} \mathrm{T}\)

Question 15.
Find out the number of protons, electrons and neutrons in the nuclide \({ }_{18}^{40} \mathrm{Ar}\).
Solution:
For the given nuclide,
Atomic number, Z = 18, Mass number, A = 40
Number of protons = Number of electrons = Z = 18
Number of neutrons (N) = A – Z = 40 – 18 = 22
Ans: Number of protons = 18, Number of electrons = 18, Number of neutrons = 22

Question 16.
How many protons, electrons and neutrons are there in the following nuclei?
i. \({ }_{8}^{17} \mathrm{O}\)
ii. \({ }_{12}^{25} \mathrm{Mg}\)
iii. \({ }_{35}^{80} \mathrm{Br}\)
Solution:
i. \({ }_{8}^{17} \mathrm{O}\)
Atomic number, Z = 8, Mass number, A = 17
Number of protons = Number of electrons = Z = 8
Number of neutrons (N) = A – Z = 17 – 8 = 9
Ans: Number of protons = 8, Number of electrons = 8, Number of neutrons = 9

ii. \({ }_{12}^{25} \mathrm{Mg}\)
Atomic number, Z = 12, Mass number, A = 25
Number of protons = Number of electrons = Z = 12
Number of neutrons (N) = A – Z = 25 – 12 = 13
Ans: Number of protons = 12, Number of electrons = 12, Number of neutrons = 13

iii. \({ }_{35}^{80} \mathrm{Br}\)
Atomic number, Z = 35, Mass number, A = 80
Number of protons = Number of electrons = Z = 35
Number of neutrons (N) = A – Z = 80 – 35 = 45
Ans: Number of protons = 35, Number of electrons = 35, Number of neutrons = 45

Question 17.
Define isotopes.
Answer:
Isotopes are defined as the atoms of an element having the same number ofprotons but different number of neutrons in their nuclei.
e.g. \({ }_{6}^{12} \mathrm{C}\), \({ }_{6}^{13} \mathrm{C}\) and \({ }_{6}^{14} \mathrm{Br}\) are isotopes.

Question 18.
Complete the information about the isotopes of carbon in the following table:

Answer:

Question 19.
Define isobars.
Answer:
Isobars are defined as the atoms of different elements having the same mass number but different atomic number.
e.g. \({ }_{6}^{14} \mathrm{C}\) and \({ }_{7}^{14} \mathrm{N}\) are isobars.

Question 20.
Complete the following table:

Answer:

Question 21.
The two natural isotopes of chlorine viz. \({ }_{17}^{35} \mathrm{Cl}\) and \({ }_{17}^{37} \mathrm{Cl}\) exist in relative abundance of 3 : 1. Find out the average atomic mass of chlorine.
Solution:
Given: Isotopes of chlorine \({ }_{17}^{35} \mathrm{Cl}\) and \({ }_{17}^{37} \mathrm{Cl}\).
Ratio of relative abundance of these isotopes is 3 : 1.
To find: Average atomic mass of chlorine
Calculation: From the relative abundance 3 : 1, it is understood that out of 4 chlorine atoms, 3 atoms have mass 35 and 1 atom has mass 37.
Therefore, the average atomic mass of chlorine = \(\frac{3 \times 35+1 \times 37}{4}\) = 35.5
∴ Average atomic mass of chlorine = 35.5 u
Ans: The average atomic mass of chlorine is 35.5 u.

Question 22.
Find out the average atomic mass of lithium (Li) from the following data:

Solution:
Given: Three isotopes of lithium along with respective atomic mass and % abundance.
To find: Average atomic mass of lithium
Calculation:

Ans: Average atomic mass of lithium is 6.940 u.

Question 23.
Certain results were obtained when scientists studied the interactions of radiation with matter. What were the two results, utilized by Neils Bohr to overcome the drawbacks of Rutherford model?
Answer:
The two results utilized by Neils Bohr to overcome the drawbacks of Rutherford model were:

• Wave particle duality of electromagnetic radiation
• Line emission spectra of hydrogen

Question 24.
Explain in short the wave particle duality of light (electromagnetic radiation).
Answer:
Wave particle duality of light (electromagnetic radiation):

• Light has both particle and wave like nature.
  Phenomena such as diffraction and interference of light could be explained by treating light as electromagnetic wave.
• However, the black-body radiation or photoelectric effect could not be explained by wave nature of light. This could be accounted for by considering particle nature of light. Thus, both phenomena could be explained only by accepting that light has dual behaviour.
• When light interacts with matter it behaves as a stream of particles (called photons) and when light propagates, it behaves as an electromagnetic wave.

Question 25.
Observe the following figure of an electromagnetic wave and answer the questions given below:

i. What does ‘x’ represent?
ii. What does ‘y’ represent?
Answer:
i. ‘x’ represents amplitude of the wave.
ii. ‘y’ represents wavelength of the wave.

Question 26.
Define and explain the following terms:
i. Wavelength (λ).
ii. Frequency (ν)
iii. Wavenumber (\(\bar{v}\))
iv. Amplitude (A)
v. Velocity (c)
Answer:
i. Wavelength (λ):

• The distance between two consecutive crests or two consecutive troughs in a wave is called wavelength.
• It is represented by Greek letter λ (lambda).
• The SI unit for wavelength is metre (m).

Note: The other units include Angstrom, nanometre, picometer (1 pm = 10^-12 m) and micron (1µ = 10^-6 m).
1Å = 10^-8 cm = 10^-10 m
1nm = 10^-9 m = 10Å

ii. Frequency (ν):

• The number of waves that pass a given point in one second is called frequency.
• It is represented by Greek letter ‘ν’ (nu).
• The SI unit of frequency is Hertz (Hz) or s^-1.

Note: 1 Hz = 1 cycle per second (1 cps)
The units, kilo Hertz (kHz) and mega Hertz (mHz) are commonly used.
1 kHz = 10³ Hz = 10³ cps
1 mHz = 10⁶ Hz = 10⁶ cps

iii. Wavenumber (\(\bar{v}\)):

• The number of wavelengths per unit length is called the wavenumber.
• It is represented by \(\bar{v}\) (nu bar).
• The commonly used unit for wavenumber is cm^-1 while its SI unit is m^-1.
• Wavenumber of a wave is related to the wavelength as follows:
  \(\bar{v}\) = \(\frac{1}{\lambda}\)

iv. Amplitude (A):

• The height of a crest or the depth of a trough from the line of propagation of the wave is called
  amplitude.
• It is represented by letter ‘A’.
• The square of the amplitude represents the intensity (brightness) of the radiation.

v. Velocity (c):

• The distance travelled by a wave in one second is called the velocity of the wave.
• It is denoted by letter c.
• It is the product of the frequency and wavelength. Hence, c = νλ
• The velocity of all types of electromagnetic radiations (in space or in vacuum) is the same and it is equal to the velocity of light (3 × 10¹⁰ cm s^-1 or 3 × 10⁸ m s^-1. However, they may have different wavelengths and frequencies.

Question 27.
Write a short note on quantum theory of radiation.
Answer:
i. Max Planck put forward a theory known as Planck’s quantum theory to explain black-body radiation.
ii. According to this theory, the energy of electromagnetic radiation depends upon the frequency and not the amplitude.
iii. The smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation is called as ‘quantum’.
iv. The energy (E) of each quantum of radiation is directly proportional to its frequency (ν).
i.e., E ∝ ν ; E = hν
where, h = Planck’s constant = 6.626 × 10^-34 J s.

Question 28.
Parameters of blue and red light are 400 nm and 750 nm respectively. Which of the two is of higher energy?
Answer:
400 nm and 750 nm are the wavelengths of blue and red light respectively. Energy of radiation is given by the expression E = hν and the frequency (ν), of radiation is related to the wavelength by the expression.
ν = \(\frac{c}{\lambda}\)
∴ E = \(\frac{\mathrm{hc}}{\lambda}\)
Therefore, shorter the wavelength, λ, larger the frequency, ν, and higher the energy, E. Thus, blue light which has shorter λ (400 nm) than red light (750 nm) has higher energy.

Question 29.
What is an emission spectrum?
Answer:

• When a substance is irradiated with electromagnetic radiation, it absorbs energy. Atoms, molecules or ions, which have absorbed radiation are said to be ‘excited’. Heating can also result in an excited state.
• The excited species emits the absorbed energy in the form of radiation. This process is called emission of radiation and the recorded spectrum of this emitted radiation is called ‘emission spectrum’.

Question 30.
Give some examples of commonly used light sources that work on atomic emission.
Answer:
Examples are fluorescent tube, sodium vapor lamp, neon sign and halogen lamp.

Question 31.
Write a short note on emission spectrum of hydrogen. Also, list all the five series of lines in the hydrogen spectrum.
Answer:
i. When electric discharge is passed through gaseous hydrogen, it emits radiation. The recorded spectrum of this emitted radiation is called hydrogen emission spectrum.

ii. This spectrum falls in different regions of electromagnetic radiation and it is comprised of a series of lines corresponding to different frequencies. That is, the spectrum was discontinuous.

iii. In the year, 1885, Balmer expressed the wave numbers of the emission lines in the visible region of electromagnetic spectrum by the formula:
\(\overline{\mathrm{v}}=109677\left[\frac{1}{2^{2}}-\frac{1}{\mathrm{n}^{2}}\right] \mathrm{cm}^{-1}\)
where, n = 3, 4, 5,…
These lines are known as Balmer series.

iv. Rydberg found that other series of lines could be described by the following formula:
\(\bar{v}=109677\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right] \mathrm{cm}^{-1}\)
where, 109677 cm^-1 is called Rydberg constant for hydrogen (R_H).

v. Different series of emission spectral lines for hydrogen are as follows:

Question 32.
Observe the emission spectrum of hydrogen and answer the following questions.

i. Is the spectrum continuous?
ii. In which region of electromagnetic radiation does the Paschen series belong?
iii. Which series falls in the visible region of electromagnetic radiation?
Answer:
i. The spectra is not continuous and comprises of a series of lines corresponding to different frequencies.
ii. Paschen series falls in the infrared region of electromagnetic radiation.
iii. Balmer series falls partly in the visible region of electromagnetic radiation.

Question 33.
Give the expression to calculate wavenumber of the emission lines in the Balmer series.
Answer:
\(\bar{v}=109677\left[\frac{1}{2^{2}}-\frac{1}{\mathrm{n}^{2}}\right] \mathrm{cm}^{-1}\)
where, n = 3, 4, 5, ….

Question 34.
Visible light has wavelengths ranging from 400 nm (violet) to 750 nm (red). Express these wavelengths in terms of frequency (Hz). (1 nm = 10^-9 m)
Solution:
Given: Wavelengths: λ₁ = 400 nm (for violet light), λ₂ = 750 nm (for red light)
To find: Frequencies: ν₁, ν₂
Formula: ν = \(\frac{c}{\lambda}\)
Calculation: i. Wavelength of violet light, λ₁ = 400 nm = 400 × 10^-9 m
Frequency, ν = \(\frac{c}{\lambda}\) where c is speed of light = 3.0 × 10⁸ ms^-1
∴ ν₁ = \(\frac{3.0 \times 10^{8} \mathrm{~ms}^{-1}}{400 \times 10^{-9} \mathrm{~m}}\) = 7.50 × 10¹⁴ Hz
ii. Wavelength of red light, λ₂ = 750 nm = 750 × 10^-9 m
Frequency, ν = \(\frac{c}{\lambda}\)
∴ ν₂ = \(\frac{3.0 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}}{750 \times 10^{-9} \mathrm{~m}}\) = 4.00 × 10¹⁴ Hz
Ans: The frequency of violet light is 7.50 × 10¹⁴ Hz and that of red light is 4.00 × 10¹⁴ Hz.

Question 35.
Yellow light emitted from a lamp has a wavelength of 580 nm. Find the frequency and wavenumber of this light.
Solution:
Given: Wavelength (λ) = 580 nm
To find: Frequency (ν), Wave number \(\bar{v}\)
Formulae : \(v=\frac{c}{\lambda}, \bar{v}=\frac{1}{\lambda}\)
Calculation: Wavelength of yellow light (λ) = 580 nm = 580 × 10^-9 m [1 nm = 10^-9 m]
We know that frequency (ν) is related to wavelength as: ν = \(\frac{c}{\lambda}\)
where, c, velocity of light = 3.0 × 10^-8 m s^-1
∴ ν = \(\frac{3.0 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}}{580 \times 10^{-9} \mathrm{~m}}=5.17 \times 10^{14} \mathrm{~s}^{-1}\)
Again, Wave number, \(\bar{v}=\frac{1}{\lambda}=\frac{1}{580 \times 10^{-9} \mathrm{~m}}=1.72 \times 10^{6} \mathrm{~m}^{-1}\)
Ans: Frequency = 5.17 × 10¹⁴ s^-1 and wave number = 1.72 × 10⁶ m^-1

Question 36.
Calculate the energy of a photon of radiation having wavelength 300 nm. [h = 6.63 × 10^-34 J s]
Solution:
Given: Wavelength (λ) = 300 nm
To find: Energy of a photon (E)
Formulae: E = \(\frac{\mathrm{hc}}{\lambda}\)
Calculation: From formula,
E = \(\frac{6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s} \times 3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}}{300 \times 10^{-9} \mathrm{~m}}=6.63 \times 10^{-19} \mathrm{~J}\)
Ans: Energy of a photon is 6.63 × 10^-19 J.

Question 37.
Explain briefly the results of Bohr’s theory for hydrogen atom.
Answer:
i. The stationary states for electrons are numbered n = 1, 2, 3……. These integers are known as principal quantum numbers.
ii. The radii of the stationary states are r_n = n² a₀, where a₀ = 52.9 pm (picometer). Thus, the radius of the first stationary state, called the Bohr radius is 52.9 pm.
iii. The most important property associated with the electron is the energy of its stationary state. It is given by the formula:
E_n = -R_H (1/n²), where n = 1, 2, 3, …..
R_H is the Rydberg constant for hydrogen and its value in joules is 2.18 × 10^-18 J.
The lowest energy state is called the ground state. Energy of the ground state (n = 1) is:
E₁ = -2.18 × 10^-18 × 1/1² = -2.18 × 10^-18 J
Energy of the stationary state corresponding to n = 2 is
E₂ = -2.18 × 10^-18 × (1/(2)²) = -0.545 × 10^-18 J.
iv. Bohr theory can be applied to hydrogen like species. For example, He⁺, Li²⁺, Be³⁺ and so on. Energies and radii of the stationary states associated with these species are given by:

where, Z is the atomic number. From the above expressions, it can be seen that the energy decreases (becomes more negative) and radius becomes smaller as the value of Z increases.
v. Velocities of electrons can also be calculated from the Bohr theory. Qualitatively, it is found that the magnitude of velocity of an electron increases with increase of Z and decreases with increase in the principal quantum number (n).

Question 38.
How many electrons are present in \({ }_{1}^{2} \mathrm{H}\), ₂He and He⁺ ? Which of these are hydrogen-like species?
Answer:
Hydrogen-like species contain only one electron.
Consider \({ }_{1}^{2} \mathrm{H}\):
Number of protons = Number of electrons = 1
Consider ₂He:
Number of protons = Number of electrons = 2
Consider He⁺:
Number of electrons = (Number of electrons in He – 1) = 2 – 1 = 1
Thus, \({ }_{1}^{2} \mathrm{H}\) and He⁺ are hydrogen-like species.
[Note: Bohr’s theory is applicable to hydrogen atom and hydrogen-like species, which contain only one electron.]

Question 39.
Describe how the line spectrum of hydrogen is explained by Bohr theory.
Answer:
i. According to second postulate of Bohr theory, radiation is emitted when an electron moves from an outer orbit of higher principal quantum number (n_i) to an inner orbit of lower principal quantum number (n_f). The energy difference (ΔE) between the initial and final orbit of the electronic transition corresponds to the energy of the emitted radiation.
ii. From the third postulate of Bohr theory, ΔE can be expressed as
ΔE = E_i – E_f …….(1)
iii. According to the results derived from Bohr theory, the energy (E_n) of an orbit is related to its principal quantum number ‘n’ by the equation:
E = \(-\mathrm{R}_{\mathrm{H}}\left(\frac{1}{\mathrm{n}^{2}}\right)\) ……(2)
iv. On combining these two equations, we get:

v. Substituting the value of R_H in joules, we get

vi. This expression can be rewritten in the terms of wavenumber of the emitted radiation in the following steps:
(ΔE) J = (h) J s × (ν) Hz
and

This equation appears like the Rydberg equation, where, n_f = n₁ and n_i = n₂.
In other words, Bohr theory successfully accounts for the empirical Rydberg equation for the line emission spectrum of hydrogen.

Question 40.
Observe the following diagram showing electronic transition in the hydrogen spectrum.

i. Electron jumps from higher energy level to n = 1. Which series does it correspond to?
ii. Electron jumps from higher energy level to n = 4. Which series does it correspond to?
iii. Which transition will give second line of Balmer series?
Answer:
i. When electron jumps from higher energy level to n = 1, it corresponds to Lyman series.
ii. When electron jumps from higher energy level to n = 1, it corresponds to Bracket series.
iii. When electron jumps from n = 4 to n = 1, the second line of Balmer series is observed.

Question 41.
Explain de Broglie’s equation.
Answer:
de Broglie’s equation:

• de Broglie proposed (in 1924) that matter should exhibit a dual behaviour. That is, every object which possesses a mass and velocity behaves both as a particle and as a wave. An electron has mass and velocity. This means that an electron should have momentum (p), a property of particle as well as wavelength (λ), a property of wave.
• According to de Broglie, the wavelength λ of a particle of mass m moving with a velocity v is
  λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) where, h is Planck’s constant.
• The quantity mv gives the momentum of the particles.
  λ = \(\frac{\mathrm{h}}{\mathrm{p}}\) where, p represents the momentum of the particle.
• de Broglie’s prediction was confirmed by diffraction experiments (a wave property).

[Note: According to de Broglie’s equation, the wavelength of a moving particle is inversely proportional to its mass. Therefore, heavier particles have much smaller wavelength than lighter particles like electrons.]

Question 42.
Write a note on Heisenberg’s uncertainty principle.
Answer:
i. Uncertainty principle was proposed by Wemer Heisenberg in 1927. It can be stated as “It is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron ”.
ii. If Δx is the uncertainty in the determination of the position of a very small moving particle and Δpx is the uncertainty in the determination of its momentum, then
Δx Δp_x ≥ \(\frac{\mathrm{h}}{4 \pi}\) …….(1) where h is Planck’s constant
iii. The above equation can alternatively be stated as,
Δx × m × Δv_x ≥ \(\frac{\mathrm{h}}{4 \pi}\), because Δp_x = m × Δv_x ……..(2)
where Δv_x is the uncertainty in the determination of velocity and m is the mass of the particle.

Question 43.
Calculate the radius and energy associated with the first orbit of He.
Solution:
Given: n = 1
To find: Radius and energy associated with the first orbit of He⁺
Formulae:

Calculation: He⁺ is a hydrogen-like species having Z = 2.
Using formula (i),
Radius of the first orbit of He⁺ = r₁ = \(\frac{52.9 \times(1)^{2}}{2} \mathrm{pm}\)
= 26.45 pm
Using formula (ii),
Energy of the first orbit of He⁺ = E₁ = -2.18 × 10^-18 \(\left(\frac{2^{2}}{1^{2}}\right) \mathrm{J}\)
= -8.72 × 10^-18 J
Ans: Radius of the first orbit of He⁺ is 26.45 pm and energy of the first orbit of He⁺ is -8.72 × 10^-18 J.

Question 44.
What is the wavelength of the photon emitted during the transition from the orbit of n = 5 to that of n = 2 in hydrogen atom?
Solution:
Given: n_i = 5, n_f = 2
To find: Wavelength of the photon emitted

Ans: Wavelength of the photon emitted is 434 nm.

Question 45.
Calculate the mass of a hypothetical particle having wavelength 5894 A and velocity 1.0 × 10⁸ ms^-1.
Solution:
Given: Wavelength (λ) = 5894 Å, Velocity (ν) = 1.0 × 10⁸ ms^-1
To find: Mass of a particle
Formula: λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) or m = \(\frac{\mathrm{h}}{\lambda \mathrm{v}}\) (according to de-Broglie equation)
Calculation: m = \(\frac{\mathrm{h}}{\lambda \mathrm{v}}\)
∴ m = \(\frac{6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}}{\left(5894 \times 10^{-10} \mathrm{~m}\right) \times\left(1.0 \times 10^{8} \mathrm{~ms}^{-1}\right)}\)
= 1.124 × 10^-35 kg
Ans: Mass of a particle is 1.124 × 10^-35 kg.

[Calculation using log table:
\(\frac{6.626 \times 10^{-34}}{5894 \times 10^{-10} \times 1.0 \times 10^{8}}=\frac{6.626}{5894} \times 10^{-32}\)
= Antilog₁₀ [log₁₀ 6.626 – log₁₀ 5894] × 10^-32
= Antilog₁₀ [0.8213 – 3.7704] × 10^-32
= Antilog₁₀ [latex]\overline{3} .0509[/latex] × 10^-32 = 1.124 × 10^-35]

Question 46.
Write a short note on Schrodinger equation.
Answer:
Schrodinger equation or wave equation:
i. Schrodinger developed the fundamental equation of quantum mechanics which incorporates wave particle duality of matter. The Schrodinger equation or wave equation is written as:
It \(\hat{\mathbf{H}}\)ψ = Eψ
Here \(\hat{\mathbf{H}}\) is a mathematical operator called Hamiltonian, ψ (psi) is the wave function and E is the total energy of the system.

ii. When Schrodinger equation is solved for an electron in hydrogen atom, the possible values of energy states (E) that the electron may have along with the corresponding wave function (ψ) are obtained. As a consequence of solving this equation, a set of three quantum numbers characteristic of the quantized energy levels and the corresponding wave functions are obtained. These are: Principal quantum number (n), azimuthal quantum number (l) and magnetic quantum number (m_l).

iii. The solution of Schrodinger wave equation led to three quantum numbers and successfully predicted features of hydrogen atom emission spectrum.

iv. Splitting of spectral lines in multi-electron atomic emission spectra could not be explained through such model. These were explained by George Uhlenbeck and Samuel Goudsmit (1925) who proposed the presence of the fourth quantum number called electron spin quantum number, m_s.

Question 47.
Write a short note on magnetic orbital quantum number (m_l).
Answer:
Magnetic orbital quantum number (m_l):

• Magnetic orbital quantum number describes the relative spatial orientation of the orbitals in a given subshell.
• It is denoted by m; and it has values from -l to +l through zero, giving total values or total orientations equal to (2l + 1).
• For s-subshell, 1 = 0, hence, m_l = 0. Thus, s-subshell contains only one orbital.
• For p-subshell, l = 1, hence, m_l = +1, 0, -1. Thus, p-subshell contains three orbitals having distinct orientations.

Question 48.
If n = 2, what are the values of quantum number l and m_l ?
Answer:
For a given n, l = 0 to (n – 1) and for given l, m_l = -l…., 0…. + l
Therefore, the possible values of l and ml for n = 2 are:

Question 49.
What are the values of m_l for f-subshell?
Answer:
For f-subshell, l = 3. Therefore, m_l has seven values: + 3, + 2, + 1, 0, -1, -2, -3.

Question 50.
How many orbitals make the N-shell? What is the subshell wise distribution of orbitals in the N-shell?
Answer:
For N-shell principal quantum number n = 4
∴ Total number of orbitals in N-shell = n² = 4² = 16. The total number of subshells in N-shell = n = 4.
The four subshells with their azimuthal quantum numbers and the constituent number of orbitals are as shown below:

Question 51.
Complete the following flow chart:

Answer:

Question 52.
Write a short note on electron spin quantum number.
Answer:
Electron spin quantum number (m_s):

• Electron spin quantum number describes the spin state of the electron in an orbital. It is designated as ms.
• An electron spins around its axis and this imparts spin angular momentum to it.
• The two orientations which the spin angular momentum of an electron can take up give rise to the spin states which can be distinguished from each other by the spin quantum number, m_s, which can be either +1/2 or -1/2.
• The two spin states are represented by two arrows, ↑ (pointing up) and ↓ (pointing down) and thus have opposite spins.

Question 53.
An atom has two electrons in its 4s orbital. Write the values of the four quantum numbers for each of them.
Answer:
For the 4s orbital, 4 stands for the principal quantum number n; s stands for the subshell s having the value of azimuthal quantum number, l = 0. In the ‘s’ subshell, there is only one orbital and has magnetic quantum number, ml = 0. The two electrons in this orbital have opposite spins. Thus, the four quantum numbers of two electrons in 4s orbital are:

Question 54.
Write a short note on probability density of electron.
Answer:
i. The probability of finding an electron at a given point in an atom is proportional to the square of the wave function at that point (ψ²).
ii. According to Max Born, the square of wave function at a point in an atom is the probability density of the electron at that point.
The following figure shows the probability density diagrams of Is and 2s atomic orbitals. These diagrams appear like a cloud.
The electron cloud of 2s orbital shows one node, which is a region with nearly zero probability density and displays the change of sign for its corresponding wavefunction.
Diagram:

Question 55.
What is meant by the term ‘boundary surface diagram’?
Answer:
A boundary surface is drawn in space for an orbital such that the value of probability density (ψ²) is constant and encloses a region where the probability of finding electron is typically more than 90%. Such a boundary surface diagram is a good representation of shape of an orbital.
e.g. Boundary surface diagram of Is and 2s orbitals are spherical in shape.

Question 56.
Describe the shape of s orbital.
Answer:
Shape of s orbital:

• For each value of principal quantum number ‘n’, there is only one s-orbital.
• For s-orbital, l = 0 and m_l = 0, hence s-orbital has only one orientation i.e., the probability of finding the electrons is same in all directions. Thus, s-orbital is spherically symmetrical around the nucleus.
• The value of n determines the size of an orbital. With increase in the value of n, the size of the s-orbital increases.

Diagram:

Question 57.
Describe the shape of 2p orbitals.
Answer:
Shape of 2p orbitals:
i. For p orbital, l = 1. For l = 1, m_l = +1, 0, -1. Thus, p orbitals have three orientations.
ii. Each orbital has two lobes on the two sides of a nodal plane passing through the nucleus.
iii. Shape of 2p orbitals resembles a dumbbell.
iv. The size and energy of the three 2p orbitals are the same. However, their orientations in space are different. The lobes of the three 2p orbitals are along the x, y and z axes. Accordingly, the corresponding orbitals are designated as 2p_x, 2p_y and 2p_z. The size and energy of the orbitals in p subshell increase with the increase of principal quantum number.
Diagram:

Question 58.
Describe the shape of 3d orbitals.
Answer:
Shape of 3d orbitals:

• For d orbital, l = 2. For l = 2, m_l = +2, +1, 0, -1, -2. Thus, d orbitals have five orientations.
• They are designated as d_xy, d_yz, d_zx, d_x²-y² and d_z.
• Shape of 3d orbitals are shown in the following figure. The first three have double dumb-bell shape. They lie in xy, yz and xz plane, respectively. The d_x²-y² is also dumb-bell shaped and lies along the x and y axes. d_z² is dumb-bell shaped along z axis with a dough-nut shaped ring of high electron density around the nucleus in xy plane.
• In spite of difference in their shapes, the five d orbitals are equivalent in energy. The shapes of 4d, 5d, 6d…….. orbitals are similar to those of 3d orbitals, but their respective size and energies are large or they are said to be more diffused.

Diagram:

Question 59.
Explain (n + l) rule with respect to energies of orbitals.
Answer:
The lower the sum (n + l) for an orbital, the lower is its energy. If two orbitals have the same (n + l) values, then the orbital with the lower value of n is of lower energy. This is called the (n + l) rule.

Question 60.
What are the two methods of representing electronic configuration of an atom?
Answer:
The two methods of representing electronic configuration of an atom are:
i. Orbital notation: ns^a np^b nd^c …..
In the orbital notation method, a shell is represented by the principal quantum number (n) followed by respective symbol of the subshell. The number of electrons occupying that subshell being written as superscript on right side of the symbol.

ii. Orbital diagram:
In the orbital diagram method, each orbital in a subshell is represented by a box and the electrons represented by an arrow (↑ for up spin and ↓ for down spin) are placed in the respective boxes. In this method, all the four quantum numbers of electron are accounted for.

Note: Consider two electrons in 3s orbital:

Question 61.
Complete the following table:

Answer:

Question 62.
Explain condensed orbital notation of electronic configuration of an atom.
Answer:

• The orbital notation of electronic configuration of an element with high atomic number comprises a long train of symbols of orbitals with an increasing order of energy.
• It can be condensed by dividing it into two parts: Inner or core part of electronic configuration and outer electronic configuration.
• Electronic configuration of the preceding inert gas is a part of the electronic configuration of any element. In the condensed orbital notation, it is implied by writing symbol of that inert gas in a square bracket. It is core part of the electronic configuration of that element. The outer electronic configuration is specific to a particular element and written immediately after the bracket.
• For example, the orbital notation of potassium ‘K (Z = 19) is Is² 2s² 2p⁶ 3s² 3p⁶ 4s¹’. Its core part is the electronic configuration of the preceding inert gas argon ‘Ar: 1s² 2s² 2p⁶ 3s² 3p⁶, while ‘4s¹’ is an outer part. Therefore, the condensed orbital notation of electronic configuration of potassium is ‘K: [Ar] 4s².’

Note: Electronic configuration of the elements with atomic numbers 1 to 30 is as follows:

Question 63.
Write electronic configuration of ₁₈Ar and ₁₉K using orbital notation and orbital diagram method.
Answer:
From the atomic numbers, it is understood that 18 electrons are to be filled in Ar atom and 19 electrons are to be filled in K atom. These are to be filled in the orbitals according to the Aufbau principle. The electronic configuration of these atoms can be represented as:

Question 64.
Write condensed orbital notation of electronic configuration of the following elements:
i. Fluorine (Z = 9)
ii. Scandium (Z = 21)
iii. Cobalt (Z = 27)
iv. Zinc (Z = 30)
Answer:

Question 65.
Find out one dinegative anion and one unipositive cation which are isoelectronic with Ne atom. Write their electronic configuration using orbital notations and orbital diagram method.
Answer:
Atomic number (Z) of Ne is 10. Therefore, Ne and its isoelectronic species contain 10 electrons each. The dinegative anionic species, isoelectronic with Ne is obtained by adding two electrons to the atom with Z = 8. This is O^2- ion.
The unipositive cationic species, isoelectronic with Ne is obtained by removing one electron from an atom with Z = 11. It is Na⁺ ion.
These species and their electronic configurations are shown below:

Question 66.
A student pictorially represented the electronic configuration of cobalt (Z = 27) in ground state as shown in the following figure.

i. Is this the correct representation?
ii. Identify the rules of electron filling that are violated (if any) in the above answer and give the correct representation.
Answer:
i. No, the electronic configuration of cobalt in ground state is incorrectly represented.
ii. The mles of electron filling that are violated in the above diagram are Pauli’s exclusion principle and Hund’s rule. The correct electronic configuration is:

Question 67.
With reference to the representative model of the gold foil experiment, answer the following questions:

i. What evidence regarding an atom do lines A, B and C provide?
ii. How does Rutherford’s model contradict Thomson’s plum-pudding model?
iii. What results would you expect from the experiment if Thomson’s plum-pudding model was correct?
Answer:
i. Line A – Most of the space inside the atom is empty because most of the α-particles passed through the gold foil without getting deflected.
Line B – The nucleus is positively charged since some α-particles were deflected at small angles.
Line C – The nucleus contains most of the atoms mass since few α-particles were deflected backward, i.e. toward the radioactive source.

ii. According to plum-pudding model, an atom was considered a positively charged sphere with negatively charged electrons embedded in it. However, Rutherford’s gold foil experiment proved that an atom consists of large empty space with positive charge concentrated only at the centre (nucleus) and negatively charged electrons revolve around the nucleus in various orbits.

iii. If Thomson’s plum-pudding model was correct, then in the gold foil experiment we would not expect to see any significant deflection of the α-particles, i.e., Most α-particles would pass through the foil with very small or no deflections.

Question 68.
Complete the following table:

Answer:

Multiple Choice Questions

1. Which of the following statements about the electron is INCORRECT?
(A) It is a negatively charged particle.
(B) The mass of electron is equal to the mass of neutron.
(C) It is a basic constituent of all atoms.
(D) It is a constituent of cathode rays.
Answer:
(B) The mass of electron is equal to the mass of neutron.

2. The isotopes of an element differ in
(A) the number of neutrons in the nucleus
(B) the charge on the nucleus
(C) the number of extra-nuclear electrons
(D) both the nuclear charge and the number of extra-nuclear electrons
Answer:
(A) the number of neutrons in the nucleus

3. The difference between U²³⁵ and U²³⁸ atoms is that U²³⁸ contains ………….
(A) 3 more protons
(B) 3 more protons and 3 more electrons
(C) 3 more neutrons and 3 more electrons
(D) 3 more neutrons
Answer:
(D) 3 more neutrons

4. The number of electrons, protons and neutrons in ³¹P^3- ion is respectively ……………
(A) 15, 15, 16
(B) 15, 16, 15
(C) 18, 15, 16
(D) 15, 16, 18
Answer:
(C) 18, 15, 16

5. In vacuum, the speed of all types of electromagnetic radiation is equal to ………….
(A) 3.0 × 10⁶ m s^-1
(B) 3.0 × 10⁸ m s^-1
(C) 3.0 × 10¹⁰ m s^-1
(D) 3.0 × 10¹² m s^-1
Answer:
(B) 3.0 × 10⁸ m s^-1

6. In the electromagnetic spectrum, the ultraviolet region is around ………….. Hz.
(A) 10⁶
(B) 10¹⁰
(C) 10¹⁶
(D) 10²⁶
Answer:
(C) 10¹⁶

7. In hydrogen spectrum, the series of lines appearing in ultraviolet region of electromagnetic spectrum are called ………….
(A) Lyman series
(B) Balmer series
(C) Pfund series
(D) Brackett series
Answer:
(A) Lyman series

8. The energy of electron in the n^th Bohr orbit of H-atom is …………

Answer:
(C) \(-\frac{2.18 \times 10^{-18}}{\mathrm{n}^{2}} \mathrm{~J}\)

9. Which of the following is a hydrogen-like species?
(A) He²⁺
(B) Be³⁺
(C) Li⁺
(D) H⁺
Answer:
(B) Be³⁺

10. The de Broglie wavelength associated with a particle of mass 10^-6 kg moving with a velocity of 10 m s^-1 is ………..
(A) 6.63 × 10^-22 m
(B) 6.63 × 10^-29 m
(C) 6.63 × 10^-31 m
(D) 6.63 × 10^-34 m
Answer:
(B) 6.63 × 10^-29 m

11. An orbital is designated by …………. quantum numbers while an electron in an atom is designated by …………. quantum numbers.
(A) two, three
(B) three, two
(C) four, two
(D) three, four
Answer:
(D) three, four

12. In a multi-electron atom, the energy of the orbital depends on two quantum numbers: ……….
(A) n and ms
(B) n and ml
(C) ml and ms
(D) n and l
Answer:
(D) n and l

13. The number of subshells in a shell is equal to ………..
(A) n
(B) n²
(C) n – 1
(D) l + 1
Answer:
(A) n

14. The maximum number of electrons in a subshell for which l = 3 is …………..
(A) 14
(B) 10
(C) 8
(D) 4
Answer:
(A) 14

15. Which of the following is INCORRECT?
(A) A nodal plane has ψ² very close to zero.
(B) The value of ψ² at any finite distance from the nucleus is always zero.
(C) A boundary surface diagram enclosing 100 % probability density cannot be drawn.
(D) A boundary surface diagram is a good representation of shape of an orbtial.
Answer:
(B) The value of ψ² at any finite distance from the nucleus is always zero.

16. p_z-Orbital has ……….. nodal plane/planes.
(A) zero
(B) one
(C) two
(D) three
Answer:
(B) one

17. Which of the following pairs of d-orbitals will have electron density along the axis?
(A) d_xy, d_x²-y²
(B) d_z², d_xz
(C) d_xz, d_yz
(D) d_z², d_x²-y²
Answer:
(D) d_z², d_x²-y²

18. The two electrons have the following set of quantum numbers:
P = 3, 2, -2, +\(\frac {1}{2}\)
Q = 3, 0, 0, +\(\frac {1}{2}\)
Which of the following statement is TRUE?
(A) P and Q have same energy
(B) P has greater energy than Q
(C) P has lesser energy than Q
(D) P and Q represent same electron
Answer:
(B) P has greater energy than Q

19. For 3d orbital, the values of n and l are ………… respectively.
(A) 0, 3
(B) 3, 2
(C) 3, 0
(D) 3, 3
Answer:
(B) 3, 2

20. For the electron present in 1 s orbital of helium atom, the correct set of values of quantum numbers is ………..
(A) 1, 0, 0, +1/2
(B) 1, 1, 0, +1/2
(C) 1, 1, 1, +1/2
(D) 2, 0, 0, +1/2
Answer:
(A) 1, 0, 0, +1/2

21. The ground state electronic configuration for chromium atom (Z = 24) is …………
(A) [Ar] 3d⁵ 4s¹
(B) [Ar] 3d⁴ 4s²
(C) [Ar] 3d⁸
(D) [Ar] 4s¹ 4p⁵
Answer:
(A) [Ar] 3d⁵ 4s¹

22. The electronic configuration of Ni²⁺ is ……….. (Atomic number of Ni = 28)
(A) [Ar] 4s² 3d⁶
(B) [Ar] 4s¹ 3d⁸
(C) [Ar] 3d⁸
(D) [Ar] 4s² 3d⁸
Answer:
(C) [Ar] 3d⁸

Balbharti Maharashtra State Board 11th Commerce Book Keeping & Accountancy Important Questions Chapter 9 Final Accounts of a Proprietary Concern Important Questions and Answers.

Maharashtra State Board 11th Commerce BK Important Questions Chapter 9 Final Accounts of a Proprietary Concern

1. Answer in One Sentence.

Question 1.
What do you mean by pre-received income?
Answer:
Income that is received before it is due for receipt is called pre-received income.

Question 2.
What do you mean by bad debts?
Answer:
The debts which are not recoverable in spite of repeated efforts to collect the same are called bad debts.

Question 3.
What is Capital?
Answer:
Amount invested by the proprietor from time to time in business is known as capital.

Question 4.
What are the adjustments?
Answer:
Adjustments are additional information given below the trial balance and are to be considered for arriving at the correct profit or loss.

Question 5.
State the meaning of Current assets.
Answer:
Assets that are purchased with the intention of converting them into cash during the operating year are called current assets.
E.g. stock of goods.

2. Give a word, term, or phrase which can substitute each of the following statements:

Question 1.
Debit balance of Trading Account.
Answer:
Gross Loss

Question 2.
The credit balance of the Trading Account.
Answer:
Gross Profit

Question 3.
Debit balance of Profit and Loss Account.
Answer:
Net Loss

Question 4.
The credit balance of Profit and Loss Account.
Answer:
Net Profit

Question 5.
A debt that cannot be recovered.
Answer:
Bad debts

Question 6.
Reduction in the value of fixed assets due to its continuous use.
Answer:
Depreciation

Question 7.
Carriage paid on the purchase of goods.
Answer:
Carriage Inwards

Question 8.
Statement of balances of various ledger accounts.
Answer:
Trial Balance

Question 9.
An amount withdraws by a proprietor from a business in cash or kind.
Answer:
Drawings

Question 10.
Account prepared to find out gross profit or gross loss.
Answer:
Trading A/c

Question 11.
Income received before it is due.
Answer:
Pre-received Income

Question 12.
Unpaid expenses of a business.
Answer:
Outstanding Expenses

Question 13.
Account prepared on the basis of direct expenses and direct income of the business.
Answer:
Trading A/c

Question 14.
Account prepared on the basis of indirect expenses and indirect incomes of the business.
Answer:
Profit and Loss A/c

Question 15.
Group of accounts which gives the result of business activities.
Answer:
Final Accounts

3. Select the most appropriate alternatives given below and rewrite the sentence:

Question 1.
A list of balances of all the accounts in ledger is called _______________
(a) Balance Sheet
(b) Profit and Loss A/c
(c) Trading A/c
(d) Trial Balance
Answer:
(d) Trial Balance

Question 2.
Opening stock is entered in a Trading Account on the _______________ side.
(a) credit
(b) debit
(c) asset
(d) liabilities
Answer:
(b) debit

Question 3.
Drawing account is closed by transferring the balance to the _______________ account.
(a) Drawing
(b) Liabilities
(c) Assets
(d) Capital
Answer:
(d) Capital

Question 4.
Outstanding expenses is a _______________ account.
(a) Real
(b) Personal
(c) Nominal
(d) None of them
Answer:
(b) Personal

Question 5.
Depreciation is always charged on _______________ assets.
(a) Current
(b) Fixed
(c) Fictitious
(d) Intangible
Answer:
(b) Fixed

Question 6.
Pre-received income is shown on _______________ side of Balance sheet.
(a) Assets
(b) Liabilities
(c) Credit
(d) Debit
Answer:
(b) Liabilities

Question 7.
Royalty on production is a _______________ expenses.
(a) direct
(b) indirect
(c) capital
(d) none of them
Answer:
(a) direct

Question 8.
Interest on investment is _______________ of business concern.
(a) a profit
(b) a loss
(c) an expense
(d) an income
Answer:
(d) an income

Question 9.
All items of indirect income are shown on the credit side of the _______________ Account.
(a) Balance Sheet
(b) Profit and Loss
(c) Manufacturing
(d) None of them
Answer:
(b) Profit and Loss

Question 10.
Reserve for discount on debtor has a _______________ balance.
(a) credit
(b) debit
(c) nil
(d) positive
Answer:
(a) credit

4. State True or False with reasons:

Question 1.
Closing Stock is valued at cost or market price whichever is more.
Answer:
This statement is False.
Closing stock is valued at cost or market price whichever is less. It is based on the theory of anticipated profit is not brought with the account before actual realization or the Principle of conservatism.

Question 2.
Income received in advance is a liability.
Answer:
This statement is True.
Income received in advance is the liability to the business because it has not yet earned the money and the business has an obligation to deliver the goods or services to the customer.

Question 3.
Prepaid expenses are a liability.
Answer:
This statement is False.
Prepaid means paid in advance for the future period. It is the amount paid but has not yet been used up or has not yet expired. So prepaid expenses are an asset.

Question 4.
A balance Sheet is a real account.
Answer:
This statement is False.
The balance sheet is a statement prepare at the end of the financial or accounting year. It is not an Account. It gives an idea of Total Assets and liabilities on a particular day.

Question 5.
Depreciation need not be provided if the asset is not in use.
Answer:
This statement is False.
The working life of fixed assets decreases with passes of time. The value of these assets decreases every year as new technology introduced in the market old becomes outdated so it is necessary to depreciate an asset even it is not in use.

5. Fill in the blanks:

Question 1.
A Copy Right is _______________ Asset.
Answer:
An intangible

Question 2.
Wages paid for installation of Machinery should be debited to _______________ A/c.
Answer:
Machinery

Question 3.
A provision made for debts irrecoverable from debtors is called _______________
Answer:
Reserve for doubtful debts

Question 4.
In the absence of information interest on Drawings is charged for _______________ months.
Answer:
six

Question 5.
Return outward are deducted from _______________
Answer:
purchases

Question 6.
Net profit is transferred to _______________
Answer:
Balance Sheet Capital A/c

Question 7.
If cash/goods withdrawn by proprietor for domestic use, it is called _______________
Answer:
Drawings

Question 8.
Payment made in advance are shown on _______________ side of Balance Sheet.
Answer:
Asset

Question 9.
Royalty on production is debited to _______________ A/c of final A/c.
Answer:
Trading

Question 10.
General Expenses are recolored to the debit side of _______________ A/c when Office expenses, Sundry expenses, or General expenses are given in the trial balance.
Answer:
Trading

6. Find the odd one:

Question 1.
Salary, Sundry expenses, General Expenses.
Answer:
General Expenses

Question 2.
Creditors, Bank Loan, Investment.
Answer:
Investment

Question 3.
Royalty on purchases, Octroi, Discount Allowed.
Answer:
Discount Allowed

Question 4.
Rent, Factory Lighting, Freight.
Answer:
Rent

Question 5.
Outstanding Salary, Accrued Interest, Outstanding Rent.
Answer:
Accrued Interest

7. Do you agree or disagree with the following statement:

Question 1.
Software expenses paid for the installation of the computer should be debited to Software A/c.
Answer:
Disagree

Question 2.
In absence of information interest on the drawing is charged for twelve months.
Answer:
Disagree

Question 3.
Only carriage means carriage on sales.
Answer:
Disagree

Question 4.
Final accounts are prepared at the end of the month.
Answer:
Disagree

Question 5.
Return Inwards means purchase return.
Answer:
Disagree

8. Correct and Rewrite the following statements:

Question 1.
Bank overdraft is an Asset of the business concern.
Answer:
Bank overdraft is a liability of business concern.

Question 2.
Discount allowed is an income for business
Answer:
Discount allowed is an expense for the business.

Question 3.
To Goods withdrawn by proprietor A/c
Capital A/c……………Dr
(Being goods withdrawn by the proprietor for personal use)
Answer:
To Goods withdrawn by proprietor A/c
To Capital A/c………….Dr
(Being goods withdrawn by the proprietor for use)

Question 4.
Sundry Debtors A/c……………Dr
To Bad debts A/c
(Being Bad debts written off)
Answer:
Bad Debts A/c……………Dr
To Sundry Debtors A/c
(Being Bad debts written off)

Question 5.
Opening stock A/c…………….Dr
Direct expenses A/c…………….Dr
Purchase A/c………………Dr
To Trading A/c
(Being Opening Stock, Direct expenses, and purchase transferred to Trading A/c)
Answer:
Trading A/c……….Dr
To Opening stock A/c
To Direct expenses A/c
To Purchase A/c
(Being opening stock, Direct expenses and purchase transferred to Trading A/c)

Balbharti Maharashtra State Board 11th Commerce Book Keeping & Accountancy Important Questions Chapter 10 Single Entry System Important Questions and Answers.

Maharashtra State Board 11th Commerce BK Important Questions Chapter 10 Single Entry System

1. Answer in One sentence only.

Question 1.
In which method statement of affairs is prepared?
Answer:
Under the Net worth method of a single entry, a statement of affairs is prepared.

Question 2.
How is closing capital calculated under a single entry system?
Answer:
Under single entry system closing capital is calculated by deducting the total closing value of liabilities from the total closing value of assets.

Question 3.
Which statement is prepared under the single entry system to ascertain profit?
Answer:
A statement of profit or loss is prepared to ascertain profit under a single entry system.

Question 4.
What is a Statement of Profit or Loss?
Answer:
A statement that is prepared under a single entry system to calculate profit or loss is called a statement of profit or loss.

2. Write a word, term, or phrase which can substitute each of the following statements.

Question 1.
A system of book-keeping in which both the aspects of transactions are recorded.
Answer:
Double Entry System

Question 2.
Name the method of accounting in which only cash and personal transactions are recorded.
Answer:
Single Entry System

Question 3.
A statement is similar to the Balance sheet was prepared to ascertain the amounts of closing capital.
Answer:
Closing Statement of Affairs

Question 4.
The system of accounting is most scientific and reliable.
Answer:
Double Entry System

Question 5.
Name the statement prepared to find out profit or loss under a single entry system.
Answer:
Statement of Profit or Loss

Question 6.
Excess of opening capital over closing capital of proprietor under single entry system.
Answer:
Loss

Question 7.
Method of accounting in which real accounts and nominal accounts are not maintained.
Answer:
Single Entry System

Question 8.
A statement that shows profit or loss of business under a single entry system.
Answer:
Statement of Profit or Loss

Question 9.
An accounting system where rules of debit and credit are not followed.
Answer:
Single Entry System

Question 10.
The incomplete method of the accounting system.
Answer:
Single Entry System

Question 11.
A system of book-keeping which records only one aspect of business transactions and ignores other aspects.
Answer:
Single Entry System

Question 12.
A statement that shows the balances of various assets and liabilities at their approximate or estimated values as on a particular date.
Answer:
Statement of Affairs

3. Select the most appropriate answer from the alternatives given below and rewrite the sentence.

Question 1.
The difference between the capital at the end of the year and capital at the beginning of the year is called _____________
(a) Profit
(b) Income
(c) Drawings
(d) Expenses
Answer:
(a) Profit

Question 2.
A statement of _____________ is to be prepared in under to find out profit or loss under a single entry system.
(a) Income
(b) Affairs
(c) Revenue
(d) Profit or Loss
Answer:
(d) Profit or Loss

Question 3.
A statement of affairs is a summarised statement of an estimated _____________
(a) Financial Position
(b) Profit
(c) Income
(d) Loss
Answer:
(a) Financial Position

Question 4.
If closing capital is ₹ 30,000 and profit is ₹ 5,000 opening capital was _____________
(a) ₹ 35,000
(b) ₹ 30,000
(c) ₹ 25,000
(d) ₹ 15,000
Answer:
(c) ₹ 25,000

Question 5.
Under single Entry system, Profit = Closing Capital less _____________
(a) Opening Capital
(b) Opening Assets
(c) Opening Liabilities
(d) Drawings
Answer:
(a) Opening Capital

Question 6.
The capital at the end of the accounting year is ascertained by preparing _____________
(a) Cash Account
(b) Closing Statement of Affairs
(c) Total Debtors Account
(d) Opening Statement of Affairs
Answer:
(b) Closing Statement of Affairs

Question 7.
The capital at the beginning of the accounting year is ascertained by preparing _____________
(a) Receipt and Payment Account
(b) Cash Account
(c) Opening Statement of Affairs
(d) Closing Statement of Affairs
Answer:
(c) Opening Statement of Affairs

Question 8.
Under Single Entry System only _____________ are opened.
(a) Cash and Personal Accounts
(b) Real Accounts
(c) Nominal Accounts
(d) Real and Nominal Accounts
Answer:
(a) Cash and Personal Accounts

Question 9.
Statement of Affairs is just like _____________
(a) Profit and Loss A/c
(b) Real A/c
(c) Trading A/c
(d) Balance Sheet
Answer:
(d) Balance Sheet

Question 10.
Under the Net worth method, the basis for ascertaining profit or loss is the difference between _____________
(a) Capital on two dates
(b) Gross assets on two dates
(c) Liabilities on two dates
(d) Net assets on two dates
Answer:
(a) Capital on two dates

4. State True or False with reasons:

Question 1.
Statement of profit is just like Profit and Loss Account.
Answer:
This statement is False.
The profit and loss account has the debit and credit side which shows all expenses on the debit side and all incomes on the credit side and the differences are profit and loss for the year. Whereas statement of profit just adds and less income and expenses to find profit or loss.

Question 2.
The single Entry System is based on certain rules and principles.
Answer:
This statement is False.
A single Entry System is an ancient method of recording business transactions. It is a simple method of book-keeping. It is not a scientific and accurate system of Accounting. This system has no proper set of rules to be followed.

Question 3.
All transactions are recorded in the Single Entry System.
Answer:
This statement is False.
All transactions are not recorded in a single entry system. Only cash books and personal accounts of debtors and creditors are maintained. All transactions are recorded in the Double Entry System of Book-Keeping and Accountancy.

Question 4.
Arithmetical accuracy cannot be checked in Single Entry.
Answer:
This statement is True.
All transactions and accounts are not recorded in the Single Entry System. So it is impossible to prepare a Trial balance under this system without which Arithmetical accuracy cannot be checked.

Question 5.
Drawings made during the year decrease the profit under the Single Entry System.
Answer:
This statement is False.
Drawings made during the year are added to the closing capital in the statement of profit, it increases the profit under the Single Entry System.

5. Do you agree with the following statements?

Question 1.
The single Entry System of Book-keeping is a scientific method of books of accounts.
Answer:
Disagree

Question 2.
The single Entry System is useful only for large organizations.
Answer:
Disagree

Question 3.
Statement of Affairs is just like a profit and loss account.
Answer:
Disagree

Question 4.
The difference between Assets and Liabilities is called net profit.
Answer:
Disagree

Question 5.
The single Entry System follows the golden rules of accounts.
Answer:
Disagree

6. Fill in the Blanks.

Question 1.
In _____________ Book-keeping system, only Cash/Bank A/c and Personal accounts of Debtors and Creditors are opened.
Answer:
Single Entry

Question 2.
Capital is the difference between _____________ and _____________
Answer:
Assets, Liabilities

Question 3.
Single Entry System of Book-keeping is _____________ system of books of accounts.
Answer:
Conventional Accounting

Question 4.
_____________ accuracy is not guaranteed under Single Entry System.
Answer:
Arithmetical

Question 5.
In statement of profit or loss, profit on sale of assets are _____________ to closing capital.
Answer:
Added

Question 6.
Bad debts are _____________ from closing capital in statement of profit or loss.
Answer:
Deducted

Question 7.
_____________ unscientific system of Book-keeping.
Answer:
Single Entry System

Question 8.
Under the Single Entry System, profit or loss is calculated by deducting the opening capital balance from _____________ at the end of the year.
Answer:
the closing capital balance

7. Find the odd one.

Question 1.
Stock in trade, Bank overdraft, Bills receivable.
Answer:
Bank overdraft

Question 2.
Interest on Loan, Interest on Investment, Income receivable.
Answer:
Interest on Loan

Question 3.
Bad debts, Reserve for Bad debts, Reserve for a discount on creditors.
Answer:
Reserve for a discount on creditors

Question 4.
Income received in advance, Prepaid Expenses, Outstanding Expenses.
Answer:
Prepaid Expenses

8. Complete the following table:

Question 1.

Answer:
₹ 13,000

Question 2.

Answer:
₹ 45,000

Question 3.

Answer:
₹ 85,000

Question 4.

Answer:
₹ 40,000, ₹ 18,000

Question 5.

Answer:
₹ 6,000

Balbharti Maharashtra State Board 11th Commerce Book Keeping & Accountancy Important Questions Chapter 8 Rectification of Errors Important Questions and Answers.

Maharashtra State Board 11th Commerce BK Important Questions Chapter 8 Rectification of Errors

1. Answer in One Sentence:

Question 1.
What is an Arithmetical error?
Answer:
An error committed to totaling the amount columns of journal and ledger is called arithmetical error.

Question 2.
What do you mean by one-sided errors?
Answer:
One-sided errors are those errors that either affect the debit or credit aspect of the transaction. It affects the agreement of the trial balance.

Question 3.
What do you mean by two-sided errors?
Answer:
Two-sided errors are those errors that affect both the aspects of a transaction i.e. debit and credit. It does not affect the agreement of the trial balance.

Question 4.
What is a suspense account?
Answer:
An account that is opened to transfer the difference in the totals of the trial balance is known as a suspense account.

Question 5.
What are accounting errors?
Answer:
Mistakes or errors committed while writing the books of accounts are known as accounting errors.

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

Question 1.
Transactions remained to be recorded at all in the books of account.
Answer:
Errors of omission

Question 2.
Errors are always rectified by passing rectification entries.
Answer:
Two-sided errors

Question 3.
Errors that affect the debit and credit side unequally.
Answer:
One-sided errors

Question 4.
Errors that affect the debit and credit side equally.
Answer:
Two-sided errors

Question 5.
An error in which the transaction is entered in the original book but not posted into the ledger.
Answer:
The error of posting/partial omission

Question 6.
Error in the process of transferring the entry from original books into the ledger.
Answer:
Error of posting

Question 7.
Errors can be rectified without passing rectification entries.
Answer:
One-sided errors

3. Select the most appropriate alternative from those given below and rewrite the sentence.

Question 1.
Wages paid for the installation of machinery wrongly debited to wages account is an error of ___________
(a) omission
(b) principle
(c) commission
(d) duplication
Answer:
(b) principle

Question 2.
If the trial balance shows a short credit the suspense account will have a ___________ balance.
(a) debit
(b) zero
(c) credit
(d) nil
Answer:
(c) credit

Question 3.
If the trial balance does not agree the difference of the trial balance is placed in ___________ account.
(a) Personal
(b) Suspense
(c) Rectification
(d) Real
Answer:
(b) Suspense

Question 4.
Errors which compensate the effect of each other are called ___________ errors.
(a) compensating
(b) one-sided
(c) two sided
(d) clerical
Answer:
(a) compensating

Question 5.
One sided errors are disclosed by ___________
(a) Trial Balance
(b) Suspense Account
(c) Journal
(d) Ledger Account
Answer:
(a) Trial Balance

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

Question 1.
Errors of principle are not disclosed by the Trial Balance.
Answer:
This statement is True.
The error of Principle is those where some basic principles of bookkeeping and accountancy are not Properly Followed while recording a business transaction. Such errors can be rectified by passing journal entries.
Eg.: Capital expender showed as revenue expenditure or vice versa. It won’t affect the Trial balance. If will agreed.

Question 2.
Transaction not recorded in the books is an error of principle.
Answer:
This statement is False.
Transaction not recorded in the books is an error of omission and not an error of principle.

Question 3.
If the purchase hook is undermasted, the purchase account is debited in rectification.
Answer:
This statement is True.
Purchases are expenses. All expenses are debited as per the rule of nominal account. When purchases are underacted they can be corrected by debiting in rectification.

Question 4.
When a transaction is not recorded according to the principles of book-keeping the error is said to be an error of principle.
Answer:
This statement is True.
The error of Principle is those where some basic principles of bookkeeping and accountancy are not properly followed while recording a business transaction.
Eg.: Capital expenditure is shown as revenue expenditure or Vice-a-Varsa and they are called an error of principle.

Question 5.
The error of omission is disclosed by the Trial Balance.
Answer:
This statement is False.
The complete omission of a transaction will bot disclosed by a trial balance. Trial balance, balances will be agreed with such errors. So the omission of transaction or an error of omission will not be disclosed by Trial Balance.

5. Do you agree or disagree with the following statements.

Question 1.
A temporary account opened to rectify the entry is known as suspense A/c.
Answer:
Agree

Question 2.
Rectified entries are passed in Ledger.
Answer:
Disagree

Question 3.
Compensating errors affect the agreement of Trial Balance.
Answer:
Disagree

Question 4.
There is only one type of error.
Answer:
Disagree

Question 5.
Transaction recorded without following the accounting principles and rules are known as Errors of Principle.
Answer:
Agree

6. Complete the following sentences.

Question 1.
___________ sided errors affect the total of Trial Balance.
Answer:
One

Question 2.
___________ sided errors do not affect the Trial Balance.
Answer:
Two

Question 3.
One sided error do not require ___________ entry.
Answer:
Rectification

Question 4.
Errors which are committed in writing the accounts are called error of ___________
Answer:
Posting

Question 5.
Under casting of Sales book is corrected by ___________ Sales Account.
Answer:
Crediting

Question 6.
The disagreement of Trial Balance indicates that an ___________ has been committed.
Answer:
Error

Question 7.
An asset has been purchased for the firm. However, the amount was debited to the purchase account. It is an error of ___________
Answer:
Principle

Question 8.
___________ account is necessary for rectification of one-sided error.
Answer:
Suspense A/c

Question 9.
An item of ₹ 95 has been debited to a personal account as ₹ 59. It is an error of ___________
Answer:
Commission

Question 10.
Rectification entries are passed in ___________
Answer:
Journal Proper

Question 11.
Two sided errors are rectified by passing ___________ entry.
Answer:
Rectification

Question 12.
If the Trial Balance does not tally, its difference is transferred to ___________ Account.
Answer:
Suspense

Question 13.
Under casting of an account is ___________ sided error.
Answer:
One

Question 14.
If the transaction is not at all recorded in the books, it is called an error of ___________
Answer:
Complete Omission

Question 15.
If the total balance shows short a credit, the suspense account will have a ___________ balance.
Answer:
Credit

Question 16.
An error that affects debit as well as credit side, it is called as ___________ errors.
Answer:
Two-Sided

Question 17.
Wages paid for installation of machinery debited to Wages Account is an error of ___________
Answer:
Principle

Question 18.
Errors which cancel out the effect of one another is called ___________ errors.
Answer:
Compensating

Question 19.
One sided errors are disclosed by ___________
Answer:
Trial Balance

Question 20.
In an error of omission, the debit and credit are ___________
Answer:
Equal

Practical Problems

Question 1.
The trial balance of Sagar did not agree. It showed an excess credit of ₹ 7,550. Sagar put the difference to the suspense account. He located the following errors:
1. Sales return book was overcast by ₹ 1,200.
2. Purchase book was undercast by ₹ 750.
3. Goods returned to Mahesh ₹ 1,000 were recorded through the sales book.
4. Credit purchases from Mahadev ₹ 6,000 were recorded through the sales book.
5. Credit purchases from Damodhar ₹ 4,000 were recorded through the sales book. However, Damodhar’s account was correctly credited.
6. Salary paid ₹ 3,500 was debited to the employee’s personal account.
Give journal entries to rectify the above errors and prepare Suspense Account.
Solution:
Journal Proper

Question 2.
The trial balance of Radhika did not agree. Radhika put the difference to the suspense account. Subsequently, she located the following errors.
1. Furniture purchased for ₹ 6,000 was posted to the purchases account as ₹ 600.
2. Repairs to Machinery ₹ 500 were debited to the Machinery account.
3. Wages paid for the installation of Machinery ₹ 750 was posted to wages account.
4. Purchased material ₹ 8,000 and Wages ₹ 2,000 were used for construction of the building. No adjustment was made in the books.
5. Total of sales returns book ₹ 2,000 was not posted to the ledger.
6. Old Furniture sold to Dinesh at its book value of ₹ 2,500 was recorded through sales book.
Give journal entries and prepare Suspense Account.
Solution:
Journal Proper