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Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 13 Nuclear Chemistry and Radioactivity Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 13 Nuclear Chemistry and Radioactivity

Question 1.
Explain the term nuclear chemistry. Give few examples of nuclear reactions.
Answer:
Nuclear chemistry is a branch of physical chemistry and it deals with the study of reactions involving changes in atomic nuclei. This branch started with the discovery of natural radioactivity by physicist Antoine Henri Becquerel.

Examples of nuclear reactions are as follows:

• Radioactive decay
• Artificial transmutation
• Nuclear fission
• Nuclear fusion

Question 2.
Write a short note on similarity between the solar system and structure of atom.
Answer:
Solar system: It consists of the Sun and planets in which Sun is at the centre of solar system and planets move around it under the force of gravity.

Atomic system: It consists of tiny central core called as nucleus at the centre of atom around which electrons are present. Like in solar system, electrostatic attractions hold subatomic particles in a structure of atom. The nucleus consists of protons and neutrons.

Question 3.
Answer the following.
i. Give the symbolic representation for calcium, (no. of protons = 20, mass number = 40)
ii. Calculate the number of neutrons for calcium.
Answer:
i. \({ }_{20}^{40} \mathrm{Ca}\), in which Z = 20 and A = 40.
ii. Number of neutrons: It can be calculated from formula (A = Z + N).
For calcium, N = A – Z = 40 – 20 = 20
Nucleus of the calcium atom contains 20 neutrons.

Question 4.
Explain the term nucleons with examples.
Answer:
The term nucleon refers to the sum of protons (p) and neutrons (n) present in atom, e.g. Number of nucleons present in \({ }_{20}^{40} \mathrm{Ca}\) are 40 (i.e., 20 protons and 20 neutrons). Number of nucleons present in \({ }_{11}^{23} \mathrm{Na}\) are 23 (i.e., 11 protons and 12 neutrons).

Question 5.
Define: Nuclide
Answer:
The nucleus of a specific isotope is called as nuclide.

Question 6.
Atom as a whole is electrically neutral. Justify.
Answer:

• The magnitude of electronic charge (e) on the nucleus is +Ze and that of outer sphere is -Ze. Number of protons and number of electrons are always equal in an atom.
• As a result of this, the charges get nullified, therefore, the atom as a whole is electrically neutral.

Question 7.
Define:
i. Isotopes
ii. Isobars
Answer:
i. Isotopes: Nuclides which contain same number of protons but different number of neutrons in their nuclei are called as isotopes. e.g. \({ }_{11}^{22} \mathrm{Na}\), \({ }_{11}^{23} \mathrm{Na}\) and \({ }_{11}^{24} \mathrm{Na}\)
ii. Isobars: Nuclides (of different element) which have same mass number but have different number of protons and neutrons in their nuclei are called as isobars.
OR
The atoms of different elements having the same mass number but different atomic numbers are called isobars.
e.g. \({ }_{6}^{14} \mathrm{C}\) and \({ }_{7}^{14} \mathrm{N}\)

Question 8.
Define mirror nuclei and isotones.
Answer:

• Isobars in which the number of protons and neutrons differ by 1 unit and are interchanged are called as mirror nuclei.
• Isotones are defined as nuclides having the same number of neutrons but different number of protons and hence, different mass numbers.

Question 9.
Name the following.
i. Nuclides in which number of protons and neutrons differ by 1 and are interchanged.
ii. Nuclides having the same number of neutrons but different number of protons.
iii. Nuclides with the same mass number which differ in energy states.
Answer:
i. Mirror nuclei
ii. Isotones
iii. Nuclear isomers

Question 10.
Explain the term nuclear isomers?
Answer:

• The nuclides with the same number of protons (Z) and neutrons (N) or the same mass number (A) which differ in energy states are called nuclear isomers.
• In this, the isomer of higher energy is said to be in the metastable state which is represented by writing “m” after the mass number.
  e.g. Nuclear isomers of cobalt can be represented as, ^60mCo and ⁶⁰Co.

Question 11.
State true or false. Correct the false statement.
i. The number of nucleons in C-12 atom is 6.
ii. N-13 and C-13 are mirror nuclei.
iii. Nuclear isomers have same number of protons and neutrons.
Answer:
i. False,
The number of nucleons in C-12 atom is 12.
ii. True
iii. True

Question 12.
Give classification of nuclides on the basis of nuclear stability.
Answer:
Nuclides can be classified into stable and unstable/radioactive nuclides on the basis of nuclear stability.

• Stable nuclides: In this type of nuclides, the number of electrons and the location of nuclei may change in outer sphere but the number of protons and neutrons remain unchanged.
• Radioactive (unstable) nuclides: These nuclides undergo spontaneous change forming new nuclides.

Question 13.

	Number of protons (Z)	Number of neutrons (N)	Number of such nuclides
i.	Even	Even	165
ii.	Even	Odd	55

What conclusion can be drawn from the above given data?
Answer:

• Number of nuclides with even ‘Z’ and even ‘N’ are higher in number as compared to nuclides with even ‘Z’ and odd ‘N’
• Nuclides with even number of ‘Z’ and odd number of ‘N’ are about 1/3rd of nuclides where both ‘Z’ and ‘N’ are even.
• Nuclides with even number of protons (Z) and even number of neutrons (N) are most stable. These nuclides tend to fonn proton-proton and neutron-neutron pairs. This impart stability to the nucleus.

Question 14.
Write a note on naturally occurring nuclides with either odd number of protons or odd number of neutrons.
Answer:
i. The number of stable nuclides with either Z or N odd is about one third of nuclides where both are even.
ii. These nuclides are less stable than those having even number of protons and neutrons.
iii. In these nuclides one nucleon has no partner and therefore, these nuclides are less stable.
iv. Further the number of nuclides with odd A are nearly the same, irrespective of Z or N is odd. This indicates that protons and neutrons behave similarly in the respect of stability.
v. Following table gives the estimate of such nuclides occurring in nature.

	Number of protons (Z)	Number of neutrons (N)	Number of such nuclides
i.	Even	Odd	55
ii.	Odd	Even	50

Question 15.
State true or false. Correct the false statements.
i. The nuclides with even Z and even N constitute 85% of earth crust.
ii. Nuclides with either ‘Z’ or ‘N’ odd are more stable than those having even number of both ‘Z’ and ‘N’
iii. The number of nuclides with odd number of ‘Z’ and odd number of ‘N’ are only four.
Answer:
i. True
ii. False
Nuclides with either ‘Z’ or ‘N’ odd are less stable than nuclides having even number of both ‘Z’ and ‘N’.
iii. True

Question 16.
Heavier nuclides require greater number of neutrons (than protons) to attain stability. Justify.
Answer:

• The heavier nuclides with the increasing number of protons lead to large coulombic repulsions.
• Increased number of neutrons will separate the protons within the nuclei, which will impart stability. Thus, in order to attain stability heavier nuclide need more number of neutrons.

Question 17.
Consider the graph of neutron (N) plotted against proton number (Z). How will you identify radioactive nuclides from the graph?
Answer:
Nuclides which fall outside the belt or stability zone are radioactive nuclides.

Question 18.
Write a note Magic numbers.
Answer:
Magic numbers: The nuclei with 2, 8, 20, 28, 50, 82 and 126 neutrons or protons are particularly stable and abundant in nature. These numbers are known as magic numbers.
e.g. Lead (\({ }_{82}^{208} \mathrm{~Pb}\)) has two magic numbers, 82 protons and 126 neutrons.

Question 19.
What is the order of distance between two protons present in the nucleus?
Answer:
The order of distance between two protons present in the nucleus is typically of order of 10^-15 m.

Question 20.
Which factor is responsible for nuclear stability?
Answer:
Nuclear forces of attractions exist within nuclei. These are attractions between proton-proton (p-p), neutron-neutron(n-n) and proton-neutron(p-n). They constitute or give rise to nuclear potential which is responsible for nuclear stability.

Question 21.
Write short notes on: nuclear potential.
Answer:

• Nuclear potential is the attraction between p-p, n-n and p-n.
• These attractive forces are independent of the charge on nucleons or attraction between p-p, n-n and p-n are equal.
• These attractive forces operate over short range within the nucleus.
• Nuclear potential is responsible for the nuclear stability.

Question 22.
State true or false. Correct the false statement.
i. The nuclear forces of attractions are dependent on the charge on the nucleons.
ii. The actual mass of an atom is observed to be more than sum of the masses of its constituents.
Answer:
i. False
The nuclear forces of attractions are independent of the charge on the nucleons.
ii. False
The actual mass of an atom is observed to be less than sum of the masses of its constituents.

Question 23.
Define: Nuclear binding energy
Answer:
An energy equivalent to the mass lost is released during the formation of nucleus. This is called the nuclear binding energy.
OR
The energy requiredfor holding the nucleons together within the nucleus of an atom is called as the nuclear binding energy.

Question 24.
Explain the term: mass defect.
Answer:
During the formation of nucleus, certain mass is lost. This phenomenon is known as mass defect (Δm).
The exact mass of nucleus is slightly less than sum of the exact masses of the constituent nucleons. This difference is called as mass defect. It is represented by symbol Δm.
Formulae: Δm = calculated mass – observed mass

Question 25.
Explain the relation between nuclear mass and energy? Also give the energy released in the conversion of one atomic mass unit into energy.
Answer:
i. The nuclear mass is expressed in atomic mass unit (u) which is exactly 1/12^th of the mass of ¹²C atom. Thus, u = 1/12^th mass of C-12 atom = 1.66 × 10^-2 kg.
ii. The conversion of mass into energy is established through Einstein’s equation, E = mc².
Where m is the mass of matter converted into energy (E) and velocity of light (c).
iii. The energy released in the conversion of one u mass into energy is given by:
E = mc² = (1.66 × 10^-27kg) × (3 × 10⁸ m s^-1)²

Question 26.
Derive the expression for nuclear binding energy for a nuclide.
Answer:
Expression for nuclear binding energy:
i. Consider a nuclide \({ }_{z}^{A} X\) that contains Z protons and (A – Z) neutrons. Suppose the mass of the nuclide is m. The mass of proton is m_p and that of neutron is mn.

ii. Total mass = (A – Z)m_n + Zm_p + Zm_e …..(1)
Δm = [(A – Z)m_n + Zm_p + Zm_e] – m
= [(A – Z)m_n + Z(m_p + m_e] – m
= [(A – Z)m_n + Zm_H] – m …..(2)
Where (m_p + m_e) = m_H = mass of H atom.
Thus, (Δm) = [Zm_p + (A – Z)m_n] – m
Where Z = atomic number
A = mass number
(A – Z) = neutron number
m_p and m_n = masses of proton and neutron, respectively
m = mass of nuclide

iii. The mass defect, Δm is related to binding energy of nucleus by Einstein’s equation,
ΔE = Δm × c²
Where, ΔE = Binding energy, Δm = mass defect.
iv. Nuclear energy is measured in million electro volt (MeV).
v. The total binding energy is then given by,
B.E. = Δm (u) × 931.4
Where 1.00 u = 931.4 MeV
B.E. = 931.4 [Zm_H + (A – Z)m_n – m] ……(3)
Total binding energy of nucleus containing A nucleons is the B.E.
vi. The binding energy per nucleon is then given by,
\(\bar{B}\) = B.E./A

Question 27.
Calculate the mean binding energy per nucleon for the formation of \({ }_{8}^{16} \mathrm{O}\) nucleus. The mass of oxygen atom is 15.994 u. The masses of H atom and neutron are 1.0078 u and 1.0087 u, respectively.
Solution:
Given: m_H = 1.0078 u
m_n= 1.0087 u
m= 15.994 u
Z = 8, A= 16
To find: Mean binding energy per nucleon (\(\bar{B}\))
Formulae: i. Δm = Zm_H + (A – Z)m_n – m
ii. B.E. = Δm × 931.4 MeV
iii. \(\overline{\mathrm{B}}=\frac{\mathrm{B} . \mathrm{E} .}{\mathrm{A}}\)
Calculation: i. The mass defect, Δm = Zm_H + (A – Z)m_n – m
Δm = 8 × 1.0078 u + 8 × 1.0087 u – 15.994 u = 0.138 u
ii. Total binding energy, B.E. (MeV) = Δm (amu) × 931.4
Hence, B.E. = 0.138 × 931.4 = 128.533 MeV
iii. Binding energy per nucleon, \(\overline{\mathrm{B}}=\frac{\mathrm{B} . \mathrm{E} .}{\mathrm{A}}\)
Hence, \(\bar{B}\) = \(\frac{128.533}{16}\) = 8.033 MeV/nucleon
Ans: Binding energy per nucleon for the formation of \({ }_{8}^{16} \mathrm{O}\) nucleus = 8.033 MeV/nucleon

Question 28.
Calculate the binding energy per nucleon for the formation of \({ }_{2}^{4} \mathrm{He}\) nucleus. Mass of \({ }_{2}^{4} \mathrm{He}\) atom = 4.0026 u.
Solution:
Given: m = 4.0026 u
Z = 2, A = 4
To find: Binding energy per nucleon (\(\bar{B}\))
Formulae: i. Δm = Zm_H + (A – Z)m_n – m
ii. B.E. = Δm × 931.4 MeV
iii. \(\overline{\mathrm{B}}=\frac{\mathrm{B} . \mathrm{E} .}{\mathrm{A}}\)
The mass defect, Δm = [Zm_H + (A – Z)m_n] – m
Δm = [(2 × 1.0078) + (2 × 1.0087)] – 4.0026 = 0.0304 u
Total binding energy, B.E. (MeV) = Δm (amu) × 931.4
= 0.0304 × 931.4
= 28.315 MeV
iii. B.E. per nucleon, \(\overline{\mathrm{B}}=\frac{\mathrm{B} . \mathrm{E} .}{\mathrm{A}}\)
\(\bar{B}\) = \(\frac{28.315}{4}\) = 7.079 Mey/nucleon
Ans: Binding energy per nucleon for formation of \({ }_{2}^{4} \mathrm{He}\) nucleus = 7.079 MeV/nucleon

Question 29.
Define radioactivity and give examples of two radioactive elements.
Answer:
Radioactivity is a phenomenon in which the nuclei spontaneously emit a nuclear particle and gamma radiation transforming to a different nuclide. e.g. Uranium and radium
[Note: Radioactivity is the phenomenon related to the nucleus.]

Question 30.
What is the criteria for an element to be known as radioactive element?
Answer:

• An element is considered to be radioactive if the nuclei of its atoms are unstable.
• That is, when element undergoes nuclear changes (i.e., emission of nuclear particles and gamma radiation), it is said to be radioactive.

Question 31.
What are the different types of radiations emitted by radioactive element?
Answer:
The radiations emitted by radioactive elements are as follows:

• Alpha (α) radiations
• Beta (β) radiations
• Gamma (γ) radiations

Question 32.
Write the unit of rate of decay.
Answer:
The rate of decay is expressed in the form of disintegrations per second (dps).

Question 33.
Derive the equation λ = \(\frac{\left(-\frac{\mathbf{d} \mathbf{N}}{\mathbf{d} \mathbf{t}}\right)}{\mathbf{N}}\) and write what does λ denotes.
Answer:
The rate of decay of a radioelement at any instant is proportional to the number of nuclei (atoms) present at that instant. It can be represented as,
\(-\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}} \propto \mathrm{N} \quad \text { or }-\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=\lambda \mathrm{N}\) …….(i)
Where, \(-\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}\) = Rate of decay at any time, t
λ = Decay constant
N = Number of nuclei (atoms) present at time, t
From equation (i),

Decay constant (λ) is the fraction of nuclei decaying in unit time.
OR
It is the ratio of the amount of substance disintegrated per unit time to the amount of substance present at that time.

Question 34.
Derive the expression for decay constant.
Answer:
Decay constant (λ) is the fraction of nuclei decaying in unit time.
Thus,
λ = \(-\frac{d N}{d t} \times \frac{1}{N}\) …(i)
Rearranging equation (i) we get,
\(\frac{\mathrm{d} \mathrm{N}}{\mathrm{N}}\) = -λ dt
On integrating above equation, we get
∫\(\frac{\mathrm{d} \mathrm{N}}{\mathrm{N}}\) = -∫ λ dt …(ii)
On performing the integration, we get lnN = -λt + C ……(iii)
where C is the constant of integration whose value is obtained as follows:
Let N₀ be the number of nuclei present at some arbitrary zero time. At time t, the number of nuclei is N. So, at t = 0, N = N₀, substituting in equation (iii), we get
lnN₀ = C
With this value of C, equation (iii) becomes
lnN = -λt + lnN₀
or λt = lnN₀ – InN = ln \(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\) ……(iv)
Hence, λ = \(\frac{1}{t} \ln \frac{N_{0}}{N}\) …….(v)
Converting natural logarithm (ln) to logarithm to the base 10, equation (v) becomes
λ = \(\frac{2.303}{t} \log _{10} \frac{N_{0}}{N}\) ………(vi)
The equation (iv) can be expressed as ln \(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\) = -λt. Taking antilog of both sides, we get
\(\frac{\mathrm{N}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \mathrm{t}} \text { or } \mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}\) …….(vii)
The equation (vi) and equation (vii) give the decay constant.

Question 35.
Write a note on half-life of a radioelement.
Answer:

• Half-life of a radioelement (t_1/2): It is the time needed for a given number of nuclei (atoms) of radioelement to decay exactly to half of its initial value.
• Each radio isotope has its own half-life.
• Half-life of a radioelement can be expressed in seconds, minutes, hours, days or years.
• Mathematical expression for half-life of a radioelement can be given as,
  \(t_{1 / 2}=\frac{0.693}{\lambda}\)

Question 36.
Complete the following statements based on the given graph.
i. As decay progresses, the number of radioactive atoms will ……….. with time.
ii. As decay progresses, the rate of decay will …………..
iii. Rate of radioactive decay at any instant is ………… to the number of atoms of the radioactive element present at that instant.
Answer:
i. decrease
ii. decrease
iii. proportional

Question 37.
²¹⁸Po decays initially at a rate of 816 dps. The rate falls to 408 dps after 24 min. Calculate the decay constant.
Solution:

Question 38.
After how many seconds will the concentration of radioactive element X will be halved, if the decay constant is 1.155 × 10^-3 s^-1?
Solution:

Ans: Concentration of radioactive element (X) will be halved in 600 s.

Question 39.
⁴¹Ar decays initially at a rate of 575 Bq. The rate falls to 358 dps after 75 minutes. What is the half-life of ⁴¹Ar?
Solution:


Ans: The half-life of Ar is 109.7 min.

Question 40.
The half-life of ³²P is 14.26 d. What percentage of ³²P sample will remain after 40 d?
Solution:
Given: t_1/2 = 14.26 d,
N₀ = 100,
t = 40 d
To find: Percentage of ³²P sample remaining after 40 d

Question 41.
The half-life of ³⁴Cl is 1.53 s. How long does it take for 99.9 % of sample of ³⁴Cl to decay?
Solution:
Given: t_1/2 = = 1.53 s,
N₀ = 100,
N = 100 – 99.9 = 0.1,
To find: Time (t)

Question 42.
The half-life of ²⁰⁹Po is 102 y. How much of 1 mg sample of polonium decays in 62 y?
Solution:
Given: t_1/2 = 102y,
t = 62 y,
N₀ = 1 mg
To find: Amount of polonium that decayed in 62 y

Taking antilog of both sides we get,
\(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\) = antilog (0.1829) = 1.524
N = \(\frac{\mathrm{N}_{0}}{1.524}=\frac{1 \mathrm{mg}}{1.524}\) = 0.656 mg
N is the amount that remains after 62 y.
Hence, the amount decayed in 62 y = 1 mg – 0.656 mg = 0.344 mg
Ans: The amount decayed in 62 y is 0.344 mg

Question 43.
What will be the approximate time taken for 90 % decay of ¹⁷⁴Ir in terms of its half-life?
Solution:
Given: N₀ = 100
N = 100 – 90 = 10
To find: Time (t)

Ans: Thus, the approximate time required for 90 % decay of ¹⁷⁴Ir in terms of its half-life is 3.3t_1/2.

Question 44.
A radioactive decay of element X (Z = 35) is 30 % complete in 2 hours. Calculate its half-life period.
Solution:
Given: t = 2 hrs,
N₀ = 100
N= 100 – 30 = 70
To find: t_1/2

Question 45.
What are the different modes by which radio elements decay?
Answer:
There are 3 modes by which radio elements decay: α-decay, β-decay and γ-emission.

Question 46.
What is α-decay?
Answer:
Radioactive isotope/radioelement when undergoes decay by the emission of α-particle from the nuclei then the process involved is referred to as α-decay.

Question 47.
Give equation for radium-222 when it undergoes decay by emission of an α-particle.
\({ }_{88}^{226} \mathrm{Ra} \longrightarrow{ }_{86}^{222} \mathrm{X}+{ }_{2}^{4} \mathrm{He}\)
Answer:
\({ }_{88}^{226} \mathrm{Ra} \longrightarrow{ }_{86}^{222} \mathrm{X}+{ }_{2}^{4} \mathrm{He}\)
Thus, atomic number of element ‘X’ will be 86 and atomic mass number will be 222.

Question 48.
Identify the mode of decay and state whether following equation is CORRECT or NOT. Justify.
\({ }_{92}^{238} \mathbf{U} \longrightarrow{ }_{90}^{234} \mathrm{Th}+{ }_{2}^{4} \mathbf{H e}\)
Ans:
i. It involves α-decay process.
ii. As uranium undergoes decay by emission of an α-particle (i.e., \({ }_{2}^{4} \mathrm{He}\)), daughter nuclei (in this case thorium) ‘will observe the decrease in atomic number by 2 units and decrease in atomic mass number by 4 units.
Hence, the given equation is correct.

Question 49.
If radioactive element ‘X’ undergoes α-emission then what will be the position of daughter nuclei in the periodic table with respect to element ‘X’.
Answer:
If radioactive element ‘X’ undergoes α-emission, then corresponding daughter nuclei formed will occupy two places to the left of the periodic table with respect to element ‘X’.

Question 50.
What is β^– – decay? Also explain the changes that occur in the parent nuclei due to β-emission with one example.
Answer:
β^– – decay: The emission of negatively charged stream of β^– particles from the nucleus is called β^– – decay.
i. β^– – Particles are electrons with a charge and mass of an electron, mass being negligible as compared to the nuclei.
ii. When a nucleus decays by emitting a high-speed electron called a beta particle (β), a new nucleus is formed with the same mass number as the original nucleus and with an atomic number that is one unit greater than the parent nuclei.
General equation:

Note: The mass number A does not change, the atomic number changes when a nuclei undergoes β-decay. e.g. Neptunium-238 decays to form plutonium-238:

Question 51.
Mention the atomic number and atomic mass number of the parent radioelement ‘X’ in the following case if parent nuclei undergo β-emission.
i. \(\mathrm{X} \longrightarrow{ }_{94}^{238} \mathrm{Pu}\)
ii. \(\mathrm{X} \longrightarrow{ }_{95}^{241} \mathrm{Am}\)
Answer:

Question 52.
How many α and β-particles are emitted in the following?
\({ }_{93}^{237} \mathrm{~Np} \longrightarrow{ }_{83}^{209} \mathrm{Bi}\)
Answer:
The emission of one α-particle decreases the mass number by 4 whereas the emission of β particles has no effect on mass number.
Net decrease in mass number = 237 – 209 = 28. This decrease is only due to α- particles. Hence, number of α- particles emitted = \(\frac {28}{4}\) = 7
Now, the emission of one α-particle decreases the atomic number by 2 and one β-particle emission increases it by 1.
The net decrease in atomic number = 93 – 83 = 10
The emission of 7 α-particles causes decrease in atomic number by 14. However, the actual decrease is only 10. It means atomic number increases by 4. This increase is due to emission of 4 β-particles.
Thus, 7 α and 4 β- particles are emitted.

Question 53.
Explain the process of γ-decay in detail with a suitable example.
Answer:
γ-decay:
i. γ-Radiation is always accompanied with α and β^– decay processes.
ii. During γ-radiation, the daughter nucleus is left in energetically excited state which decays to the ground state of product with emission of γ-rays.
For example, \({ }_{92}^{238} \mathrm{U} \longrightarrow{ }_{90}^{234} \mathrm{Th}+{ }_{2}^{4} \mathrm{He}+\gamma\)
iii. \({ }_{92}^{238} \mathrm{U}\) emits α-particles of two different energies, 4.147 MeV (23%) and 4.195 MeV (77%).
iv. When α-particles of energy 4.147 MeV are emitted, ²³⁴Th is left in an excited state which de-excites to the ground state with emission of γ-ray photons with energy 0.0048 MeV.

Question 54.
Half-life of ²⁰⁹Po is 102 y. How many α-particles are emitted in 1 s from 2 mg sample of Po?
Solution:
Given: t_1/2 = 102 y,
t = 1 s,
Amount of sample = 2 mg
To find: Number of α-particles emitted

Question 55.
Nuclear transmutation is a spontaneous or non-spontaneous process?
Answer:
Nuclear transmutation is a non-spontaneous (man-made) process.

Question 56.
What is nuclear transmutation?
Answer:
Nuclear transmutation:

• It is the process of transformation of a stable nucleus into another nucleus which can be stable or unstable.
• It can occur by the radioactive decay of a nucleus or the reaction of a nucleus with another particle.

Question 57.
Differentiate between chemical reactions and nuclear reactions.
Answer:
Chemical reactions:

• Rearrangement of atoms by breaking and forming of chemical bonds.
• Different isotopes of an element have same behaviour.
• Only outer shell electrons take part in the chemical reaction.
• The chemical reaction is accompanied by relatively small amounts of energy.
  e.g. chemical combustion of 1.0 g methane releases only 56 kJ energy.
• The rates of reaction are influenced by the temperature, pressure, concentration and catalyst.

Nuclear reactions:

• Elements or isotopes of one element are converted into another element in a nuclear reaction.
• Isotopes of an element behave differently.
• In addition to electrons, protons, neutrons, other elementary particles may be involved.
• The nuclear reaction is accompanied by a large amount of energy change, e.g. The nuclear transformation of 1 g of Uranium – 235 release 8.2 × 10⁷ kJ
• The rate of nuclear reactions is unaffected by temperature, pressure and catalyst.

Question 58.
What will happen when a nucleus of J’B is bombarded with α-particle? Identify the process involved.
Answer:
i. When a stable nucleus of \({ }_{5}^{10} \mathrm{~B}\) is is bombarded with α-particle, it transforms into \({ }_{7}^{13} \mathrm{~N}\), which is radioactive and spontaneously emits positrons to produces \({ }_{6}^{13} \mathrm{C}\).
This can be represented as,

ii. The process involved is known as induced radioactivity or artificial radioactivity.

Question 59.
Define: Nuclear fission
Answer:
Nuclear fission is defined as a process which involves splitting of the heavy nucleus of an atom into two nearly equal fragments accompanied by release of the large amount of energy.

Question 60.
Nuclear fission of ²³⁵U is a chain process. Justify.
Answer:

• Nuclear fission of ²³⁵U occurs when nucleus absorbs neutron. When a uranium nucleus absorbs neutron, it breaks into two lighter fragments and releases energy (heat), more neutrons, and other radiation.
• When one uranium 235 nucleus undergoes fission, three neutrons are emitted.
• These neutrons emitted in fission cause more fission of the uranium nuclei which yield more neutrons. These neutrons again bring forth fission producing further neutrons.
• The process continues indefinitely leading to chain reaction which continues even after the removal of bombarding neutrons.

Question 61.
Explain the term: Nuclear fusion and give one example.
Answer:
Nuclear fusion: In this process, the lighter nuclei combine (fuse) together and form a heavy nucleus which is accompanied by an enormous amount of energy.
e. g. The energy received by earth from the sun is due to the nuclear fusion reactions.

Question 62.
Which will produce more energy: Nuclear fission or fusion?
Answer:
Nuclear fusion will produce relatively more energy per given mass of fuel.

Question 63.
What is the range of temperature required to carry out nuclear fusion reaction?
Answer:
Nuclear fusion reaction requires extremely high temperature typically of the order of 108 K.

Question 64.
Distinguish between nuclear fission and nuclear fusion.
Answer:
Nuclear fission:

• It is the process in which a heavy nucleus splits up into two lighter nuclei of nearly equal masses.
• About 200 MeV of energy is available per fission in case of \({ }_{92}^{235} \mathrm{U}\).
• The products of nuclear fission are, in general, radioactive.

Nuclear fusion:

• It is the process in which two lighter nuclei combine together to form a heavy nucleus.
• Energy available per fusion is much less but the energy per unit mass of material is much greater than that for fission of heavy nuclei.
• The products of fusion are, in general, non-radioactive.

Question 65.
Estimate the energy released in the fusion reaction.
\({ }_{1}^{2} \mathbf{H}+{ }_{2}^{3} \mathbf{H e} \longrightarrow{ }_{2}^{4} \mathbf{H e}+{ }_{1}^{1} \mathbf{H}\)
(Given atomic masses: ²H = 2.0141 u. ³He = 3.0160 u, ⁴He = 4.0026 u, ¹H = 1.0078 u)
Answer:

Question 66.
Explain the term: Radiocarbon dating in detail.
Answer:
Radiocarbon dating: The technique is used to find the age of historic and archaeological organic samples such as old wood samples and animal or human fossils.
Radioisotope used for carbon dating is ¹⁴C.
i. Radioactive ¹⁴C is formed in the upper atmosphere by bombardment of neutrons from cosmic ray on ¹⁴N.
\({ }_{7}^{14} \mathrm{~N}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{6}^{14} \mathrm{C}+{ }_{1}^{1} \mathrm{H}\)
ii. ¹⁴C combines with atmospheric oxygen to form ¹⁴CO₂ which mixes with ordinary ¹²CO₂.
iii. This carbon dioxide is absorbed by plants during photosynthesis.
iv. Animals eat plants which have absorbed a carbon dioxide (¹⁴CO₂ + ¹²CO₂). Hence, ¹⁴C becomes a part of plant and animal bodies.
v. As long as the plant is alive, the ratio ¹⁴C/¹²C remains constant.
vi. When the plant dies, photosynthesis will not occur and the ratio ¹⁴C/¹²C decreases with the decay of radioactive ¹⁴C which has a half-life 5730 years.
vii. The decay process of ¹⁴C is given below:
\({ }_{6}^{14} \mathrm{C} \longrightarrow{ }_{7}^{14} \mathrm{~N}+{ }_{-1}^{0} \mathrm{e}\)
viii. The activity (N) of given wood sample and that of fresh sample of live plant (N0) is measured, where, N0 denotes the activity of the given sample at the time of death.
ix. The age of the given wood sample, can be determined by applying following Formulae:

Question 67.
What is nuclear power?
Answer:
Nuclear power is the electricity generated from the fission of uranium and plutonium.

Question 68.
Nuclear power is a clean source of energy. Justify.
Answer:
Nuclear power offers huge environmental benefits in producing electricity because,

• it releases zero carbon dioxide.
• it releases zero sulphur and nitrogen oxides.
• these are atmospheric pollutants which pollute the air.

Thus, nuclear power is a clean source of energy.

Question 69.
How much energy will be produced by fission of 1 gram of ²³⁵U?
Answer:
Fission of 1 gram of uranium-235 produces about 24,000 kW/h of energy.

Question 70.
Nuclear fission is an alternative energy source. Explain.
Answer:

• Fission of 1 gram of uranium-235 produces about 24,000 kW/h of energy.
• This is the same amount of energy produced by burning 3 tons of coal or 12 barrels of oil, or nearly 5000 m³ of natural gas.
• The sources like coal, oil, natural gas are depleting very fast.
• Also, the costs of petrol and other products from petroleum industry is increasing.
• Thus, we need to depend on the nuclear fission as an alternative source of energy for electricity.

Question 71.
Label the follow ing diagram of simplified nuclear reactor.

Answer:

Question 72.
Explain in brief: Nuclear reactor
Answer:
Nuclear reactor: Nuclear reactor is a device for using atomic energy in controlled manner for peaceful purposes. During nuclear fission energy is released. The released energy can be utilized to generate electricity in a nuclear reactor.

Working of a nuclear reactor:

• In a nuclear reactor, U²³⁵ or U²³⁹, a fissionable material is stacked with heavy water (D₂O deuterium oxide) or graphite called moderator.
• The neutrons produced in the fission pass through the moderator and lose a part of their energy. The slow neutrons produced during the process are captured which initiate new fission.
• Cadmium rods are inserted in the moderator as they have ability to absorb neutrons. This controls the rate of chain reaction.
• The energy released during the reaction appears as heat and removed by circulating a liquid (coolant). The coolant which has absorbed excess of heat from the reactor is passed over a heat exchanger for producing steam.
• Steam is then passed through the turbines to produce electricity. Thus, the atomic energy produced with the use of fission reaction can be controlled in the nuclear reactor.
• This process can be explored for peaceful purpose such as conversion of atomic energy into electrical energy which can be used for civilian purposes, ships, submarines, etc.

Note: Schematic diagram of nuclear power plant:

Question 73.
Why cadmium rods are used in nuclear reactor?
Answer:
Cadmium rods are inserted in the moderator as they have ability to absorb neutrons which help to control the rate of chain reaction.

Question 74.
Why short-lived isotopes are used for diagnostic purposes?
Answer:
For diagnostic purpose, short-lived isotopes are used in order to limit the exposure time to radiation. Note: Diagnostic Radioisotopes are listed below:

Question 75.
Give one application of therapeutic radioisotopes.
Answer:
Therapeutic radioisotopes are used to destroy abnormal cell growth in the body, e.g. cancerous cells.
Note: Therapeutic Radioisotopes are listed below:

Question 76.
Give example of isotopes used in following.
i. Isotope used in the treatment of leukaemia.
ii. Isotopes used in the preservation of agricultural products by irradiation.
Answer:
i. Isotope of phosphorus, \({ }_{15}^{35} \mathrm{P}\).
ii. ⁶⁰Co or ¹³⁷Cs

Question 77.
At which places has BARC Mumbai set up irradiation plants for preservation of agricultural produce?
Answer:
Bhabha Atomic Research Centre (BARC) Mumbai has set up irradiation plants for preservation of agricultural produce such as mangoes, onion and potatoes at Vashi (Navi Mumbai) and Lasalgaon (Nashik).

Question 78.
Why radiotracer technique is used in chemistry?
Answer:
Radiotracer technique is used to trace the path/mechanism followed by a reaction in the system.

Question 79.
The half-life for radioactive decay of an element X is 140 days. Complete the following flow chart showing decay of 1 g of X.

Answer:

Shortcut method:
Amount of the element X left after n half-lives is given as [X] = \(\frac{[\mathrm{X}]_{0}}{2^{n}}\)
e.g. \(\frac{1}{2^{4}} \mathrm{~g}=\frac{1}{16} \mathrm{~g}\)

Question 80.
A sample of 35S complete its 10% decay in 20 min, then calculate the time required to complete decay by 19%.
Answer:
When decay is 10 % complete, if N₀ = 100 , then N = 100 – 10 = 90 and t = 20 minutes
When decay is 19 % complete, N = 100 – 19 = 81
Substituting these values in formula we get,

Multiple Choice Questions

1. Radius of the nucleus is related to the mass number A by ………….
(A) R = R₀A^1/2
(B) R = R₀A
(C) R = R₀A²
(D) R = R₀A^1/3
Answer:
(D) R = R₀A^1/3

2. Which of the following nuclides has the magic number of both protons and neutrons?
(A) \({ }_{50}^{115} \mathrm{Sn}\)
(B) \({ }_{81}^{206} \mathrm{Pb}\)
(C) \({ }_{82}^{208} \mathrm{Pb}\)
(D) \({ }_{50}^{118} \mathrm{Pb}\)
Answer:
(C) \({ }_{82}^{208} \mathrm{Pb}\)

3. The probability of decay of a radioactive element depends on …………..
i. the age of nucleus
ii. the presence of catalyst
iii. pressure
iv. temperature
(A) only i. and iv
(B) all of these
(C) only ii. And iii.
(D) none of these
Answer:
(D) none of these

4. The decay constant for 67Ga is 7.0 × 10^-4 s^-1. If initial concentration of is 0.07 g, what is the half-life of ⁶⁷Ga?
(A) 990 s
(B) 79.2 s
(C) 12375 s
(D) 10.10 × 10^-4 s
Answer:
(A) 990 s

5. The half-life of radioactive element X having decay constant of 1.7 × 10^-5 s^-1 is …………
(A) 21.5 h
(B) 19.7 h
(C) 11.3 h
(D) 2.8 h
Answer:
(C) 11.3 h

6. A radioactive decay of element X (Z = 90) is 30 % complete in 30 minutes. It has a half-life period of ……………
(A) 24.3 min
(B) 58.3 min
(C) 102.3 min
(D) 120.3 min
Answer:
(B) 58.3 min

7. The half-life of radium is 1600 years. The fraction of a sample of radium that would remain after 6400 year is ……….
(A) \(\frac {1}{2}\)
(B) \(\frac {1}{4}\)
(C) \(\frac {1}{8}\)
(D) \(\frac {1}{16}\)
Answer:
(D) \(\frac {1}{16}\)

8. The half-life of an element is 5 d. How much time is required for the decay of 7/8^th of the sample?
(A) 5 d
(B) 10 d
(C) 15 d
(D) 35/8 d
Answer:
(C) 15 d

9. The composition of an α-particle can be expressed as ……………….
(A) 1p + 1n
(B) 1p + 2n
(C) 2p + 1n
(D) 2p + 2n
Answer:
(D) 2p + 2n

10. If a radioactive nuclide of group 15 element undergoes β-particle emission, the daughter element will be found in ………………..
(A) 16 group
(B) 14 group
(C) 13 group
(D) same group
Answer:
(A) 16 group

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 9 Elements of Group 13, 14 and 15 Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 9 Elements of Group 13, 14 and 15

Question 1.
What are p-block elements?
Answer:

• Elements in which the differentiating electron (the last filling electron) enters the outermost p orbital are p-block elements.
• Since maximum six electrons can be accommodated in p-subshell i.e., three p-orbitals, the p-block contains six groups numbered from 13 to 18 in the modem periodic table.
• The p-block elements show greater variation in the properties than s-block elements.

Question 2.
Write the names of the elements present in groups 13, 14 and 15.
Answer:

Group	Name of family	Name of the elements
13	Boron family	Boron (5B), aluminium (13Al), gallium (31Ga), indium (49In), thallium (81Tl)
14	Carbon family	Carbon (6C), silicon (14Si), germanium (32Ge), tin (50Sn), lead (82Pb)
15	Nitrogen family	Nitrogen (7N), phosphorus (15P), arsenic (33AS), antimony (51Sb), bismuth (83Bi)

Question 3.
i. Write the general outer electronic configuration of the elements of group 13, group 14 and group 15.
ii. By how many electrons do their outer electronic configurations differ from their nearest inert gas?
Answer:
i.

Group	General outer electronic configuration
13	ns2 np1
14	ns2 np2
15	ns2 np3

ii. The outer electronic configurations of the elements group 13, group 14 and group 15 differ from their nearest inert gas by 5, 4 and 3 electrons, respectively.

Question 4.
In which form do the elements of groups 13,14 and 15 occur in nature?
Answer:

• The elements of groups 13, 14 and 15 do not occur in free monoatomic state and are found as compounds with other elements.
• They also occur in the form of polyatomic molecules (such as N₂, P₄, C₆₀) or polyatomic covalent arrays (such as graphite, diamond).

Question 5.
Write condensed electronic configurations of the following elements.
₁₃Al, ₄₉In, ₁₄Si, ₅₀Sn, ₁₅P, ₃₃As
Answer:
Condensed electronic configurations of
i. ₁₃Al: [Ne]3s² 3p¹
ii. ₄₉In: [Kr]4d¹⁰5s²5p¹
iii. ₁₄Si: [Ne]3s²3p²
iv. ₅₀Sn: [Kr]4d¹⁰5s²5p²
v. ₁₅P: [Ne]3s²3p³
vi. ₃₃As: [Ar]3d¹⁰4s²4p³

Note: Condensed electronic configurations of elements of groups 13, 14 and 15 are given in the table below.

Question 6.
Name the following.
i. A metalloid present in group 13.
ii. A group 13 element which is the third most abundant element in the earth’s crust.
Answer:
i. Boron
ii. Aluminium

Question 7.
Why boron is classified as a metalloid?
Answer:
Boron is glossy and hard solid like metals but a poor conductor of electricity like nonmetals. Since it exhibits properties of both metals and nonmetals, boron is classified as a metalloid.

Question 8.
Describe the variation in the electronegativity of group 13 elements.
Answer:

• In group 13, on moving down the group, the electronegativity decreases from B to Al.
• However, there is a marginal increase in the electronegativity from Al to Tl.
• This trend is a result of the irregularities observed in atomic size of elements.

Question 9.
Atomic numbers of the group 13 elements are in the order B < Al < Ga < In < Tl. Arrange these elements in increasing order of ionic radii of M³⁺.
Answer:

• The given elements are in an increasing order of their atomic number.
• The general outer electronic configuration of group 13 elements is ns²np¹.
• M³⁺ ion is formed by the removal of three electrons from the outermost shell ‘n’.
• In the M³⁺ ions, the ‘n-1’ shell becomes the outermost shell. Size of the ‘n-1’ shell increases down the group.

Therefore, the ionic radii of M³⁺ ion increases down the group in the following order:
B³⁺ < Al³⁺ < Ga³⁺ < In³⁺ < Tl³⁺

Question 10.
Why the atomic radius of Gallium is less than that of aluminium?
Answer:

• Atomic radius of the elements increases down the group due to addition of new shells.
• Electronic configuration of Al is [Ne]3s²3p¹ while that of Ga is [Ar]3d¹⁰4s²4p¹.
• As Al does not have d-electrons, it offers an exception to this trend.
• As we go from Al down to Ga the nuclear charge increases by 18 units. Out of the 18 electrons added, 10 electrons are in the inner 3d subshell of Ga. These d-electrons offer poor shielding effect.
• Therefore, the effect of attraction due to increased nuclear charge is experienced prominently by the outer electrons of Ga and thus, its atomic radius becomes smaller than that of Al.

Hence, the atomic radius of gallium is less than that of aluminium.

Question 11.
The values of the first ionization enthalpy of Al, Si and P are 577, 786 and 1012 kJ mol^-1 respectively. Explain the observed trend.
Answer:

• The trend shows increasing first ionization enthalpy from Al to Si to P.
• Al, Si and P belong to the third period in the periodic table and hence, they have same valence shell.
• As we move across a period from left to right, the nuclear charge increases. Due to this, electrons in the valence shell are held more tightly by the nucleus as we go from Al to Si to P.
• Therefore, more energy is required to remove an electron from its outermost shell.

Hence, the value of first ionization enthalpy increases from Al to Si to P.

Note: Atomic and physical properties of group 13 elements.

Question 12.
Name metal(s), nonmetal(s) and metalloid(s) of group 14.
Answer:
i. Metal: Tin, lead
ii. Nonmetal: Carbon
iii. Metalloid: Silicon, germanium

Question 13.
Explain the variation in the following properties of the group 14 elements,
i. Atomic radii
ii. Ionization enthalpy
iii. Electronegativity
Answer:
i. Atomic radii (Covalent radii):

• In the periodic table as we move down the group 14 from C to Pb, the atomic radii increases due to the addition of new shell at each succeeding element.
• However, the increase is comparatively less after silicon due to poor shielding by inner d- and f-electrons in the atoms.

ii. Ionization enthalpy:

• Due to increased effective nuclear charge, group 14 elements have higher value of ionization enthalpy than corresponding group 13 elements.
• In the periodic table, as we move down the group 14 from C to Sn, the ionization enthalpy decreases.
• From Si to Sn, the ionization enthalpy decreases slightly.
• However, from Sn to Pb, the ionization enthalpy increases slightly. It is due to the poor shielding effect of intervening d and f orbitals and increase in the size of the atoms.

iii. Electronegativity:

• Due to small atomic size, group 14 elements are slightly more electronegative than the corresponding group 13 elements.
• As we move down the group 14 from C to Si in the periodic table, the electronegativity decreases.
• The electronegativity values for elements from Si to Pb are almost the same.
• Among group 14 elements, carbon is the most electronegative with electronegativity of 2.5.

Question 14.
Explain why there is a phenomenal decrease in ionization enthalpy from carbon to silicon.
Answer:

• Carbon is the first element of group 14 and thus, it has the smallest atomic size.
• The ionization enthalpy of carbon (1086 kJ mol^-1) is very high due to its small atomic size (77 pm) and high electronegativity (2.5).
• However, the ionization enthalpy of silicon (786 kJ mol^-1) decreases phenomenally due to the increase in its atomic size (118 pm) and low electronegativity (1.8).

Note: Atomic and physical properties of group 14 elements.

Question 15.
Which type of elements are present in group 15? Mention their physical state.
Answer:

• Group 15 includes all the three traditional types of elements i.e., metals, nonmetals and metalloids.
• Nitrogen is a gas whereas the remaining group 15 elements are solids.
• The gaseous nitrogen and brittle phosphorus are nonmetals.
• Arsenic and antimony are metalloids while bismuth is moderately reactive metal.

Question 16.
Explain the trends in physical properties of group 15 elements.
i. Atomic and ionic radii
ii. Ionization enthalpy
iii. Electronegativity
Answer:
i. Atomic and ionic radii:

• Atomic size increases down the group with increasing atomic number.
• The effective nuclear charge in case of group 15 elements is larger than that of group 14 elements. Due to the increased effective nuclear charge, electrons are strongly attracted by the nucleus. Thus, the atomic and ionic radii of group 15 elements are smaller than the atomic and ionic radii of the corresponding group 14 elements.
• On moving down the group, number of shells increases which leads to increased shielding effect and as a result atomic radii and ionic radii increases.

ii. Ionization enthalpy:

• Due to extra stability of half-filled p-orbitals and relatively smaller size of group 15 elements, ionization enthalpy of group 15 elements is much greater than that of the group 14 elements in the corresponding periods.
• On moving down the group, increase in atomic size and screening effect overcome the effective nuclear charge and thus, ionization enthalpy decreases.

iii. Electronegativity:

• Due to smaller size and greater effective nuclear charge of atoms, group 15 elements have higher electronegativity values than group 14 elements.
• On moving down the group, electronegativity values decreases due to increase in the size of the atoms and shielding effect.
• Nitrogen is the most electronegative element among group 15 elements. However, there is not much of a difference between the electronegativity values of other elements of group 15.

Note: Atomic and physical properties of group 15 elements.

Question 17.
Write a note on the oxidation state of p-block elements with respect to groups 13, 14 and 15 elements.
Answer:

• Oxidation state is the primary chemical property of all elements.
• The highest oxidation state exhibited by the p-block elements is equal to the total number of valence electrons i.e., the sum of s- and p-electrons present in the valence shell. This is sometimes called the group oxidation state.
• In boron, carbon and nitrogen families, the group oxidation state is the most stable oxidation state for the lighter elements.
• Besides, the elements of groups 13, 14 and 15 exhibit other oxidation states which are lower than the group oxidation state by two units.
• The lower oxidation states become increasingly stable as we move down to heavier elements in the groups.

Note: Group oxidation states and common oxidation states with examples for groups 13, 14 and 15.

Question 18.
What are general oxidation states of group 13 elements? Explain.
Answer:

• The general oxidation states of group 13 are +1 and +3.
• The group 13 elements have the outermost electronic configuration ns² np¹.
• If only np¹ electron takes part in bonding, the oxidation state is +1 and if all the three electrons i.e., ns² np¹ take part in bonding, the oxidation state is +3. Hence, the expected oxidation states are +1 and +3.

Question 19.
Give reason: The increased stability of the oxidation state is lowered by 2 units than the group oxidation state in heavier p-block elements.
Answer:

• The increased stability of the oxidation state lowered by 2 units than the group oxidation state in heavier p-block elements is due to inert pair effect.
• In these elements, the two s-electrons are involved less readily in chemical reactions.
• This is because, in heavier p-block elements, the s-electrons of valence shell experience poor shielding than valence p-electrons due to ten inner d-electrons.

Hence, the increased stability of the oxidation state is lowered by 2 units than the group oxidation state in heavier p-block elements.

Question 20.
Why Tl¹⁺ ion is more stable than Tl³⁺?
Answer:

• Tl is a heavy element which belongs to group 13 of the p-block.
• The common oxidation state for this group is +3.
• In p-block, the lower oxidation state is more stable for heavier elements due to inert pair effect.

Hence, Tl¹⁺ ion is more stable than Tl³⁺ ion.

Question 21.
How can you explain the higher stability of BCl₃ as compared to TlCl₃?
Answer:

• Boron is a light element in group 13 and has outermost electronic configuration 2s² 2p¹ whereas thallium is a heavy element in group 13 and has outermost electronic configuration 6s² 6p¹.
• Because of its small ionic radius, boron forms stable compounds in +3 oxidation state.
• Thallium has a large atomic size and due to the inert pair effect forms more stable compounds with lower oxidation state +1 than compounds with +3 oxidation state.

Therefore, BCl₃ has higher stability than TlCl₃.

Question 22.
State the oxidation state for the following:
i. The group oxidation state of group 14 elements.
ii. The stable oxidation state for lead.
iii. Oxidation state of carbon in CH₄.
Answer:
i. +4
ii. +2
iii. -4

Question 23.
GeCl4 is more stable than GeCl₂ while PbCl₂ is more stable than PbCl₄. Explain.
Answer:

• Elements Ge and Pb belong to 4th and 6th period in the group 14.
• The group oxidation state of group 14 elements is +4.
• However, the stability of other oxidation state which is lower by 2 units i.e., +2, increases down the group due to inert pair effect.
• Therefore, the stability of the oxidation state +4 is more in Ge than in Pb while the stability of the oxidation state +2 is more in Pb than in Ge.

Hence, GeCl₄ is more stable than GeCl₂ while PbCl₂ is more stable than PbCl₄.

Question 24.
Name the elements of group 14 which are generally occur in +2 oxidation state.
Answer:
The elements of group 14 that are generally occur in +2 oxidation state are tin (Sn) and lead (Pb).

Question 25.
Discuss the nature of bonding in compounds of group 13, 14 and 15 elements.
Answer:

• The lighter elements in groups 13, 14 and 15 have small atomic radii and high ionization enthalpy values. They form covalent bonds with other atoms by overlapping of valence shell orbitals.
• As we move down the group, the value of ionization enthalpy decreases. The atomic radius increases since the valence shell orbitals are more diffused.
• The heavier elements in these groups tend to form ionic bonds. The first member of these groups belongs to second period and do not have d orbitals and hence, B, C and N cannot expand their octet.
• The subsequent elements in the group possess vacant d orbital in their valence shell, which can expand their octet forming a variety of compounds.

Question 26.
Explain the reactivity of groups 13, 14 and 15 elements towards air.
Answer:
i. Group 13 elements:
a. On heating with air or oxygen, group 13 elements form oxide of the type E₂O₃.
\(4 \mathrm{E}_{(\mathrm{s})}+3 \mathrm{O}_{2(\mathrm{~g})} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{E}_{2} \mathrm{O}_{3(\mathrm{~s})}\) (where, E = B, Al, Ga, In, Tl)

b. At high temperature, group 13 elements also react with nitrogen present in the air to form corresponding nitrides.
\(2 \mathrm{E}_{(\mathrm{s})}+\mathrm{N}_{2(\mathrm{~g})} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{EN}_{(\mathrm{s})}\) (where, E = B, Al, Ga, In, Tl)

ii. Group 14 elements: The elements of group 14 on heating in air or oxygen form oxide of the type EO and EO₂ in accordance with the stable oxidation state and availability of oxygen.

iii. Group 15 elements: The elements of group 15 on heating in air or oxygen forms two types of oxide i.e., E₂O₃ and E₂O₅.

Due to increase in metallic character down the groups 13, 14 and 15, the nature of their oxides gradually varies from acidic through amphoteric to basic.
[Note: The temperature required for the reaction of nitrogen with oxygen is very high. This is produced by striking an electric arc.]

Question 27.
Classify the following oxides into acidic, basic or amphoteric.
B₂O₃, Ga₂O₃, Tl₂O₃, In₂O₃, Al₂O₃
Answer:

Acidic oxide	B2O3
Basic oxides	In2O3, Tl2O3
Amphoteric oxides	Al2O3, Ga2O3

Question 28.
Match the following.

	Column A		Column B
i.	N2O5	a.	Amphoteric
ii.	Bi2O3	b.	Acidic
iii.	Sb2O3	c.	Basic

Answer:
i – b,
ii – c,
iii – a

Note: Nature of stable oxides of groups 13, 14 and 15 elements

Question 29.
State TRUE or FALSE. Correct the false statement.
i. Sb is more stable in +3 oxidation state.
ii. Oxides of the type E₂O₅ are formed by group 15 elements.
iii. As₄O₆ is an acidic oxide.
Answer:
i. True
ii. True
iii. False
As₄O₆ is an amphoteric oxide.

Question 30.
What happens when the elements of groups 13, 14 and 15 react with water?
Answer:
i. Most of the elements of groups 13, 14 and 15 are unaffected by water.
ii. Aluminium reacts with water on heating and forms hydroxide while tin reacts with steam to form oxide.

iii. Lead is unaffected by water due to the formation of a protective film of oxide.

Question 31.
Why is phosphorus stored under water?
Answer:
Phosphorus is highly reactive and hence, it is stored under water to prevent its reaction with air as it catches fire on being exposed to air.

Question 32.
Explain the reactivity of group 13 elements towards halogens.
Answer:
i. All the elements of group 13 react directly with halogens to form trihalides (EX3).
2E_(S) + 3X_2(g) → 2EX_3(s) (where, E = B, Al, Ga, In and X = F, Cl, Br, I)
ii. Thallium is an exception as it forms monohalides (TlX).

Question 33.
Describe the reactivity of group 14 elements with halogens.
Answer:

• All the elements of group 14 (except carbon) react directly with halogens to form tetrahalides (EX₄).
• The heavy elements Ge and Pb form dihalides as well.
• Stability of dihalides increases down the group due to inert pair effect.
• The ionic character of halides also increases steadily down the group.

Question 34.
Discuss the reactivity of group 15 elements with halogens.
Answer:

• Elements of the group 15 reacts with halogens to form two series of halides i.e., trihalides (EX₃) and pentahalides (EX₅).
• The pentahalides possess more covalent character due to availability of vacant d orbitals of the valence shell for bonding.
• Nitrogen being second period element, does not have d orbitals in its valence shell, and therefore, does not form pentahalides.
• Trihalides of the group 15 elements are predominantly covalent except BiF₃. The only stable trihalide of nitrogen is NF₃.

Question 35.
Nitrogen does not form pentahalides. Give reason.
Answer:

• The electronic configuration of 7N is 1s² 2s² 2p³. It has 3 unpaired electrons which can form 3 covalent bonds, thus forming NX₃ molecule.
• Valence shell of nitrogen (n = 2) contains only s and p orbitals.
• Thus, due to the absence of d orbitals in the valence shell, nitrogen cannot expand its octet, therefore, it cannot form compounds like NCl₅ and NF₅.

Hence, nitrogen does not form pentahalides.

Question 36.
Nitrogen does not form NCl₅ or NF₅ but phosphorus can. Explain.
Answer:

• The electronic configuration of 7N is 1s2 2s2 2p3 while that of 15P is 1s² 2s² 2p⁶ 3s² 3p³.
• As phosphorus contains d orbitals, it can expand its octet to form MX₃ as well MX₅ compounds.
• However, due to absence of d orbitals, nitrogen cannot form MX₃ or MX₅.

Hence, Nitrogen does not form NCl₅ or NF₅ but phosphoms can form compounds like PCl₅ or PF₅.

Question 37.
Define catenation.
Answer:
The property of self-linking of atoms of an element by covalent bonds to form chains and rings is called catenation.

Question 38.
Explain catenation of group 14 elements.
Answer:
i. The property of self-linking of atoms of an element by covalent bonds to form chains and rings is called catenation.
ii. The strength of the element-element bond determines the tendency of an element to form a chain.
iii. Among the elements of group 14, the bond strength is maximum for C-C bond (348 kJ mol^-1). Hence, carbon has maximum tendency for catenation.

Bond	Bond strength (Bond enthalpy kJ mol-1)
C-C	348
Si-Si	297
Ge-Ge	260
Sn-Sn	240

iv. From the values of bond enthalpy, it can be concluded that the tendency to form chains is maximum for carbon and much lesser for silicon. Germanium has still lesser tendency and tin has hardly any tendency for catenation. Lead does not show catenation.
Therefore, the order of catenation of group 14 elements is C >> Si > Ge = Sn.

Question 39.
State TRUE or FALSE. Correct the false statement.
i. Among the group 14 elements, Ge does not show the property of catenation.
Answer:
i. False
Among the group 14 elements, Pb (lead) does not shows the property of catenation.

Question 40.
Define allotropy.
Answer:
When a solid element exists in different crystalline forms with different physical properties such as colour, density, melting point, etc. the phenomenon is called allotropy.

Question 41.
i. What are allotropes?
ii. Name various allotropes of carbon.
Answer:
i. When a solid element exists in different crystalline forms with different physical properties such as colour, density, melting point, etc. the phenomenon is called allotropy and the individual crystalline forms are called allotropes.
ii. Diamond, graphite, fiillerenes, graphene and carbon nanotubes are various allotropes of carbon.

Question 42.
Explain the structure of diamond.
Answer:
Structure of diamond:

• In diamond, each carbon atom undergoes sp3 hybridization and is linked to four other carbon atoms in tetrahedral manner.
• The C – C bond length is 154 pm.
• The tetrahedra are linked together forming a three-dimensional network structure involving strong C-C single bonds which makes diamond the hardest natural substance.

Question 43.
Write physical properties of diamond. Also, state its uses.
Answer:
i. Physical properties

• Diamond is the hardest natural substance.
• It has abnormally high melting point (3930 °C).
• It is a bad conductor of electricity.

ii. Uses: Diamond is used

• for cutting glass and in drilling tools.
• for making dies for drawing thin wire from metal.
• for making jewellery.

Question 44.
Describe the structure of graphite.
Answer:

• Graphite is composed of layers of two-dimensional sheets of carbon atoms.
• Each sheet is made up of hexagonal net of sp² carbons bonded to three neighbours forming three bonds.
• The fourth electron in the unhybrid p-orbital of each carbon is shared by all carbon atoms resulting in a π bond. These it electrons are delocalized over the whole layer.
• The C – C bond length in graphite is 141.5 pm.
• The individual layers are held by weak van der Waals forces and separated by 335 pm.
• Graphite is soft and slippery and is thermodynamically most stable allotrope of carbon.

Question 45.
Diamond is very hard whereas graphite is soft. Explain.
Answer:

• Diamond has three-dimensional network of sp³ hybridized carbon atoms joined by extended covalent bonds which are difficult to break. Therefore, diamond is hard.
• Graphite has two-dimensional sheet like structure, like layers of hexagonal rings formed from sp² hybridized carbon atoms. These layers are held by weak van der Waals forces, which can be broken easily. Therefore, graphite is soft and slippery.

Hence, diamond is very hard whereas graphite is soft.

Question 46.
i. What are fullerenes?
ii. How are they prepared?
Answer:
i. Fullerenes are allotropes of carbon in which carbon molecules are linked by a definite numbers of carbon atoms, for example as in C₆₀.
ii. Fullerenes are produced when an electric arc is struck between the graphite electrodes in an inert atmosphere of argon or helium. The soot formed contains significant amount of C₆₀ fullerene and smaller amounts of other fullerenes C₃₂, C₅₀, C₇₀ and C₈₄.

Question 47.
Discuss the structure and properties of fullerene (C6o).
Answer:

• C₆₀ has a shape like soccer ball and called Buckminsterfullerene or bucky ball.
• It contains 20 hexagonal and 12 pentagonal fused rings of carbon.
• The C₆₀ fullerene structure exhibit separations between the neighbouring carbons as 143.5 pm and 138.3 pm.
• Fullerenes are covalent and soluble in organic solvents.
• Fullerene C₆₀ reacts with group 1 metals forming solids such as K₃C₆₀.
• The compound K₃C₆₀ behaves as a superconductor below 18 K, which means that its carries electric current with zero resistance.

Question 48.
Explain the structure of carbon nanotubes.
Answer:

• Carbon nanotubes are cylindrical in shape consisting of rolled-up graphite sheet.
• Nanotubes can be single-walled (SWNTs) with a diameter of less than 1 nm or multi-walled (MWNTs) with diameter reaching more than 100 nm.
• Their lengths range from several micrometres to millimetres.

Question 49.
Describe the properties of carbon nanotubes.
Answer:

• Carbon nanotubes are robust. They can be bent, and when released, they will spring back to the original shape.
• Carbon nanotubes have high electrical or heat conductivities and highest strength-to-weight ratio for any known material to date.
• The researchers of NASA are combining carbon nanotubes with other materials into composites that can be used to build lightweight spacecraft.

Question 50.
What is graphene?
Answer:

• Isolated layer of graphite is called graphene.
• Graphene sheet is a two dimensional solid.
• It has unique electronic properties.

Question 51.
Explain the structure of various allotropes of phosphorus.
Answer:
Phosphorus is found in different allotropic forms. White and red phosphorus are important allotropes of phosphorus.
i. White (yellow) phosphorus:

1. White (yellow) phosphorus consists of discrete tetrahedral P₄ molecules.
2. The P – P – P bond angle is 60°.
3. White phosphorus is less stable and hence more reactive, because of angular strain in the P₄ molecule.

ii. Red phosphorus:

• Red phosphorus consists of chains of P₄ linked together by covalent bonds.
• Thus, it is polymeric in nature.

Question 52.
Enlist properties of
i. white phosphorus.
ii. red phosphorus.
Answer:
i. Properties of white phosphorus:

• It is translucent white waxy solid.
• It glows in the dark (chemiluminescence).
• It is insoluble in water but dissolves in boiling NaOH solution.
• It is poisonous.

ii. Properties of red phosphorus:

• It is stable and less reactive.
• It is odourless and possess iron grey lustre.
• It does not glow in the dark.
• It is insoluble in water.
• It is nonpoisonous.

Question 53.
How is red phosphorus prepared?
Answer:
Red phosphorus is prepared by heating white phosphorus at 573 K in an inert atmosphere.

Question 54.
State whether the following statement is TRUE or FALSE. Correct if false.
i. Covalent molecules have irregular shape described with the help of bond lengths and bond angles.
ii. It is difficult to understand the reactivity of covalent inorganic compounds from their structures.
iii. Inorganic molecules are often represented by molecular formulae indicating their elemental composition.
Answer:
i. False
Covalent molecules have definite shape described with the help of bond lengths and bond angles.
ii. False
The reactivity of covalent inorganic compounds is better understood from their structures.
iii. True

Question 55.
Describe structure of the following molecules.
i. Boron trichloride
ii. Aluminium chloride
iii. Orthoboric acid
Answer:
i. Structure of boron trichloride (BCl₃) molecule:

• Boron trichloride (BCl₃) is a covalent compound.
• In BCl₃ molecule, boron atom is sp² hybridized having one vacant unhybridized p orbital.
• B in BCl₃ has incomplete octet.
• BCl₃ is a nonpolar trigonal planar molecule.
• Each Cl – B – Cl bond angle is 120°.

ii. Structure of aluminium chloride (AlCl₃) molecule:

• Aluminium atom in aluminium chloride is sp² hybridized, with one vacant unhybrid p-orbital.
• Aluminium chloride exists as the dimer (Al₂Cl₆) formed by overlap of vacant 3d orbital of Al with a lone pair of electrons of Cl.

iii. Structure of orthoboric or boric acid (H₃BO₃) molecule:

• Orthoboric acid has central boron atom bound to three -OH groups.
• The solid orthoboric acid has layered crystal structure in which trigonal planar B(OH)₃ units are joined together by hydrogen bonds.

Question 56.
Which are the different crystalline forms of silica?
Answer:
Quartz, cristobalite and tridymite are the different crystalline forms of silica.
[Note: These crystalline forms are inter-convertible at a suitable temperature.]

Question 57.
Explain the structure of silicon dioxide.
Answer:

• Silicon dioxide (SiO₂), is also known as silica.
• It is a covalent three-dimensional network solid.
• In SiO₂, each silicon atom is covalently bound in tetrahedral manner to four oxygen atoms.
• The crystal contains eight membered rings having alternate silicon and oxygen atoms.

Question 58.
Discuss the nature and structure of the following compounds.
i. Nitric acid
ii. Phosphoric acid
Answer:
i. Nitric acid:

• Nitric acid (HNO₃) is a strong, oxidizing mineral acid.
• The central nitrogen atom is sp² hybridized.
• HNO₃ exhibits resonance phenomenon.
• Figure (a) represents resonating structures of HNO₃ while figure (b) represents resonance hybrid of HNO₃.

ii. Phosphoric acid (Orthophosphoric acid):

• Phosphorus forms number of oxyacids. Orthophosphoric acid (H₃PO₄) is a strong nontoxic mineral acid.
• It contains three ionizable acidic hydrogens.
• The central phosphorus atom is tetrahedral.

Question 59.
Give the molecular formula of crystalline borax.
Answer:
The crystalline borax has formula Na₂B₄O₇.10H₂O or Na₂[B₄O₅(OH)₄].8H₂O.

Question 60.
How is borax obtained from its mineral colemanite?
Answer:
Borax is obtained from its mineral colemanite by boiling it with a solution of sodium carbonate.

Question 61.
Why is the aqueous solution of borax alkaline?
Answer:
On hydrolysis, borax forms a strong base (NaOH) and a weak acid (H₃BO₃). The presence of the strong base makes borax solution alkaline.

Question 62.
What happens when borax is heated strongly?
Answer:
Borax is a white crystalline solid. On heating, borax loses water molecules and swells up. On further heating, it melts and forms a transparent liquid, which solidifies into a glass like material known as borax bead.

Question 63.
Explain borax bead test.
Answer:
i. Borax bead test is used to detect coloured transition metal ions.
ii. On heating, borax loses water molecules and swells up. On further heating, it melts and forms a transparent liquid, which solidifies into a glass like material known as borax bead.

iii. The borax bead consists of sodium metaborate and boric anhydride, which reacts with metals salts to form coloured bead.
e.g. When borax is heated in a Bunsen burner flame with CoO on a loop of platinum wire, a blue coloured Co(BO₂)₂ bead is formed.

Question 64.
Write the uses of borax.
Answer:
Borax is used

• to manufacture optical and hard borosilicate glasses.
• as a flux for soldering and welding.
• as a mild antiseptic in the preparation of medical soaps.
• in qualitative analysis for borax bead test.
• as a brightener in washing powder.

Question 65.
How are silicones prepared? Write their properties.
Answer:
i. Preparation of silicones:
a. Alkyl or aryl substituted silicon chlorides having general formula R_nSiCl_(4-n) (R = alkyl or aryl group) are used as the starting materials for manufacture of silicones.
b. When methyl chloride reacts with silicon in the presence of copper catalyst at a temperature 573 K, various types of methyl substituted chlorosilane of formula MeSiCl₃, Me₂SiCl₂, Me₃SiCl with small amounts of Me₄Si are formed.

c. Hydrolysis of dimethyldichlorosilane, (CH₃)₂SiCl₂ followed by condensation polymerisation yields straight chain silicone polymers.

d. The chain length of polymer can be controlled by adding (CH₃)₃SiCl at the end.

ii. Properties:

• Silicones are water repellent.
• They have high thermal stability.
• They are good electrical insulators.
• They are resistant to oxidation and chemicals.

Question 66.
Explain the preparation of ammonia from nitrogeneous organic matter.
Answer:
Ammonia is formed by the decomposition of nitrogeneous organic matter such as urea. It is therefore, present naturally in small quantities in air and soil.

Question 67.
Describe laboratory method for preparation of ammonia.
Answer:
Ammonia is prepared on laboratory scale by decomposition of the ammonium salts with calcium hydroxide or caustic soda.

Question 68.
How is ammonia manufactured by Haber process?
Answer:

• On the large scale, ammonia is prepared by direct combination of dinitrogen and dihydrogen by Haber process.
• In this process, dinitrogen reacts with dihydrogen under high pressure of 200 × 10⁵ Pa (200 atm) and temperature around 700 K to produce ammonia.
  N_2(g) + 2H_2(g) ⇌ 2NH_3(g); Δ_fH° = -46.1 kJ mol^-1
• Iron oxide with trace amounts of K₂O and Al₂O₃ is used as catalyst in Haber process.
• High pressure favours the formation of ammonia as equilibrium is attained rapidly under these conditions.

Question 69.
State the physical properties of ammonia.
Answer:

• Ammonia is a colourless gas with pungent odour.
• It has freezing point of 198.4 K and boiling point of 239.7 K.
• It is highly soluble in water.

Question 70.
What is liquor ammonia?
Answer:
The concentrated aqueous solution of ammonia (NH₃) is called liquor ammonia.

Question 71.
Give reason: Ammonia has higher melting and boiling points.
Answer:

• In solid and liquid state, NH₃ molecules get associated together through hydrogen bonding.
• As a result, extra amount of energy is required to break such intermolecular hydrogen bonds. Hence, ammonia has higher melting and boiling points.

Question 72.
Why is ammonia basic in aqueous solution?
Answer:
i. As ammonia is highly soluble in water, it readily forms OH^– ions in its aqueous solution.
\(\mathrm{NH}_{3(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(l)} \rightleftharpoons \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{OH}_{(a q)}^{-}\)
ii. Thus, due to the formation of OH^– ions, aqueous solution of ammonia is basic in nature.

Question 73.
How does the aqueous solution of ammonia react with the following salt solutions?
i. ZnSO₄
ii. FeCl₃
Answer:
Aqueous solution of ammonia precipitates out as hydroxides (or hydrated oxides) of metals solutions.

Question 74.
Write applications of ammonia.
Answer:
Ammonia is used in

• manufacture of fertilizers such as urea, diammonium phosphate, ammonium nitrate, ammonium sulphate etc.
• manufacture of some inorganic compounds like nitric acid.
• refrigerant (liq. ammonia).
• laboratory reagent in qualitative and quantitative analysis (aq. solution of ammonia).

Question 75.
Give reactions involved in the formation of Nessler’s reagent.
Answer:

Question 76.
How does ammonia react with Nessler’s reagent?
Answer:
Ammonia react with Nessler’s reagent (an alkaline solution of K₂HgI₄) to form a brown precipitate (Millon’s base).

Question 77.
Complete and write the balanced chemical equations for:
i. Ca₂B₆O₁₁ + Na₂CO₃ →
ii. CoO + B₂O₃ →
iii. AgCl + NH₃ →
iv. ZnSO₄ + 2NH₄OH →
v. a. 2KI + HgCl₂ →
b. 2KI + HgI₂ →
Answer:

Question 78.
Naina was preparing a compound in the laboratory. She added compound ‘A’ to (CaOH)₂ solution. As a result of this, a compound ‘B’ was obtained which had a pungent smell. On adding Nessler’s reagent to the compound ‘B’, a brown precipitate of compound ‘C’ was obtained.
Write the chemical reactions involved and identify ‘A’, ‘B’ and ‘C’.
Answer:
i. When ammonium chloride is mixed with (CaOH)₂ solution, ammonia is formed which has a pungent odour.

ii. Ammonia react with Nessler’ s reagent (an alkaline solution of K₂Hgl₄) to form a brown precipitate (Millon’ s base).

Multiple Choice Questions

1. The electronic configuration of boron family is ……………
(A) ns² np²
(B) ns² np⁵
(C) ns² np⁶
(D) ns² np¹
Answer:
(D) ns² np¹

2. ………… has noble gas core plus 14 f-electrons and 10 d-electrons.
(A) Gallium
(B) Indium
(C) Thallium
(D) Boron
Answer:
(C) Thallium

3. The group 15 element having inner electronic configuration as of argon is …………..
(A) Phosphorus (Z = 15)
(B) Antimony (Z = 51)
(C) Arsenic (Z = 33)
(D) Nitrogen (Z = 7)
Answer:
(C) Arsenic (Z = 33)

4. Which of the following is NOT a metalloid?
(A) B
(B) Sn
(C) Ge
(D) Sb
Answer:
(B) Sn

5. Among the group 13 elements, melting point is highest for …………..
(A) B
(B) Al
(C) Ga
(D) In
Answer:
(A) B

6. On moving down the group 14, the ionization enthalpy
(A) increases slightly from Si to Sn and decreases slightly from Sn to Pb
(B) increases throughout uniformly from Si to Pb
(C) decreases throughout uniformly from Si to Pb
(D) decreases slightly from Si to Sn and increases slightly from Sn to Pb
Answer:
(D) decreases slightly from Si to Sn and increases slightly from Sn to Pb

7. ………… is the most electronegative element of group 14.
(A) Carbon
(B) Silicon
(C) Germanium
(D) Tin
Answer:
(A) Carbon

8. The stability of +3 oxidation state in aqueous solution is in order ……………
(A) Al > Ga > In > Tl
(B) Tl > In > Ga > Al
(C) Al > Tl > Ga > In
(D) Tl > Al > Ga > In
Answer:
(A) Al > Ga > In > Tl

9. Group oxidation state of group 15 elements is ……………
(A) +4
(B) +1
(C) +3
(D) +5
Answer:
(D) +5

10. …………. cannot expand its octet due to absence of d orbital in its valence shell.
(A) Ga
(B) C
(C) As
(D) Ge
Answer:
(B) C

11. Which one of the following statements about boron and aluminium is INCORRECT?
(A) Both exhibit oxidation state of +3.
(B) Both form oxides of the formula M₂O₃.
(C) Both form trihalides, MX₃.
(D) Both form amphoteric oxides.
Answer:
(D) Both form amphoteric oxides.

12. Which of the following is basic oxide?
(A) Bi₂O₃
(B) CO₂
(C) B₂O₃
(D) SiO₂
Answer:
(A) Bi₂O₃

13. The reaction of Al with H₂O produces ……………
(A) Al₂O₃
(B) AlH₃
(C) Al(OH)₃
(D) Al₂H₆
Answer:
(C) Al(OH)₃

14. Which of the following is a stable halide of nitrogen?
(A) NF₃
(B) NCl₅
(C) NF₅
(D) NBr₅
Answer:
(A) NF₃

15. Catenation is the ability of …………..
(A) atoms to form strong bonds with similar atoms
(B) elements to form giant molecules
(C) an element to form multiple bonds
(D) an element to form long chains of identical atoms
Answer:
(D) an element to form long chains of identical atoms

16. Among the group 13 elements, the property of allotropy is shown by ………………
(A) indium
(B) aluminium
(C) thallium
(D) boron
Answer:
(D) boron

17. Thermodynamically stable allotrope of carbon is …………..
(A) diamond
(B) graphite
(C) buckyball
(D) all of these
Answer:
(B) graphite

18. White phosphorus contains discrete …………… molecules.
(A) P₅
(B) P₄
(C) P₆
(D) P₅₂
Answer:
(B) P₄

19. In white phosphorus, the P-P-P bond angle is ……………
(A) 60°
(B) 90°
(C) 109.5
(D) 120°
Answer:
(A) 60°

20. 3c-2e bonds are present in ………………
(A) NH₃
(B) B₂H₆
(C) H₃BO₃
(D) SiCl₄
Answer:
(B) B₂H₆

21. Which of the following is borax?
(A) Na₂B₄O₇.4H₂O
(B) Na₂B₄O₇.10H₂O
(C) H₃BO₃
(D) NaBO₂
Answer:
(B) Na₂B₄O₇.10H₂O

22. In Borax bead test, the coloured ions give characteristic coloured beads due to formation of …………….
(A) metal borates
(B) metal metaborates
(C) metal phosphates
(D) metal tetraborates
Answer:
(B) metal metaborates

23. The catalyst used in Haber process contains …………..
(A) nickel
(B) palladium
(C) iron
(D) platinum
Answer:
(C) iron

24. Which of the following is used as refrigerant?
(A) Nessler’s reagent
(B) Liq. ammonia
(C) Borax
(D) Diborane
Answer:
(B) Liq. ammonia