JEE Chemistry Classification of Elements and Periodicity in Properties MCQs Set 13

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Chapter-wise Objective Questions: Classification Of Elements and Periodicity In Properties

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Multiple Choice Questions

Question 1: Incorrect order of ionic radius is:
(a) La³⁺ > Gd³⁺ > Eu³⁺ > Lu³⁺
(b) V²⁺ > V³⁺ > V⁴⁺ > V⁵⁺
(c) In⁺ > Sn²⁺ > Sb³⁺
(d) K⁺ > Sc³⁺ > V⁵⁺ > Mn⁷⁺
Show Answer & Explanation

Answer: (a) La³⁺ > Gd³⁺ > Eu³⁺ > Lu³⁺

Explanation:
1. Lanthanoid contraction makes Gd³⁺ (94 pm) smaller than Eu³⁺ (95 pm), so the order La > Gd > Eu > Lu is wrong.
2. The others are correct.

Teacher's Note:
1. Compare shells first, then nuclear charge. For isoelectronic ions, the larger nuclear charge gives the smaller radius.
2. Revise the trend with two or three examples.

Question 2: According to Slater's rule, order of effective nuclear charge for last electron in case of Li, Na and K:
(a) Li > Na > K
(b) K > Na > Li
(c) Na > Li > K
(d) Li < Na = K
Show Answer & Explanation

Answer: (d) Li < Na = K

Explanation:
1. Slater's rule gives Zeff of the outer electron as about 1.3 for Li and about 2.2 for both Na and K.
2. So Li < Na = K.

Teacher's Note:
1. Compare shells first, then nuclear charge. For isoelectronic ions, the larger nuclear charge gives the smaller radius.
2. Revise the trend with two or three examples.

Question 3: The ionisation potential of isotopes of an element will be:
(a) Same
(b) Different
(c) Depends on atomic masses
(d) Depends on number of neutrons
Show Answer & Explanation

Answer: (a) Same

Explanation:
1. Isotopes have the same electronic structure and nuclear charge, so their ionisation potentials are the same.

Teacher's Note:
1. Check half filled and filled subshells before applying the general trend. A big jump between successive ionisation energies shows the number of valence electrons.
2. Revise the trend with two or three examples.

Question 4: In which of the following pairs, the ionisation energy of the first species is less than that of the second:
(a) O⁻, O²⁻
(b) S, P
(c) N, P
(d) Be⁺, Be
Show Answer & Explanation

Answer: (b) S, P

Explanation:
1. S (1000) is less than P (1012) in first IE.
2. In the other pairs the first species has the higher value: N > P, Be⁺ > Be, O⁻ > O²⁻.

Teacher's Note:
1. Check half filled and filled subshells before applying the general trend. A big jump between successive ionisation energies shows the number of valence electrons.
2. Revise the trend with two or three examples.

Question 5: Ionisation energy increases in the order:
(a) Be, B, C, N
(b) B, Be, C, N
(c) C, N, Be, B
(d) N, C, Be, B
Show Answer & Explanation

Answer: (b) B, Be, C, N

Explanation:
1. First IE: B 801 < Be 899 < C 1086 < N 1402.
2. Be is higher than B because of its filled 2s subshell.

Teacher's Note:
1. Check half filled and filled subshells before applying the general trend. A big jump between successive ionisation energies shows the number of valence electrons.
2. Revise the trend with two or three examples.

Question 6: Mg forms Mg(II) because of:
(a) The oxidation state of Mg is + 2
(b) Difference between I.P₁ and I.P₂ is greater than 16.0 eV
(c) There is only one electron in the outermost energy level of Mg
(d) Difference between I.P₁ and I.P₂ is less than 11 eV
Show Answer & Explanation

Answer: (d) Difference between I.P₁ and I.P₂ is less than 11 eV

Explanation:
1. Mg(II) forms because the energy gap between IP1 and IP2 (7.6 and 15.0 eV, a difference of about 7.4 eV) is small, so it is recovered through lattice energy, while IP3 is very high.

Teacher's Note:
1. Check half filled and filled subshells before applying the general trend. A big jump between successive ionisation energies shows the number of valence electrons.
2. Revise the trend with two or three examples.

Question 7: Minimum first ionisation energy is shown by which electronic configuration:
(a) 1s² 2s² 2p⁵
(b) 1s² 2s² 2p⁶ 3s² 3p²
(c) 1s² 2s² 2p⁶ 3s¹
(d) 1s² 2s² 2p⁶
Show Answer & Explanation

Answer: (c) 1s² 2s² 2p⁶ 3s¹

Explanation:
1. The 3s¹ configuration (Na) has the largest atom with the outer electron least held, so its IE is the lowest.

Teacher's Note:
1. Check half filled and filled subshells before applying the general trend. A big jump between successive ionisation energies shows the number of valence electrons.
2. Revise the trend with two or three examples.

Question 8: With reference to ionisation potential which one of the following set is correct:
(a) Li > K > B
(b) B > Li > K
(c) Cs > Li > K
(d) Cs < Li < K
Show Answer & Explanation

Answer: (b) B > Li > K

Explanation:
1. First IE: B 801 > Li 520 > K 419.
2. So B > Li > K.

Teacher's Note:
1. Check half filled and filled subshells before applying the general trend. A big jump between successive ionisation energies shows the number of valence electrons.
2. Revise the trend with two or three examples.

Question 9: The successive ionisation energies of an element X (in kcal) are IP₁ = 165, IP₂ = 195, IP₃ = 556 and IP₄ = 595. The electronic configuration of X is:
(a) 1s² , 2s² 2p⁶ , 3s² 3p²
(b) 1s² , 2s¹
(c) 1s² , 2s² 2p²
(d) 1s² , 2s² 2p⁶ , 3s²
Show Answer & Explanation

Answer: (d) 1s² , 2s² 2p⁶ , 3s²

Explanation:
1. The jump comes after IP2 (195 to 556), so the element has two valence electrons.
2. So the configuration is 1s² 2s² 2p⁶ 3s².

Teacher's Note:
1. Check half filled and filled subshells before applying the general trend. A big jump between successive ionisation energies shows the number of valence electrons.
2. Revise the trend with two or three examples.

Question 10: Second IP of which of the element is maximum–:
(a) Lithium
(b) Oxygen
(c) Nitrogen
(d) Fluorine
Show Answer & Explanation

Answer: (a) Lithium

Explanation:
1. Li⁺ has the stable helium configuration, so its second IP (7298 kJ per mole) is the highest.
2. O, N and F give 3388, 2856 and 3374.

Teacher's Note:
1. Check half filled and filled subshells before applying the general trend. A big jump between successive ionisation energies shows the number of valence electrons.
2. Revise the trend with two or three examples.

Question 11: Out of Na⁺, Mg²⁺, O²⁻ and N³⁻, the pair of species showing minimum and maximum IP would be.
(a) Na⁺, Mg²⁺
(b) Mg²⁺, N³⁻
(c) N³⁻, Mg²⁺
(d) O²⁻, N³⁻
Show Answer & Explanation

Answer: (c) N³⁻, Mg²⁺

Explanation:
1. Anions have the lowest IP, and N³⁻ is the most unstable.
2. Mg²⁺ has a very high third IE (7733), so it has the maximum.

Teacher's Note:
1. Check half filled and filled subshells before applying the general trend. A big jump between successive ionisation energies shows the number of valence electrons.
2. Revise the trend with two or three examples.

Question 12: The element having highest I.P. is the from the two series C, N, O and Si, P, S:
(a) P
(b) N
(c) S
(d) O
Show Answer & Explanation

Answer: (b) N

Explanation:
1. N (1402) is the highest.
2. O 1314 and P 1012 are lower.

Teacher's Note:
1. Check half filled and filled subshells before applying the general trend. A big jump between successive ionisation energies shows the number of valence electrons.
2. Revise the trend with two or three examples.

Question 13: The strongest reducing agent among the following is:
(a) Na
(b) Mg
(c) Al
(d) K
Show Answer & Explanation

Answer: (d) K

Explanation:
1. Reducing power rises with ease of losing electrons.
2. K has the lowest ionisation energy among Na, Mg, Al and K, so it is the strongest reducing agent.

Teacher's Note:
1. Check half filled and filled subshells before applying the general trend. A big jump between successive ionisation energies shows the number of valence electrons.
2. Revise the trend with two or three examples.

Question 14: Consider (a) M⁻(g) → M(g), (b) M(g) → M⁺(g), (c) M⁺(g) → M²⁺(g), (d) M²⁺(g) → M³⁺(g), each with loss of one electron. The minimum and maximum ionisation potentials respectively are for:
(a) a, d
(b) b, c
(c) c, d
(d) d, a
Show Answer & Explanation

Answer: (a) a, d

Explanation:
1. Minimum IP is of anions (a) and maximum is of the highest charged cation (d).
2. So a and d.

Teacher's Note:
1. Check half filled and filled subshells before applying the general trend. A big jump between successive ionisation energies shows the number of valence electrons.
2. Revise the trend with two or three examples.

Question 15: Which of the following electronic configurations belong to the least and most metallic character respectively? (a) 1s² 2s¹ (b) 5s² 5p⁵ (c) 3s² 3p⁶ 4s¹ (d) 1s² 2s² 2p⁵?
(a) a, b
(b) d, c
(c) b, a
(d) c, d
Show Answer & Explanation

Answer: (b) d, c

Explanation:
1. Least metallic is F (d) and most metallic is K (c).
2. So d and c.

Teacher's Note:
1. Check half filled and filled subshells before applying the general trend. A big jump between successive ionisation energies shows the number of valence electrons.
2. Revise the trend with two or three examples.

Question 16: Triad I [N³⁻, O²⁻, Na⁺] and Triad II [N⁺, C⁺, O⁺]. Choose the species of lowest IP from triad I and highest IP from triad II respectively:
(a) N³⁻, O⁺
(b) Na⁺, C⁺
(c) N³⁻, N⁺
(d) O²⁻, C⁺
Show Answer & Explanation

Answer: (a) N³⁻, O⁺

Explanation:
1. N³⁻ is the most unstable of the ions in triad I, so its IP is lowest.
2. For triad II the second ionisation energies are O 3388, N 2856 and C 2353 kJ per mole, so O⁺ has the highest IP.

Teacher's Note:
1. Check half filled and filled subshells before applying the general trend. A big jump between successive ionisation energies shows the number of valence electrons.
2. Revise the trend with two or three examples.

Question 17: Across the second period the first ionisation energies follow Li < B < Be < C < O < N < F < Ne. The first ionisation energy of sodium (Z = 11) will be:
(a) Above Ne
(b) Below Ne but above O
(c) Below Li
(d) Between N and O
Show Answer & Explanation

Answer: (c) Below Li

Explanation:
1. After removing 1s² 2s² 2p⁶ 3s² electrons the Na atom is easier to ionise (496) than Li (520), so Na lies below Li.

Teacher's Note:
1. Check half filled and filled subshells before applying the general trend. A big jump between successive ionisation energies shows the number of valence electrons.
2. Revise the trend with two or three examples.

Question 18: In which of the following pairs, the ionisation energy of the first species is less than that of the second:
(a) N, P
(b) Be⁺, Be
(c) N, N⁻
(d) Ne, Ne⁺
Show Answer & Explanation

Answer: (d) Ne, Ne⁺

Explanation:
1. Ne⁺ has a much higher IE than Ne, since a positive charge holds the electrons tightly.
2. Ne (2081) is less than Ne⁺ (3952). Other pairs do not fit.

Teacher's Note:
1. Check half filled and filled subshells before applying the general trend. A big jump between successive ionisation energies shows the number of valence electrons.
2. Revise the trend with two or three examples.

Question 19: Select the correct order of I.E.:
(a) Cl⁻ > Cl > Cl⁺
(b) Cl⁺ > Cl > Cl⁻
(c) Cl > Cl⁺ > Cl⁻
(d) Cl⁻ > Cl⁺ > Cl
Show Answer & Explanation

Answer: (b) Cl⁺ > Cl > Cl⁻

Explanation:
1. More positive charge means the electron is held tighter.
2. So IE: Cl⁺ > Cl > Cl⁻.

Teacher's Note:
1. Check half filled and filled subshells before applying the general trend. A big jump between successive ionisation energies shows the number of valence electrons.
2. Revise the trend with two or three examples.

Question 20: Process in which maximum energy is released:
(a) O → O²⁻
(b) Mg⁺ → Mg²⁺
(c) Cl → Cl⁻
(d) F → F⁻
Show Answer & Explanation

Answer: (c) Cl → Cl⁻

Explanation:
1. Cl takes up an electron with the release of about 349 kJ per mole, the most of these.
2. O to O²⁻ and Mg⁺ to Mg²⁺ need energy and F gives only 328 kJ per mole.

Teacher's Note:
1. Electron affinity is highest for chlorine, not fluorine. Atoms with filled or half filled subshells have very low values.
2. Revise the trend with two or three examples.

Question 21: Which of the following is energy releasing process?
(a) X⁻ → X(g) + e⁻
(b) O⁻(g) + e⁻ → O²⁻(g)
(c) O(g) → O⁺(g) + e⁻
(d) O(g) + e⁻ → O⁻(g)
Show Answer & Explanation

Answer: (d) O(g) + e⁻ → O⁻(g)

Explanation:
1. O⁺ e gives O⁻ releasing 141 kJ per mole.
2. The other steps need energy.

Teacher's Note:
1. Electron affinity is highest for chlorine, not fluorine. Atoms with filled or half filled subshells have very low values.
2. Revise the trend with two or three examples.

Question 22: In which of the following processes is energy liberated?
(a) Cl → Cl⁺ e
(b) HCl → H⁺ + Cl
(c) Cl⁺ e → Cl⁻
(d) O⁻ + e → O²⁻
Show Answer & Explanation

Answer: (c) Cl⁺ e → Cl⁻

Explanation:
1. Cl + e⁻ gives Cl⁻ and releases energy, 349 kJ per mole.
2. The other steps absorb energy.

Teacher's Note:
1. Electron affinity is highest for chlorine, not fluorine. Atoms with filled or half filled subshells have very low values.
2. Revise the trend with two or three examples.

Question 23: Element of which atomic number has highest electron affinity:
(a) 35
(b) 17
(c) 9
(d) 53
Show Answer & Explanation

Answer: (b) 17

Explanation:
1. Chlorine has the highest electron affinity (349 kJ per mole), more than F (328) because of the small size of F and electron repulsion.

Teacher's Note:
1. Electron affinity is highest for chlorine, not fluorine. Atoms with filled or half filled subshells have very low values.
2. Revise the trend with two or three examples.

Question 24: The electron affinity:
(a) Of carbon is greater than oxygen
(b) Of fluorine is less than iodine
(c) Of fluorine is less than chlorine
(d) Of sulphur is less than oxygen
Show Answer & Explanation

Answer: (c) Of fluorine is less than chlorine

Explanation:
1. F has a lower electron affinity than Cl.
2. C is lower than O is false, C < O, and S > O. So only F < Cl is correct.

Teacher's Note:
1. Electron affinity is highest for chlorine, not fluorine. Atoms with filled or half filled subshells have very low values.
2. Revise the trend with two or three examples.

Question 25: The energy absorbed when a second electron is added to an atom is called:
(a) 1st IP
(b) 2nd EA
(c) 1st EA
(d) 2nd IP
Show Answer & Explanation

Answer: (b) 2nd EA

Explanation:
1. The energy absorbed when a second electron is added is the second electron affinity.

Teacher's Note:
1. Electron affinity is highest for chlorine, not fluorine. Atoms with filled or half filled subshells have very low values.
2. Revise the trend with two or three examples.

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