JEE Chemistry Classification of Elements and Periodicity in Properties MCQs Set 15

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Multiple Choice Questions

Question 1: Be²⁺ is isoelectronic with which of the following ions?
(a) H⁺
(b) Li⁺
(c) Na⁺
(d) Mg²⁺
Show Answer & Explanation

Answer: (b) Li⁺

Explanation:
1. Be²⁺ has 2 electrons, as does Li⁺.
2. H⁺ has none, Na⁺ has 10 and Mg²⁺ has 10.

Teacher's Note:
1. Compare size, ionisation energy and electron gain enthalpy trends, then check the exceptions such as Be/B, N/O and F/Cl.
2. Revise the trend with two or three examples.

Question 2: Acidity of diprotic acids in aqueous solutions increases in the order:
(a) H₂S < H₂Se < H₂Te
(b) H₂Se < H₂S < H₂Te
(c) H₂Te < H₂S < H₂Se
(d) H₂Se < H₂Te < H₂S
Show Answer & Explanation

Answer: (a) H₂S < H₂Se < H₂Te

Explanation:
1. Acidity of H2S, H2Se and H2Te rises as the H-E bond weakens down the group.
2. So H2S < H2Se < H2Te.

Teacher's Note:
1. Compare size, ionisation energy and electron gain enthalpy trends, then check the exceptions such as Be/B, N/O and F/Cl.
2. Revise the trend with two or three examples.

Question 3: Correct order of atomic radius is:
(a) V > Ti
(b) Cl > S
(c) Rb > Cs
(d) Ne > Be
Show Answer & Explanation

Answer: (d) Ne > Be

Explanation:
1. Ne (a van der Waals radius of 154 pm) is larger than Be (112 pm).
2. V < Ti, Cl < S and Cs > Rb.

Teacher's Note:
1. Compare size, ionisation energy and electron gain enthalpy trends, then check the exceptions such as Be/B, N/O and F/Cl.
2. Revise the trend with two or three examples.

Question 4: The number of d-electrons in Fe²⁺ (Z = 26) is not equal to the number of electrons in which one of the following?
(a) p-electrons in Cl (Z = 17)
(b) d-electrons in Fe (Z = 26)
(c) p-electrons in Ne (Z = 10)
(d) s-electrons in Mg (Z = 12)
Show Answer & Explanation

Answer: (a) p-electrons in Cl (Z = 17)

Explanation:
1. Fe²⁺ has six 3d electrons.
2. Cl has 11 p electrons, while Fe has 6 d, Ne has 6 p and Mg has 6 s electrons.

Teacher's Note:
1. Compare size, ionisation energy and electron gain enthalpy trends, then check the exceptions such as Be/B, N/O and F/Cl.
2. Revise the trend with two or three examples.

Question 5: The formation of the oxide ion O²⁻(g) from an oxygen atom needs first an exothermic and then an endothermic step: O(g) + e⁻ → O⁻(g), ΔfH = −141 kJ mol⁻¹; O⁻(g) + e⁻ → O²⁻(g), ΔfH = +780 kJ mol⁻¹. Thus formation of O²⁻ in the gas phase is unfavourable even though O²⁻ is isoelectronic with neon. This is due to the fact that:
(a) Oxygen is more electronegative
(b) Addition of electron in oxygen results in larger size of the ion
(c) Electron repulsion outweighs the stability gained by achieving noble gas configuration
(d) O⁻ ion has comparatively smaller size than oxygen atom
Show Answer & Explanation

Answer: (c) Electron repulsion outweighs the stability gained by achieving noble gas configuration

Explanation:
1. Adding a second electron to the small O⁻ ion causes strong electron-electron repulsion.
2. This outweighs the stability of the neon configuration, so it is endothermic.

Teacher's Note:
1. Compare size, ionisation energy and electron gain enthalpy trends, then check the exceptions such as Be/B, N/O and F/Cl.
2. Revise the trend with two or three examples.

Question 6: Which is the correct order of increasing energy of the listed orbitals in the atom of titanium ? (At. no. Z = 22)?
(a) 3s 3p 3d 4s
(b) 3s 3p 4s 3d
(c) 3s 4s 3p 3d
(d) 4s 3s 3p 3d
Show Answer & Explanation

Answer: (b) 3s 3p 4s 3d

Explanation:
1. By the (n + l) rule in Ti the order is 3s < 3p < 4s < 3d.

Teacher's Note:
1. Compare size, ionisation energy and electron gain enthalpy trends, then check the exceptions such as Be/B, N/O and F/Cl.
2. Revise the trend with two or three examples.

Question 7: The electronic configuration of Al³⁺, excluding the electrons used in bonding in the aluminate ion, is:
(a) [Ne]
(b) [Ar]
(c) [Ne]3s²
(d) [Ar]4s²
Show Answer & Explanation

Answer: (a) [Ne]

Explanation:
1. Al³⁺ has lost its three valence electrons, giving the neon configuration [Ne].

Teacher's Note:
1. Compare size, ionisation energy and electron gain enthalpy trends, then check the exceptions such as Be/B, N/O and F/Cl.
2. Revise the trend with two or three examples.

Question 8: If the I.P. of Na, Mg & Si are 496, 737 & 786 kJ/mole respectively then I.P. of Al is:
(a) 760 kJ/mole
(b) 756 kJ/mole
(c) 577 kJ/mole
(d) 986 kJ/mole
Show Answer & Explanation

Answer: (c) 577 kJ/mole

Explanation:
1. Al has a lower IP than Mg because its outermost electron is in 3p.
2. The value is 577 kJ per mole, which is also below Mg and Si.

Teacher's Note:
1. Compare size, ionisation energy and electron gain enthalpy trends, then check the exceptions such as Be/B, N/O and F/Cl.
2. Revise the trend with two or three examples.

Question 9: Which of the following not reacts with NaOH?
(a) As₂O₃
(b) Bi₂O₃
(c) Sb₂O₃
(d) SeO₂
Show Answer & Explanation

Answer: (b) Bi₂O₃

Explanation:
1. Bi2O3 is basic and does not react with NaOH.
2. As2O3, Sb2O3 (amphoteric) and SeO2 (acidic) react.

Teacher's Note:
1. Compare size, ionisation energy and electron gain enthalpy trends, then check the exceptions such as Be/B, N/O and F/Cl.
2. Revise the trend with two or three examples.

Question 10: The correct order of atomic radii in group 13 elements is:
(a) B < Al < In < Ga < Tl
(b) B < Al < Ga < In < Tl
(c) B < Ga < Al < Tl < In
(d) B < Ga < Al < In < Tl
Show Answer & Explanation

Answer: (d) B < Ga < Al < In < Tl

Explanation:
1. Group 13 radii: B 85, Ga 135, Al 143, In 167, Tl 170 pm.
2. Ga is smaller than Al because of poor shielding by the 3d electrons.

Teacher's Note:
1. Compare size, ionisation energy and electron gain enthalpy trends, then check the exceptions such as Be/B, N/O and F/Cl.
2. Revise the trend with two or three examples.

Question 11: In which of the following elements d-orbitals do not have any electrons in their outer electronic configuration:
(a) Am
(b) Lr
(c) Th
(d) Lu
Show Answer & Explanation

Answer: (a) Am

Explanation:
1. Am is [Rn] 5f⁷ 7s² with no d electron.
2. Lu (5d¹), Th (6d²) and Lr (6d¹ in the usual table) have d electrons.

Teacher's Note:
1. Compare size, ionisation energy and electron gain enthalpy trends, then check the exceptions such as Be/B, N/O and F/Cl.
2. Revise the trend with two or three examples.

Question 12: Which of the following is the correct match?
Atomic numberGroup numberPeriod number
(A)46106
(B)5836
(C)5626
(D)4265

(a) Only B, C, D
(b) Only A, B, C
(c) Only B, C
(d) Only A, C, D
Show Answer & Explanation

Answer: (a) Only B, C, D

Explanation:
1. Pd (46) is group 10 but in the 5th period.
2. Ce (58), Ba (56) and Mo (42) are correct as listed.

Teacher's Note:
1. Break each statement into one trend at a time and test it with real values. Watch for exceptions in the p-block.
2. Revise the trend with two or three examples.

Question 13: In the general electronic configuration (n–2)f¹⁻¹⁴ (n–1)d⁰⁻¹ ns², if n = 7 the configuration belongs to:
(a) Lanthanides
(b) Actinides
(c) Transition elements
(d) None
Show Answer & Explanation

Answer: (b) Actinides

Explanation:
1. (n - 2)f¹-14 with n = 7 gives the 5f series, the actinides.

Teacher's Note:
1. Break each statement into one trend at a time and test it with real values. Watch for exceptions in the p-block.
2. Revise the trend with two or three examples.

Question 14: If total 110 elements are present in periodic table then how many of them contain e– in f subshell:
(a) 28
(b) 57
(c) 58
(d) 53 Check Your Understanding
Show Answer & Explanation

Answer: (d) 53 Check Your Understanding

Explanation:
1. Electrons enter 4f from Ce (58) onwards, and Z from 58 to 110 gives 53 elements.
2. La (57) has no f electron.

Teacher's Note:
1. Break each statement into one trend at a time and test it with real values. Watch for exceptions in the p-block.
2. Revise the trend with two or three examples.

Question 15: Which of the following statements is wrong:
(a) van der Waal's radius of iodine is more than its covalent radius
(b) All isoelectronic ions belong to same period of the periodic table
(c) IE₁ of N is higher than that of O while IE₂ of O is higher than that of N
(d) The electron affinity of N is less than that of P
Show Answer & Explanation

Answer: (b) All isoelectronic ions belong to same period of the periodic table

Explanation:
1. Isoelectronic ions do not all belong to the same period, e.g. Na⁺ and F⁻ .
2. The other statements are correct.

Teacher's Note:
1. Break each statement into one trend at a time and test it with real values. Watch for exceptions in the p-block.
2. Revise the trend with two or three examples.

Question 16: Three elements A, B and C have atomic numbers Z₁, Z₂ and Z₃ respectively. If Z₃ − Z₁ = 2, (Z₁ + Z₃)/2 = Z₂ and the electronic configuration of element C is [Ar] 3d² 4s², then the correct order of atomic radius is:
(a) A²⁺ < B³⁺ < C⁴⁺
(b) A²⁺ = B³⁺ = C⁴⁺
(c) A²⁺ > B³⁺ > C⁴⁺
(d) B³⁺ < A²⁺ = C⁴⁺
Show Answer & Explanation

Answer: (c) A²⁺ > B³⁺ > C⁴⁺

Explanation:
1. C is Ti (Z = 22), so Z1 = 20 (Ca) and Z2 = 21 (Sc).
2. Ca²⁺, Sc³⁺ and Ti⁴⁺ are isoelectronic and the size falls as charge rises: Ca²⁺ > Sc³⁺ > Ti⁴⁺.

Teacher's Note:
1. Break each statement into one trend at a time and test it with real values. Watch for exceptions in the p-block.
2. Revise the trend with two or three examples.

Question 17: For M(g) → M⁺(g) + e⁻, ΔH = 100 eV and for M(g) → M²⁺(g) + 2e⁻, ΔH = 250 eV. Which is the incorrect statement?
(a) I₁ of M(g) is 100 eV
(b) I₂ of M(g) is 150 eV
(c) I₂ of M(g) is 250 eV
(d) none
Show Answer & Explanation

Answer: (c) I₂ of M(g) is 250 eV

Explanation:
1. I1 = 100 eV, so I2 = 250 - 100 = 150 eV.
2. So the statement that I2 is 250 eV is incorrect.

Teacher's Note:
1. Break each statement into one trend at a time and test it with real values. Watch for exceptions in the p-block.
2. Revise the trend with two or three examples.

Question 18: Successive ionisation energies of an element A are 100 eV, 150 eV, 181 eV, 2000 eV, 2200 eV correct statement regarding A is:
(a) Element 'A' may be metal
(b) Formula of oxide of A may be A₂O₃
(c) Oxide of element A may be amphoteric
(d) All are correct
Show Answer & Explanation

Answer: (d) All are correct

Explanation:
1. A big jump after the third IE shows three valence electrons.
2. A is a metal like Al, A2O3 is possible, and it could be amphoteric.

Teacher's Note:
1. Break each statement into one trend at a time and test it with real values. Watch for exceptions in the p-block.
2. Revise the trend with two or three examples.

Question 19: Identify the incorrect are:
(a) Shielding constant (s) : Li < Na < K < Rb
(b) Zeff : Li > Na > K > Rb
(c) Ionic radius : O²⁻ > F⁻ > Na⁺ > Mg²⁺
(d) Atomic size : Li < Na < K < Rb
Show Answer & Explanation

Answer: (b) Zeff : Li > Na > K > Rb

Explanation:
1. By Slater's rules Zeff rises from Li to Na and is about the same from Na to Rb.
2. So Li > Na > K > Rb is incorrect.

Teacher's Note:
1. Break each statement into one trend at a time and test it with real values. Watch for exceptions in the p-block.
2. Revise the trend with two or three examples.

Question 20: If electronegativity values of element X and Y are 3.8 and 1.8 respectively, then percentage of ionic character in compound XY is :
(a) 50
(b) 46
(c) 64
(d) 36 –
Show Answer & Explanation

Answer: (b) 46

Explanation:
1. % ionic character = 16 (difference) + 3.5 (difference)2.
2. With a difference of 2.0 this is 32 + 14 = 46.

Teacher's Note:
1. Break each statement into one trend at a time and test it with real values. Watch for exceptions in the p-block.
2. Revise the trend with two or three examples.

Question 21: The order of ionisation potential between He⁺ ion and H-atom (both species are in gaseous state) is:
(a) I.P. (He⁺) = I.P. (H)
(b) I.P. (He⁺) < I.P. (H)
(c) I.P. (He⁺) > I.P. (H)
(d) Cannot be compared
Show Answer & Explanation

Answer: (c) I.P. (He⁺) > I.P. (H)

Explanation:
1. He⁺ (Z = 2) has a much higher IP (54.4 eV) than H (13.6 eV), as it has a higher nuclear charge with one electron.

Teacher's Note:
1. Break each statement into one trend at a time and test it with real values. Watch for exceptions in the p-block.
2. Revise the trend with two or three examples.

Question 22: Electronic configurations: A: 1s² 2s² 2p¹; B: 1s² 2s² 2p⁶ 3s¹ 3p²; C: 1s² 2s¹ 2p¹; D: 1s² 2s² 2p⁵ 3s¹. Which among these belong to the same group in the periodic table?
(a) A & B
(b) A, B, C
(c) A, B, D
(d) A,
Show Answer & Explanation

Answer: (a) A & B

Explanation:
1. A is boron, B (excited) is aluminium, C (excited) is beryllium and D (excited) is neon.
2. A and B are both in group 13.

Teacher's Note:
1. Break each statement into one trend at a time and test it with real values. Watch for exceptions in the p-block.
2. Revise the trend with two or three examples.

Question 23: The IE₁ and IE₂ of three elements A, B and C are given (in kJ/mol): A: 400, 2650; B: 550, 1070; C: 1150, 2090. Identify the element which represents a non-metal:
(a) A
(b) B
(c) Both A & B
(d) C
Show Answer & Explanation

Answer: (d) C

Explanation:
1. Element C has high IE1 and IE2, which is typical of a non-metal.
2. A is an alkali metal and B is an alkaline earth metal.

Teacher's Note:
1. Break each statement into one trend at a time and test it with real values. Watch for exceptions in the p-block.
2. Revise the trend with two or three examples.

Question 24: In which of the following arrangements is the order NOT according to the property indicated against it?
(a) Al³⁺ < Mg²⁺ < Na⁺ < F⁻ (increasing ionic size)
(b) B < C < N < O (increasing first ionisation energy)
(c) I < Br < F < Cl (increasing electron gain enthalpy with negative sign)
(d) Li < Na < K < Rb (increasing metallic radius)
Show Answer & Explanation

Answer: (b) B < C < N < O (increasing first ionisation energy)

Explanation:
1. B < C < N < O is not the order of IE1, because O (1314) is less than N (1402).
2. The other orders are right.

Teacher's Note:
1. Break each statement into one trend at a time and test it with real values. Watch for exceptions in the p-block.
2. Revise the trend with two or three examples.

Question 25: Which one of the following orders represents the correct sequence of the increasing basic nature of the given oxides ?
(a) Na₂O < K₂O < MgO < Al₂O₃
(b) K₂O < Na₂O < Al₂O₃ < MgO
(c) Al₂O₃ < MgO < Na₂O < K₂O
(d) MgO < K₂O < Al₂O₃ < Na₂O
Show Answer & Explanation

Answer: (c) Al₂O₃ < MgO < Na₂O < K₂O

Explanation:
1. Basic character increases with metallic character: Al2O3 < MgO < Na2O < K2O.

Teacher's Note:
1. Break each statement into one trend at a time and test it with real values. Watch for exceptions in the p-block.
2. Revise the trend with two or three examples.

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