JEE Chemistry Classification of Elements and Periodicity in Properties MCQs Set 06

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Multiple Choice Questions

Question 1: The correct order of decreasing second ionisation enthalpy of Ti (22), V (23), Cr (24) and Mn (25) is
(a) Mn > Cr > Ti > V
(b) Ti > V > Cr > Mn
(c) Cr > Mn > V > Ti
(d) V > Mn > Cr > Ti
Show Answer & Explanation

Answer: (c) Cr > Mn > V > Ti

Explanation:
1. IE2 (kJ/mol): Cr 1591 (3d5 after loss of 4s1), Mn 1509, V 1414, Ti 1310.

Teacher's Note:
1. Cr+ has the very stable 3d5.

Question 2: Among Ca, Mg, P and Cl, the order of increasing atomic radii is
(a) Cl < P < Mg < Ca
(b) P < Cl < Ca < Mg
(c) Ca < Mg < P < Cl
(d) Mg < Ca < Cl < P
Show Answer & Explanation

Answer: (a) Cl < P < Mg < Ca

Explanation:
1. Mg and Ca are metals; Ca is bigger because it is in the 4th period. In period 3, radius decreases from Mg to Cl.

Teacher's Note:
1. Radius: Cl 99, P 110, Mg 160, Ca 197 pm.

Question 3: The correct order of decreasing ionic radii among the isoelectronic species is
(a) K+ > Ca2+ > Cl- > S2-
(b) Ca2+ > K+ > S2- > Cl-
(c) Cl- > S2- > Ca2+ > K+
(d) S2- > Cl- > K+ > Ca2+
Show Answer & Explanation

Answer: (d) S2- > Cl- > K+ > Ca2+

Explanation:
1. All have 18 electrons; more protons contract the electron cloud.

Teacher's Note:
1. S2- (16) > Cl- (17) > K+ (19) > Ca2+ (20).

Question 4: Which is the correct order of first ionisation potential (IP1)?
(a) Na > Al
(b) Mg > Al
(c) Ga > Ca
(d) Mg > Be
Show Answer & Explanation

Answer: (b) Mg > Al

Explanation:
1. IP1 (kJ/mol): Mg 737 > Al 577; Na 496 < Al 577; Ga 579 < Ca 590; Mg 737 < Be 899.

Teacher's Note:
1. Mg has a filled 3s2 subshell, making Al lower.

Question 5: What is the value of electron gain enthalpy of Na+ if IE1 of Na = 5.1 eV?
(a) +0.2 eV
(b) -5.1 eV
(c) -10.2 eV
(d) +2.55 eV
Show Answer & Explanation

Answer: (b) -5.1 eV

Explanation:
1. Na+ + e- gives Na is the reverse of ionisation, so delta H = -IE1 = -5.1 eV.

Teacher's Note:
1. Electron gain by a cation equals the negative of the ionisation of the atom.

Question 6: The first ionisation enthalpies of Na, Mg and Si are 496, 737 and 786 kJ/mol respectively. The first ionisation enthalpy of Al will be
(a) More than 766 kJ/mol
(b) More than 496 and less than 737 kJ/mol
(c) More than 737 and less than 766 kJ/mol
(d) Less than 496 kJ/mol
Show Answer & Explanation

Answer: (b) More than 496 and less than 737 kJ/mol

Explanation:
1. Al is lower than Mg because of the filled 3s2 in Mg and ejection from the higher-energy 3p; so Na < Al < Mg.

Teacher's Note:
1. Actual value is 577 kJ/mol.

Question 7: 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+.

Teacher's Note:
1. H+ has none; Na+ and Mg2+ have 10.

Question 8: Acidity of diprotic hydrides in aqueous solution 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. Down the group, bond strength to H falls, so the acid strength rises: H2S < H2Se < H2Te.

Teacher's Note:
1. The M-H bond becomes longer and weaker.

Question 9: Which of the following is the correct order of atomic radius?
(a) V > Ti
(b) Cl > S
(c) Rb > Cs
(d) Ne > Be
Show Answer & Explanation

Answer: (d) Ne > Be

Explanation:
1. Using van der Waals radius for Ne (154 pm) it is larger than Be (112 pm); the other orders are wrong (Ti > V, S > Cl, Cs > Rb).

Teacher's Note:
1. Noble gas radii are van der Waals radii.

Question 10: Which is an incorrect order of acidic strength?
(a) H2S > H2Se
(b) HI > HBr
(c) HBr > HCl
(d) H2Te > H2S
Show Answer & Explanation

Answer: (a) H2S > H2Se

Explanation:
1. H2Se is a stronger acid than H2S; so 'H2S > H2Se' is incorrect.

Teacher's Note:
1. Acidity rises down a group.

Question 11: The number of d-electrons in Fe2+ (Z = 26) is not equal to the number of electrons in which 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+: 3d6 (6 d-electrons). Fe: 3d6; Ne: 2p6; Mg: 1s2 2s2 3s2 = 6 s-electrons.
2. Cl has 11 p-electrons (2p6 3p5).

Teacher's Note:
1. Write each configuration.

Question 12: Because of lanthanoid contraction, which of the following pairs have nearly the same atomic radii?
(a) Zr (40) and Nb (41)
(b) Zr (40) and Hf (72)
(c) Zr (40) and Ta (73)
(d) Ti (22) and Zr (40)
Show Answer & Explanation

Answer: (b) Zr (40) and Hf (72)

Explanation:
1. Zr and Hf are of nearly equal size (160 and 159 pm).

Teacher's Note:
1. Hence their chemical properties are very similar.

Question 13: The correct order of increasing energy of the listed orbitals in the titanium atom (Z = 22) is
(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. For a many-electron atom the order is 3s < 3p < 4s < 3d.

Teacher's Note:
1. Use the (n + l) rule.

Question 14: Smallest ionic radius is shown by
(a) La3+
(b) U3+
(c) Yb3+
(d) Cf3+
Show Answer & Explanation

Answer: (c) Yb3+

Explanation:
1. Yb3+ lies late in the lanthanoids, so lanthanoid contraction makes it the smallest among the options.

Teacher's Note:
1. Ionic radii decrease across the lanthanoid series.

Question 15: The element with Z = 114 recently discovered would belong to which family and configuration?
(a) Carbon family, [Rn] 5f14 6d10 7s2 7p2
(b) Oxygen family, [Rn] 5f14 6d10 7s2 7p4
(c) Nitrogen family, [Rn] 5f14 6d10 7s2 7p6
(d) Halogen family, [Rn] 5f14 6d10 7s2 7p5
Show Answer & Explanation

Answer: (a) Carbon family, [Rn] 5f14 6d10 7s2 7p2

Explanation:
1. Z = 114: [Rn] 5f14 6d10 7s2 7p2, giving 4 valence electrons, group 14 (carbon family).

Teacher's Note:
1. Flerovium.

Question 16: The ionisation energies of Na, Mg and Si are 496, 737 and 786 kJ/mol. The ionisation energy 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 IE lower than Mg (737) and higher than Na (496); 577 kJ/mol fits.

Teacher's Note:
1. The p electron in Al is easier to remove than the s electron of Mg.

Question 17: Which of the following oxides is most acidic in nature?
(a) MgO
(b) BeO
(c) BaO
(d) CaO
Show Answer & Explanation

Answer: (b) BeO

Explanation:
1. BeO is amphoteric and the others are basic, so BeO is the most acidic of these.

Teacher's Note:
1. Basicity of group 2 oxides increases down the group.

Question 18: The correct order of atomic radii in group 13 elements is
(a) B < Al < In < 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. Radii (pm): B 85, Ga 135, Al 143, In 167, Tl 170.
2. Ga is smaller than Al due to poor screening by 3d electrons.

Teacher's Note:
1. d-block contraction affects Ga.

Question 19: The atomic number of Ag is 47. In the same group, the atomic numbers of the elements placed above and below Ag are
(a) 37, 67
(b) 29, 79
(c) 39, 69
(d) 29, 65
Show Answer & Explanation

Answer: (b) 29, 79

Explanation:
1. Group 11: Cu (29), Ag (47), Au (79).

Teacher's Note:
1. Move up 18 (period 4) and down 32 (period 6).

Question 20: In the general electronic configuration (n-2)f(1-14) (n-1)d(0-1) ns2, if n = 7 the configuration is of
(a) Lanthanoids
(b) Actinoids
(c) Transition elements
(d) None
Show Answer & Explanation

Answer: (b) Actinoids

Explanation:
1. With n = 7 the filling is of 5f, so these are the actinoids.

Teacher's Note:
1. For n = 6 it describes the lanthanoids.

Question 21: Three elements A, B and C have atomic numbers Z1, Z2, Z3. If Z3 - Z1 = 2 and (Z1 + Z3)/2 = Z2, and the electronic configuration of C is [Ar] 3d5 4s2, then the correct order of radii is (ions A2+, B3+, C4+)
(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 Mn (25), so Z1 = 23 (V) and Z2 = 24 (Cr).
2. V2+, Cr3+ and Mn4+ are all d3: higher charge and nuclear charge mean smaller size: V2+ > Cr3+ > Mn4+.

Teacher's Note:
1. These are isoelectronic (3d3).

Question 22: M(g) to M+(g) + e-, delta H = 100 eV; M(g) to M2+(g) + 2e-, delta H = 250 eV. Which statements are incorrect?
(a) I1 of M is 100 eV
(b) I2 of M is 150 eV
(c) I2 of M is 250 eV
(d) None
Show Answer & Explanation

Answer: (c) I2 of M is 250 eV

Explanation:
1. I1 = 100 eV and I2 = 250 - 100 = 150 eV, so the statement 'I2 = 250 eV' is incorrect.

Teacher's Note:
1. The second ionisation energy is the difference.

Question 23: Given K to K+ (a), K+ to K2+ (b), Ca to Ca+ (c) and Ca+ to Ca2+ (d) with ionisation energies a, b, c and d, which order is NOT correct?
(a) c > a
(b) b > a
(c) d > c
(d) b < d
Show Answer & Explanation

Answer: (d) b < d

Explanation:
1. IE (kJ/mol): a 419, b 3052, c 590, d 1145. So b is greater than d; 'b < d' is not correct.

Teacher's Note:
1. K+ has a noble gas configuration.

Question 24: Which of the following is the weakest base among NaOH, Ca(OH)2, KOH and Zn(OH)2?
(a) NaOH
(b) KOH
(c) Ca(OH)2
(d) Zn(OH)2
Show Answer & Explanation

Answer: (d) Zn(OH)2

Explanation:
1. Zn(OH)2 is amphoteric and the weakest base; the others are strongly basic alkali or alkaline earth hydroxides.

Teacher's Note:
1. Basicity increases as electronegativity of the metal falls.

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