Download JEE MCQs for JEE Chemistry: Classification Of Elements and Periodicity In Properties
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Chapter-wise Objective Questions: Classification Of Elements and Periodicity In Properties
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Multiple Choice Questions
(a) 290
(b) 72
(c) 190
(d) 144
Show Answer & Explanation
Answer: (a) 290
Explanation:
1. Period lengths would be 2, 8, 8, 18, 18, 32, 32, 50, 50 and 72, which add up to 290.
Teacher's Note:
1. Each new period length follows the number of orbitals available: 2, 8, 18, 32, 50, 72.
(a) f, s, d and p
(b) s, d, f and p
(c) d, f, s and p
(d) p, s, f and d
Show Answer & Explanation
Answer: (a) f, s, d and p
Explanation:
1. Z = 102 (No) is an actinoid (f-block); Z = 120 would be an alkaline earth (s-block); Z = 107 (Bh) is d-block and Z = 82 (Pb) is p-block.
Teacher's Note:
1. Identify the last subshell filled.
(a) C, B
(b) B, Be
(c) P, S
(d) Mg, Na
Show Answer & Explanation
Answer: (b) B, Be
Explanation:
1. For B and Be, IE1 of B (801) < IE1 of Be (900), so the first condition fails. C/B, P/S and Mg/Na satisfy both conditions.
Teacher's Note:
1. Compare IE1 and IE2 for each pair.
(a) \( \ce{P^{3+}} \) > \( \ce{P^{5+}} \)
(b) \( \ce{Ti^{4+}} \) < \( \ce{Mn^{7+}} \)
(c) \( \ce{Cr^{6+}} \) > \( \ce{Cr^{3+}} \)
(d) \( \ce{K+} \) > \( \ce{Cl-} \)
Show Answer & Explanation
Answer: (a) \( \ce{P^{3+}} \) > \( \ce{P^{5+}} \)
Explanation:
1. Higher positive charge on the same element gives a smaller ion, so P\(^{3+}\) > P\(^{5+}\). In the others, Ti4+ (60 pm) > Mn7+ (46 pm), Cr3+ > Cr6+ and Cl- (181 pm) > K+ (138 pm).
Teacher's Note:
1. Same element: more positive charge, smaller radius.
(a) Y and La
(b) Zr and Hf
(c) Mo and W
(d) Tc and Re
Show Answer & Explanation
Answer: (a) Y and La
Explanation:
1. Both Zr/Hf, Mo/W and Tc/Re are 4d/5d pairs separated by the lanthanoids. Y and La are in group 3 before the lanthanoids start, so La is simply larger due to the extra shell.
Teacher's Note:
1. Contraction acts on elements after the 4f series.
(a) [Xe] 5f\(^{14}\) 6d\(^{10}\) 6p\(^{1-6}\)
(b) [Rn] 5f\(^{14}\) 5d\(^{10}\) 6p\(^{1-6}\)
(c) [Xe] 4f\(^{14}\) 5d\(^{10}\) 6p\(^{1-6}\) (with 6s\(^2\))
(d) [Xe] 4f\(^{0}\) 5d\(^{10}\) 6p\(^{1-6}\)
Show Answer & Explanation
Answer: (c) [Xe] 4f\(^{14}\) 5d\(^{10}\) 6p\(^{1-6}\) (with 6s\(^2\))
Explanation:
1. In the 6th period after Hg the 4f and 5d subshells are full, and 6p fills from Tl to Rn with xenon core.
Teacher's Note:
1. The p-block of the 6th period begins after the 5d series.
(a) \( \ce{Li+} \) and \( \ce{Na+} \)
(b) \( \ce{Be^{2+}} \) and \( \ce{Mg^{2+}} \)
(c) \( \ce{Li+} \) and \( \ce{Mg^{2+}} \)
(d) \( \ce{Be^{2+}} \) and \( \ce{Li+} \)
Show Answer & Explanation
Answer: (c) \( \ce{Li+} \) and \( \ce{Mg^{2+}} \)
Explanation:
1. Li\(^+\) (76 pm) and Mg\(^{2+}\) (72 pm) are diagonally related and similar in size.
Teacher's Note:
1. Diagonal relationship pairs have similar charge-to-radius ratios.
(a) I > Li > F < Cl > Br
(b) I > Br > Cl > F > Li
(c) Li > I > F < Cl < Br
(d) Li > F < I < Br
Show Answer & Explanation
Answer: (c) Li > I > F < Cl < Br
Explanation:
1. Radii (pm): Li 152, I 133, F 64, Cl 99, Br 114; thus Li > I > F < Cl < Br.
Teacher's Note:
1. Use covalent radii for halogens.
(a) Cu\(^+\) > Zn\(^+\)
(b) Tl < Al
(c) Cu < Zn
(d) Pd > Ag
Show Answer & Explanation
Answer: (b) Tl < Al
Explanation:
1. Tl (589) has higher IE1 than Al (577) because of poor shielding by 4f and 5d electrons, so Tl < Al is wrong.
Teacher's Note:
1. The lanthanoid contraction raises IE of Tl and Pb.
(a) C > N
(b) C > Br
(c) Sr > Ca
(d) P > Si
Show Answer & Explanation
Answer: (d) P > Si
Explanation:
1. P (2.1) > Si (1.8) is correct; N (3.0) > C (2.5), Br (2.8) > C (2.5) and Ca (1.0) > Sr (0.95).
Teacher's Note:
1. Across a period EN increases.
(a) Cl > O > N > C
(b) Cl > N > C > O
(c) Cl > C > O > N
(d) Cl > O > C > N
Show Answer & Explanation
Answer: (d) Cl > O > C > N
Explanation:
1. EA (kJ per mol): Cl 349, O 141, C 122, N about zero.
Teacher's Note:
1. Nitrogen has an extremely low electron gain enthalpy.
(a) 575
(b) 760
(c) 390
(d) 1120
Show Answer & Explanation
Answer: (a) 575
Explanation:
1. The order is Na (496) < Al < Mg (737) < Si (786). Al is lower than Mg because its 3p electron is easier to remove than a 3s\(^2\) electron, and the actual value is about 577 kJ per mol.
Teacher's Note:
1. Al and Ga show the dip in IE in group 13.
(a) Chloride ion is more electronegative than fluorine atom
(b) The size of \( \ce{H-} \) is larger than that of \( \ce{F-} \)
(c) Rb is more electropositive than Ca
(d) \( \ce{Na+} \) is more electronegative than Na
Show Answer & Explanation
Answer: (a) Chloride ion is more electronegative than fluorine atom
Explanation:
1. Cl\(^-\) is an anion and holds no tendency to attract further electrons; fluorine is the most electronegative element. The other three statements are correct.
Teacher's Note:
1. Cations are more electronegative than their atoms.
(a) Beryllium
(b) Magnesium
(c) Aluminium
(d) Boron
Show Answer & Explanation
Answer: (d) Boron
Explanation:
1. IE2 values (kJ per mol): B 2427, Be 1757, Al 1816, Mg 1451. Boron is the highest because B\(^+\) has a stable 2s\(^2\) configuration.
Teacher's Note:
1. Removal from a filled s subshell needs more energy.
(a) \( \ce{Ni^{2+}} \)
(b) \( \ce{Co^{3+}} \)
(c) \( \ce{Fe^{2+}} \)
(d) \( \ce{Cu^{2+}} \)
Show Answer & Explanation
Answer: (d) \( \ce{Cu^{2+}} \)
Explanation:
1. Unpaired electrons: Cu2+ (3d9) 1, Ni2+ (3d8) 2, Co3+ (3d6) 4 and Fe2+ (3d6) 4 (high-spin). Cu2+ has the lowest moment (1.73 BM).
Teacher's Note:
1. Magnetic moment = root of n(n+2).
(a) 8
(b) 58
(c) 38
(d) 29
Show Answer & Explanation
Answer: (d) 29
Explanation:
1. Z = 78 is Pt: [Xe] 4f\(^{14}\) 5d\(^9\) 6s\(^1\). d electrons: 3d 10 + 4d 10 + 5d 9 = 29.
Teacher's Note:
1. Count d electrons of all shells, not just the valence one.
(a) Na, Mg
(b) Al, Be
(c) Li+, Be2+
(d) \( \ce{O^{2-}} \), \( \ce{N^{3-}} \)
Show Answer & Explanation
Answer: (d) \( \ce{O^{2-}} \), \( \ce{N^{3-}} \)
Explanation:
1. Both ions are isoelectronic (10 electrons). N\(^{3-}\) has fewer protons, so it is larger than O\(^{2-}\). In the other pairs the second species is smaller.
Teacher's Note:
1. Among isoelectronic species, lower nuclear charge means larger radius.
(a) +580, +780
(b) +580, -780
(c) +780, +590
(d) -780, -580
Show Answer & Explanation
Answer: (c) +780, +590
Explanation:
1. O\(^-\) to O\(^{2-}\) has +780 kJ per mol (large repulsion in the small ion) and S\(^-\) to S\(^{2-}\) about +590 kJ per mol (the source value 580 is approximate).
Teacher's Note:
1. Second electron gain enthalpies are always positive.
(a) 2s\(^2\)2p\(^4\)
(b) 2s\(^2\)2p\(^5\)
(c) 2s\(^2\)2p\(^3\)
(d) 2s\(^2\)2p\(^1\)
Show Answer & Explanation
Answer: (b) 2s\(^2\)2p\(^5\)
Explanation:
1. 2s\(^2\)2p\(^5\) is fluorine, which strongly gains one electron to reach neon configuration (-328 kJ per mol), more than O, N or B.
Teacher's Note:
1. Compare among elements of the same period.
(a) Be would have been in the first period
(b) The fourth period would have contained 36 elements
(c) The number of periods would have been fewer than in the modern table
(d) The second period would have contained 12 elements
Show Answer & Explanation
Answer: (d) The second period would have contained 12 elements
Explanation:
1. Each orbital would hold 4 electrons.
2. First period: 1s holds 4, so Be (Z = 4) closes period 1. Second period: 2s and 2p = 4 orbitals, i.e. 16 elements (not 12). Fourth period: 4s, 3d and 4p = 9 orbitals, i.e. 36 elements.
Teacher's Note:
1. Count orbitals in each period and multiply by the electron capacity of an orbital.
(a) IE\(_2\)(Mg) > IE\(_2\)(Na)
(b) IE\(_1\)(Al) < IE\(_1\)(Mg)
(c) IE\(_1\)(Na) < IE\(_1\)(Mg)
(d) IE\(_3\)(Mg) > IE\(_3\)(Al)
Show Answer & Explanation
Answer: (a) IE\(_2\)(Mg) > IE\(_2\)(Na)
Explanation:
1. IE2 of Na (4562) is far greater than IE2 of Mg (1451), so the given order is wrong.
Teacher's Note:
1. IE2 of an alkali metal is exceptionally large.
(a) Pb
(b) Sn
(c) Si
(d) C
Show Answer & Explanation
Answer: (b) Sn
Explanation:
1. IE1 (kJ per mol): Sn 709, Pb 716, Si 787 and C 1086. Sn is the lowest.
Teacher's Note:
1. In group 14, Pb has slightly higher IE than Sn.
(a) zero
(b) 10
(c) 16
(d) 30
Show Answer & Explanation
Answer: (a) zero
Explanation:
1. In the 5th period the filling is 5s, 4d, 5p. The 5d subshell is not occupied in any ground state, so the answer is zero.
Teacher's Note:
1. The 5d subshell begins to fill only in the 6th period (La).
(a) Ionisation enthalpy increases for each successive electron
(b) Removal of an electron from an orbital with a lower n value is easier than from an orbital with a higher n value
(c) The greatest increase is on removal of an electron from a core noble gas configuration
(d) End of valence electrons is marked by a big jump in ionisation enthalpy
Show Answer & Explanation
Answer: (b) Removal of an electron from an orbital with a lower n value is easier than from an orbital with a higher n value
Explanation:
1. Electrons in orbitals with lower n are closer to the nucleus and harder to remove, so the given statement is the reverse of the truth.
Teacher's Note:
1. Inner electrons are held tightly.
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