JEE Chemistry Classification Of Elements and Periodicity In Properties MCQs Set 25

Chemistry Objective Questions and Answers: Classification Of Elements and Periodicity In Properties

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Multiple Choice Questions

Question 1: The successive ionisation energies (kcal per mol) of five elements are: P: 300, 549, 920; Q: 99, 734, 1100; R: 118, 1091, 1652; S: 176, 347, 1848; T: 497, 947, 1500. Which element is a noble gas?
(a) P
(b) R
(c) T
(d) S
Show Answer & Explanation

Answer: (c) T

Explanation:
1. T has by far the highest first ionisation energy (497) with no sudden jump after the first value, characteristic of a noble gas such as He.

Teacher's Note:
1. Noble gases have the highest first ionisation energies.

Question 2: The correct order of ionic radii is:
(a) \( \ce{Ce^{3+}} \) < \( \ce{Yb^{3+}} \) < \( \ce{Pm^{3+}} \) < \( \ce{La^{3+}} \)
(b) \( \ce{Yb^{3+}} \) < \( \ce{Pm^{3+}} \) < \( \ce{La^{3+}} \) < \( \ce{Ce^{3+}} \)
(c) \( \ce{Yb^{3+}} \) < \( \ce{Pm^{3+}} \) < \( \ce{Ce^{3+}} \) < \( \ce{La^{3+}} \)
(d) \( \ce{Pm^{3+}} \) < \( \ce{La^{3+}} \) < \( \ce{Ce^{3+}} \) < \( \ce{Yb^{3+}} \)
Show Answer & Explanation

Answer: (c) \( \ce{Yb^{3+}} \) < \( \ce{Pm^{3+}} \) < \( \ce{Ce^{3+}} \) < \( \ce{La^{3+}} \)

Explanation:
1. Because of lanthanoid contraction, the radius of \( \ce{Ln^{3+}} \) decreases steadily with atomic number: La (103 pm) > Ce (101) > Pm (97) > Yb (87).

Teacher's Note:
1. Across the lanthanoids the 4f electrons shield poorly, so size shrinks from La to Lu.

Question 3: The successive ionisation energies (kcal per mol) of five elements are: P: 300, 549, 920; Q: 99, 734, 1100; R: 118, 1091, 1652; S: 176, 347, 1848; T: 497, 947, 1500. Which element is a non-metal (excluding the noble gas)?
(a) Q
(b) S
(c) R
(d) P
Show Answer & Explanation

Answer: (d) P

Explanation:
1. P has a high first ionisation energy (300) with gradually rising IE2 and IE3, typical of a non-metal; Q, R and S are metals (alkali or alkaline earth).

Teacher's Note:
1. Non-metals have high IE with no big jump among small values.

Question 4: Which of the following properties of the halogens decreases from fluorine to iodine?
(a) Electronegativity only
(b) Bond energy only
(c) Electronegativity and ionisation energy (both)
(d) Atomic size
Show Answer & Explanation

Answer: (c) Electronegativity and ionisation energy (both)

Explanation:
1. Electronegativity falls from F (4.0) to I (2.5) and ionisation energy falls from F to I. The bond energy is not monotonic since F-F is weaker than Cl-F; atomic size increases.

Teacher's Note:
1. Check each property separately.

Question 5: Which statement is correct for the processes (i) Cl + e- to Cl- (electron affinity); (ii) Cl- to Cl + e-; (iii) Cl to Cl+ + e-; (iv) Cl+ to Cl2+ + e-?
(a) |I.E. of process (ii)| = |E.A. of process (i)|
(b) |I.E. of process (iii)| = |I.E. of process (ii)|
(c) |I.E. of process (iv)| = |E.A. of process (i)|
(d) |I.E. of process (iv)| = |I.E. of process (iii)|
Show Answer & Explanation

Answer: (a) |I.E. of process (ii)| = |E.A. of process (i)|

Explanation:
1. Process (ii) is simply the reverse of process (i), so their energies have equal magnitude and opposite sign.

Teacher's Note:
1. Reverse processes have equal and opposite energy changes.

Question 6: Regarding electron affinity, which statement is correct?
(a) Carbon is greater than oxygen
(b) Sulphur is lesser than oxygen
(c) Iodine is higher than bromine
(d) Bromine has a lower electron affinity than chlorine
Show Answer & Explanation

Answer: (d) Bromine has a lower electron affinity than chlorine

Explanation:
1. EA (kJ per mol): Cl 349 > Br 325 > I 295; O 141 < S 200; C 122 < O 141.

Teacher's Note:
1. Second-period elements are exceptions only for F and O versus Cl and S.

Question 7: In which pair is the first atom or ion not larger than the second?
(a) N, F
(b) O, S
(c) \( \ce{Cl-} \), Cl
(d) \( \ce{Fe^{2+}} \), \( \ce{Fe^{3+}} \)
Show Answer & Explanation

Answer: (b) O, S

Explanation:
1. O is smaller than S (the group trend). N > F, Cl- > Cl and Fe2+ > Fe3+ are all correct for 'first larger than second'.

Teacher's Note:
1. Read the stem carefully: it asks for the pair where the first is not larger.

Question 8: If the atomic number of an inert gas is Z, an element with which atomic number will have the highest ionisation energy among the options?
(a) Z - 2
(b) Z + 1
(c) Z - 1
(d) Z + 2
Show Answer & Explanation

Answer: (c) Z - 1

Explanation:
1. Among the options, Z - 1 is a halogen (e.g. F, Cl) with a nearly complete shell, and has the highest IE; Z + 1 is an alkali metal and Z - 2 is a chalcogen.

Teacher's Note:
1. Elements just before a noble gas have high IE.

Question 9: According to Slater's rules, the set of elements that shows an incorrect order of \( Z_{eff} \) is:
(a) Al > Mg
(b) Na < Al
(c) K > Na
(d) Sc < Ti
Show Answer & Explanation

Answer: (c) K > Na

Explanation:
1. K and Na both have 2.20 as the Z_eff of the valence s electron (19 - 16.8 and 11 - 8.8), so K > Na is incorrect. Across a period Z_eff increases, so the other pairs are correct.

Teacher's Note:
1. Valence s electrons of alkali metals have the same Z_eff.

Question 10: Which of the following statements is incorrect?
(a) The first ionisation energy of calcium is more than that of gallium
(b) The third ionisation energy of Mg is greater than that of Al
(c) The second ionisation energy of copper is greater than that of potassium
(d) The IE\(_1\) of \( \ce{Mg+} \) is less than the IE\(_1\) of \( \ce{Na+} \)
Show Answer & Explanation

Answer: (c) The second ionisation energy of copper is greater than that of potassium

Explanation:
1. IE2 of K (3052) is greater than IE2 of Cu (1958) since K\(^+\) has a noble gas core. The other three are correct.

Teacher's Note:
1. Compare the configuration after losing the first electron.

Question 11: Which of the following ions has the highest ionic radius?
(a) \( \ce{Sc^{3+}} \)
(b) \( \ce{Ce^{3+}} \)
(c) \( \ce{La^{3+}} \)
(d) \( \ce{Lu^{3+}} \)
Show Answer & Explanation

Answer: (c) \( \ce{La^{3+}} \)

Explanation:
1. Among group 3 and lanthanoid trivalent ions, La\(^{3+}\) (103 pm) is largest; Ce3+ is 101 pm, Lu3+ 86 pm and Sc3+ 75 pm.

Teacher's Note:
1. Lanthanoid ions shrink along the series.

Question 12: Which one of the following hydroxides is least basic?
(a) \( \ce{Tb(OH)3} \)
(b) \( \ce{Yb(OH)3} \)
(c) \( \ce{Gd(OH)3} \)
(d) \( \ce{Eu(OH)3} \)
Show Answer & Explanation

Answer: (b) \( \ce{Yb(OH)3} \)

Explanation:
1. Basic character of \( \ce{Ln(OH)3} \) decreases with decreasing ionic radius (increasing Z); Yb3+ is the smallest here.

Teacher's Note:
1. Smaller cation, more covalent M-OH bond, lower basicity.

Question 13: The correct values of ionisation enthalpies (kJ per mol) of Si, P, Cl and S respectively are:
(a) 1012, 786, 999, 1256
(b) 786, 1012, 999, 1256
(c) 786, 1012, 1256, 999
(d) 786, 999, 1012, 1256
Show Answer & Explanation

Answer: (c) 786, 1012, 1256, 999

Explanation:
1. Order: Si (786) < S (999) < P (1012) < Cl (1256); S is lower than P because of half-filled 3p\(^3\) in P.

Teacher's Note:
1. Half-filled configurations raise ionisation energy.

Question 14: The correct order of size is:
(a) Ge > Br > C > F
(b) F > C > Br > Ge
(c) Ge > C > F > Br
(d) Br > Ge > C > F
Show Answer & Explanation

Answer: (a) Ge > Br > C > F

Explanation:
1. Covalent radii: Ge 122 pm, Br 114 pm, C 77 pm, F 64 pm.

Teacher's Note:
1. Compare periods first, then groups.

Question 15: Select the incorrect statement:
(a) In general, more ionisation energy means more electronegativity
(b) Electronegativity increases means metallic character increases
(c) Lower ionisation energy means higher reducing property
(d) Cl has higher electron affinity than F
Show Answer & Explanation

Answer: (b) Electronegativity increases means metallic character increases

Explanation:
1. Higher electronegativity means more non-metallic character, so the first statement is false.

Teacher's Note:
1. Metallic character is inversely related to electronegativity.

Question 16: Read the statements: (I) Atomic sizes of Fe, Co; Al, Ga; Zr, Hf are almost the same. (II) Electron affinity order Be < N < B < Li < C < O. (III) Ionisation energy: Al < Ga and Zr < Hf. (IV) Ionic radius: \( \ce{O^{2-}} < \ce{F-} < \ce{Na+} < \ce{Mg^{2+}} \). The correct statements are:
(a) I, II and III
(b) I, III and IV
(c) II, III and IV
(d) All the four
Show Answer & Explanation

Answer: (a) I, II and III

Explanation:
1. IV is wrong: O\(^{2-}\) is larger than F\(^-\) which is larger than Na\(^+\) and Mg\(^{2+}\) (isoelectronic with decreasing size as nuclear charge rises). I, II and III are correct.

Teacher's Note:
1. For isoelectronic ions the more negative one is the larger.

Question 17: The correct order of increasing size is:
(a) S < O < Se < C
(b) O < C < S < Se
(c) O < Se < S < C
(d) C < O < S < Se
Show Answer & Explanation

Answer: (b) O < C < S < Se

Explanation:
1. Covalent radii: O 66 pm, C 77 pm, S 104 pm, Se 117 pm.

Teacher's Note:
1. Size decreases across a period and increases down a group.

Question 18: When one mole of F atoms is converted to F\(^-\), the energy released is X joules. Then:
(a) X joules is sufficient to ionise 1 mole of gaseous Cl into Cl\(^+\)
(b) X joules is sufficient to ionise 1 mole of gaseous F into F\(^+\)
(c) X joules can ionise 1 mole each of F and Cl
(d) less than 1 mole of gaseous F or Cl atoms would be ionised to F\(^+\) or Cl\(^+\) with X joules
Show Answer & Explanation

Answer: (d) less than 1 mole of gaseous F or Cl atoms would be ionised to F\(^+\) or Cl\(^+\) with X joules

Explanation:
1. Electron gain enthalpy of F is -328 kJ per mol while ionisation enthalpies of F (1681) and Cl (1251) are far larger, so X can ionise only a small fraction of a mole.

Teacher's Note:
1. Ionisation energies are much larger than electron affinities.

Question 19: Which of the following facts is not explained by lanthanoid contraction?
(a) The ionisation energy of Ag is less than that of Au
(b) The first ionisation energy of Hg is greater than that of Cd
(c) The size of Hf is almost equal to that of Zr
(d) The size of Y is less than that of La
Show Answer & Explanation

Answer: (d) The size of Y is less than that of La

Explanation:
1. Y lies above La in group 3, so Y < La is the normal group trend caused by one extra shell. The other three facts arise because the 4f electrons shield poorly in the 6th-period elements.

Teacher's Note:
1. Lanthanoid contraction affects only the 5d series versus the 4d series.

Question 20: Select the correct combination of correct statements: (a) The effective nuclear charge for a 2p electron of nitrogen is 3.90. (b) The ionisation potential of Ne is more than that of Na\(^+\). (c) Electronegativity order is sp > sp\(^2\) > sp\(^3\). (d) Acidic character order is \( \ce{NH3} \) < \( \ce{PH3} \) < \( \ce{AsH3} \).
(a) a, b, d
(b) b, c
(c) a, b, c, d
(d) a, c, d
Show Answer & Explanation

Answer: (d) a, c, d

Explanation:
1. (a) Z_eff = 7 - (0.35 x 4 + 0.85 x 2) = 3.90: correct.
2. (b) Na\(^+\) is isoelectronic with Ne but has higher nuclear charge, so its IE (4562) exceeds Ne (2081): false.
3. (c) More s character means higher electronegativity: correct.
4. (d) Bond strength of E-H falls down the group, so acidity rises: correct.

Teacher's Note:
1. Isoelectronic species: higher nuclear charge, higher ionisation energy.

Question 21: The incorrect order of radius is:
(a) \( \ce{Sr^{2+}} \) < \( \ce{Rb+} \) < \( \ce{Br-} \) < \( \ce{Se^{2-}} \)
(b) Co > \( \ce{Co^{2+}} \) > \( \ce{Co^{3+}} \) > \( \ce{Co^{4+}} \)
(c) \( \ce{Ba^{2+}} \) < \( \ce{Cs+} \) < \( \ce{Se^{2-}} \) < \( \ce{As^{3-}} \)
(d) Ni > Cu > Zn
Show Answer & Explanation

Answer: (d) Ni > Cu > Zn

Explanation:
1. Ni (124 pm) < Cu (128 pm) < Zn (134 pm), so the order given in the first option is incorrect; the rest follow the expected trends.

Teacher's Note:
1. Across the 3d series the radius first falls and then rises after Ni.

Question 22: The difference between atomic radii is maximum in which one of the following pairs?
(a) Rb, Cs
(b) Na, Ca
(c) Li, Na
(d) Na, K
Show Answer & Explanation

Answer: (d) Na, K

Explanation:
1. Radii (pm): Li 152, Na 186, K 227, Rb 248, Cs 265. The Na to K difference is 41 pm, the largest of the pairs given (Li-Na 34, Rb-Cs 17).

Teacher's Note:
1. The increment is large where the number of inner electrons jumps.

Question 23: Which of the following electron affinity orders is not correct?
(a) Cl > O > N > C
(b) N < O < S
(c) O < S < F < Cl
(d) B < C < Si < S
Show Answer & Explanation

Answer: (a) Cl > O > N > C

Explanation:
1. EA values: C 122, N about 0, O 141: so the order is O > C > N, not O > N > C. The other orders hold.

Teacher's Note:
1. N has almost zero electron affinity.

Question 24: The correct order of first ionisation energy is:
(a) Li < Be < N
(b) C > N > O
(c) Mg > Al > P
(d) Tl > B > In
Show Answer & Explanation

Answer: (a) Li < Be < N

Explanation:
1. Li (520) < Be (900) < N (1402) is correct. The others fail: actual orders are N (1402) > O (1314) > C (1086); P (1012) > Mg (738) > Al (577); B (801) > Tl (589) > In (558).

Teacher's Note:
1. Check anomalies at group 2/13 and group 15/16.

Question 25: An element Q reacts with fluorine and oxygen to give compounds with molecular formulae respectively:
(a) \( \ce{QF3} \) and \( \ce{Q2O3} \)
(b) QF and \( \ce{Q2O} \)
(c) \( \ce{QF2} \) and QO
(d) none of these
Show Answer & Explanation

Answer: (b) QF and \( \ce{Q2O} \)

Explanation:
1. Q is an alkali metal (IE jumps after the first value, Q: 99, 734, 1100), so it forms the +1 compounds QF and \( \ce{Q2O} \).

Teacher's Note:
1. Alkali metals have valency 1.

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