JEE Chemistry Classification Of Elements and Periodicity In Properties MCQs Set 27

Multiple Choice Questions (MCQs) for JEE Chemistry: Classification Of Elements and Periodicity In Properties

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Practice Classification Of Elements and Periodicity In Properties MCQs for JEE Chemistry

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Multiple Choice Questions

Question 1: If elements with atomic numbers 117 and 120 are discovered, their blocks will be respectively:
(a) s, p
(b) p, s
(c) p, d
(d) d, p
Show Answer & Explanation

Answer: (b) p, s

Explanation:
1. Z = 117 would be a halogen of period 7 (7s27p5), a p-block element.
2. Z = 120 would be 8s2, an s-block element.
3. Answer: (b)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 2: The correct order of metallic character is:
(a) P < Si < Be < Mg < Na
(b) P > Si > Be > Mg > Na
(c) P < Si < Be > Mg > Na
(d) P > Si > Be < Mg < Na
Show Answer & Explanation

Answer: (a) P < Si < Be < Mg < Na

Explanation:
1. Metallic character falls from left to right in a period and rises down a group.
2. Na > Mg > Be > Si > P.
3. Answer: (a)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 3: Which statement concerning the element with atomic number 10 is false?
(a) the element is monoatomic
(b) it has almost zero electron affinity
(c) it forms a covalent network solid
(d) it has an extremely high value of I.E. in its own period
Show Answer & Explanation

Answer: (c) it forms a covalent network solid

Explanation:
1. Z = 10 is neon, a noble gas existing as single atoms with a completed octet.
2. It forms no covalent network solid, so statement (3) is false.
3. Answer: (c)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 4: The following are configurations of 4 atoms: P = [Ne]3s23p3; Q = [Ar]3d104s24p3; R = [He]2s22p5; S = [Ne]3s1. The incorrect statement is:
(a) EA: Q > R
(b) EN: R > P
(c) IP: P > S
(d) atomic radius: S > R
Show Answer & Explanation

Answer: (a) EA: Q > R

Explanation:
1. The atoms are P, As, F and Na.
2. Electron affinity of F (R) is much more negative than that of As (Q), so 'Q > R' is incorrect. The other statements hold.
3. Answer: (a)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 5: Among P, S, Cl and F, the elements with the most and the least negative ΔegH will be respectively:
(a) Cl, P
(b) P, Cl
(c) Cl, F
(d) F, Cl
Show Answer & Explanation

Answer: (a) Cl, P

Explanation:
1. ΔegH: Cl (−349) is the most negative; F (−328) is lower because of its small size.
2. P (−72) is the least negative of the four.
3. Answer: (a)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 6: Which of the following is valid for the order Li < B < Be < C < O < N < F < Ne?
(a) ionisation enthalpy
(b) electron gain enthalpy
(c) electronegativity
(d) size
Show Answer & Explanation

Answer: (a) ionisation enthalpy

Explanation:
1. First ionisation enthalpies (kJ/mol): Li 520, B 801, Be 899, C 1086, O 1314, N 1402, F 1681, Ne 2081.
2. The order follows this sequence, with dips at B and O.
3. Answer: (a)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 7: Match Column-I (elements) with Column-II (periodic properties). (A) F, (B) Cl, (C) Fe, (D) He. (P) maximum ionisation energy, (Q) maximum electronegativity, (R) maximum electron affinity, (S) variable oxidation state.
(a) (A)→(P), (B)→(Q), (C)→(R), (D)→(S)
(b) (A)→(R), (B)→(Q), (C)→(P), (D)→(S)
(c) (A)→(Q), (B)→(R), (C)→(S), (D)→(P)
(d) (A)→(Q), (B)→(P), (C)→(S), (D)→(R)
Show Answer & Explanation

Answer: (c) (A)→(Q), (B)→(R), (C)→(S), (D)→(P)

Explanation:
1. Fluorine has the highest electronegativity; chlorine the highest electron gain enthalpy (more negative than F).
2. Iron shows variable oxidation states and helium has the highest ionisation energy.
3. Answer: (c)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 8: The highest fluoroanion of boron and aluminium is:
(a) BF63−, AlF63−
(b) BF4−, AlF6−
(c) BF4−, AlF63−
(d) BF4−, AlF4−
Show Answer & Explanation

Answer: (c) BF4−, AlF63−

Explanation:
1. Boron has no d-orbitals and a small size, so its maximum covalency is 4 (BF4−).
2. Aluminium can expand to 6 (AlF63−).
3. Answer: (c)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 9: Which of the following orders is wrong?
(a) NH3 < PH3 < AsH3 (acidic)
(b) Li < Be < B < C (first IP)
(c) Al2O3 < MgO < Na2O < K2O (basic)
(d) Li+ < Na+ < K+ < Cs+ (ionic radius)
Show Answer & Explanation

Answer: (b) Li < Be < B < C (first IP)

Explanation:
1. First IP: Li 520 < B 801 < Be 899 < C 1086, so Be > B and the stated order is wrong.
2. The other three orders are correct.
3. Answer: (b)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 10: The electronic configuration of 90Th is:
(a) (n−2)f2, (n−1)d1 ns1
(b) (n−2)f1, (n−1)d1 ns2
(c) (n−2)f0, (n−1)d2 ns2
(d) (n−2)f2, (n−1)d0 ns2
Show Answer & Explanation

Answer: (c) (n−2)f0, (n−1)d2 ns2

Explanation:
1. Thorium is [Rn]6d27s2.
2. So (n−2)f is empty, (n−1)d has 2 electrons and ns has 2.
3. Answer: (c)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 11: Which property decreases from left to right across the periodic table and increases from top to bottom? (i) atomic radius, (ii) electronegativity, (iii) ionisation energy, (iv) metallic character
(a) (i) only
(b) (i), (ii) and (iii)
(c) (i), (iii) and (iv)
(d) (i) and (iv)
Show Answer & Explanation

Answer: (d) (i) and (iv)

Explanation:
1. Atomic radius and metallic character both fall across a period and rise down a group.
2. Electronegativity and ionisation energy do the opposite.
3. Answer: (d)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 12: Which of the following statements is not correct?
(a) the electron affinity of Si is greater than that of C
(b) BeO is amphoteric while B2O3 is acidic
(c) the ionisation energy of Tl is less than that of Al
(d) the ionisation energy of elements of the Cu group is less than that of the respective elements of the Zn group
Show Answer & Explanation

Answer: (c) the ionisation energy of Tl is less than that of Al

Explanation:
1. Tl (589 kJ/mol) has a higher first IE than Al (577) because of poor shielding by 4f and 5d electrons.
2. So 'Tl less than Al' is incorrect.
3. Answer: (c)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 13: Which order for atomic radii is incorrect?
(a) H− > Li+ ≥ Mg2+ > Al3+
(b) MnO2 > KMnO4 (radius of Mn)
(c) O2− > F− > Na+ > Mg2+
(d) B > Al ≃ Ga < In ≃ Tl
Show Answer & Explanation

Answer: (d) B > Al ≃ Ga < In ≃ Tl

Explanation:
1. B (85 pm) is much smaller than Al (143 pm), so 'B > Al' is wrong.
2. The other orders follow isoelectronic and oxidation-state trends.
3. Answer: (d)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 14: Which of the following is not incorrect for acidic strength?
(a) H2S < H2Se < H2Te
(b) HClO4 > HClO3 > HClO2 > HClO
(c) P4O10 > SiO2
(d) all are correct
Show Answer & Explanation

Answer: (d) all are correct

Explanation:
1. Acidity of hydrides increases down a group, oxoacid strength rises with the oxidation state of Cl, and P4O10 is more acidic than SiO2.
2. All three statements are correct.
3. Answer: (d)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 15: Which is correct for the magnitude of ΔegH?
(a) Cl > F > Br > I > S
(b) O− > O > O+
(c) S− > O−
(d) Cl > F > S > O > N > P
Show Answer & Explanation

Answer: (a) Cl > F > Br > I > S

Explanation:
1. Magnitudes (kJ/mol): Cl 349, F 328, Br 325, I 295, S 200.
2. The other orders are wrong (e.g. O− repels an added electron, and N and P order is reversed).
3. Answer: (a)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 16: Match Column I with Column II. (a) N2O, (b) Na2O, (c) Ga2O3, (d) C3O2, (e) V3O4 with (P) normal oxide, (Q) neutral oxide, (R) sub oxide, (S) basic oxide, (T) amphoteric oxide, (U) mixed oxide.
(a) (a)-Q, (b)-P,S, (c)-R, (d)-T, (e)-U
(b) (a)-Q, (b)-P,S, (c)-T, (d)-R, (e)-U
(c) (a)-R,P, (b)-Q,S, (c)-U, (d)-T, (e)-R
(d) (a)-P,Q, (b)-R,S, (c)-T, (d)-U, (e)-R
Show Answer & Explanation

Answer: (b) (a)-Q, (b)-P,S, (c)-T, (d)-R, (e)-U

Explanation:
1. N2O is neutral; Na2O is a normal, basic oxide.
2. Ga2O3 is amphoteric, C3O2 a suboxide and V3O4 a mixed oxide.
3. Answer: (b)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 17: Which has the maximum number of unpaired electrons?
(a) Na
(b) Mn
(c) Cr
(d) Fe2+
Show Answer & Explanation

Answer: (c) Cr

Explanation:
1. Unpaired electrons: Na 1, Mn (3d54s2) 5, Cr (3d54s1) 6, Fe2+ (3d6) 4.
2. Cr has the most.
3. Answer: (c)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 18: Among the following statements, identify the one that is NOT correct.
(a) oxygen has the least electron affinity in its group
(b) iodine has the largest electron affinity among the halogens
(c) S− has a higher ionisation energy than O−
(d) none of these
Show Answer & Explanation

Answer: (b) iodine has the largest electron affinity among the halogens

Explanation:
1. Chlorine has the largest electron gain enthalpy among the halogens, not iodine (349 for Cl against 295 for I).
2. The other statements are correct.
3. Answer: (b)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 19: Which among the following statements is wrong?
(a) the electronic configuration of Gd (64) is 4f75d16s2
(b) Ce4+ is a good reductant
(c) actinoids exhibit higher oxidation states than lanthanoids
(d) actinoid contraction is greater from element to element than lanthanoid contraction
Show Answer & Explanation

Answer: (b) Ce4+ is a good reductant

Explanation:
1. Ce4+ readily gains an electron to become Ce3+, so it is a good oxidant, not a reductant.
2. The other statements are true.
3. Answer: (b)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 20: Which of the following orders of ionisation potential is incorrect?
(a) Na+ > Mg+
(b) Mg2+ > Mg+
(c) He > Li+
(d) Be > B
Show Answer & Explanation

Answer: (c) He > Li+

Explanation:
1. IP of Li+ (7298 kJ/mol) is much higher than that of He (2372) because Li+ has a higher nuclear charge on the same 1s2 core.
2. So 'He > Li+' is incorrect.
3. Answer: (c)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 21: Consider the IE (eV) values of elements W and X. W: 10.5, 15.5, 24.9, 79.8; X: 8, 14.8, 78.9, 105.8 (IE1 to IE4). Other elements Y and Z have outer configurations ns2np4 and ns2np5. Which of the compounds is/are not possible? (a) W2Y3 (b) X2Y3 (c) WZ2 (d) XZ2
(a) a, b
(b) b, c
(c) c, d
(d) a, d
Show Answer & Explanation

Answer: (b) b, c

Explanation:
1. W has a big jump after IE3 (valency 3) and X after IE2 (valency 2); Y is −2 and Z is −1.
2. W2Y3 and XZ2 are possible, while X2Y3 and WZ2 are not.
3. Answer: (b)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 22: Which represents the electronic configuration of the most electropositive element?
(a) [He]2s1
(b) [Xe]6s1
(c) [He]2s2
(d) [Xe]6s2
Show Answer & Explanation

Answer: (b) [Xe]6s1

Explanation:
1. Electropositive character rises down a group and is highest for alkali metals.
2. [Xe]6s1 is caesium.
3. Answer: (b)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 23: The IP and EA of F are 17.42 and 3.45 eV/atom respectively. The electronegativity of F on the Pauling scale (Mulliken value divided by 2.8) will be:
(a) 2.7
(b) 3.7
(c) 4.07
(d) 6.7
Show Answer & Explanation

Answer: (b) 3.7

Explanation:
1. Mulliken EN = (IP + EA)/2 = (17.42 + 3.45)/2 = 10.4 eV.
2. Pauling EN = 10.4/2.8 = 3.7.
3. Answer: (b)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 24: A = 1s22s22p4, B = 1s22s22p5, C = 1s22s22p6 (one of A/B/C is a neutral atom). A, B and C are atoms or ions of the same element. Then:
(a) B(g) + e− → C(g) is exothermic
(b) A(g) → A+(g) + e− is exothermic
(c) C(g) → A(g) + 2e is exothermic
(d) B(g) → B+(g) + e− is exothermic
Show Answer & Explanation

Answer: (a) B(g) + e− → C(g) is exothermic

Explanation:
1. The element is fluorine: A = F+, B = F (neutral), C = F−.
2. B + e− → C is the electron gain of F, which is exothermic; removal of electrons (the other options) needs energy.
3. Answer: (a)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 25: The correct order of increasing atomic size is:
(a) N < F < Si < P
(b) F > N > P > Si
(c) F < N < P < Si
(d) F < N < Si < P
Show Answer & Explanation

Answer: (c) F < N < P < Si

Explanation:
1. Atomic radii (pm): F 64, N 75, P 110, Si 118.
2. So F < N < P < Si.
3. Answer: (c)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 26: The correct order of second IP is:
(a) Mg > Na > Si > Al
(b) Mg < Si < Al < Na
(c) Na < Mg < Al < Si
(d) Na > Mg > Al > Si
Show Answer & Explanation

Answer: (b) Mg < Si < Al < Na

Explanation:
1. IE2 (kJ/mol): Mg 1451, Si 1577, Al 1817, Na 4562.
2. So Mg < Si < Al < Na.
3. Answer: (b)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 27: The required energy will be maximum for the process:
(a) Ba → Ba2+
(b) Be → Be2+
(c) Cs → Cs+
(d) Li → Li+
Show Answer & Explanation

Answer: (b) Be → Be2+

Explanation:
1. Total energies (IE1 + IE2 where needed, kJ/mol): Be 899 + 1757 = 2656; Ba 503 + 965 = 1468; Li 520; Cs 376.
2. Be → Be2+ needs the most.
3. Answer: (b)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 28: Which of the following molecules has the highest electronegativity of Xe?
(a) XeF2
(b) XeF4
(c) XeO3
(d) none of these
Show Answer & Explanation

Answer: (c) XeO3

Explanation:
1. Electronegativity rises with the oxidation state of the element.
2. Xe is +6 in XeO3, higher than +2 and +4.
3. Answer: (c)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 29: The compound X–O–H is likely to act as a base compared with hydrogen if X has:
(a) higher ionisation potential
(b) lower electronegativity
(c) higher electronegativity
(d) lower atomic radius
Show Answer & Explanation

Answer: (b) lower electronegativity

Explanation:
1. If X is less electronegative than H, the O–X bond breaks heterolytically, giving OH−.
2. So the compound behaves as a base.
3. Answer: (b)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 30: Among the following hydroxides, which is most basic?
(a) Lu(OH)3
(b) La(OH)3
(c) Pm(OH)3
(d) Yb(OH)3
Show Answer & Explanation

Answer: (b) La(OH)3

Explanation:
1. Basic strength of lanthanoid hydroxides falls with lanthanoid contraction as the cation shrinks.
2. La(OH)3 has the largest cation and is the most basic.
3. Answer: (b)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 31: Amongst the following elements whose electronic configurations are given, the one having the highest ionisation enthalpy is:
(a) [Ne]3s23p1
(b) [Ne]3s23p3
(c) [Ne]3s23p2
(d) [Ar]3d104s24p3
Show Answer & Explanation

Answer: (b) [Ne]3s23p3

Explanation:
1. The elements are Al, P, Si and As.
2. P (1012 kJ/mol) has a half-filled 3p subshell, which gives the highest IE.
3. Answer: (b)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 32: The block of the element A, when its electronic configuration is [Rn]6d27s2, along with its group and period is:
(a) d-block, IV B, 7th
(b) d-block, II B, 7th
(c) f-block, II B, 6th
(d) f-block, III B, 7th
Show Answer & Explanation

Answer: (d) f-block, III B, 7th

Explanation:
1. [Rn]6d27s2 is thorium (Z = 90), an actinoid.
2. It is placed in the f-block, group III B, 7th period.
3. Answer: (d)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 33: Match Column-I (type of elements) with Column-II (outer electronic configuration). (A) inert gas elements, (B) representative elements, (C) transition elements, (D) inner transition elements. (i) ns1–2np0–5, (ii) 1s2 and ns2np6, (iii) (n−2)f1–14(n−1)d0–1ns2, (iv) (n−1)d1–10ns1 or 2.
(a) A–i, B–ii, C–iii, D–iv
(b) A–ii, B–i, C–iii, D–iv
(c) A–ii, B–i, C–iv, D–iii
(d) A–ii, B–i, C–iv, D–iii (with A–i)
Show Answer & Explanation

Answer: (c) A–ii, B–i, C–iv, D–iii

Explanation:
1. Noble gases are 1s2 or ns2np6 and representative elements are ns1–2np0–5.
2. Transition elements fill (n−1)d and inner transition elements fill (n−2)f.
3. Answer: (c)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 34: The correct order of spin magnetic moment of trivalent lanthanoid ions is:
(a) Gd3+ < Ce3+ < Lu3+
(b) Lu3+ < Ho3+ < Gd3+
(c) Gd3+ < Yb3+ < Tb3+
(d) Pr3+ > Sm3+ > Eu3+
Show Answer & Explanation

Answer: (b) Lu3+ < Ho3+ < Gd3+

Explanation:
1. Unpaired 4f electrons: Lu3+ 0, Ho3+ 4, Gd3+ 7.
2. Magnetic moment rises with unpaired electrons: Lu3+ < Ho3+ < Gd3+.
3. Answer: (b)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 35: Incorrect order of radius is:
(a) H− > Li+ > Mg2+
(b) P3− > S2− > K+
(c) Br− > S2− > Cl− > F−
(d) Ni > Cu > Zn
Show Answer & Explanation

Answer: (d) Ni > Cu > Zn

Explanation:
1. Metallic radii (pm): Zn 134, Cu 128, Ni 124.
2. So Zn > Cu > Ni, and the stated order is incorrect.
3. Answer: (d)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 36: Incorrect order of ionic radius is:
(a) La3+ > Gd3+ > Eu3+ > Lu3+
(b) Na+ > Li+ > Mg2+ > Al3+ > Be2+
(c) In+ > Sn2+ > Sb3+
(d) K+ > Sc3+ > V5+ > Mn7+
Show Answer & Explanation

Answer: (a) La3+ > Gd3+ > Eu3+ > Lu3+

Explanation:
1. Across the lanthanoids radius falls with atomic number, so Eu3+ (95 pm) > Gd3+ (94 pm).
2. The stated Gd3+ > Eu3+ is incorrect.
3. Answer: (a)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 37: Which of the following orders of I.P. is incorrect?
(a) Al > Mg > Be > B
(b) S < P < O < N
(c) Sc > Y > La
(d) Ni < Pd < Pt
Show Answer & Explanation

Answer: (a) Al > Mg > Be > B

Explanation:
1. IE of Al (577) is lower than that of Mg (738), so Al > Mg is wrong.
2. The other three orders are correct.
3. Answer: (a)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 38: The correct trend of first ionisation energy in group 13 is:
(a) B > Al > Ga > In > Tl
(b) B > Al > Ga > Tl > In
(c) B > Tl > Ga > Al > In
(d) B > Ga > Al > In > Tl
Show Answer & Explanation

Answer: (c) B > Tl > Ga > Al > In

Explanation:
1. IE (kJ/mol): B 801, Tl 589, Ga 579, Al 577, In 558.
2. Poor shielding by d- and f-electrons raises Ga and Tl above Al and In.
3. Answer: (c)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 39: Which of the following electron affinity orders is not correct?
(a) B < Al < C < Si
(b) Mg < C < S < F
(c) O < F < S < Cl
(d) N < C < Si < S
Show Answer & Explanation

Answer: (c) O < F < S < Cl

Explanation:
1. Electron gain magnitudes: F (328) is smaller than Cl (349) but larger than S (200).
2. So the order O < F < S < Cl is incorrect, while the others hold.
3. Answer: (c)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 40: Considering the elements B, C, N, F and Si, the correct order of their non-metallic character is:
(a) B > C > Si > N > F
(b) Si > C > B > N > F
(c) F > N > C > B > Si
(d) F > N > C > Si > B
Show Answer & Explanation

Answer: (c) F > N > C > B > Si

Explanation:
1. Non-metallic character increases across a period (B < C < N < F) and decreases down a group (Si < C).
2. Boron is more non-metallic than silicon, so F > N > C > B > Si.
3. Answer: (c)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 41: Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidising property is:
(a) F > Cl > O > N
(b) F > O > Cl > N
(c) Cl > F > O > N
(d) O > F > N > Cl
Show Answer & Explanation

Answer: (b) F > O > Cl > N

Explanation:
1. Oxidising power follows electronegativity and ease of gaining electrons.
2. F (4.0) > O (3.5) > Cl (3.0) > N (3.0, with a stable bond).
3. Answer: (b)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 42: Match the electronic configuration with the electron gain enthalpy (kJ mol−1): (i) 1s22s22p6, (ii) 1s22s22p63s1, (iii) 1s22s22p5, (iv) 1s22s22p4 with (A) −53, (B) −328, (C) −141, (D) +48.
(a) i→D, ii→A, iii→B, iv→C
(b) i→A, ii→B, iii→C, iv→D
(c) i→D, ii→B, iii→A, iv→C
(d) i→D, ii→C, iii→A, iv→B
Show Answer & Explanation

Answer: (a) i→D, ii→A, iii→B, iv→C

Explanation:
1. (i) is neon, a noble gas with positive EA (+48); (ii) is sodium (−53).
2. (iii) is fluorine (−328) and (iv) is oxygen (−141).
3. Answer: (a)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 43: In which of the following options does the order of arrangement not agree with the variation of the property indicated against it?
(a) Al3+ < Mg2+ < Na+ < F− (increasing ionic size)
(b) B < C < N < O (increasing first ionisation enthalpy)
(c) I < Br < F < Cl (increasing electron gain enthalpy)
(d) Li < Na < K < Rb (increasing metallic radius)
Show Answer & Explanation

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

Explanation:
1. First IE: N (1402) is greater than O (1314) because of the half-filled 2p3.
2. So B < C < N < O is not correct.
3. Answer: (b)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 44: The process requiring the absorption of energy is:
(a) F → F−
(b) H → H−
(c) Cl → Cl−
(d) O → O2−
Show Answer & Explanation

Answer: (d) O → O2−

Explanation:
1. The first three are exothermic electron gains.
2. The second electron gain of oxygen (O− → O2−) is endothermic, so the overall process absorbs energy.
3. Answer: (d)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 45: Select the correct order of ionic radii:
(a) Ti2+ > Ti3+ > Ti4+
(b) Ti4+ > Ti2+
(c) Ti3+ > Ti2+ > Ti4+
(d) Ti4+ > Ti3+ > Ti2+
Show Answer & Explanation

Answer: (a) Ti2+ > Ti3+ > Ti4+

Explanation:
1. For the same element, radius falls as positive charge rises.
2. Ti2+ > Ti3+ > Ti4+.
3. Answer: (a)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 46: Europium belongs to:
(a) s-block
(b) p-block
(c) d-block
(d) f-block
Show Answer & Explanation

Answer: (d) f-block

Explanation:
1. Europium (Z = 63) is a lanthanoid, filling the 4f subshell.
2. Answer: (d)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 47: The ionic radius of Cr is minimum in which of the following compounds?
(a) CrF3
(b) CrCl3
(c) Cr2O3
(d) K2CrO4
Show Answer & Explanation

Answer: (d) K2CrO4

Explanation:
1. Radius falls as the oxidation state rises.
2. Cr is +3 in the first three and +6 in K2CrO4, which has the smallest radius.
3. Answer: (d)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

Question 48: Which of the following is/are Döbereiner triads? (i) P, As, Sb; (ii) Cu, Ag, Au; (iii) Fe, Co, Ni; (iv) S, Se, Te
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iv)
(d) all
Show Answer & Explanation

Answer: (c) (i) and (iv)

Explanation:
1. In a Döbereiner triad the atomic mass of the middle element is about the mean of the other two.
2. This holds for P, As, Sb (31, 75, 122) and S, Se, Te (32, 79, 128) but not for Cu, Ag, Au or Fe, Co, Ni.
3. Answer: (c)

Teacher's Note:
1. Recall the periodic trend and its exceptions before choosing. Check each option against a known example.

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