JEE Chemistry The D and F Block Elements MCQs Set 08

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Multiple Choice Questions

Question 1: The highly unstable species is
(a) Sc(II)
(b) Ti(II)
(c) Co(II)
(d) V(II)
Show Answer & Explanation

Answer: (a) Sc(II)

Explanation:
1. Sc(II) would be 3d1 4s0 and loses its electron to give the stable noble gas Sc3+.

Teacher's Note:
1. Sc is always +3.

Question 2: Which of the following is diamagnetic?
(a) Aurum
(b) Tungsten
(c) Palladium
(d) Niobium
Show Answer & Explanation

Answer: (c) Palladium

Explanation:
1. Palladium has the configuration 4d10 5s0.

Teacher's Note:
1. Pd: all electrons paired.

Question 3: Which statement is correct when SO2 is passed through acidified K2Cr2O7 solution?
(a) The solution turns blue
(b) The solution is decolourised
(c) SO2 is reduced
(d) Green Cr2(SO4)3 is formed
Show Answer & Explanation

Answer: (d) Green Cr2(SO4)3 is formed

Explanation:
1. SO2 reduces dichromate to green Cr3+.

Teacher's Note:
1. Orange to green.

Question 4: The electronic configurations of Eu (63), Gd (64) and Tb (65) are
(a) [Xe]4f6 6s2, [Xe]4f8 6s2 and [Xe]4f8 5d1 6s2
(b) [Xe]4f7 5d1 6s2, [Xe]4f7 5d1 6s2 and [Xe]4f9 6s2
(c) [Xe]4f6 5d1 6s2, [Xe]4f7 5d1 6s2 and [Xe]4f8 5d1 6s2
(d) [Xe]4f7 6s2, [Xe]4f7 5d1 6s2 and [Xe]4f9 6s2
Show Answer & Explanation

Answer: (d) [Xe]4f7 6s2, [Xe]4f7 5d1 6s2 and [Xe]4f9 6s2

Explanation:
1. Half-filled 4f7 occurs for Eu (no 5d electron) and Gd (one 5d electron); Tb is 4f9 6s2.

Teacher's Note:
1. Half-filled shells are favoured.

Question 5: Which of the following is ferromagnetic?
(a) U
(b) Co
(c) Cr
(d) Cu
Show Answer & Explanation

Answer: (b) Co

Explanation:
1. Iron, cobalt and nickel are the ferromagnetic metals.

Teacher's Note:
1. Fe, Co, Ni.

Question 6: Which can convert Cl- into Cl2?
(a) KMnO4
(b) H2SO3
(c) Fe2(SO4)3
(d) HIO3
Show Answer & Explanation

Answer: (a) KMnO4

Explanation:
1. Permanganate in acid oxidises chloride.

Teacher's Note:
1. Strong oxidant needed.

Question 7: Which lanthanoid ion shows yellow colour?
(a) La3+
(b) Yb2+
(c) Lu3+
(d) Ce4+
Show Answer & Explanation

Answer: (d) Ce4+

Explanation:
1. Ce4+ (f0) is orange-yellow due to charge-transfer.

Teacher's Note:
1. Charge transfer colour.

Question 8: Which gas readily decolourises acidified KMnO4 solution?
(a) SO2
(b) NO2
(c) P2O5
(d) CO2
Show Answer & Explanation

Answer: (a) SO2

Explanation:
1. SO2 is a reducing gas.

Teacher's Note:
1. SO2 reduces permanganate.

Question 9: HgCl2 and I2 both dissolve in water containing I- ions. The pair of species formed is
(a) HgI2, I-
(b) HgI4 2-, I3-
(c) Hg2I2, I-
(d) HgI2, I3-
Show Answer & Explanation

Answer: (b) HgI4 2-, I3-

Explanation:
1. Iodide gives the complex tetraiodomercurate(II), and iodine forms triiodide.

Teacher's Note:
1. Excess iodide.

Question 10: The reason for the greater range of oxidation states in actinoids is attributed to
(a) Actinoid contraction
(b) 5f, 6d and 7s levels having comparable energies
(c) 4f and 5d levels being close in energy
(d) The radioactive nature of actinoids
Show Answer & Explanation

Answer: (b) 5f, 6d and 7s levels having comparable energies

Explanation:
1. All three levels are close in energy, so electrons from each take part.

Teacher's Note:
1. Comparable energies.

Question 11: Gadolinium is
(a) Ferromagnetic
(b) Anti ferromagnetic
(c) Ferrimagnetic
(d) Diamagnetic
Show Answer & Explanation

Answer: (a) Ferromagnetic

Explanation:
1. Gd is ferromagnetic below about 293 K.

Teacher's Note:
1. Seven unpaired f electrons.

Question 12: Which pair contains all diamagnetic ions?
(a) Yb2+, Lu3+
(b) Ce4+, Pr4+
(c) Dy3+, Lu3+
(d) Lu3+, Gd3+
Show Answer & Explanation

Answer: (a) Yb2+, Lu3+

Explanation:
1. Yb2+ is 4f14 and Lu3+ is 4f14.

Teacher's Note:
1. Fully filled f shells.

Question 13: Which statement about interstitial compounds is incorrect?
(a) They have high melting points in comparison to pure metals
(b) They are very hard
(c) They retain metallic conductivity
(d) They are chemically very reactive
Show Answer & Explanation

Answer: (d) They are chemically very reactive

Explanation:
1. Interstitial compounds are chemically inert.

Teacher's Note:
1. Hard, inert, high melting.

Question 14: Which pair does not decolourise acidified KMnO4?
(a) KI and NaNO2
(b) NaNO3 and Na2C2O4
(c) Na2SO4 and Na2C2O4
(d) I2 and NaNO3
Show Answer & Explanation

Answer: (d) I2 and NaNO3

Explanation:
1. Iodine and nitrate are not oxidisable by permanganate.

Teacher's Note:
1. Reducing agents decolourise permanganate.

Question 15: Which ion exhibits d-d transition and is paramagnetic as well?
(a) CrO4 2-
(b) Cr2O7 2-
(c) MnO4-
(d) MnO4 2-
Show Answer & Explanation

Answer: (d) MnO4 2-

Explanation:
1. MnO4 2- is d1, paramagnetic with one d electron available for d-d transition.

Teacher's Note:
1. The other three are d0.

Question 16: Match the metal ions in Column I with the spin-only magnetic moments in Column II: a. Co3+ b. Cr3+ c. Fe3+ d. Ni2+; i. sqrt8, ii. sqrt35, iii. sqrt3, iv. sqrt24, v. sqrt15 B.M.
(a) a-iv, b-v, c-ii, d-i
(b) a-i, b-ii, c-iii, d-iv
(c) a-iv, b-i, c-ii, d-iii
(d) a-iii, b-v, c-i, d-ii
Show Answer & Explanation

Answer: (a) a-iv, b-v, c-ii, d-i

Explanation:
1. Co3+ d6 gives 4 unpaired (sqrt24), Cr3+ d3 gives 3 (sqrt15), Fe3+ d5 gives 5 (sqrt35), Ni2+ d8 gives 2 (sqrt8).

Teacher's Note:
1. Spin-only formula.

Question 17: Which pair of elements has the least difference in E0 (M2+/M) values?
(a) V and Mn
(b) Mn and Fe
(c) Fe and Co
(d) Co and Ni
Show Answer & Explanation

Answer: (a) V and Mn

Explanation:
1. V (-1.18 V) and Mn (-1.18 V) have almost identical values.

Teacher's Note:
1. Values are scattered across the series.

Question 18: The violet colour of glass is due to
(a) K(I)
(b) Mn(IV)
(c) I2
(d) Cr(III)
Show Answer & Explanation

Answer: (b) Mn(IV)

Explanation:
1. Mn(IV) in the form of MnO2 gives violet glass; with decolouriser it is Mn(III).

Teacher's Note:
1. Glass makers use MnO2.

Question 19: In which compound does the oxidation state not change in acidic and basic media?
(a) MnO4-
(b) CrO4 2-
(c) FeO4 2-
(d) MnO4 2-
Show Answer & Explanation

Answer: (b) CrO4 2-

Explanation:
1. Chromate interconverts with dichromate by pH alone, with no redox change.

Teacher's Note:
1. Cr(VI) stays Cr(VI).

Question 20: The spin-only magnetic moment of a divalent metal ion with 23 electrons is
(a) 5.9
(b) 5.2
(c) 4.0
(d) 5.0
Show Answer & Explanation

Answer: (a) 5.9

Explanation:
1. M2+ with 23 electrons is Mn2+ (3d5) with five unpaired electrons, giving 5.92 B.M.

Teacher's Note:
1. n(n+2) with n = 5.

Question 21: In salts of Co, Sb, Sn and Cd, H2S was added. The colours of the precipitates (black, yellow, orange, brown from A to D) are correct for
(a) A-Co
(b) B-Sb
(c) C-Sn
(d) D-Cd
Show Answer & Explanation

Answer: (a) A-Co

Explanation:
1. CoS is black, CdS is yellow, Sb2S3 is orange and SnS is brown.

Teacher's Note:
1. Learn the sulphide colours.

Question 22: The colour of the compound formed when KMnO4 is fused with KOH is
(a) Green
(b) Black
(c) Brown
(d) Pale pink
Show Answer & Explanation

Answer: (a) Green

Explanation:
1. The product is green potassium manganate, K2MnO4.

Teacher's Note:
1. Mn(VI) is green.

Question 23: In which conversion is blue colour observed?
(a) MnO4- + SO2
(b) Cr2O7 2- + KI
(c) K2Cr2O7 + NaCl + conc. H2SO4
(d) CuSO4 + KI + starch
Show Answer & Explanation

Answer: (d) CuSO4 + KI + starch

Explanation:
1. The liberated iodine gives the blue starch-iodine complex.

Teacher's Note:
1. Iodine plus starch.

Question 24: Which pair is soluble in excess NH4OH?
(a) Zn(OH)2, Pb(OH)2
(b) Zn(OH)2, Sn(OH)2
(c) Cd(OH)2, Zn(OH)2
(d) Be(OH)2, Cd(OH)2
Show Answer & Explanation

Answer: (c) Cd(OH)2, Zn(OH)2

Explanation:
1. Zn(OH)2 and Cd(OH)2 form ammine complexes.

Teacher's Note:
1. Zn2+ and Cd2+ form complexes.

Question 25: The lanthanide contraction is caused by
(a) The same effective nuclear charge from Ce to Lu
(b) The imperfect shielding on outer electrons by 4f electrons from the nuclear charge
(c) The appreciable shielding on outer electrons by 4f electrons
(d) The appreciable shielding on outer electrons by 5d electrons
Show Answer & Explanation

Answer: (b) The imperfect shielding on outer electrons by 4f electrons from the nuclear charge

Explanation:
1. 4f electrons shield poorly, so the effective nuclear charge rises.

Teacher's Note:
1. Poor shielding.

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