Chemistry Objective Questions and Answers: Unit 04 The d- and f-Block Elements
Review structured MCQ sets for Class 12 Chemistry Unit 04 The d- and f-Block Elements. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.
Download Unit 04 The d- and f-Block Elements MCQs with Answers
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Multiple Choice Questions
(a) (n − 1)d1−10 ns0−2
(b) nd10 ns2
(c) (n − 1)d10 ns2
(d) (n − 1)d1−5 ns2
Show Answer & Explanation
Answer: (a) (n − 1)d1−10 ns0−2
Explanation:
1. General electronic configuration: (n − 1)d1–10 ns0–2
Teacher's Note:
1. The ns0 case covers Pd (4d10 5s0); the d subshell is always the (n − 1) shell.
I. La(OH)3 is least basic among hydroxides of lanthanides.
II. Zr4+ and Hf4+ possess almost the same ionic radii.
III. Ce4+ can act as an oxidizing agent.
Which of the above is/are true?
(a) I and III
(b) II and III
(c) II only
(d) I and II
Show Answer & Explanation
Answer: (b) II and III
Explanation:
1. La(OH)3 is most basic. Hence, (I) is wrong.
2. (II) is correct due to lanthanoid contraction.
3. (III) is correct because Ce4+ tends to change to stable Ce3+.
Teacher's Note:
1. Lu(OH)3 is the least basic lanthanoid hydroxide; La(OH)3 is the most basic.
(i) [Fe(CN)6]4− (ii) [Fe(CN)6]3−
(iii) [Cr(NH3)6]3+ (iv) [Ni(H2O)4]2+
(a) (i) < (ii) < (iii) < (iv)
(b) (i) < (ii) < (iv) < (iii)
(c) (ii) < (iii) < (i) < (iv)
(d) (iii) < (i) < (ii) < (iv)
Show Answer & Explanation
Answer: (b) (i) < (ii) < (iv) < (iii)
Explanation:
1. [Fe(CN)6]4−: No. of unpaired electrons = 0
2. [Fe(CN)6]3−: No. of unpaired electrons = 1
3. [Ni(H2O)4]2+: No. of unpaired electrons = 2
4. [Cr(NH3)6]3+: No. of unpaired electrons = 3
Teacher's Note:
1. CN− is a strong field ligand and pairs up the 3d electrons, which is why both cyano complexes have the fewest unpaired electrons.
(a) Fe3+ has smaller size than Fe2+
(b) Fe3+ has 3d5 configuration (half-filled)
(c) Fe3+ has higher oxidation state
(d) Fe3+ is paramagnetic in nature.
Show Answer & Explanation
Answer: (b) Fe3+ has 3d5 configuration (half-filled)
Explanation:
1. 3d5 configuration is more stable due to singly occupied half-filled orbitals.
Teacher's Note:
1. Exactly half-filled and completely filled subshells have extra exchange energy, which gives the extra stability.
VO2+ < Cr2O72− < MnO4−
The reason for this increasing order of oxidising power is
(a) increasing stability of the lower species to which they are reduced
(b) increasing stability of the higher species to which they are oxidised
(c) increasing stability of the higher species to which they are reduced
(d) increasing stability of the lower species to which they are oxidised.
Show Answer & Explanation
Answer: (a) increasing stability of the lower species to which they are reduced
Explanation:
1. The increasing order of oxidising power is due to increasing stability of the lower species to which they are reduced.
Teacher's Note:
1. An oxidising agent is itself reduced, so options that talk of the species being “oxidised” can be ruled out at once.
(a) Cu2+
(b) Ni2+
(c) Co2+
(d) Fe2+
Show Answer & Explanation
Answer: (d) Fe2+
Explanation:
1. More the number of unpaired d-electrons, more is the magnetic moment.
2. Cu2+ – 3d9; No. of unpaired electrons = 1
3. Ni2+ – 3d8; No. of unpaired electrons = 2
4. Co2+ – 3d7; No. of unpaired electrons = 3
5. Fe2+ – 3d6; No. of unpaired electrons = 4
Teacher's Note:
1. Among M2+ ions of the 3d series, Mn2+ (d5) has the maximum; on either side of it the number of unpaired electrons falls.
(i) Mn2O7 (ii) CrO3 (iii) Cr2O3 (iv) CrO (v) V2O5 (vi) V2O4
(a) (i) and (iv)
(b) (ii) and (iii)
(c) (iii) and (v)
(d) (ii) and (vi)
Show Answer & Explanation
Answer: (c) (iii) and (v)
Explanation:
1. Cr2O3 and V2O5 are amphoteric oxides.
Teacher's Note:
1. Mn2O7 and CrO3 are acidic, CrO is basic; Cr2O3 and V2O5 are the standard amphoteric examples.
(a) Gd
(b) Hs
(c) Es
(d) Cs
Show Answer & Explanation
Answer: (b) Hs
Explanation:
1. Hs (Z = 108) belongs to 6d series with electronic configuration – [Rn]5f14 6d6 7s2.
Teacher's Note:
1. Gd is a lanthanoid, Es an actinoid (both f-block) and Cs an s-block element.
(a) Lanthanum
(b) Nobelium
(c) Thorium
(d) Cerium
Show Answer & Explanation
Answer: (d) Cerium
Explanation:
1. Ce is most commonly used lanthanoid, nobelium (No) and Th (thorium) are actinoids.
Teacher's Note:
1. Cerium is the most abundant lanthanoid and is the main metal in misch metal used in lighter flints.
(a) Cr
(b) Cu
(c) Ni
(d) All of these
Show Answer & Explanation
Answer: (a) Cr
Explanation:
1. Cr forms CrF3 whereas Cu and Ni do not form CuF3 and NiF3.
Teacher's Note:
1. +3 is a common, stable state for Cr (3d3); for Cu and Ni the +2 state is the stable one.
(a) belong to d-block
(b) have same number of electrons
(c) have similar atomic radius
(d) belongs to the same group of the periodic table.
Show Answer & Explanation
Answer: (c) have similar atomic radius
Explanation:
1. Due to lanthanoid contraction, Zr and Hf have nearly equal size.
Teacher's Note:
1. Being in the same group is not enough (Ti differs a lot from Zr); the near-equal radii are what make Zr and Hf so alike.
(i) paramagnetic behaviour
(ii) diamagnetic behaviour
(iii) variable oxidation states
(iv) formation of coloured ions
(a) (ii), (iii) and (iv)
(b) (i), (iii) and (iv)
(c) (i), (ii) and (iii)
(d) (i), (ii) and (iv)
Show Answer & Explanation
Answer: (b) (i), (iii) and (iv)
Explanation:
1. Due to presence of unpaired electrons in (n − 1)d orbitals, most of the transition metal ions and their compounds are paramagnetic. They form coloured ions and show variable oxidation states due to presence of vacant d-orbitals.
Teacher's Note:
1. Paramagnetism and colour both come from partly filled d orbitals, so they usually go together.
(a) K2Cr2O7
(b) K2MnO4
(c) CuSO4·5H2O
(d) MnCl2
Show Answer & Explanation
Answer: (a) K2Cr2O7
Explanation:
1. The electronic configuration of Cr is 24Cr → [Ar] 3d5 4s1 (six unpaired electrons)
2. In K2Cr2O7, the oxidation number of Cr is +6. So it has no unpaired electron.
Teacher's Note:
1. The orange colour of dichromate (Cr6+, d0) is due to charge transfer, not d–d transition; MnO4− is coloured for the same reason.
(a) lanthanoid contraction which results in almost same radii of second and third row metals
(b) diagonal relationship between second and third row elements
(c) similar ionisation enthalpy of second and third row elements
(d) similar oxidation states of second and third row metals.
Show Answer & Explanation
Answer: (a) lanthanoid contraction which results in almost same radii of second and third row metals
Explanation:
1. Due to lanthanoid contraction, the atomic radii of second and third row transition elements is almost same. Hence, they resemble each other much more as compared to first row elements.
Teacher's Note:
1. The poor shielding by 4f electrons cancels the expected size increase from the 4d to the 5d series.
(a) Na2[CuCl4]
(b) Na2[CdCl4]
(c) K4[Fe(CN)6]
(d) K3[Fe(CN)6]
Show Answer & Explanation
Answer: (b) Na2[CdCl4]
Explanation:
1. Na2[CuCl4]; Cu = +2 or Cu2+ → 3d9
2. Na2[CdCl4]; Cd = +2 or Cd2+ → 4d10
3. K4[Fe(CN)6]; Fe = +2 or Fe2+ → 3d6
4. K3[Fe(CN)6]; Fe = +3 or Fe3+ → 3d5
5. Since Cd2+ has completely filled d-subshell hence it is colourless.
Teacher's Note:
1. Group 12 ions (Zn2+, Cd2+, Hg2+) are d10 and their simple compounds are white or colourless.
(a) Cu+ has 3d10 configuration
(b) Cu+ disproportionates easily to Cu(0) and Cu2+
(c) Cu+ disproportionates easily to Cu2+ and Cu3+
(d) Cu+ is easily reduced to Cu2+.
Show Answer & Explanation
Answer: (b) Cu+ disproportionates easily to Cu(0) and Cu2+
Explanation:
1. Cu+ ions undergo disproportionation,
2. 2Cu+ → Cu2+ + Cu
Teacher's Note:
1. Option (d) is self-contradictory: going from Cu+ to Cu2+ is oxidation, not reduction.
(a) d-s transition
(b) s-s transition
(c) s-d transition
(d) d-d transition.
Show Answer & Explanation
Answer: (d) d-d transition.
Explanation:
1. The colour of transition metal ions is due to d-d transitions.
Teacher's Note:
1. The ligands split the d orbitals; absorption of visible light promotes an electron between these split levels and we see the complementary colour.
Which of the following facts cannot be explained on the basis of the given graph?
(a) Inability of Cu to liberate H2 from acids
(b) Extra stability of d5 (Mn2+) and d10 (Zn2+) configuration
(c) Extra stability of Ni2+ due to d10 configuration
(d) All of these.
Show Answer & Explanation
Answer: (c) Extra stability of Ni2+ due to d10 configuration
Explanation:
1. Ni2+ has 3d8 configuration, not d10.
2. Negative value of E° for Ni2+/Ni is related to the highest negative ΔH°hyd.
Teacher's Note:
1. Cu is the only 3d metal with a positive E°(M2+/M), which is why it cannot liberate H2 from dilute acids.
[Fe(H2O)6]2+ + NO3− + H+ → X + Y + H2O
[Fe(H2O)6]2+ + X → Z + H2O
The incorrect statement about X, Y and Z is
(a) Magnetic moment of Y is 5.9 B.M.
(b) Oxidation state of Fe in Z is +1.
(c) Complex Z is reddish-brown in colour.
(d) X is an acidic oxide of nitrogen.
Show Answer & Explanation
Answer: (d) X is an acidic oxide of nitrogen.
Explanation:
1. 3[Fe(H2O)6]2+ + NO3− + 4H+ → NO (X) + 3[Fe(H2O)6]3+ (Y) + 2H2O
2. [Fe(H2O)6]2+ + NO → [Fe(H2O)5NO]2+ (Z, brown) + H2O
3. In Y, Fe3+ is present: [Ar]3d5; \( \mu = \sqrt{n(n+2)} \) = 5.9 B.M.
4. NO is a neutral oxide.
Teacher's Note:
1. This is the brown ring test for nitrate; in the ring complex NO is present as NO+, which makes Fe +1.
Correct statement about W, X, Y and Z is
(a) magnetic moment of X in its +2 oxidation state is 2.83 B.M.
(b) W3+ ion is green in colour.
(c) Y3+ catalyses reaction between iodide and persulphate ions.
(d) stable oxidation states of Z are +1, +2 and +6.
Show Answer & Explanation
Answer: (c) Y3+ catalyses reaction between iodide and persulphate ions.
Explanation:
1. From the melting point trend it is clear that W → Cr, X → Mn, Y → Fe, Z → Co
2. Mn2+ ⇒ [Ar]3d5; \( \mu = \sqrt{5(5+2)} \) = 5.9 B.M.
3. Cr3+ → Violet
4. Fe3+ catalyses the given reaction: 2I− + S2O82− → I2 + 2SO42−
5. Co shows +2 and +3 oxidation states.
Teacher's Note:
1. The dip in melting point at Mn (weak metallic bonding due to the stable 3d5 4s2 configuration) is the clue to identify the four metals.
Cr3+/Cr2+ = −0.41 V
Cr2+/Cr = −0.90 V
Mn3+/Mn2+ = +1.57 V
Mn2+/Mn = −1.18 V
On the basis of these values, a student concluded the following statements:
I. Cr2+ is a reducing agent
II. Mn3+ is an oxidizing agent
III. both Cr2+ and Mn3+ exhibit d4 electronic configuration
IV. when Cr2+ is used as a reducing agent, the chromium ion attains d5 electronic configuration.
The incorrect conclusion made by him is
(a) I
(b) II
(c) III
(d) IV
Show Answer & Explanation
Answer: (d) IV
Explanation:
1. (I) Cr2+ is a reducing agent, it gets oxidised to Cr3+ (3d3 or t2g3, stable half-filled configuration).
2. (II) Mn3+ is an oxidizing agent, it gets reduced to Mn2+ (3d5, most stable, half-filled configuration).
3. (III) Cr (24): 3d5 4s1; Mn (25): 3d5 4s2; Cr2+: 3d4; Mn3+: 3d4. Both Cr2+ and Mn3+ exhibit d4 electronic configuration.
4. (IV) When Cr2+ is used as a reducing agent, the chromium ion attains d3 electronic configuration.
Teacher's Note:
1. Cr3+ is d3, which is stable because the three electrons half-fill the lower t2g set, not because it is d5.
(a) Good oxidising agent: Ce4+
(b) Paramagnetic lanthanide ion: Yb2+
(c) Ions that can exist in aqueous solution: Eu2+, Yb2+
(d) Colourless ions: Ce3+, Yb3+
Show Answer & Explanation
Answer: (b) Paramagnetic lanthanide ion: Yb2+
Explanation:
1. Ce3+ and Yb3+ are colourless despite having one unpaired electron.
2. The lanthanide ions, other than 4f0 type (La3+ and Ce4+) and the 4f14 type (Yb2+ and Lu3+) are all paramagnetic.
Teacher's Note:
1. Yb2+ is 4f14, a completely filled subshell, so it has no unpaired electron and is diamagnetic.
(a) 25
(b) 26
(c) 27
(d) 24
Show Answer & Explanation
Answer: (b) 26
Explanation:
1. It is formed by the loss of 3 electrons, so the configuration of element X is [Ar] 3d64s2.
2. Therefore, atomic number = number of electrons in a neutral atom of the element = 18 + 6 + 2 = 26.
Teacher's Note:
1. Add back the lost electrons to the 4s first and then the 3d: the ion has 23 electrons, so the atom has 26.
(a) 0
(b) 2
(c) 3
(d) 4
Show Answer & Explanation
Answer: (a) 0
Explanation:
1. Let oxidation state of Ni = x
2. x + 0 = 0 (as CO is neutral ligand.)
3. x = 0
Teacher's Note:
1. Metal carbonyls such as Ni(CO)4, Fe(CO)5 and Cr(CO)6 all have the metal in the zero oxidation state.
(a) Inert pair effect
(b) Completely filled 3d subshell
(c) Completely filled 4s subshell
(d) Common ion effect
Show Answer & Explanation
Answer: (b) Completely filled 3d subshell
Explanation:
1. Zinc does not exhibit variable oxidation states as it has completely filled 3d subshell.
2. Zn — [Ar] 3d10 4s2 (Atomic number 30)
3. It only exhibits +2 oxidation state.
Teacher's Note:
1. Zn loses only its two 4s electrons; removing a 3d electron from the stable d10 set needs too much energy.
(a) V2+
(b) Sc3+
(c) Cu2+
(d) Mn3+
Show Answer & Explanation
Answer: (b) Sc3+
Explanation:
1. The electronic configuration of Sc3+ is [Ar]3d0. Since its d subshell is empty, it is diamagnetic as it has attained noble gas configuration.
Teacher's Note:
1. Diamagnetic means no unpaired electrons: V2+ (3d3), Cu2+ (3d9) and Mn3+ (3d4) all have unpaired electrons.
(a) Sc3+, Ti4+, Mn3+
(b) Sc3+, Zn2+, Ni2+
(c) V3+, V2+, Fe3+
(d) Ti3+, Ti4+, Ni2+
Show Answer & Explanation
Answer: (c) V3+, V2+, Fe3+
Explanation:
1. V3+, V2+, Fe3+ ions exhibit specific colours.
2. Atomic number of V = 23,
3. Electronic configuration of V - [Ar]3d3 4s2
4. Electronic configuration of V2+ - [Ar]3d3
5. Electronic configuration of V3+ - [Ar] 3d2
6. Atomic number of Fe = 26
7. Electronic configuration of Fe - [Ar]3d6 4s2
8. Electronic configuration of Fe3+ - [Ar]3d5
9. Since these ions have partially filled d-subshells, they exhibit colour because unpaired electrons of d-orbital of lower energy excites to the higher energy level, and absorbs radiations from visible region and exhibits complementary colours.
10. As for other ions,
11. Atomic number of Sc = 21
12. Electronic configuration of Sc - [Ar]3d1 4s2
13. Electronic configuration of Sc3+ - [Ar]3d0
14. Since d subshell is empty, it shows no colour.
15. Atomic number of Ti = 22
16. Electronic configuration of Ti - [Ar]3d2 4s2
17. Electronic configuration of Ti4+ - [Ar]3d0
18. Since d subshell is empty, it shows no colour.
19. Atomic number of Mn = 25
20. Electronic configuration of Mn - [Ar]3d5 4s2
21. Electronic configuration of Mn2+ - [Ar]3d5
22. Since d subshell is partially filled, it shows colour.
23. Atomic number of Ni = 28
24. Electronic configuration of Ni - [Ar]3d8 4s2
25. Electronic configuration of Ni2+ - [Ar]3d8
26. Since d subshell is partially filled, it shows colour.
27. Atomic number of Zn = 30
28. Electronic configuration of Zn - [Ar]3d10 4s2
29. Electronic configuration of Zn2+ - [Ar]3d10
30. Since d subshell is fully filled, it shows no colour.
Teacher's Note:
1. Quick test: an ion is coloured only if it is d1 to d9; any set containing a d0 (Sc3+, Ti4+) or d10 (Zn2+) ion is ruled out.
(a) +2
(b) +3
(c) +4
(d) +5
Show Answer & Explanation
Answer: (b) +3
Explanation:
1. Most common oxidation state for lanthanoids is +3.
Teacher's Note:
1. +2 (Eu, Yb) and +4 (Ce, Tb) appear only where they give an empty, half-filled or filled 4f shell; +3 is shown by every lanthanoid.
(a) Ce
(b) Sm
(c) Tm
(d) All of the above
Show Answer & Explanation
Answer: (d) All of the above
Explanation:
1. The lanthanoids are the 14 elements from cerium (Ce, Z = 58) to lutetium (Lu, Z = 71).
2. Ce (58), Sm (62) and Tm (69) all lie in this range.
Teacher's Note:
1. Remember the range Z = 58 to 71 (Ce to Lu); lanthanum itself (Z = 57) is usually included only for comparison.
(a) [Xe] 4f75d16s2
(b) [Xe] 4f65d26s2
(c) [Xe] 4f86d2
(d) [Xe] 4f95s1
Show Answer & Explanation
Answer: (a) [Xe] 4f75d16s2
Explanation:
1. 64Gd : [Xe] 4f7 5d1 6s2
2. Rest of the configurations violates the Aufbau principle to fill electrons in the orbitals.
Teacher's Note:
1. Gd keeps a half-filled 4f7 set and puts the eighth electron in 5d; the same idea explains why Gd3+ (4f7) is very stable.
(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. The non-stoichiometric interstitial compounds have voids present in their crystal lattice. That is occupied by small atoms and hence remain chemically inert.
Teacher's Note:
1. Typical examples are TiC, Mn4N, Fe3H and VH0.56: high melting, very hard, metallic conductivity, chemically inert.
Assertion–Reason Questions
Assertion (A): When Zn is placed in a magnetic field, it is feebly magnetised in a direction opposite to that of the magnetising field.
Reason (R): Zn has completely filled atomic orbitals.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Zinc has all electrons paired [Ar]3d104s2. So, it is diamagnetic in nature.
Teacher's Note:
1. Being feebly magnetised opposite to the field is the definition of diamagnetism.
Assertion (A): The correct order of oxidising power is: VO2+ < VO < VO2+.
Reason (R): The oxidation state of Mn in MnO4− is +7.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. The oxidation states of the given compounds are the following,
2. VO2+: x + 2(−2) = +1 ⇒ x = +5
3. VO: x − 2 = 0 ⇒ x = +2
4. VO2+: x + 1(−2) = +2 ⇒ x = +4
5. The correct order of oxidising power is: VO < VO2+ < VO2+
6. In MnO4−: x + 4(−2) = −1 ⇒ x − 8 = −1 ⇒ x = −1 + 8 = +7
Teacher's Note:
1. For oxo-species of the same metal, the oxidising power increases with the oxidation state of the metal.
Assertion (A): Transition metals form a large number of interstitial compounds.
Reason (R): They have high melting point and boiling point.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. Some non-metallic atoms (e.g., H, B, C, N, etc.) are able to fit in the interstitial sites of transition metals lattices to form interstitial compounds.
Teacher's Note:
1. Interstitial compounds form because small atoms fit in the voids of the metal lattice; high melting point is a separate property.
Assertion (A): Members of 4d and 5d series of transition elements have nearly same atomic radii.
Reason (R): Atomic and ionic radii for transition elements are smaller than their corresponding s-block elements.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. It is due to lanthanide contraction.
Teacher's Note:
1. The near-equal size of 4d and 5d members is explained only by lanthanoid contraction, which the reason does not mention.
Assertion (A): In transition elements, ns orbital is filled up first and (n − 1)d afterwards, during ionization ns electrons are lost prior to (n − 1)d electrons.
Reason (R): The effective nuclear charge felt by (n − 1)d electrons is higher as compared to that by ns electrons.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Once the (n − 1)d orbitals are occupied, they lie lower in energy than ns because the d electrons feel a higher effective nuclear charge. Hence the outer ns electrons are removed first during ionisation.
Teacher's Note:
1. This is why Fe2+ is [Ar]3d6 and not [Ar]3d44s2: the 4s electrons leave first.
Assertion (A): The maximum oxidation state of chromium in its compounds is +6.
Reason (R): Chromium has only six electrons in ns and (n − 1)d orbitals.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Cr: [Ar]3d5 4s1 has six electrons in the 3d and 4s orbitals; all six can be used in bonding, so the maximum oxidation state is +6 (as in CrO3 and Cr2O72−).
Teacher's Note:
1. In the first half of the 3d series the maximum oxidation state equals the total number of 3d + 4s electrons.
Assertion (A): Transition metals are poor reducing agents.
Reason (R): Transition metals form numerous alloys with other metals.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. In actual practice transition metals react with acid very slowly and act as poor reducing agents. This is due to the protection of metal as a result of formation of thin oxide protective film. Further, their poor tendency as reducing agent is due to high ionisation energy, high heat of vapourization and low heat of hydration.
Teacher's Note:
1. Alloy formation is unrelated to reducing power; look for energy terms (ionisation, atomisation, hydration) to explain reducing behaviour.
Assertion (A): Cu2+ iodide is not known.
Reason (R): Cu2+ oxidises I− to iodine.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Cu2+ oxidises iodide to iodine hence cupric iodide is converted to cuprous iodide.
2. 2CuI2 (unstable cupric iodide) → Cu2I2 (cuprous iodide stable) + I2
Teacher's Note:
1. In the lab this shows up as: 2Cu2+ + 4I− → Cu2I2 + I2, which is the basis of the iodometric estimation of copper.
Assertion (A): The highest oxidation state of osmium is +8.
Reason (R): Osmium is a 5d-block element.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. The highest oxidation state of osmium (Os) is +8. It is due to its ability to expand octet by using its all 8 electrons.
Teacher's Note:
1. Os (5d66s2) can use all 8 valence electrons, as in OsO4; being a 5d element alone does not explain the +8 state.
Assertion (A): Separation of Zr and Hf is difficult.
Reason (R): Zr and Hf lie in the same group of the periodic table.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. Separation of Zr and Hf is difficult as both have same atomic size due to lanthanoid contraction. Thus, these are occurred in nature altogether.
Teacher's Note:
1. Zr (160 pm) and Hf (159 pm) are almost the same size because of lanthanoid contraction; merely being in the same group is not enough, since Ti is easily separated from them.
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