Vidaara.orgClass 12 · Chemistry
CodeVID-C12-08-CH-01
Chapter Assignment — The d- and f-Block Elements
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
The general outer electronic configuration of d-block elements is:
- A.$(n-1)d^{1-10}\,ns^{0-2}$
- B.$ns^{1-2}\,np^{1-6}$
- C.$(n-2)f^{1-14}$
- D.$ns^2$
2.
Which ion is colourless?
- A.$\text{Cu}^{2+}$
- B.$\text{Ti}^{3+}$
- C.$\text{Sc}^{3+}$
- D.$\text{Ni}^{2+}$
3.
The highest oxidation state of manganese is shown in:
- A.$\text{MnO}_2$
- B.$\text{KMnO}_4$
- C.$\text{MnCl}_2$
- D.$\text{Mn}_2\text{O}_3$
4.
Lanthanoid contraction is mainly due to:
- A.Good shielding by 4f electrons
- B.Poor shielding by 4f electrons
- C.Increase in atomic number only
- D.Larger 5d orbitals
5.
The colour of $\text{Cr}_2\text{O}_7^{2-}$ ion is:
- A.Green
- B.Yellow
- C.Orange
- D.Purple
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
6.
Why are Zn, Cd and Hg not regarded as typical transition elements?
7.
Calculate the spin-only magnetic moment of $\text{Fe}^{2+}$ ($d^6$).
8.
Write the ionic equation for the oxidation of iodide by acidified dichromate.
9.
Why do transition metals form coloured compounds?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
10.
List three consequences of lanthanoid contraction.
11.
Explain why $\text{Mn}^{2+}$ is more stable than $\text{Mn}^{3+}$ while $\text{Fe}^{3+}$ is more stable than $\text{Fe}^{2+}$.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
12.
Describe the preparation of potassium permanganate from pyrolusite and give its reaction with acidified ferrous sulphate and with oxalic acid.
Answer Key
Section A — Multiple Choice Questions
- (A) $(n-1)d^{1-10}\,ns^{0-2}$
- (C) $\text{Sc}^{3+}$
- (B) $\text{KMnO}_4$
- (B) Poor shielding by 4f electrons
- (C) Orange
Section B — Short Answer (2 marks)
- Their $d$ orbitals are completely filled ($d^{10}$) in the ground state and in their common oxidation states, so they show none of the characteristic transition-metal properties.
- $\text{Fe}^{2+}$ has 4 unpaired electrons; $\mu=\sqrt{4(4+2)}=\sqrt{24}=4.90\ \text{BM}$.
- $\text{Cr}_2\text{O}_7^{2-}+14\text{H}^++6\text{I}^-\rightarrow2\text{Cr}^{3+}+3\text{I}_2+7\text{H}_2\text{O}$.
- Their partially filled $d$ orbitals allow d–d electronic transitions; the energy absorbed lies in the visible region, so the complementary colour is transmitted.
Section C — Short Answer (3 marks)
- Similar sizes of 4d and 5d elements (e.g. Zr ≈ Hf) making their separation difficult; a steady decrease in size and basicity of lanthanoid hydroxides along the series; the high density of third-row transition elements.
- $\text{Mn}^{2+}$ is $3d^5$ (half-filled, extra-stable), so further oxidation is unfavourable. $\text{Fe}^{3+}$ is also $3d^5$ (half-filled), so $\text{Fe}^{2+}$ ($3d^6$) readily loses one electron to attain this stable configuration.
Section D — Long Answer (5 marks)
- Pyrolusite is fused with KOH and an oxidant (KNO3/air) to give green $\text{K}_2\text{MnO}_4$, which on electrolytic (or disproportionation) oxidation gives purple $\text{KMnO}_4$. In acidic medium: with $\text{Fe}^{2+}$, $\text{MnO}_4^-+5\text{Fe}^{2+}+8\text{H}^+\rightarrow\text{Mn}^{2+}+5\text{Fe}^{3+}+4\text{H}_2\text{O}$; with oxalic acid, $2\text{MnO}_4^-+5\text{C}_2\text{O}_4^{2-}+16\text{H}^+\rightarrow2\text{Mn}^{2+}+10\text{CO}_2+8\text{H}_2\text{O}$.
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