Level 1 — Recognitiond-Block (Transition Metals) & f-Block

d-Block (Transition Metals) & f-Block

30 marksprintable — key stays hidden on paper

Level 1 — Recognition Test

Time: 20 minutes Total Marks: 30

Section A — Multiple Choice Questions (1 mark each)

Choose the single best answer.

Q1. The general outer electronic configuration of transition elements is: (a) $ns^{1-2}\,np^{1-6}$ (b) $(n-1)d^{1-10}\,ns^{0-2}$ (c) $(n-1)f^{1-14}\,ns^2$ (d) $nd^{1-10}\,ns^2$

Q2. The spin-only magnetic moment of $\mathrm{Mn^{2+}}$ (in BM) is approximately: (a) 1.73 (b) 3.87 (c) 4.90 (d) 5.92

Q3. Which of the following ions is colourless in aqueous solution? (a) $\mathrm{Cu^{2+}}$ (b) $\mathrm{Ti^{3+}}$ (c) $\mathrm{Sc^{3+}}$ (d) $\mathrm{Cr^{3+}}$

Q4. In the manufacture of $\mathrm{H_2SO_4}$ by the Contact process, the catalyst used is: (a) Fe (b) Ni (c) $\mathrm{V_2O_5}$ (d) Pt

Q5. In acidic medium, $\mathrm{KMnO_4}$ is reduced to: (a) $\mathrm{MnO_2}$ (b) $\mathrm{Mn^{2+}}$ (c) $\mathrm{MnO_4^{2-}}$ (d) Mn metal

Q6. The most common oxidation state of lanthanides is: (a) +2 (b) +3 (c) +4 (d) +6

Q7. The lanthanide contraction is primarily due to: (a) increasing shielding of 4f electrons (b) poor shielding of 4f electrons causing increased effective nuclear charge (c) increasing atomic radius across the series (d) filling of 5d orbitals

Q8. The oxidation state of chromium in $\mathrm{K_2Cr_2O_7}$ is: (a) +3 (b) +5 (c) +6 (d) +7

Q9. Which pair of elements has almost identical atomic radii due to lanthanide contraction? (a) Ti and Zr (b) Zr and Hf (c) V and Nb (d) Cr and Mo

Q10. Actinides differ from lanthanides mainly because actinides: (a) show only the +3 oxidation state (b) do not include radioactive elements (c) show a wider range of oxidation states and are all radioactive (d) have completely filled 5f orbitals

Section B — Matching (1 mark per correct match, 5 marks)

Q11. Match Column I with Column II.

Column I (Species/Element)	Column II (Property)
(i) $\mathrm{Zn^{2+}}$	(P) catalyst in Haber process
(ii) Fe	(Q) $d^0$ configuration, colourless
(iii) $\mathrm{Sc^{3+}}$	(R) $d^{10}$ , diamagnetic
(iv) Ni	(S) catalyst in hydrogenation of oils
(v) $\mathrm{Ti^{3+}}$	(T) $d^1$ , coloured (violet)

Section C — True / False WITH Justification

State True or False and give a one-line reason. (2 marks each: 1 for T/F, 1 for justification)

Q12. All transition metal ions are coloured.

Q13. $\mathrm{Sc^{3+}}$ is paramagnetic.

Q14. $\mathrm{K_2Cr_2O_7}$ acts as an oxidising agent in acidic medium and is itself reduced to $\mathrm{Cr^{3+}}$ .

Q15. The variable oxidation states of transition metals arise because the $(n-1)d$ and $ns$ orbitals have nearly equal energies.

Q16. Cu shows an anomalous electronic configuration $[\mathrm{Ar}]3d^{10}4s^1$ .

Q17. Atomic radii of the second and third transition series are nearly the same.

Q18. Promethium (Pm) is the only radioactive lanthanide.

End of Paper

Answer keyMark scheme & solutions

Section A (1 mark each)

Q1. (b) $(n-1)d^{1-10}\,ns^{0-2}$ — the defining d-block configuration; d fills after s. [1]

Q2. (d) 5.92. $\mathrm{Mn^{2+}}$ is $d^5$ , so $n=5$ unpaired electrons. $\mu=\sqrt{5(5+2)}=\sqrt{35}=5.92$ BM. [1]

Q3. (c) $\mathrm{Sc^{3+}}$ — configuration $[\mathrm{Ar}]3d^0$ ; no d-electrons ⇒ no d–d transition ⇒ colourless. The others have partly filled d subshells. [1]

Q4. (c) $\mathrm{V_2O_5}$ — oxidises $\mathrm{SO_2}\to\mathrm{SO_3}$ in the Contact process. [1]

Q5. (b) $\mathrm{Mn^{2+}}$ — Mn goes from +7 to +2 (gain of 5 electrons) in acidic medium. [1]

Q6. (b) +3 — most stable/common lanthanide oxidation state. [1]

Q7. (b) poor shielding of 4f electrons → increased effective nuclear charge → steady size decrease. [1]

Q8. (c) +6. Check: $2(+1)+2x+7(-2)=0 \Rightarrow 2x=12 \Rightarrow x=+6$ . [1]

Q9. (b) Zr and Hf — lanthanide contraction offsets the expected size increase. [1]

Q10. (c) wider range of oxidation states and all radioactive. [1]

Section B

Q11. (5 marks, 1 each)

(i) $\mathrm{Zn^{2+}}$ → (R) $d^{10}$ , diamagnetic
(ii) Fe → (P) catalyst in Haber process
(iii) $\mathrm{Sc^{3+}}$ → (Q) $d^0$ , colourless
(iv) Ni → (S) hydrogenation of oils
(v) $\mathrm{Ti^{3+}}$ → (T) $d^1$ , violet coloured [5]

Section C (2 marks each: 1 T/F + 1 reason)

Q12. False. Ions with $d^0$ (e.g. $\mathrm{Sc^{3+}}$ , $\mathrm{Ti^{4+}}$ ) or $d^{10}$ (e.g. $\mathrm{Zn^{2+}}$ ) are colourless — no d–d transition possible. [2]

Q13. False. $\mathrm{Sc^{3+}}$ is $d^0$ ; zero unpaired electrons ⇒ diamagnetic. [2]

Q14. True. $\mathrm{Cr_2O_7^{2-}}+14H^++6e^-\to 2Cr^{3+}+7H_2O$ ; Cr is reduced +6→+3. [2]

Q15. True. Small energy gap between $(n-1)d$ and $ns$ allows both to participate in bonding, giving variable oxidation states. [2]

Q16. True. A fully filled $3d^{10}$ subshell gives extra stability, so one $4s$ electron shifts into $3d$ . [2]

Q17. True. Lanthanide contraction cancels the expected radius increase, so 2nd and 3rd series radii are nearly equal (e.g. Zr≈Hf). [2]

Q18. True. Pm (Z=61) has no stable isotope; it is the only radioactive lanthanide. [2]

[
  {"claim":"Spin-only moment of Mn2+ (d5, n=5) ≈ 5.92 BM","code":"n=5; mu=sqrt(n*(n+2)); result = abs(float(mu)-5.916)<0.01"},
  {"claim":"Cr oxidation state in K2Cr2O7 is +6","code":"x=symbols('x'); sol=solve(2*1+2*x+7*(-2),x); result = sol[0]==6"},
  {"claim":"Mn changes by 5 electrons from +7 to +2","code":"result = (7-2)==5"},
  {"claim":"Cr2O7 half reaction uses 6 electrons (2*(6-3))","code":"result = 2*(6-3)==6"}
]