Question

Given below are two statements:
Statement I: The number of pairs among [Ti⁴⁺, V²⁺], [V²⁺, Mn²⁺], [Mn²⁺, Fe³⁺] and [V²⁺, Cr²⁺] in which both ions are coloured is 3.
Statement II: The number of pairs among [La³⁺, Yb²⁺], [Lu³⁺, Ce⁴⁺] and [Ac³⁺, Lr³⁺] ions in which both are diamagnetic is 3.

• A. Both Statement I and Statement II are correct
• B. Both Statement I and Statement II are incorrect
• C. Statement I is correct but Statement II is incorrect
• D. Statement I is incorrect but Statement II is correct

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
To determine if transition metal ions are coloured or possess magnetic properties, we must look at their electronic configurations. Ions with partially filled $d$ or $f$ orbitals (unpaired electrons) exhibit colour (via $d-d$ or $f-f$ transitions) and are paramagnetic. Ions with empty ($d^0, f^0$) or completely full ($d^{10}, f^{14}$) orbitals are colourless and diamagnetic.
Step 2: Key Formula or Approach:
Evaluate the electronic configurations for the transition metal ions and the f-block ions to count the number of unpaired electrons ($n$).
Coloured/Paramagnetic $\implies n>0$.
Colourless/Diamagnetic $\implies n = 0$.
Step 3: Detailed Explanation:
Evaluate Statement I (d-block ions):
- $Ti^{4+}$: [Ar] $3d^0 \implies$ 0 unpaired $e^- \implies$ Colourless.
- $V^{2+}$: [Ar] $3d^3 \implies$ 3 unpaired $e^- \implies$ Coloured.
- $Mn^{2+}$: [Ar] $3d^5 \implies$ 5 unpaired $e^- \implies$ Coloured.
- $Fe^{3+}$: [Ar] $3d^5 \implies$ 5 unpaired $e^- \implies$ Coloured.
- $Cr^{2+}$: [Ar] $3d^4 \implies$ 4 unpaired $e^- \implies$ Coloured.
Now let's check the given pairs:
1. $[Ti^{4+}, V^{2+}]$: $Ti^{4+}$ is colourless, so the pair is not "both coloured".
2. $[V^{2+}, Mn^{2+}]$: Both are coloured.
3. $[Mn^{2+}, Fe^{3+}]$: Both are coloured.
4. $[V^{2+}, Cr^{2+}]$: Both are coloured.
There are precisely 3 pairs where BOTH ions are coloured. Statement I is correct.
Evaluate Statement II (f-block ions):
- $La^{3+}$: [Xe] $4f^0 \implies$ 0 unpaired $e^- \implies$ Diamagnetic.
- $Yb^{2+}$: [Xe] $4f^{14} \implies$ 0 unpaired $e^- \implies$ Diamagnetic.
- $Lu^{3+}$: [Xe] $4f^{14} \implies$ 0 unpaired $e^- \implies$ Diamagnetic.
- $Ce^{4+}$: [Xe] $4f^0 \implies$ 0 unpaired $e^- \implies$ Diamagnetic.
- $Ac^{3+}$: [Rn] $5f^0 \implies$ 0 unpaired $e^- \implies$ Diamagnetic.
- $Lr^{3+}$: [Rn] $5f^{14} \implies$ 0 unpaired $e^- \implies$ Diamagnetic.
Now let's check the given pairs:
1. $[La^{3+}, Yb^{2+}]$: Both are diamagnetic.
2. $[Lu^{3+}, Ce^{4+}]$: Both are diamagnetic.
3. $[Ac^{3+}, Lr^{3+}]$: Both are diamagnetic.
There are precisely 3 pairs where BOTH ions are diamagnetic. Statement II is correct.
Step 4: Final Answer:
Both Statement I and Statement II are correct.