Q53 / 180 · KCET 2024 · Chemistry

Match the following. Figure
Codes

  • A
    $\text{I} \to \text{(i) II} \to \text{(ii) III} \to \text{(iii)}$
  • Correct answer
    B
    $\text{I} \to \text{(ii) II} \to \text{(iii) III} \to \text{(i)}$
  • C
    $\text{I} \to \text{(ii) II} \to \text{(i) III} \to \text{(iii)}$
  • D
    $\text{I} \to \text{(i) II} \to \text{(iii) III} \to \text{(ii)}$

Solution

The correct match is I-(ii), II-(iii), III-(i) - $\mathrm{Zn}^{2+}(Z=30)$ The outer electronic configuration of $\mathrm{Zn}^{2+}$ is $3 d^{10}$. Thus, it has no unpaired electrons. Hence, it is colourless. - $\mathrm{Cu}^{2+}(Z=29)$ The outer electronic configuration of $\mathrm{Cu}^{2+}$ is $3 d^9$. Hence, it contains one unpaired electron. So, $\mu=\sqrt{1(1+2)}=1.73 \mathrm{BM}$ - $\mathrm{Ni}^{2+}(Z=28)$ The outer electronic configuration of $4 d^8 \mathrm{Ni}^{2+}$ is $3 d^8$. Hence, its configuration is $d^8$.

Hence, the correct answer is option: 'B'.

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