KCET · Chemistry · Coordination Compounds
Match the following.
Codes
- A \(\text{I} \to \text{(i) II} \to \text{(ii) III} \to \text{(iii)}\)
- 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 \(\text{I} \to \text{(i) II} \to \text{(iii) III} \to \text{(ii)}\)
Answer & Solution
Correct Answer
(B) \(\text{I} \to \text{(ii) II} \to \text{(iii) III} \to \text{(i)}\)
Step-by-step Solution
Detailed explanation
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 \(d^8\).
- \(\mathrm{Zn}^{2+}(Z=30)\) 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 \(d^8\).
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