KCET · Chemistry · Coordination Compounds
The "spin only" magnetic moment of \(\mathrm{Ni}^{2+}\) in aqueous solution would be [At. no. of \(\mathrm{Ni}=28\) ]
- A \(\sqrt{15} \mathrm{BM}\)
- B \(\sqrt{2} \mathrm{BM}\)
- C \(\sqrt{8} \mathrm{BM}\)
- D \(\sqrt{6} \mathrm{BM}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{8} \mathrm{BM}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{Ni}^{2+}\) in aqueous solution refers to
\({\left.\left[\mathrm{Ni}_{2} \mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \cdot \mathrm{Ni}^{2+}\) has following electronic \(}
\) orientation.
\(\therefore\) Number of unpaired electrons \(=2\)
Spin only magnetic moment \(=\sqrt{n(n+2)}\)
\[
=\sqrt{2(2+2)}=\sqrt{8} \mathrm{BM}
\]
\({\left.\left[\mathrm{Ni}_{2} \mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \cdot \mathrm{Ni}^{2+}\) has following electronic \(}
\) orientation.
\(\therefore\) Number of unpaired electrons \(=2\)
Spin only magnetic moment \(=\sqrt{n(n+2)}\)
\[
=\sqrt{2(2+2)}=\sqrt{8} \mathrm{BM}
\]
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