The pair having the same magnetic moment is :
[At. No.: Cr = 24, Mn = 25, Fe = 26, Co = 27]
[At. No.: Cr = 24, Mn = 25, Fe = 26, Co = 27]
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- \(\mathrm{[Cr(H_2O)_6]^{2+}}\) and \(\mathrm{[Fe(H_2O)_6]^{2+}}\)
- \(\mathrm{[Mn(H_2O)_6]^{2+}}\) and \(\mathrm{[Cr(H_2O)_6]^{2+}}\)
- \(\mathrm{[CoCl_4]^{2-}}\) and \(\mathrm{[Fe(H_2O)_6]^{2+}}\)
- \(\mathrm{[Cr(H_2O)_6]^{2+}}\) and \(\mathrm{[CoCl_4]^{2-}}\)
The Correct Option is A
Solution and Explanation
\(\mathrm{Fe^{2+}}\): \(d^6\); high spin with \(\mathrm{H_2O}\), 4 unpaired electrons.
\(\mathrm{Mn^{2+}}\): \(d^5\), 5 unpaired electrons.
\(\mathrm{Co^{2+}}\) in \(\mathrm{[CoCl_4]^{2-}}\): \(d^7\), tetrahedral, 3 unpaired electrons.
\(\mu = \sqrt{n(n+2)}\): n = 4 gives 4.9 BM for the Cr and Fe complexes, 5.9 BM for Mn, 3.87 BM for Co.
So the pair with equal magnetic moment is \(\mathrm{[Cr(H_2O)_6]^{2+}}\) and \(\mathrm{[Fe(H_2O)_6]^{2+}}\).
Answer: (1) \(\mathrm{[Cr(H_2O)_6]^{2+}}\) and \(\mathrm{[Fe(H_2O)_6]^{2+}}\)
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