The magnetic moment of the complex anion \([\mathrm{Cr(NO)(NH_3)(CN)_4}]^{2-}\) is :
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- 5.91 BM
- 3.87 BM
- 1.73 BM
- 2.82 BM
The Correct Option is D
Solution and Explanation
Let the oxidation state of Cr be x: \(x + 0 + 0 + 4(-1) = -2\), so \(x = +2\).
\(\mathrm{Cr^{2+}}\): \([\mathrm{Ar}]3d^4\). CN- and NO are strong-field ligands, so the complex is low spin and the four d electrons occupy only the \(t_{2g}\) orbitals: \(t_{2g}^{2,1,1}\).
Number of unpaired electrons \(n = 2\).
\(\mu = \sqrt{n(n+2)} = \sqrt{2 imes 4} = \sqrt{8} pprox 2.83\) BM, i.e. about 2.82 BM.
Answer: (4) 2.82 BM
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