JEE Advance - Chemistry (2009 - Paper 2 Offline - No. 2)
The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)$$_6$$ is
0
2.84
4.90
5.92
Explanation
In [Cr(CO)$$_6$$]: Cr(24) = [Ar] 3$$d^5$$ 4$$s^1$$
Since CO is a strong field ligand, so pairing of electrons will take place and the configuration will become $$3d^6$$. There will be no unpaired electrons, so, the spin only magnetic moment is zero.
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