1.

The spin only magnetic moment value (in Bohr magnetion units) of Cr(CO)_(6) is

Answer»

0
2.84
`4.90`
`5.92`

Solution :In `[CR(CO)_(6)],Cr` is in zero oxidation state
`._(24)Cr=[Ar] 3d^(5)4s^(1)`
As CO is a STRONG ligand, all the six unpaired electrons will PAIR up, i.e., there will be no unpaired electron. HENCE, `mu=0`


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