Hybridization shape and magnetic moment of `K_(3)[Co(CO)_(6)]` is

Options

A. `d^(2)sp^(3)`, octahedral, 4.9 B.M.
B. `sp^(3)d^(2)`, octahedral, 4.9 B. M.
C. `dsp^(2)`, square planar, 4.9 B.M.
D. `sp^(3)`, tetrahedral, 4.9 B.M.

Answer

B

Explanation

In each case, magnetic moment is 4.9 B.M. which shows that each of them has 4 unpaired electrons. Further, in the given complex, oxidation state of cobalt is +3. Hence, it will have `3d^(6)` configuration, i.e.,

Thus, hybridisation possible is `sp^(3)d^(2)`, i.e., outer orbital (high spin) octahedral complex.