The cobalt atom in $Hg \left[ Co ( SCN )_{4}\right]$ carries +2 charge and has $3 d ^{7}$ as the outer electronic configuration.
Here Hg shows +2 oxidation state and the compound is known as Mercuric tetrathiocyanatocobaltate(II).
It contains three unpaired electrons.
For three unpaired electrons, the expression becomes $\mu_{e f f}=\sqrt{3(3+2)} B \cdot M .=\sqrt{15} B \cdot M$
Therefore, option $C$ is correct.
| સૂચી $- I$ | સૂચી $- II$ |
| $(A)$ $\left[ PtCl _{4}\right]^{2-}$ | $(I)$ $sp ^{3} d$ |
| $(B)$ $BrF _{5}$ | $(II)$ $d ^{2} sp ^{3}$ |
| $(C)$ $PCl _{5}$ | $(III)$ $dsp ^{2}$ |
| $(D)$ $\left[ Co \left( NH _{3}\right)_{6}\right]^{3+}$ | $(IV)$ $sp ^{3} d ^{2}$ |
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