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Q.

The number of electrons in $t_{2 g}$ orbitals in $K_{4} [Fe {(CN)}_{6}]$ is

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a

2

b

6

c

3

d

5

answer is C.

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Detailed Solution

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$\begin{array}{l} {Fe}^{2 +} : 3 d^{6} \\ {(t_{2 g})}^{6} {(eg)}^{0} \end{array}$

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