Color of $C d_{2} [F e (C N)_{6}]$ is bluish white.

True

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False

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Solution

The correct option is B False
The given statement is False.
$C d_{2} [F e (C N)_{6}]$ is white or colorless.
$2 C d^{2 +} + K_{4} [F e (C N)_{6}] \to C d_{2} [F e (C N)_{6}] + 4 K^{+}$
Cyanide ion is strong field ligand. Hence, it forms low spin complex.
$F e^{2 +}$ has $3 d^{6}$ outer electronic configuration with all electrons paired. Hence, it is colorless.

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