Taking $H X_{2} O$ as weak field (high spin) and $CN X^{-}$ as strong field (low spin), and ignoring pairing energy, the ratio of CFSE of $[Fe (H X_{2} O) X_{6}] X^{2 +}$ to that of $[Fe (CN) X_{6}] X^{4 -}$ is

Answer: D

Both contain Fe(II),

d^{6}

. High spin:

t_{2 g}^{4} e_{g}^{2}

gives

4 (- 0.4) + 2 (0.6) = - 0.4 Δ_{o}

. Low spin:

t_{2 g}^{6}

gives

- 2.4 Δ_{o}

. The ratio is

0.4 : 2.4 = 1 : 6