Q. In which of the following octahedral complexes of Co (at. No. 27), will be magnitude of $\mathrm{\Delta}$_{0} be the highest?
A) $\left\lbrack Co(CN)_{6} \right\rbrack^{3 -}$
B) $\left\lbrack Co\left( C_{2}O_{4} \right)_{3} \right\rbrack^{3 -}$
C) $\left\lbrack Co\left( H_{2}O \right)_{6} \right\rbrack^{3 +}$
D) $\left\lbrack Co\left( {NH}_{3} \right)_{6} \right\rbrack^{3 +}$
Explanation
(a)
CFSE (crystal field splitting energy) for octahedral complex, $\mathrm{\Delta}_{0}$ depends on the strength of negative ligand. Spectrochemically it has been found that the strength of splitting is as follows
$CO > \ \underline{CN^{-}}\ > NO_{2}^{-} > en > \ \underline{NH_{3}}\ > py >$
$NCS^{-} > H_{2}O > O^{2 -} > OX^{2 -} > OH^{-} > F^{-} >$
$Cl^{-} > SCN^{-} > S^{2 -} > Br^{-} > I^{-}\$
Therefore, magnitude of $\mathrm{\Delta}_{0}$ will be highest in case of $\left\lbrack Co(CN)_{6} \right\rbrack^{3 -}$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.