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(A) The central metal ion in all three complexes is the same,$Ni^{2+}$. Therefore,the crystal field splitting energy,$\Delta$,depends on the strength

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$[Ni(H_{2}O)_{6}]^{2+} > [Ni(NH_{3})_{6}]^{2+} > [Ni(NO_{2})_{6}]^{4-}$

Vedclass · Answer

(A) The central metal ion in all three complexes is the same,$Ni^{2+}$. Therefore,the crystal field splitting energy,$\Delta$,depends on the strength of the ligands. According to the spectrochemical series,the order of ligand field strength is: $H_{2}O < NH_{3} < NO_{2}^{-}$. Since $\Delta = \frac{hc}{\lambda}$,the magnitude of crystal field splitting is inversely proportional to the wavelength of absorption,$\lambda$. The order of crystal field splitting,$\Delta$,is: $[Ni(H_{2}O)_{6}]^{2+} < [Ni(NH_{3})_{6}]^{2+} < [Ni(NO_{2})_{6}]^{4-}$. Therefore,the order of wavelengths of absorption,$\lambda$,is: $[Ni(H_{2}O)_{6}]^{2+} > [Ni(NH_{3})_{6}]^{2+} > [Ni(NO_{2})_{6}]^{4-}$.

$I$. $NCS^{-}$	$II$. $S^{2-}$
$III$. $en$	$IV$. $SCN^{-}$

What will be the correct order for the wavelengths of absorption in the visible region for the following complexes: $[Ni(NO_{2})_{6}]^{4-}, [Ni(NH_{3})_{6}]^{2+}, [Ni(H_{2}O)_{6}]^{2+}$?

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