An ion M^{2+} forms the complexes [M(H_2O)_6] | vedclass

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(D) The crystal field splitting energy $\Delta_o$ is directly proportional to the strength of the ligand. Stronger ligands cause larger splitting,lead

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Blue,red and green

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(D) The crystal field splitting energy $\Delta_o$ is directly proportional to the strength of the ligand. Stronger ligands cause larger splitting,leading to the absorption of light with shorter wavelengths. The order of ligand field strength is $Br^{-} Consequently,the order of $\Delta_o$ is $[MBr_6]^{4-} As the energy of absorbed light increases,the wavelength of the absorbed light decreases,shifting the observed (complementary) color from green to red to blue. Therefore,$[M(H_2O)_6]^{2+}$ is blue,$[M(en)_3]^{2+}$ is red,and $[MBr_6]^{4-}$ is green.

List-$I$ (Complex)	List-$II$ (Electronic configuration of metal/ion)
$A. [Co(NH_3)_6]^{3+}$	$I. t_{2g}^5 e_g^0$
$B. [CoF_6]^{3-}$	$II. t_{2g}^6 e_g^0$
$C. [Ni(CO)_4]$	$III. t_{2g}^4 e_g^2$
$D. [Fe(CN)_6]^{3-}$	$IV. t^4 e^6$

An ion $M^{2+}$ forms the complexes $[M(H_2O)_6]^{2+}$,$[M(en)_3]^{2+}$,and $[MBr_6]^{4-}$. Match the complex with the appropriate colour.

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