Consider the following metal complexes:1. [Co | vedclass

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(B) The energy of absorbed light is inversely proportional to the wavelength: $\Delta_0 \propto \frac{1}{\lambda}$.To absorb light with the shortest w

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(B) The energy of absorbed light is inversely proportional to the wavelength: $\Delta_0 \propto \frac{1}{\lambda}$.To absorb light with the shortest wavelength,the complex must have the largest crystal field splitting energy $(\Delta_0)$.Among the ligands provided $(Cl^- Therefore,$[Co(CN)_6]^{3-}$ has the largest $\Delta_0$ and absorbs light of the shortest wavelength.In $[Co(CN)_6]^{3-}$,the oxidation state of $Co$ is $+3$,which corresponds to a $d^6$ configuration.Since $CN^-$ is a strong field ligand,it causes pairing of electrons,resulting in a low-spin $t_{2g}^6 e_g^0$ configuration.The number of unpaired electrons $(n)$ is $0$,so the spin-only magnetic moment $\mu = \sqrt{n(n+2)} = 0 \ B.M.$

Consider the following metal complexes:
$1. [Co(NH_3)_6]^{3+}$
$2. [CoCl(NH_3)_5]^{2+}$
$3. [Co(CN)_6]^{3-}$
$4. [Co(NH_3)_5(H_2O)]^{3+}$
The spin-only magnetic moment value of the complex that absorbs light with the shortest wavelength is $B.M.$ (Nearest integer).

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Consider the following metal complexes:$1. [Co(NH_3)_6]^{3+}$$2. [CoCl(NH_3)_5]^{2+}$$3. [Co(CN)_6]^{3-}$$4. [Co(NH_3)_5(H_2O)]^{3+}$The spin-only magnetic moment value of the complex that absorbs light with the shortest wavelength is $B.M.$ (Nearest integer).