Assertion : Complex ion [Co(NH_3)_6]^{2+} is | vedclass

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(B) The electronic configuration of $Co$ $(Z=27)$ is $[Ar] 3d^7 4s^2$.In $[Co(NH_3)_6]^{2+}$,the oxidation state of $Co$ is $+2$,so the configuration

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Assertion is true but Reason is false.

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(B) The electronic configuration of $Co$ $(Z=27)$ is $[Ar] 3d^7 4s^2$.In $[Co(NH_3)_6]^{2+}$,the oxidation state of $Co$ is $+2$,so the configuration is $3d^7$.$NH_3$ is a strong field ligand,but in $Co^{2+}$ complexes,it does not force pairing of all electrons to form an inner orbital complex because the resulting configuration would be highly unstable. Instead,one electron is promoted to the $4d$ orbital to allow for $d^2sp^3$ hybridization,making the complex $[Co(NH_3)_6]^{2+}$ unstable and easily oxidized to the more stable $d^6$ $[Co(NH_3)_6]^{3+}$ complex.The unpaired electron in $[Co(NH_3)_6]^{2+}$ is promoted to the $4d$ orbital,not the $4p$ orbital. Thus,the Assertion is true,but the Reason is false.

Assertion $:$ Complex ion $[Co(NH_3)_6]^{2+}$ is readily oxidized to $[Co(NH_3)_6]^{3+}$.
Reason $:$ Unpaired electron in complex ion $[Co(NH_3)_6]^{2+}$ is present in $4p$ orbital.

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Assertion $:$ Complex ion $[Co(NH_3)_6]^{2+}$ is readily oxidized to $[Co(NH_3)_6]^{3+}$.Reason $:$ Unpaired electron in complex ion $[Co(NH_3)_6]^{2+}$ is present in $4p$ orbital.