Consider the following complex [Co(NH_3)_5CO_ | vedclass

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(B) The complex is $[Co(NH_3)_5CO_3]ClO_4$. $NH_3$ is a monodentate ligand ($5$ ligands) and $CO_3^{2-}$ is a bidentate ligand ($1$ ligand). Coordinat

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$7, 2, 7, 1$

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(B) The complex is $[Co(NH_3)_5CO_3]ClO_4$. $NH_3$ is a monodentate ligand ($5$ ligands) and $CO_3^{2-}$ is a bidentate ligand ($1$ ligand). Coordination number $(C.N.)$ $= 5 	imes 1 + 1 	imes 2 = 7$. Let the oxidation state of $Co$ be $x$. $x + 5(0) + 1(-2) + 1(-1) = 0 \implies x - 3 = 0 \implies x = +3$. Electronic configuration of $Co$ $(Z=27)$ is $[Ar] 3d^7 4s^2$. Electronic configuration of $Co^{3+}$ is $[Ar] 3d^6 4s^0$. Number of $d-$ electrons is $6$. Since $NH_3$ and $CO_3^{2-}$ are ligands,in the presence of strong field ligands,the $6$ electrons in the $3d$ orbital are paired. Number of unpaired $d-$ electrons $= 0$.

Consider the following complex $[Co(NH_3)_5CO_3]ClO_4$. The coordination number,oxidation number,number of $d-$ electrons,and number of unpaired $d-$ electrons on the metal are respectively:

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