Consider the following complex: [Co(CO_3)(NH_ | vedclass

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(A) In $[Co(NH_3)_5(CO_3)]ClO_4$,the central metal atom is $Co$. The ligands are five $NH_3$ (monodentate) and one $CO_3^{2-}$ (bidentate). However,in

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$6-3-6-0$

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(A) In $[Co(NH_3)_5(CO_3)]ClO_4$,the central metal atom is $Co$. The ligands are five $NH_3$ (monodentate) and one $CO_3^{2-}$ (bidentate). However,in this specific complex,$CO_3^{2-}$ acts as a monodentate ligand to satisfy the coordination sphere. Thus,the coordination number is $5 + 1 = 6$.Let $X$ be the oxidation number of $Co$. The charge on $NH_3$ is $0$,$CO_3$ is $-2$,and $ClO_4$ is $-1$. Therefore,$X + 5(0) + (-2) + (-1) = 0$,which gives $X = +3$.The atomic number of $Co$ is $27$. The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$. For $Co^{3+}$,the configuration is $[Ar] 3d^6 4s^0$. Thus,there are $6$ $d$-electrons.$NH_3$ is a strong field ligand. In the presence of strong ligands,the $6$ electrons in the $3d$ orbitals of $Co^{3+}$ pair up,resulting in $0$ unpaired $d$-electrons.

Consider the following complex: $[Co(CO_3)(NH_3)_5]ClO_4$.
Mark the correct option for the following parameters in order:
Coordination number $-$ Oxidation number $-$ Number of $d$-electrons $-$ Number of unpaired $d$-electrons.

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Consider the following complex: $[Co(CO_3)(NH_3)_5]ClO_4$.Mark the correct option for the following parameters in order:Coordination number $-$ Oxidation number $-$ Number of $d$-electrons $-$ Number of unpaired $d$-electrons.