Consider the following complex: [Co(NH_3)_5CO | vedclass

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(D) In $[Co(NH_3)_5CO_3]ClO_4$,five $NH_3$ ligands (monodentate) and one $CO_3^{2-}$ ligand (bidentate) are attached to the central cobalt atom. Howev

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$6, 3, 6, 0$

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(D) In $[Co(NH_3)_5CO_3]ClO_4$,five $NH_3$ ligands (monodentate) and one $CO_3^{2-}$ ligand (bidentate) are attached to the central cobalt atom. However,the carbonate ion $CO_3^{2-}$ acts as a bidentate ligand,but in this specific complex,it is often considered to occupy two coordination sites. Wait,$NH_3$ is $5$ and $CO_3$ is $1$ (bidentate),so $5 + 2 = 7$ is incorrect; the coordination number is $6$ because $CO_3^{2-}$ is bidentate,but the complex is $[Co(NH_3)_5(CO_3)]^+$. Actually,$NH_3$ is $5$ and $CO_3$ is $1$ (bidentate),so $5 + 2 = 7$. Let us re-evaluate: $Co$ is bonded to $5$ $NH_3$ and $1$ $CO_3^{2-}$. $5 + 2 = 7$. Wait,the coordination number is $6$ if $CO_3$ is monodentate or if the formula implies $5+1=6$. Given the options,$6$ is the coordination number.Let $x$ be the oxidation state of $Co$. $x + 5(0) + (-2) + (-1) = 0$,so $x = +3$.The atomic number of $Co$ is $27$. Configuration: $[Ar] 3d^7 4s^2$. For $Co^{3+}$,it is $3d^6$.Number of $d-$ electrons is $6$.Since $NH_3$ is a strong field ligand,it causes pairing. For $d^6$ in an octahedral field with strong ligands,all electrons are paired. Thus,unpaired 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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