The coordination number,oxidation number,numb | vedclass

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(C) $1$. Coordination number: $NH_3$ is a monodentate ligand $(4 	imes 1 = 4)$ and $CO_3^{2-}$ is a bidentate ligand $(1 	imes 2 = 2)$. Total coordi

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$6, 3, 6, 0$

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(C) $1$. Coordination number: $NH_3$ is a monodentate ligand $(4 	imes 1 = 4)$ and $CO_3^{2-}$ is a bidentate ligand $(1 	imes 2 = 2)$. Total coordination number = $4 + 2 = 6$.$2$. Oxidation number: Let the oxidation state of $Co$ be $x$. The complex is $[Co(NH_3)_4CO_3]ClO_4$. The charge on $ClO_4$ is $-1$,so the complex ion $[Co(NH_3)_4CO_3]^+$ has a charge of $+1$. Thus,$x + 4(0) + 1(-2) = +1$,which gives $x = +3$.$3$. Number of electrons in $d$-orbital: $Co$ $(Z=27)$ has the configuration $[Ar] 3d^7 4s^2$. $Co^{3+}$ has the configuration $[Ar] 3d^6$. Thus,there are $6$ electrons in the $d$-orbital.$4$. Number of unpaired electrons: In the presence of strong field ligands like $NH_3$ and $CO_3^{2-}$,the $6$ electrons in the $3d$ orbitals of $Co^{3+}$ pair up in the $t_{2g}$ set. Therefore,the number of unpaired electrons is $0$.

The coordination number,oxidation number,number of electrons in $d$-orbital,and number of unpaired electrons respectively in the complex $[Co(NH_3)_4CO_3]ClO_4$ are . . . . . . .

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