Consider the complex [Co(NH_3)_4 CO_3]ClO_4. | vedclass

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(A) The complex is $[Co(NH_3)_4 CO_3]ClO_4$. $NH_3$ is a monodentate ligand $(4 	imes 1 = 4)$ and $CO_3^{2-}$ is a bidentate ligand $(1 	imes 2 = 2)

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$6, 3, 6$

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(A) The complex is $[Co(NH_3)_4 CO_3]ClO_4$. $NH_3$ is a monodentate ligand $(4 	imes 1 = 4)$ and $CO_3^{2-}$ is a bidentate ligand $(1 	imes 2 = 2)$. Coordination number $= 4 + 2 = 6$. Let the oxidation state of $Co$ be $x$. $x + 4(0) + (-2) = +1$ (since $ClO_4^-$ has a charge of $-1$). $x - 2 = +1 \implies x = +3$. The atomic number of $Co$ is $27$. The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$. Thus,the number of $d$-electrons is $6$.

Consider the complex $[Co(NH_3)_4 CO_3]ClO_4$. The coordination number,oxidation number,and number of $d$-electrons on the metal are respectively:

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