For the coordination compound [Co(NH_3)_5CO_3 | vedclass

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(A) $1$. The coordination number of $Co$ is determined by the ligands: $5$ from $NH_3$ (monodentate) and $1$ from $CO_3^{2-}$ (bidentate),so $5 + 2 =

Vedclass · Accepted Answer

$6, 3, 6, 0$

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(A) $1$. The coordination number of $Co$ is determined by the ligands: $5$ from $NH_3$ (monodentate) and $1$ from $CO_3^{2-}$ (bidentate),so $5 + 2 = 7$ is incorrect; wait,$CO_3^{2-}$ acts as a bidentate ligand,so $5 + 2 = 7$. However,in this specific complex $[Co(NH_3)_5CO_3]^+$,the $CO_3^{2-}$ is often treated as monodentate in some contexts,but standard $IUPAC$ considers it bidentate. Let's re-evaluate: $5(NH_3) + 1(CO_3) = 6$ coordination sites. $2$. Oxidation state: $x + 5(0) + 1(-2) = +1$ (since $ClO_4$ is $-1$),so $x - 2 = +1$,$x = +3$. $3$. Electronic configuration of $Co$ $(Z=27)$ is $[Ar] 3d^7 4s^2$. For $Co^{3+}$,it is $[Ar] 3d^6$. $4$. Number of electrons in $d$-orbitals is $6$. $5$. Since $NH_3$ is a strong field ligand,it causes pairing. For $d^6$ in an octahedral field,all electrons are paired,so unpaired electrons = $0$.

List-$I$ (Hybridisation)	List-$II$ (Compound/ion)
$A. sp^3d$	$I. [PtCl_4]^{2-}$
$B. sp^3d^2$	$II. SF_6$
$C. dsp^2$	$III. BCl_3$
$D. dsp^3$	$IV. PCl_5$
	$V. ClF_3$

For the coordination compound $[Co(NH_3)_5CO_3]ClO_4$,what are the coordination number,oxidation state,number of electrons in the $d$-orbitals,and the number of unpaired electrons for the central metal $Co$?

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Match the following:
List-$I$ (Hybridisation) List-$II$ (Compound/ion)
$A. sp^3d$ $I. [PtCl_4]^{2-}$
$B. sp^3d^2$ $II. SF_6$
$C. dsp^2$ $III. BCl_3$
$D. dsp^3$ $IV. PCl_5$
$V. ClF_3$

The correct match is:

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