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(B) For $[Co(NH_3)_6]Cl_2$:$Co$ is in $+2$ oxidation state. The electronic configuration of $Co^{2+}$ is $[Ar] 3d^7$.Since $NH_3$ is a weak field liga

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$3$

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(B) For $[Co(NH_3)_6]Cl_2$:$Co$ is in $+2$ oxidation state. The electronic configuration of $Co^{2+}$ is $[Ar] 3d^7$.Since $NH_3$ is a weak field ligand for $Co^{2+}$,no pairing occurs.The configuration is $t_{2g}^5 e_g^2$,resulting in $3$ unpaired electrons.For $[Co(NH_3)_6]Cl_3$:$Co$ is in $+3$ oxidation state. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.Since $NH_3$ is a strong field ligand for $Co^{3+}$,pairing occurs.The configuration is $t_{2g}^6 e_g^0$,resulting in $0$ unpaired electrons.Total number of unpaired electrons = $3 + 0 = 3$.

List $-I$	List $-II$
$A$. $[PtCl_4]^{2-}$	$I$. $sp^3d$
$B$. $BrF_5$	$II$. $d^2sp^3$
$C$. $PCl_5$	$III$. $dsp^2$
$D$. $[Co(NH_3)_6]^{3+}$	$IV$. $sp^3d^2$

The total number of unpaired electrons present in $[Co(NH_3)_6]Cl_2$ and $[Co(NH_3)_6]Cl_3$ is $......$

Similar Questions

From the magnetic behaviour of $[NiCl_4]^{2-}$ (paramagnetic) and $[Ni(CO)_4]$ (diamagnetic),choose the correct geometry and oxidation state.

Hybridization in $[Ni(CO)_4]$ and $[Ni(CN)_4]^{2-}$ are respectively . . . . . . .

Match List $-I$ with List $-II$:

List $-I$ List $-II$

$A$. $[PtCl_4]^{2-}$ $I$. $sp^3d$

$B$. $BrF_5$ $II$. $d^2sp^3$

$C$. $PCl_5$ $III$. $dsp^2$

$D$. $[Co(NH_3)_6]^{3+}$ $IV$. $sp^3d^2$

Among $Ni(CO)_4, [Ni(CN)_4]^{2-}$ and $[NiCl_4]^{2-}$,which of the following statements is correct?

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