Question

Give the oxidation state, d-orbital occupation and coordination number of the central metal ion in the following complexes:
$K_3\left[Co\right(C_2{O}_4{)}_3]$

$\left(N{H}_4{)}_2\right[Co{F}_4]$

$Cis-\left[Cr\right(en{)}_{2}C{l}_2]Cl$

$\left[Mn\right(H_{2}O{)}_6]S{O}_4$

Answer 1

$K_3\left[Co\right(C_2{O}_4{)}_3]$

$\begin{array}{c}(+1)3+x+(-2)\times 3=0\\ +3+x-6=0\\ x=+3\end{array}$
Thus, Co is present as $C{o}^{3+}$
$\underset{27}{Co}=\left[Ar\right]3d^{7}4s^2$
$C{o}^{3+}=\left[Ar\right]3d^6$ so d-orbital occupation is $d^{6}or{t}_{2g}^6{e}_g^0$
$\left(as{C}_2{O}_4^{2-}isstrongfieldligand\right)$
Coordination Number of Co = $3\times denticityof{C}^2{O}_4$
=$3\times 2\left(as{C}_2{O}_4^{2-}isabidentateligand\right)$= 6

$\left(N{H}_4{)}_2\right[Co{F}_4]$
$\begin{array}{c}(+1)\times 2x+(-1)\times 4=0\\ 2+x-4=0\\ x=+2\end{array}$
Thus, Co is present as $C{o}^{2+}$
$C{o}^{2+}=\left[Ar\right]3d^7$ so, d-orbital occupation is $d^{7}or{t}_{2g}^5{e}_g^2$
(as $F^{-}$ is a weak field ligand)
CN of Co = 4

$Cis-\left[Cr\right(en{)}_{2}C{l}_2]Cl$
$\begin{array}{c}x=\left(0\right)2+(-1)\times 2+(-1)=0\\ x+0-2-1=0\\ x=+3\end{array}$
Thus, Cr is present as $C{r}^{3+}.$
$C{r}^{3+}=\left[Ar\right]3d^3$ so, d-orbital occupation is $d^{3}or{t}_{2g}^3,e{g}^0$
CN of Cr =$2\times x$ denticity of en $+2=2\times 2+2=6$

$\left[Mn\right(H_{2}O{)}_6]S{O}_4$
$\begin{array}{c}x+\left(0\right)6+(-2)=0\\ x=+2\end{array}$
Thus, Mn is present as $M{n}^{2+}$.
$M{n}^{2+}=\left[Ar\right]3d^5$ So, d-orbital occupation is $d^{5}or{t}_{2g}^3{e}_g^2$
CN of Mn = 6