Question

The coordination number of $Th$ in $K_4\left[Th\right(C_2{O}_4{)}_4\left(O{H}_2{)}_2\right]$ is:
$C_2{O}_4^{2-}=\text{oxalato}$

• A. $14$
• B. $10$
• C. $8$
• D. $6$

Answer 1

The correct option is B $10$

which the metal is di rectly at tached. Here, $C_2{O}_4^{2-}$ is a bidentate ligand, and $H_{2}O$ is a monodentate ligand, So, total number of sites offered by $C_2{O}_4^{2-}$ and $H_{2}O$ ligands around $Th\left(IV\right)=\text{Coordination number of}Th\left(IV\right)$
$=4\times 2\left(by{C}_2{O}_4^{2-}\right)+2\times 1\left(by{H}_{2}O\right)=10$