The coordination number of Th in K_4[Th(C_2O_ | vedclass

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(D) In the complex $K_4[Th(C_2O_4)_4(OH_2)_2]$,the central metal atom is $Th$.The ligands present are:$1$. Four oxalate ions $(C_2O_4^{2-})$,which are

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$10$

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(D) In the complex $K_4[Th(C_2O_4)_4(OH_2)_2]$,the central metal atom is $Th$.The ligands present are:$1$. Four oxalate ions $(C_2O_4^{2-})$,which are bidentate ligands. Contribution = $4 	imes 2 = 8$.$2$. Two water molecules $(OH_2)$,which are monodentate ligands. Contribution = $2 	imes 1 = 2$.The coordination number is the sum of the contributions of all ligands attached to the central metal atom.Coordination number = $8 + 2 = 10$.

The coordination number of $Th$ in $K_4[Th(C_2O_4)_4(OH_2)_2]$ is
$(C_2O_4^{2-} = \text{Oxalato})$

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The coordination number of $Th$ in $K_4[Th(C_2O_4)_4(OH_2)_2]$ is$(C_2O_4^{2-} = \text{Oxalato})$