The oxidation state and coordination number o | vedclass

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(B) $1$. The complex is $K_3[Cr(C_2O_4)_3]$.$2$. Let the oxidation state of $Cr$ be $x$.$3$. The oxalate ion $(C_2O_4)^{2-}$ has a charge of $-2$.$4$.

Vedclass · Accepted Answer

$+3$ and $6$

Vedclass · Answer

(B) $1$. The complex is $K_3[Cr(C_2O_4)_3]$.$2$. Let the oxidation state of $Cr$ be $x$.$3$. The oxalate ion $(C_2O_4)^{2-}$ has a charge of $-2$.$4$. Potassium $(K)$ has a charge of $+1$.$5$. The sum of oxidation states in a neutral complex is zero: $3(+1) + x + 3(-2) = 0$.$6$. $3 + x - 6 = 0$,which gives $x = +3$.$7$. The oxalate ion $(C_2O_4)^{2-}$ is a bidentate ligand,meaning it forms $2$ coordinate bonds per ligand.$8$. With $3$ oxalate ligands,the coordination number is $3 	imes 2 = 6$.$9$. Thus,the oxidation state is $+3$ and the coordination number is $6$.

The oxidation state and coordination number of $Cr$ in $K_3[Cr(C_2O_4)_3]$ are respectively:

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