In the complex K[Cr(H_2O)_2(C_2O_4)_2] cdot 3 | vedclass

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(D) $1$. The complex is $K[Cr(H_2O)_2(C_2O_4)_2] \cdot 3H_2O$. $2$. Let the oxidation state of $Cr$ be $x$. $3$. The charge on $K$ is $+1$,$H_2O$ is $

Vedclass · Accepted Answer

$+3, 6$

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(D) $1$. The complex is $K[Cr(H_2O)_2(C_2O_4)_2] \cdot 3H_2O$. $2$. Let the oxidation state of $Cr$ be $x$. $3$. The charge on $K$ is $+1$,$H_2O$ is $0$,and $C_2O_4^{2-}$ is $-2$. $4$. The equation is: $1 + x + 2(0) + 2(-2) = 0$. $5$. $1 + x - 4 = 0$,which gives $x = +3$. $6$. The coordination number is determined by the ligands attached to the central metal: $2(H_2O)$ are monodentate $(2 	imes 1 = 2)$ and $2(C_2O_4^{2-})$ are bidentate $(2 	imes 2 = 4)$. $7$. Total coordination number = $2 + 4 = 6$. $8$. Thus,the oxidation state is $+3$ and the coordination number is $6$.

In the complex $K[Cr(H_2O)_2(C_2O_4)_2] \cdot 3H_2O$,the oxidation state and coordination number of the central metal ion are . . . . . . and . . . . . . respectively.

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