The formula for potassium dicyanobis(oxalato) | vedclass

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Question

(A) $1$. The central metal atom is Nickel $(Ni)$ with an oxidation state of $+2$.$2$. The ligands are two cyano groups $(CN^-)$ and two oxalato groups

Vedclass · Accepted Answer

$K_4[Ni(CN)_2(C_2O_4)_2]$

Vedclass · Answer

(A) $1$. The central metal atom is Nickel $(Ni)$ with an oxidation state of $+2$.$2$. The ligands are two cyano groups $(CN^-)$ and two oxalato groups $(C_2O_4^{2-})$.$3$. The charge on the coordination sphere is calculated as: $x + 2(-1) + 2(-2) = +2 - 2 - 4 = -4$. Wait,let's re-calculate: $Ni$ is $+2$,$2(CN^-)$ is $-2$,$2(C_2O_4^{2-})$ is $-4$. Total charge = $+2 - 2 - 4 = -4$.$4$. To balance a charge of $-4$,we need $4$ potassium ions $(K^+)$.$5$. Thus,the formula is $K_4[Ni(CN)_2(C_2O_4)_2]$.

The formula for potassium dicyanobis(oxalato)nickelate $(II)$ is:

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