The IUPAC name for K_2[Cr(CN)_2O_2(O_2)NH_3] | vedclass

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(D) To determine the $IUPAC$ name of $K_2[Cr(CN)_2O_2(O_2)NH_3]$,we first calculate the oxidation state of $Cr$.Let the oxidation state of $Cr$ be $x$

Vedclass · Accepted Answer

Potassium amminedicyanodioxoperoxo chromate $(VI)$

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(D) To determine the $IUPAC$ name of $K_2[Cr(CN)_2O_2(O_2)NH_3]$,we first calculate the oxidation state of $Cr$.Let the oxidation state of $Cr$ be $x$.The charge on $K$ is $+1$,$CN$ is $-1$,$O$ is $-2$,$O_2$ (peroxo) is $-2$,and $NH_3$ is $0$.$2(+1) + x + 2(-1) + 1(-2) + 1(-2) + 0 = 0$$2 + x - 2 - 2 - 2 = 0$$x - 4 = 0$,so $x = +6$.The complex anion is $[Cr(CN)_2O_2(O_2)NH_3]^{2-}$.Naming the ligands in alphabetical order: ammine $(NH_3)$,cyano $(CN^-)$,dioxo $(O^{2-})$,and peroxo $(O_2^{2-})$.Since there are two cyano,two oxo,and one peroxo ligand,the name is Potassium amminedicyanodioxoperoxo chromate $(VI)$.

The $IUPAC$ name for $K_2[Cr(CN)_2O_2(O_2)NH_3]$ is

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