Question

IUPAC name of \[{{K}_{2}}[Cr{{(CN)}_{2}}{{O}_{2}}({{O}_{2}})N{{H}_{3}}]\]
A. Potassium amminedicyanotetraoxo chrominum (III)
B. Potassium amminedicyanodioxygendioxo chromate (IV)
C. Potassium amminedicyanosuperoxoperoxo chromate (III)
D. Potassium amminedicyanodioxoperoxo chromate (VI)

Answer 1

Hint: At first we have to determine the oxidation state of the central metal atom that is $Cr$ . As potassium is present outside the coordination sphere so, it is an anionic complex. For anionic sphere complex like the given one the name of the cation is written first.

Complete Step by Step Solution:
The given compound is \[{{K}_{2}}[Cr{{(CN)}_{2}}{{O}_{2}}({{O}_{2}})N{{H}_{3}}]\].
We know that amine group is a neutral group so, it does not contribute any charge. Again this complex has one amine groups so, the naming for amine is ammine.
Here potassium is present as a cation and have charge of $+1$ .
The $CN$ group has $-1$ charge and the term for this group is cyano and as two groups are present so, dicyano.
One ${{O}_{2}}$ is present as oxo and one as peroxo. Here two oxo groups are present so, dioxo is written.
Here the central metal atom is chromium.
The symbol of chloride ligand is $Cl$ .
Let the oxidation state of $Cr$ is $x$ .
Again chloride is a negative ligand so, the charge of chloride ligand is $-1$ .
The charge due to amine is zero.
Thus we can write as follows-
$x+0\times 1+(-2)-1+2-2\times 2-1=0$
$x-6=0$
$x=+6$
So, the value of $x$ by solving the above equation is $+6$ .
So, the oxidation state of the central metal atom chromium is also $+6$ .
In alphabetical order first amine, then cyano followed by oxo and dioxo groups come in IUPAC naming.
Thus the IUPAC name of the given complex is \[{{K}_{2}}[Cr{{(CN)}_{2}}{{O}_{2}}({{O}_{2}})N{{H}_{3}}]\] is potassium amminedicyanodioxoperoxo chromate (VI).
Thus the correct option is D.

Note: Oxygen can bind as bridging or terminal ligand. The type of binding of oxygen decides its name.