The oxidation states of Cr in [Cr(H$$_2$$O)$$_6$$]Cl$$_3$$, [Cr(C$$_6$$H$$_6$$)$$_2$$] and K$$_2$$[Cr(CN)$$_2$$(O)$$_2$$(O$$_2$$)(NH$$_3$$)], respectively, are:

We have to find the oxidation state (O.S.) of chromium in each of the three complexes one by one. In every case we shall write:

$$\text{(Oxidation state of Cr)} + \sum \text{(charges of ligands)} = \text{charge on the whole complex species}$$

and then solve for the unknown oxidation state.

First complex  $$[\,\text{Cr(H}_2\text{O)}_6]Cl_3$$

The complete compound is electrically neutral, so the algebraic sum of all charges must be zero.

Each chloride ion $$Cl^-$$ has a charge $$-1$$ and there are three of them, so the total charge contributed by the counter-ions is $$3(-1)=-3.$$ This means the complex cation $$[\,\text{Cr(H}_2\text{O)}_6]^{3+}$$ must balance it with a charge $$+3$$.

Inside the coordination sphere every $$\text{H}_2\text{O}$$ ligand is neutral (charge $$0$$). Let the oxidation state of Cr be $$x$$. Applying the above formula,

$$x + 6(0) = +3 \;\Longrightarrow\; x = +3.$$\p>

So the oxidation state of Cr in the first complex is $$+3.$$

Second complex  $$[\,\text{Cr(C}_6\text{H}_6)_2]$$

The entire species shown inside the square brackets is the whole compound; there is no counter-ion, so its overall charge is $$0.$$

A benzene ring $$C_6H_6$$ acting as an η6-ligand is a neutral donor, hence each $$C_6H_6$$ contributes a charge $$0.$$\p>

If the oxidation state of Cr is again called $$x$$, we write

$$x + 2(0) = 0 \;\Longrightarrow\; x = 0.$$

Thus, chromium is in the zero oxidation state in $$[\,\text{Cr(C}_6\text{H}_6)_2].$$

Third complex  $$K_2[\,\text{Cr(CN)}_2(\text{O})_2(\text{O}_2)(\text{NH}_3)]$$

There are two potassium ions $$K^+$$ outside, each with charge $$+1,$$ so together they give $$+2.$$ Therefore the complex ion must have charge $$-2$$ to make the whole salt neutral:

$$[\,\text{Cr(CN)}_2(\text{O})_2(\text{O}_2)(\text{NH}_3)]^{2-}$$

Now list the charges of all ligands inside the brackets.

• $$\text{CN}^-$$ (cyano) has charge $$-1,$$ and there are two: total $$2(-1)=-2.$

• $$\text{O}^{2-}$$ (oxo) has charge $$-2,$$ and there are two: total $$2(-2)=-4.$

• $$\text{O}_2^{2-}$$ (peroxo) acts as a ligand with charge $$-2.$$ There is one such ligand: total $$-2.$$

• $$\text{NH}_3$$ (ammine) is neutral: charge $$0.$$

Let the oxidation state of chromium be $$x.$$ Substituting every charge into the master equation,

$$x + \big[-2\;(\text{from }2\,\text{CN}^-)\big] + \big[-4\;(\text{from }2\,\text{O}^{2-})\big] + \big[-2\;(\text{from }\text{O}_2^{2-})\big] + 0 = -2.$$

Simplifying the left-hand side,

$$x - 2 - 4 - 2 = -2 \;\Longrightarrow\; x - 8 = -2.$$

Adding $$8$$ to both sides gives

$$x = +6.$$

So the oxidation state of Cr in this anion is $$+6.$$

Collecting the three results, the oxidation states of Cr are $$+3,\,0,$$ and $$+6$$ respectively.

Comparing with the given options, these values match Option D.

Hence, the correct answer is Option D.