Question

The oxidation states of Cr in $\left[Cr\left(H_{2}O\right)\right]{Cl}_3,\left[{Cr\left(C_6{H}_6\right)}_2\right],$ and $K_2\left[Cr{\left(CN\right)}_2{\left(O\right)}_2{\left(O\right)}_2\left({NH}_3\right)\right]$ respectively are :

• A. +3,+4, and +6
• B. +3,+2, and +4
• C. +3, 0, and +6
• D. +3, 0, and +4

Answer 1

The correct option is C +3, 0, and +6

In the given molecules,

1. In $\left[Cr\right(H_{2}O\left)\right]C{l}_3$, let oxidation number of Cr be x, Here the molecule separates as its ions as, $\left[Cr\right(H_{2}O){]}^{3+}$ and $3C{l}^{-}$. As $H_{2}O$ is a neutral molecule its oxidation number is zero. Hence, $x=+3$
2. In $\left[Cr\right(C_6{H}_6{)}_2]$ let oxidation number of Cr be y, Benzene is a neutral molecule and hence on observing the cross multiplied factor we get the oxidation number of Cr as 0. Hence,$y=0$.
   
3. In $K_2\left[Cr\right(CN{)}_2\left(O{)}_2\right(N{H}_3\left)\right]$,let oxidation number of Cr be z, Here the ions are- $2K^{+}$ and $\left[Cr\right(CN{)}_2\left(O{)}_2\right(N{H}_3){]}^{2-}$. As $N{H}_3$ is neutral molecule and $O^{-}$and $C{N}^{-}$ is $2\times (-1)+4\times (-1)=-6$ and hence oxidation number of Cr is, z-6=-0$⟹z=+6$
   

Hence,option C is correct answer.