Coordination Compounds 1 Question 7
7. The oxidation states of $Cr$, in $\left[Cr\left(H _2 O\right) _6\right] Cl _3,\left[Cr\left(C _6 H _6\right) _2\right]$, and $K _2\left[Cr(CN) _2(O) _2\left(O _2\right)\left(NH _3\right)\right]$ respectively are
(a) $+3,+4$ and +6
(b) $+3,+2$ and +4
(c) $+3,0$ and +6
(d) $+3,0$ and +4
(2018 Main)
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Answer:
Correct Answer: 7. (c)
Solution:
- Let the oxidation state of $Cr$ in all cases is ’ $x$ '
(i) Oxidation state of $Cr$ in $\left[Cr\left(H _2 O\right) _6\right] Cl _3$
$ x+(0 \times 6)+(-1 \times 3)=0 $
$ \text { or } \quad x+0-3=0 \text { or } x=+3 $
(ii) Oxidation state of $Cr$ in $\left[Cr\left(C _6 H _6\right) _2\right]$
$ x+(2 \times 0)=0 \quad \text { or } \quad x=0 $
(iii) Oxidation state of $Cr$ in
$ \begin{gathered} K _2\left[Cr(CN) _2(O) _2\left(O _2\right)\left(NH _3\right)\right] \\ 1 \times 2+x+(-1 \times 2)+(-2 \times 2)+(-2)+0=0 \\ \text { or } \quad 2+x-2-4-2=0 \text { or } x-6=0 \end{gathered} $
hence $x=+6$
Thus, $+3,0$ and $+6$ is the answer.
BETA
