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. (a)
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.
