Q4 / 50 · Manipal (MET) 2024 · Chemistry

The oxidation state of Cr in $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3,\left[\mathrm{Cr}\left(\mathrm{C}_6 \mathrm{H}_6\right)_2\right]$, $\mathrm{K}_2\left[\mathrm{Cr}(\mathrm{CN})_2(\mathrm{O})_2\left(\mathrm{O}_2\right)\left(\mathrm{NH}_3\right)\right]$ respectively are

  • A
    $+3,+4,+6$
  • B
    $+3,+2,+4$
  • Correct answer
    C
    $+3,0,+6$
  • D
    $+3,0,+4$

Solution

Oxidation state of Cr , $\begin{aligned} & {In} \left[\mathrm{Cr}\left(\mathrm{H}_2O\right)_6\right] \mathrm{Cl}_3 \\ & x+0 \times 6-1 \times 3=0 \\ & x=3...(i) \end{aligned}$ $\ln \left[\mathrm{Cr}\left(\mathrm{C}_6 \mathrm{H}_6\right)_2\right]$ $\begin{aligned} & x+0 \times 2=0 \\ & x=0..(ii)\end{aligned}$ $\text { In } \mathrm{K}_2\left[\mathrm{Cr}(\mathrm{CN})_2(\mathrm{O})_2\left(\mathrm{O}_2\right)\left(\mathrm{NH}_3\right)\right]$ $(1 \times 2)+x-(1 \times 2)-(2 \times 2)-(1 \times 2)-(0 \times 1)=0$ $x-6=0$ $x=+6...(iii)$ Therefore, oxidation states are $+3,0,+6$.

Hence, the correct answer is option: 'C'.

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