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EEJ MAIN Chemistry QUESTION #7534
Question 1

Find the oxidation states of Cr in $[\text{Cr(H}_2\text{O)}_6]\text{Cl}_3$, $[\text{Cr(C}_6\text{H}_6)_2]$, and $\text{K}_2[\text{Cr(CN)}_2(\text{O})_2(\text{O}_2)(\text{NH}_3)]$ respectively.

  • +3, 0, and +6✔️
  • +3, 0, and +4
  • +3, +4, and +6
  • +3, +2, and +4
Correct Answer Explanation

$[\text{Cr(H}_2\text{O)}_6]\text{Cl}_3$: H$_2$O is neutral, Cl$^-$ is $-1$, complex has charge 3+. So Cr = +3.

$[\text{Cr(C}_6\text{H}_6)_2]$: Benzene is a neutral ligand (0 charge). Complex is neutral. So Cr = 0.

$\text{K}_2[\text{Cr(CN)}_2(\text{O})_2(\text{O}_2)(\text{NH}_3)]$: K$^+$ gives $+2$ outside; complex ion has $-2$ charge. CN$^-$ ($-1$ each, $2\times-1=-2$); O$^{2-}$ ($-2$ each, $2\times-2=-4$); O$_2^{2-}$ (peroxo, $-2$); NH$_3$ (0). Let Cr = $x$: $x + (-2) + (-4) + (-2) + 0 = -2 \Rightarrow x - 8 = -2 \Rightarrow x = +6$.

Oxidation states: +3, 0, +6.