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The oxidation state of $\mathrm{Cr}$ in $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}$is
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Verified Answer
The correct answer is:
$+3$
$+3$
$\left(\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right)^{+}$
$X+4 \times 0+2 \times-1=1$
$\mathrm{X}=+3$
$X+4 \times 0+2 \times-1=1$
$\mathrm{X}=+3$
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