$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ in acidic medium converts into

$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ in acidic medium converts into
  1. $\mathrm{Cr}^{2+}$
  2. $\mathrm{Cr}^{3+}$
  3. $\mathrm{Cr}^{4+}$
  4. $\mathrm{Cr}^{5+}$

Solution

$\mathrm{Cr}_2 \mathrm{O}_7^{2-+} 14 \mathrm{H}^{+}+6 e^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}$

Asked in: NEET 2015 (Phase 2)

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