At $\mathrm{pH}=4, \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ exists as

At $\mathrm{pH}=4, \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ exists as
  1. $\mathrm{CrO}_{4}^{2-}$
  2. $\mathrm{CrO}_{3}$
  3. $\mathrm{CrO}_{2}^{2+}$
  4. $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$

Solution

$\mathrm{pH}=4$ (i.e., acidic medium) no change but
at $\mathrm{pH}=12$ (i.e., basic medium)
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ changes to $\mathrm{CrO}_{4}^{2-}$
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+2 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{CrO}_{4}^{2-}+\mathrm{H}_{2} \mathrm{O}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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