Question

Chromium hydroxide on reaction with excess $NaOH$ gives a soluble compound. Guess the molecular formula.

• A. $\left[{Cr\left(OH\right)}_3\right]$
• B. ${\left[{Cr\left(OH\right)}_4\right]}^{-}$
• C. ${\left[{Cr\left(OH\right)}_5\right]}^{2-}$
• D. ${\left[{Cr\left(OH\right)}_6\right]}^{3-}$

Answer 1

The correct option is D ${\left[{Cr\left(OH\right)}_6\right]}^{3-}$

An excess of sodium hydroxide solution is added to a solution of the hexaaquachromium(III) ions to produce a solution of green hexahydroxochromate(III) ions.

$\left[Cr\right(H_{2}O{)}_6{]}^{3+}\stackrel{NaOH}{\to }\left[Cr\right(H_{2}O{)}_3\left(OH{)}_3\right]\stackrel{excessNaOH}{\to }\left[Cr\right(OH{)}_6{]}^{-3}$