In aqueous acidified solution, CrO$_4^{2-}$ ion converts to which of the following ?

Question

Read the following passage and answer the next five questions based on it.
The transition metals are very hard and have low volatility. Their melting and boiling points are high. In any row, the melting points of these metals rise to a maximum at d5 and fall regularly as atomic number increases. The high melting points of these metals are attributed to the involvement of greater number of electrons from (n– 1)d in addition to ns electrons in the interatomic metallic bonding.

Answer 1

Step 1: Chromate-dichromate equilibrium.
In aqueous solution: $2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O$. Yellow chromate and orange dichromate coexist, with the ratio governed by pH.
Step 2: Effect of acidification.
Adding acid increases $[H^+]$. By Le Chatelier's principle, the equilibrium shifts to the right, converting chromate to dichromate. The solution turns from yellow to orange.
Step 3: Product identified.
$CrO_3$ and $Cr^{3+}$ are not formed under simple acidification; the product is dichromate ion.
\[ \boxed{Cr_2O_7^{2-}} \]

List-I	Transition metal/compound
A	V₂O₅
B	Fe
C	PdCl₂
D	Ni complex

List-II	Catalytic role
I	Preparation of ammonia from N₂/H₂ mixture
II	Polymerisation of alkynes
III	Oxidation of SO₂ to SO₃
IV	Preparation of ethyne to ethene

In aqueous acidified solution, CrO$_4^{2-}$ ion converts to which of the following ?

Show Hint

The Correct Option is C

Solution and Explanation

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