JEE MAIN - Chemistry (2024 - 27th January Morning Shift - No. 15)

Yellow compound of lead chromate gets dissolved on treatment with hot $$\mathrm{NaOH}$$ solution. The product of lead formed is a :
Tetraanionic complex with coordination number six
Neutral complex with coordination number four
Dianionic complex with coordination number six
Dianionic complex with coordination number four

Explanation

Lead chromate ( $$\mathrm{PbCrO}_4$$ ) dissolves in hot NaOH solution to form products due to a chemical reaction. The reaction involves the formation of a compound where lead ( $$\mathrm{Pb}$$ ) is coordinated by hydroxide ions ( $$\mathrm{OH}^-$$ ).

The balanced chemical reaction is:

$$\mathrm{PbCrO}_4 + 4\mathrm{NaOH} (hot, excess) \rightarrow \left[\mathrm{Pb}(\mathrm{OH})_4\right]^{2-} + \mathrm{Na}_2\mathrm{CrO}_4$$

In this reaction, lead forms a dianionic complex ( $$\left[\mathrm{Pb}(\mathrm{OH})_4\right]^{2-}$$ ) with hydroxide ions. This indicates that the lead is in the center of a complex with a coordination number of four, meaning it is directly bound to four hydroxide ions. Therefore, the correct description of the product formed with lead is a dianionic complex with a coordination number of four.

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