JEE Main 2023 — Chemistry Inorganic Chemistry

2023mcqhard

When ${\mathrm{Cu}}^{2+}$ ion is treated with $\mathrm{KI}$ , a white precipitate, $X$ appears in solution. The solution is titrated with sodium thiosulphate, the compound $Y$ is formed. $X$ and $Y$ respectively are

Official previous-year question

Held on 31 Jan 2023 · Verified 6 Jul 2026.

Options

A
$X={\mathrm{Cu}}_{2}{I}_{2},Y={\mathrm{Na}}_{2}{S}_{4}{O}_{5}$
B
$X={\mathrm{Cu}}_{2}{I}_{2},Y={\mathrm{Na}}_{2}{S}_{4}{O}_{6}$
C
$X={\mathrm{CuI}}_{2},Y={\mathrm{Na}}_{2}{S}_{4}{O}_{3}$
D
$X={\mathrm{CuI}}_{2},Y={\mathrm{Na}}_{2}{S}_{4}{O}_{6}$

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Solution

The reaction between ${\mathrm{Cu}}^{2+}$ and $\mathrm{KI}$ will takes place as:

${\mathrm{Cu}}^{2+}+2\mathrm{KI}\longrightarrow \underset{\mathrm{Unstable}}{{\mathrm{CuI}}_{2}\downarrow }+2{K}^{+}$

${I}^{-}$ is a strong reducing agent, it reduces ${\mathrm{Cu}}^{2+}$ to ${\mathrm{Cu}}^{+}$ and a white precipitate of ${\mathrm{Cu}}_{2}{I}_{2}$ is formed.

$2{\mathrm{CuI}}_{2}\longrightarrow \underset{(\mathrm{White})'X'}{{\mathrm{Cu}}_{2}{I}_{2}\downarrow }+{I}_{2}$

Sodium thiosulphate is used as reducing agent to titrate iodine, with iodide being the product of the reaction.

$2{\mathrm{Na}}_{2}{S}_{2}{O}_{3}+{I}_{2}\rightarrow 2\mathrm{NaI}+\underset{(Y)}{{\mathrm{Na}}_{2}{S}_{4}{O}_{6}}$

So, the compound $X\text{and}Y$ are ${\mathrm{Cu}}_{2}{I}_{2}\text{and}{\mathrm{Na}}_{2}{S}_{4}{O}_{6}$ respectively.

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Options

$X={\mathrm{Cu}}_{2}{I}_{2},Y={\mathrm{Na}}_{2}{S}_{4}{O}_{5}$

$X={\mathrm{Cu}}_{2}{I}_{2},Y={\mathrm{Na}}_{2}{S}_{4}{O}_{6}$

$X={\mathrm{CuI}}_{2},Y={\mathrm{Na}}_{2}{S}_{4}{O}_{3}$

$X={\mathrm{CuI}}_{2},Y={\mathrm{Na}}_{2}{S}_{4}{O}_{6}$

Solution

The reaction between ${\mathrm{Cu}}^{2+}$ and $\mathrm{KI}$ will takes place as:

${\mathrm{Cu}}^{2+}+2\mathrm{KI}\longrightarrow \underset{\mathrm{Unstable}}{{\mathrm{CuI}}_{2}\downarrow }+2{K}^{+}$

${I}^{-}$ is a strong reducing agent, it reduces ${\mathrm{Cu}}^{2+}$ to ${\mathrm{Cu}}^{+}$ and a white precipitate of ${\mathrm{Cu}}_{2}{I}_{2}$ is formed.

$2{\mathrm{CuI}}_{2}\longrightarrow \underset{(\mathrm{White})'X'}{{\mathrm{Cu}}_{2}{I}_{2}\downarrow }+{I}_{2}$

Sodium thiosulphate is used as reducing agent to titrate iodine, with iodide being the product of the reaction.

$2{\mathrm{Na}}_{2}{S}_{2}{O}_{3}+{I}_{2}\rightarrow 2\mathrm{NaI}+\underset{(Y)}{{\mathrm{Na}}_{2}{S}_{4}{O}_{6}}$

So, the compound $X\text{and}Y$ are ${\mathrm{Cu}}_{2}{I}_{2}\text{and}{\mathrm{Na}}_{2}{S}_{4}{O}_{6}$ respectively.

JEE Main 2023 — Chemistry Inorganic Chemistry

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JEE Main 2023 — Chemistry Inorganic Chemistry

Options

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Solution

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JEE Main Inorganic Chemistry in other years

More JEE Main Inorganic Chemistry questions

Other Chemistry topics for JEE Main

Physical Chemistry

Organic Chemistry

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