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In the reaction
$X+\mathrm{I}_2+2 \mathrm{HCl} \longrightarrow \mathrm{SnCl}_4+2 \mathrm{HI}$
The correct option regarding $X$ is/are
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$X+\mathrm{I}_2+2 \mathrm{HCl} \longrightarrow \mathrm{SnCl}_4+2 \mathrm{HI}$
The correct option regarding $X$ is/are
Solution:
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Verified Answer
The correct answer is:
All of the above
$\mathrm{SnCl}_2+\mathrm{I}_2+2 \mathrm{HCl} \longrightarrow \mathrm{SnCl}_4+2 \mathrm{HI}$
$\mathrm{SnCl}_2$ is an strong reducing agent because it is readily oxidised into $\mathrm{Sn}^{4+}$ ion.
Since, Sn-atom in $\mathrm{SnCl}_2$ is $\mathrm{sp}^2$-hybridised.
So, it is an angular molecule.
On the other hand, it is used as a reagent in test of $\mathrm{Hg}^{2+}$ radical.
$\mathrm{SnCl}_2$ is an strong reducing agent because it is readily oxidised into $\mathrm{Sn}^{4+}$ ion.
Since, Sn-atom in $\mathrm{SnCl}_2$ is $\mathrm{sp}^2$-hybridised.
So, it is an angular molecule.
On the other hand, it is used as a reagent in test of $\mathrm{Hg}^{2+}$ radical.
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