Question

The reaction which proceeds in the forward direction is

• A. $N{H}_3+H_{2}O+NaCl\to N{H}_{4}Cl+NaOH$
• B. $2Cul+I_2+4K^{+}\to 2C{u}^{2+}+4KI$
  
• C. $F{e}_2{O}_3+6HCl\to 2FeC{l}_3+3H_{2}O$
• D. $SnC{l}_4+H{g}_{2}C{l}_2\to SnC{l}_2+2HgC{l}_2$

Answer 1

The correct option is C $F{e}_2{O}_3+6HCl\to 2FeC{l}_3+3H_{2}O$

Analysing the options

Option A:

$F{e}_2{O}_3+6HCl\to 2FeC{l}_3+3H_{2}O$

• $F{e}_2{O}_3$ is a basic oxide it will react with strong acid $HCl$ readily since acid-base reactions are spontaneous. Hence, this reaction proceeds in forward reaction.
• As complete hydrolysis of $FeC{l}_3$ will give $Fe(OH{)}_3.x{H}_{2}O$ , but $HCl$ will not be formed as a product of hydrolysis. So, reaction will not proceed in backward direction.

Option B:

$N{H}_3+H_{2}O+NaCl\to N{H}_{4}Cl+NaOH$

Since $NaOH$ is a strong base, it will react with $N{H}_{4}Cl$ (a salt of strong acid and weak base) to give $N{H}_{4}OH(N{H}_3+H_{2}O)\text{and}NaCl$

Also, since the salt of strong acid and strong base $\left(NaCl\right)$ does not undergo hydrolysis, so, reaction will proceed from strong to weak side i.e., in backward direction only.

Option C:

$SnC{l}_4+H{g}_{2}C{l}_2\to SnC{l}_2+2HgC{l}_2$

• In $SnC{l}_2$, $Sn$ exists in $+2$ oxidation state and in $SnC{l}_4$, it exists in $+4$.
• Since $S{n}^{2+}$ has reducing nature as $Sn$ is more stable in $+4$ state, reaction can go in backward direction.
• The $S{n}^{2+}$ ions convert readily to $S{n}^{4+}$ ions because the $Sn\left(IV\right)$ oxidation state is more stable than $Sn\left(II\right)$.
• Thus, this reaction proceeds in backward direction.

Option D:

$2Cul+I_2+4K^{+}\to 2C{u}^{2+}+4KI$

Because of the reducing nature of $I^{-}$(iodide ion) of strong electrolyte $KI$ (completely dissociated), $C{u}^{2+}$ gets reduced back to $CuI$. Thus, this reaction also goes in the backward direction also.

Hence, option (A) is correct.