Question

Oxidation reaction involves loss of electrons, and reduction reaction involves gain of electrons. The reaction in which a species disproportionate into two oxidation states (lower and higher) is called disproportionate reaction
Which of the following statements is wrong?

• A. An acidified $K _2 Cr _2 O _7$ paper on being exposed to $SO _2$ turns green
• B. Mercuric chloride and stannous chloride cannot exist as such
• C. Iron turning on addition to $CuSO _4$ solution decolourises the blue colour
• D. $\left[ Cul _4\right]^{2-}$ is formed but $\left[ CuCl _4\right]^{2-}$ is not

Answer 1

Answer: (D)

a. $Cr _2 O _7^{2-}$ (orange red ) oxidises $SO _2 toSO _4^{2-}$ and is itself reduced to $Cr ^{+3}$ (green)
$\underset{\text{(Orange)}}{Cr _2 O _7^{2-}}+3 H ^{\oplus}+3 SO _2 \longrightarrow \underset{\text{(Green)}}{2 Cr ^{3+}}+3 SO _4{ }^{2-}+4 H _2 O$
b. Both reacts together
$\overset{+2}{ HgCl _2}+\overset{+2}{ SnCl _2} \longrightarrow \overset{+1}{\longrightarrow} \underset{\text{(White ppt)}}{Hg _2 Cl _2 \downarrow}+ \overset{4}{Sn}Cl _4$
$\overset{+1}{ Hg _2 Cl _2}+\overset{+2}{ SnCl _2} \longrightarrow \underset{\text{(Grey)}}{2 Hg \downarrow}+ \overset{+4}{SnCl _4}$
$\overset{0}{ Fe }+ \underset{\text{(Blue)}}{\overset{+2\,\, 2}{CuSO _4 }}\longrightarrow 2$
d. $\left[ CuCl _4\right]^{2-}$ is formed but $\left[ CuI _4\right]^{2-}$ is not
$I ^{\ominus}$ ion reduces $Cu ^{2+}$ to $CuI ^{+1}$ and itself undergoes oxidation to form $I _2$. However, $Cl ^{\ominus}$ does not reduce $Cu ^{2+}$