Chemistry - Surface Chemistry Question with Solution | TestHub

A. $AgN{O}_3$ is treated with excess of $KI$:

When silver nitrate ($AgN{O}_3$) is treated with excess potassium iodide ($KI$), silver iodide ($AgI$) is formed. Since $KI$ is in excess, there is an abundance of $I^{-}$ ions in the solution. The $AgI$ colloidal particles preferentially adsorb $I^{-}$ ions from the excess $KI$ solution onto their surface.

$$AgN{O}_3+KI(excess)\to AgI(colloidalparticle)+KN{O}_3$$

The $AgI$ particles adsorb $I^{-}$ ions:

$$AgI+I^{-}\to [AgI]I^{-}$$

This results in the formation of a negatively charged $AgI$ sol.

B. $Sn{O}_2$ is treated with $NaOH$:

Tin(IV) oxide ($Sn{O}_2$) is an amphoteric oxide. When treated with a strong base like sodium hydroxide ($NaOH$), it can react to form stannate ions, or more generally, the surface of $Sn{O}_2$ can adsorb $O{H}^{-}$ ions from the $NaOH$ solution.

$$Sn{O}_2+2O{H}^{-}\to Sn{O}_3^{2-}+H_{2}O$$

Alternatively, the $Sn{O}_2$ particles can preferentially adsorb $O{H}^{-}$ ions from the $NaOH$ solution onto their surface, leading to a negatively charged sol. This is common for metal oxide sols in basic media.

C. $H_{2}S$ is passed into $A{s}_2{O}_3$:

When hydrogen sulfide ($H_{2}S$) gas is passed into a solution of arsenic(III) oxide ($A{s}_2{O}_3$), arsenic(III) sulfide ($A{s}_2{S}_3$) sol is formed. The $A{s}_2{S}_3$ particles preferentially adsorb $S^{2-}$ ions (from the dissociation of $H_{2}S$) onto their surface.

$$A{s}_2{O}_3+3H_{2}S\to A{s}_2{S}_3(colloidalparticle)+3H_{2}O$$

The $A{s}_2{S}_3$ particles adsorb $S^{2-}$ ions:

$$A{s}_2{S}_3+S^{2-}\to [A{s}_2{S}_3]S^{2-}$$

This results in the formation of a negatively charged $A{s}_2{S}_3$ sol. This is a classic example of a negatively charged sol.

D. $FeC{l}_3$ is added to a fresh precipitate of $Fe(OH{)}_3$:

When ferric chloride ($FeC{l}_3$) is added to a freshly prepared precipitate of ferric hydroxide ($Fe(OH{)}_3$), a positively charged sol is formed. The $Fe(OH{)}_3$ particles preferentially adsorb $F{e}^{3+}$ ions from the $FeC{l}_3$ solution onto their surface.

$$Fe(OH{)}_3+F{e}^{3+}\to [Fe(OH{)}_3]F{e}^{3+}$$

This results in the formation of a positively charged $Fe(OH{)}_3$ sol.

Therefore, options A, B, and C describe conditions that lead to the formation of negatively charged sols.