Question

Which of the following are the white precipitate and the soluble substance formed by the excess of the NaOH reagent, respectively ?

• A. $AsOClandAs{O}_3^{3-}$
• B. $SbOClandSb{O}_3^{3-}$
• C. $Zn\left(OH{)}_{2}and\right[Zn(N{H}_3{)}_4{]}^{2-}$
• D. $Cd\left(OH{)}_{2}and\right[Cd(OH{)}_4{]}^{2-}$

Answer 1

Answer: (D)

$Cd\left(OH{)}_{2}and\right[Cd(OH{)}_4{]}^{2-}$

Here compound contains $C{d}^{2+}$ as a cation which reacts with NaOH and form

$C{d}^{2+}+NaOH\to Cd(OH{)}_2\to Cd(OH{)}_4^{-2}$

The yellow precipitate of a sulphite could be $CdS$ (group IIA) or $A{s}_2{S}_3/A{s}_2{S}_3/Sn{S}_2$ (group IIB), but as it is insoluble in yellow ammonium polysulphide, it should be $CdS$. Thus, the cation appears to be $C{d}^{2+}$ which is confirmed by reactions (ii) to (v). Reaction (ii) to (v). Reaction (vi) indicates $S{O}_4^{2-}$. Hence , the compound B is $CdS{O}_4$.

(i) $C{d}^{2+}$ $+$ $S^{2-}$ $\to$ $CdS\downarrow$

$\left(CdS{O}_4\right)$ $\left(H_{2}S\right)$ (yellow)

(ii) $CdS\downarrow +2H^2\to C{d}^{2+}+H_{2}S\uparrow$ $CdS\stackrel{(N{H}_4{)}_2{S}_x}{\to }$ No action

$\left(HN{O}_3\right)$ A

(iii) $C{d}^{2+}$ $+$ $2O{H}^{-}$ $\to$ $Cd(OH{)}_2\downarrow \underset{excess}{\overset{4N{H}_3}{\to }}$ $Cd\left[\right(N{H}_3{)}_4{]}^{2+}+2O{H}^{-}$

$\left(CdS{O}_4\right)$ $\left(N{H}_{4}OH\right)$ (white) tetramminecadmium (II) ion

B soluble C

(iv) $C{d}^{2+}$ $+$ $2C{N}^{-}$ $\to$ $Cd(CN{)}_2\downarrow$ $\underset{\left(KCN\right)}{\overset{2C{N}^{-}}{\to }}$ $\left[Cd\right(CN{)}_4{]}^{2-}$

$\left(CdS{O}_4\right)$ $\left(KCN\right)$ (White) tetracycnocadmiate (II) ion

B (soluble) D

(v) The tetracyanocadmiate (II) ion, formed in (iv) has a low stability constant value and , therefore , furnishes $C{d}^{2+}$ ions in sufficient concentration to give a precipitate of $CdS$ (A).

$\left[Cd\right(CN{)}_4{]}^{2-}\rightleftharpoons C{d}^{2+}+4C{N}^{-}$

$\downarrow S^{2-}$

$CdS\downarrow$

A

(vi) $B{a}^{2+}$ $+$ $S{O}_4^{2-}$ $\to$ $BaS{O}_4\downarrow$

$\left(BaC{l}_2\right)$ $\left(CdS{O}_4\right)$ (White)

Insoluble in $HCl$

or $HN{O}_3$

Thus, A is $CdS$ , B is $CdS{O}_4$ , C is $\left[Cd\right(N{H}_3{)}_4{]}^{2+}$ , D is $\left[Cd\right(CN{)}_4{]}^{2-}$.