Question

A to E are
A : ${Cd\left({CH}_{3}COO\right)}_2$
B : $[Fe\left({CH}_{3}COO\right){]}^{2+}$
C : $CdS$
D : $[{Cd\left({NH}_3\right)}_4{]}^{3+}$
E : $[{Cd\left(CN\right)}_4{]}^{2-}$
If true enter 1, else enter 0.

Answer 1

The compound A is ${Cd\left({CH}_{3}COO\right)}_2$
The compoundB is $[Fe\left({CH}_{3}COO\right){]}^{2+}$
The compoundC is $CdS$
The compoundD is $[{Cd\left({NH}_3\right)}_4{]}^{2+}$
The compoundE is $[{Cd\left(CN\right)}_4{]}^{2-}$
${Cd\left({CH}_{3}COO\right)}_2$ reacts with neutral ferric chloride solution to give blood red
colour of $[Fe\left({CH}_{3}COO\right){]}^{2+}$
${Cd\left({CH}_{3}COO\right)}_2+2FeC{l}_{3}2[Fe\left({CH}_{3}COO\right){]}^{2+}+CdC{l}_2+2C{l}_2$
${Cd\left({CH}_{3}COO\right)}_2$ reacts with hydrogen sulphide in presence of HCl to form
yellow ppt of $CdS$
${Cd\left({CH}_{3}COO\right)}_2+H_{2}S\stackrel{HCl}{\to }CdS\downarrow \left(yellow\right)+2C{H}_{3}CO{O}^{-}+2H^{+}$
$CdS$ is soluble in nitric acid but insoluble in yellow ammonium sulphide (YAS).
${Cd\left({CH}_{3}COO\right)}_2$ reacts with aq ammonia to form white ppt which is soluble
in excess of ammonia to form $[{Cd\left({NH}_3\right)}_4{]}^{2+}$
${Cd\left({CH}_{3}COO\right)}_2+4N{H}_3\to [{Cd\left({NH}_3\right)}_4{]}^{2+}+2C{H}_{3}CO{O}^{-}$
${Cd\left({CH}_{3}COO\right)}_2$ reacts with aq KCN solution to form white ppt which is
soluble in excess of aq KCN to form $[{Cd\left(CN\right)}_4{]}^{2-}$.
${Cd\left({CH}_{3}COO\right)}_2+4KCN\to [{Cd\left(CN\right)}_4{]}^{2-}+2C{H}_{3}CO{O}^{-}+4K^{+}$