Copper sulphate is finally reduced to the metal on passing $\left(\mathrm{B}\right)$ through its solution. What is the $\left(\mathrm{B}\right)$?

An inorganic iodide (A) on heating with a solution of KOH gives a gas (B) and the solution of compound (C). Precipitate of compound $\left(\mathrm{E}\right)$ is formed on reaction of $\left(\mathrm{C}\right)$ with copper sulphate solution Identify the compound (E) and give chemical equations.

Consider the following sequence of reactions:

In the above sequence of reactions, $\mathrm{Y}$ and $\mathrm{A}$, respectively:

Consider the following reactions:

$\left(\mathrm{i}\right)$ ${\mathrm{NH}}_4\mathrm{Cl} \underset{∆}{\overset{\mathrm{NaOH}}{\to }}$ product (containing nitrogen).
$\left(\mathrm{ii}\right)$ ${PH}_4\mathrm{I}\underset{∆}{\overset{\mathrm{NaOH}}{\to }}$ product (containing phosphorus).
$\left(\mathrm{iii}\right)$ ${\mathrm{P}}_4\underset{∆}{\overset{\mathrm{NaOH}}{\to }}$ product (containing phosphorous in negative oxidation state).

The order of change in bond angle (for reactant $\to$ specified product) is: