Question

Hydrazine reacts with \[KI{O_3}\] in presence of $HCl$ as:
${N_2}{H_4} + IO_3^ - + 2{H^ + } + C{l^ - } \to ICl + {N_2} + 3{H_2}O$ .
The equivalent masses of ${N_2}{H_4}$ and $KI{O_3}$ respectively are:
A. $8,87$
B. $8,35.6$
C. $16,53.5$
D. $8,53.5$

Answer 1

Hint: Equivalent is defined as the mass of one equivalent which is the mass of a given substance that will combine with or displace a fixed quantity of another substance. Thus, equivalent weight can be defined, alternatively, as the division of molecular weight by the number of electrons transferred.

Complete step by step answer:
The change in oxidation takes place in a compound by the gain or loss of electrons. When the compound gains electrons, its oxidation number decreases and when the compound loses electrons, its oxidation number increases. Let us determine the oxidation number of the atoms in the compounds.
The oxidation number of some elements is fixed beforehand. They are:
$H = + 1$
$Cl = - 1$
$I = - 1$
$O = - 2$
Now let us find the number of electrons transferred or the n-factor in the reaction:
(i) ${N_2}{H_4}$ : Let the oxidation number of nitrogen atoms be $x$ .
$2x + 4( + 1) = 0$
$ \Rightarrow x = \dfrac{{ - 4}}{2} = - 2$
As the oxidation number changes from $ - 2$ in ${N_2}{H_4}$ to $0$ in ${N_2}$ , this means that there is a transfer of two electrons during the oxidation of nitrogen. Thus, the equivalent weight of ${N_2}{H_4}$ is:
${E_w} = \dfrac{M}{n}$
Where, $M = $ molar mass of ${N_2}{H_4} = 2 \times 14 + 1 \times 4 = 32$
${E_w} = \dfrac{{32}}{{2 \times 2}} = 8$
In the denominator, the n-factor is made twice because there are two atoms of nitrogen in the compound and the equivalent determined is for a single atom.
(ii) $KI{O_3}$ : Let the oxidation number of iodine in the compound be $y$ .
$1( + 1) + y + 3( - 2) = 0$
$ \Rightarrow y = + 5$
Let the oxidation number of iodine in $ICl$ be $z$ .
$z + 1( - 1) = 0$
$z = + 1$
As the oxidation number changes from $ + 5$ in $KI{O_3}$ to $ + 1$ in $ICl$ , this means that there is a transfer of two electrons during the oxidation of nitrogen. Thus, the equivalent weight of $KI{O_3}$ is:
${E_w} = \dfrac{M}{n}$
Where, $M = $ molar mass of $KI{O_3} = 39 + 127 + 16 \times 3 = 214$
Thus, ${E_w} = \dfrac{{214}}{4} = 53.5$

Thus, the correct option is D. $8,53.5$ .

Note:
During the determination of the equivalent weight of any atom in a compound, the number of equivalents is determined per atom of the element and if there is more than one atom, the n-factor is multiplied by the number of atoms present.