Question

1. The oxidation state of Boron is $+ 3$ except in
(1) ${(BN)_1}$
(2) ${B_3}{N_3}{H_6}$
(3) ${B_2}C{l_4}$
(4) $Mg{(B{O_2})_2}$
2. $F{e_{0.95}}O \to F{e_2}{O_3}$
In the above reaction equivalent mass of reactant is-
(1) $\dfrac{M}{{0.95}}$
(2) $0.85 \times M$
(3) $\dfrac{{0.95M}}{{0.85}}$
(4) $\dfrac{M}{{0.85}}$

Answer 1

As we are aware with the oxidation states of group thirteen elements that is generally $+ 3$ and $+ 1$ and Boron possess the three valence electron hence show only $+ 3$ oxidation state which decreases down the group.

Complete Step by step answer:
- As we know very well about the oxidation states of group thirteen elements that are generally $+ 3$ and $+ 1$ and Boron possesses the three valence electrons hence show only $+ 3$ oxidation state which decreases down the group. We can easily calculate the oxidation state of these compounds like in the first compound the nitrogen is present in an oxidation state of $- 3$ so the boron will be present in $+ 3$ oxidation state.
- In the second option, boron is present in $+ 1$ oxidation state as nitrogen is in $- 3$ and hydrogen is in $+ 1$oxidation state. So by calculating the oxidation number for boron we get $+ 1$. Similarly in the third option chlorine is present in $- 1$ oxidation state and on solving we get boron in oxidation state of $+ 2$. In the last option magnesium is present in $+ 2$ oxidation state and oxygen is in $- 2$ oxidation state on solving for boron we get $+ 3$ oxidation state.

Therefore the correct answer is (2) and (3).

2. as we know that equivalent weight of an element or compound is that weight which is either reacted or displaced from $1g$ hydrogen, $8g$ oxygen or $35.5g$ chlorine and it basically calculated using the formula:
$Eq.\;wt. = \dfrac{{molecular\;weight}}{{n - factor}}$ where n-factor is calculated by change in oxidation state.
Thus, when $F{e_{0.95}}O \to F{e_2}{O_3}$ then the oxidation state of iron changes from $+ 2$ to $+ 3$. But iron in reactant is present in $0.95$ quantity so the n-factor can be calculated here as:
$n - factor = \left( {3 - \dfrac{2}{{0.95}}} \right) \times 0.95 = 0.85$
So the equivalent weight will be:
$Eq.\;wt. = \dfrac{{molecular\;weight}}{{n - factor}}$
$\Rightarrow Eq.\;wt. = \dfrac{M}{{0.85}}$

Hence the correct answer is (4).

Note: Equivalent weight of acids and base can also be calculated using the same formula where n-factor is simply replaced by the acidity and basicity of the acid and base respectively. basicity is the number of replaceable hydrogen present in acid and acidity is the number of hydroxide ions produced in solution by base.