Question

What happens when:
(A) Boric acid is added to water.
(B) Diborane reacts with ammonia
(C) $B{F_3}$ reacts with ammonia

Answer 1

This question is based on the concepts of chemical reactions of p-block elements. We need to understand the electron accepting and electron releasing abilities of p-block elements. Also, a lewis acid acts as electron acceptor while a lewis base donates electrons.

Complete answer:
(A) Boric acid is represented by the formula $B{(OH)_3}$ . In cold water, boric acid is slightly soluble, while it is readily soluble in hot water. The chemical reaction can be represented as follows:
$B{(OH)_3} + 2{H_2}O \to {[B{(OH)_4}]^ - } + {H_3}{0^ + }$
(B) Diborane can be represented by the formula ${B_2}{H_6}$ . It is also known as diboron hexahydride. Diborane is used as a reducing agent, a rubber vulcanizer, a catalyst for hydrocarbon polymerization, a flame-speed accelerator, and a doping agent in rocket propellants. It's also used in electronics to give pure crystals electrical properties.
The reaction of diborane with ammonia is shown below:
$3{B_2}{H_6} + 6N{H_3} \to 2{B_3}{N_3}{H_6} + 12{H_2}$
Diborane reacts with ammonia to produce borazine and hydrogen.
(C) $B{F_3}$ is a lewis acid, so it accepts electrons from a donor. $N{H_3}$ is a lewis base, so it donates electrons. The reaction between these two compounds results in the formation of a complete octet around $B$ in $B{F_3}$ . The reaction can be represented as:
$B{F_3} + :N{H_3} \to {F_3}B \leftarrow :N{H_3}$.

Note:
Since they have a tendency to lose an electron, P block elements are lustrous and typically good conductors of electricity and heat. In a P-block element like gallium, you'll find some incredible properties of elements. It's a metal that can easily melt in your palm. Silicon is also a key component of glass, making it one of the most common metalloids in the p-block community.