The decreasing order of the boiling points of the following hydrides is: (i) \[N{H_3}\] (ii) \[P{H_3}\] (iii) \[As{H_3}\] (iv) \[Sb{H_3}\] (v) \[{H_2}O\]
A. (v) > (iv) > (i) > (iii) > (ii)
B. (v) > (i) > (ii) > (iii) > (iv)
C. (ii) > (iv) > (iii) > (i) > (v)
D. (iv) > (iii) > (i) > (ii) > (v)

Hint: Here, in this question, five hydrides are given. We have to arrange them in the decreasing order of boiling point. The temperature at which a liquid's vapour pressure equals the liquid's atmospheric pressure is known as its boiling point. The liquid converts into a vapour at this temperature. The liquid changes into vapour without increasing the temperature at this stage, thanks to the input of heat.

Complete Step by Step Solution:
An anion for the hydrogen atom is hydride. Hydrogen can react with a variety of elements on the periodic table. Chemical compounds containing hydrogen and hydride have nucleophilic, basic, and reducing characteristics. Except for a few noble gases, all elements of the periodic table are used to make hydride compounds.

The boiling point of a chemical is determined by its molecular mass. As the molecular mass of the group decreases, so does the boiling point.

However, the presence of hydrogen bonding in \[{H_2}O\] and \[N{H_3}\] gives them a very high boiling point. As a result, \[N{H_3}\] has a greater boiling point than \[P{H_3}\] and \[As{H_3}\].

\[{H_2}O\] has the greatest boiling point of all because it can create four hydrogen bonds with surrounding water molecules, whereas \[N{H_3}\] can only form one.
As a result, the correct boiling point order is:
\[{H_2}O > Sb{H_3} > N{H_3} > As{H_3} > P{H_3}\]

Therefore, the correct answer is option A (v)>(iv)>(i)>(iii)>(ii).

Note: Hydrogen bonding: It is a special type of dipole-dipole interaction between the highly electronegative elements (fluorine, oxygen, and nitrogen) and the hydrogen atom via covalent bonding. There are two types of hydrogen bonding: