Question: The conjugate acid of \[{\text{NH}}_{\text{2}}^{\text{ - }}\] is A.\[{\text{N}}{{\text{H}}_3}\] ...

Question

The conjugate acid of ${\text{NH}}_{\text{2}}^{\text{ - }}$ is
A. ${\text{N}}{{\text{H}}_3}$
B. ${\text{N}}{{\text{H}}^{2 - }}$
C. ${\text{NH}}_4^ +$
D. ${\text{N}}_3^ +$

Answer 1

Solution

If conjugate acid is present for any species then the given species will be base. A base is that species which gains protons or ${{\text{H}}^{\text{ + }}}$ .

Complete step by step answer:
An acid is a substance which loses ${{\text{H}}^{\text{ + }}}$ ion or gains electrons. A base is a substance which either gains ${{\text{H}}^{\text{ + }}}$ ions or lose electrons. In this question we are only concerned with loss and gain of ${{\text{H}}^{\text{ + }}}$ ions.
Conjugate pairs are those pairs which differ by one ${{\text{H}}^{\text{ + }}}$ ion. Since acid is a proton donor, whenever an acid loses its proton one anionic species and one ${{\text{H}}^{\text{ + }}}$ will form. This anionic species will be the conjugate base of that acid. For example, let us consider any acid HA, it will dissociate as:
$HA \to {H^ + } + {A^ - }$ the ${A^ - }$ is termed as conjugate base.
Similarly for a base, if it takes ${{\text{H}}^{\text{ + }}}$ the respective cationic species formed will be termed as conjugate acid.
In the above question it is clear that ${\text{NH}}_{\text{2}}^{\text{ - }}$ is a base, because the respective species that is being formed is an acid. A base never donates ${{\text{H}}^{\text{ + }}}$ . So option B and D are eliminated. Now since the conjugate pairs differ by one ${{\text{H}}^{\text{ + }}}$ , option C has a difference of 2 hydrogen. Hence the only option left is option A, which is the correct option.

The correct option is A.

Note:
The IUPAC name of ${\text{NH}}_{\text{2}}^{\text{ - }}$ is Azanide. It is commonly called amide ion. Though there is also an amide functional group present in organic chemistry. When ammonia reacts with alkali metals ${\text{NH}}_{\text{2}}^{\text{ - }}$ is formed. Alkali metals reacts with ${\text{N}}{{\text{H}}_3}$ to form their respective amides such as sodium amide ${\text{(NaN}}{{\text{H}}_2})$ , potassium amide or lithium amide.