Question

Maximum bond angle at nitrogen is present in which of the following?
(A)- $N{{O}_{2}}^{+}$
(B)- $N{{O}_{3}}^{-}$
(C)- $N{{O}_{2}}$
(D)- $N{{O}_{2}}^{-}$

Answer 1

The angle between the two bonds originating from the same atom in a covalent species is known as the bond angle. In case of the molecules made up of three or more atoms, the bond angle is the average angle between the bonded orbitals (that is between the two covalent bonds).

Complete answer:
-There are many factors which affect the bond angle in different ways and are given below-
(i) Repulsion present between the atoms or groups attached to the central atoms can increase or decrease the bond angle. Larger the repulsion, the greater will be the bond angle. More the attractions, lesser the bond angle.
(ii) When the hybridization is considered, the bond angle increases with the increase in the s-character of the s-hybrid bond. For example,
(a) A molecule with a central atom in $s{{p}^{3}}$has $25$ s-character and thus has a lesser bond angle $109{}^\circ 28'$.
(b) A molecule with a central atom in $s{{p}^{2}}$has $33.33$ s-character, and thus have a bond angle a bit greater than $s{{p}^{3}}$, that is $120{}^\circ$.
(c) A molecule with a central atom in sp has the maximum s-character as $50$ and thus have maximum
bond angle as $180{}^\circ$
(iii) The bond angle decreases with the decrease in the electronegativity of the central atom. For example, the bond angle decreases as${{H}_{2}}O>{{H}_{2}}S>{{H}_{2}}Se>{{H}_{2}}Te$ , because the electronegativity of the central atom decreases. Oxygen is the most electronegative has the highest bond angle as $104.5{}^\circ$followed by sulphur and serum containing compounds which have a bond angle as $92.2{}^\circ \text{ and }91.2{}^\circ$ respectively. The tellurium being the least electronegative has the least bond angle $89.5{}^\circ$.
-In case, if the central metal atom remains the same, the bond angle increases with the decrease in the electronegativity of atoms surrounding the central atom. For example in $PC{{l}_{3}},PB{{r}_{3}},P{{I}_{3}}$ the bond angle of these compounds are $100{}^\circ ,101.5{}^\circ \text{ and 102}{}^\circ$ respectively.
-Following the above knowledge let us now try to validate each option given in the question:
-In option A, that is nitronium ion, the nitrogen has 0 lone pairs. Nitrogen in nitronium ion is in sp hybridization with predicted bond angle as $180{}^\circ$. Nitronium ion has 2 numbers of domains.
-In option B, that is nitrate ion, the nitrogen has 0 lone pairs. Nitrogen in nitrate is in $s{{p}^{2}}$ hybridization with predicted bond angle as $120{}^\circ$. Nitrate ion has 3 number of bonding domains.
-In option C, that is nitrogen dioxide, the nitrogen has 1 lone pair. Nitrogen is nitrogen dioxide is in $s{{p}^{2}}$ hybridization with predicted bond angle as $120{}^\circ$. Nitrogen dioxide has 2 numbers of bonding domains.
-In option D, that is nitrite, the nitrogen has 1 lone pair. Nitrogen in nitrite is in $s{{p}^{2}}$ hybridization with predicted bond angle as $120{}^\circ$. Nitrogen has 2 numbers of bonding domains.
Since nitrite has a maximum bond angle, therefore, the correct answer is option D.

Note:
As of now, we read everything about the bond angle, its definition and influencing factors, but what is the significance of the bond angle? Determining the bond angle of a molecule helps in determining the structure of the molecule. The bond angle helps us to differentiate between the various geometries of compounds such as linear, trigonal planar, tetrahedral, trigonal bi-pyramidal or any other geometry.