Question

The bond angles of $\mathrm{N}1{\mathrm{I}}_3,{\mathrm{NlI}}_4$ and ${\mathrm{NII}}_2$ are in the order (1) ${\mathrm{NII}}_2^{-}>{\mathrm{NII}}_3>{\mathrm{NII}}_4^{+}$ (2) ${\mathrm{NII}}_4^{+}>{\mathrm{NII}}_3>{\mathrm{NII}}_2^{-}$ (3) ${\mathrm{NII}}_3>{\mathrm{NII}}_2>{\mathrm{NII}}_4^{-}$ (4) ${\mathrm{NII}}_3>{\mathrm{NII}}_4^{+}>{\mathrm{NII}}_2^{-}$

Short answer

Option (2): ${\text{Nl}}_4^{+}>{\text{Nl}}_3>{\text{NlI}}_2^{-}$.

Step-by-step solution

- Understanding the Molecules

Identify the molecular geometries of ${\text{Nl}}_3$, ${\text{Nl}}_4^{+}$, and ${\text{NlI}}_2^{-}$. For ${\text{Nl}}_3$, the central atom is nitrogen with three iodine atoms and one lone pair, making it a trigonal pyramidal shape. ${\text{NlI}}_4^{+}$ has nitrogen with four iodine atoms and no lone pair, forming a tetrahedral shape. ${\text{NlI}}_2^{-}$ has nitrogen with two iodine atoms and two lone pairs, which gives it a bent geometry.

- Determining Bond Angles

Recognize that bond angles are influenced by lone pairs on the central atom. More lone pairs result in smaller bond angles due to increased electron repulsion. In a trigonal pyramidal shape like ${\text{Nl}}_3$, there are fewer lone pairs compared to the bent shape of ${\text{NlI}}_2^{-}$, but more than the tetrahedral ${\text{NlI}}_4^{+}$ with no lone pairs.

- Bond Angle Comparison

Compare the bond angles based on the VSEPR (Valence Shell Electron Pair Repulsion) theory. Larger angles are found in shapes with fewer lone pairs: ${\text{Nl}}_4^{+}$ (tetrahedral, ~109.5°) > ${\text{Nl}}_3$ (trigonal pyramidal, ~107°) > ${\text{NlI}}_2^{-}$ (bent, ~104.5°).

- Selecting the Correct Order

Choose the option that matches the bond angle comparisons: ${\text{NlI}}_4^{+}>{\text{NlI}}_3>{\text{NlI}}_2^{-}$. This matches option (2).

Key concepts

molecular geometry

Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. The shape of a molecule is determined by the number of bonding pairs and lone pairs of electrons around the central atom.
The key idea behind molecular geometry is the Valence Shell Electron Pair Repulsion (VSEPR) theory. This theory states that electron pairs around a central atom will position themselves as far apart as possible to minimize repulsion. This affects the molecular shape.
For example, ${\mathrm{NlI}}_3$ has a trigonal pyramidal shape because the nitrogen atom is bonded to three iodine atoms and has one lone pair. This lone pair pushes the bonds closer together.
In ${\mathrm{NlI}}_4^{+}$, there are no lone pairs and four bonding pairs, forming a perfect tetrahedral shape. Lastly, ${\mathrm{NlI}}_2^{-}$ has two bonding pairs and two lone pairs, resulting in a bent geometry.

bond angles

Bond angles are the angles between adjacent bonds at an atom. They are influenced by both bonding pairs and lone pairs of electrons.
In a molecule with no lone pairs, like a tetrahedral ${\mathrm{NlI}}_4^{+}$, the bond angles are about 109.5°.
In a trigonal pyramidal shape such as ${\mathrm{NlI}}_3$, the presence of one lone pair reduces the bond angles to about 107°.
When there are two lone pairs, as in a bent shape like ${\mathrm{NlI}}_2^{-}$, the bond angles are even smaller, closely around 104.5°.

electron pair repulsion

Electron pair repulsion is a core concept in understanding molecular shapes and bond angles. According to VSEPR theory, electron pairs repel each other and arrange themselves to be as far apart as possible around the central atom.
This principle explains why molecules with lone pairs have smaller bond angles compared to those without lone pairs.
Lone pairs occupy more space than bonding pairs, leading to increased repulsion and decreasing the bond angles between the atoms.
For example, in ${\mathrm{NlI}}_2^{-}$, the repulsion caused by two lone pairs makes the bond angles smaller compared to ${\mathrm{NlI}}_3$ and ${\mathrm{NlI}}_4^{+}$.