Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The hybridisations of the atomic orbitals of nitrogen in ${\text{NO}}_2^{-}$, ${\text{NO}}_2^{+}$ and ${\text{NH}}_4^{+}$ respectively are

Welcome to a fascinating journey into the microscopic world of molecules! Today, we are going to decode the hybridization of the central nitrogen atom in three distinct polyatomic ions: the nitrite ion (${\text{NO}}_2^{-}$), the nitronium ion (${\text{NO}}_2^{+}$), and the ammonium ion (${\text{NH}}_4^{+}$).

Understanding hybridization is like having a master key to unlock the 3D geometry and shape of any molecule. Let's dive in and see how it works.

Steric Number

To find the hybridization of a central atom, we rely on a very simple yet powerful concept called the Steric Number ($H$). The steric number tells us how many hybrid orbitals are needed to accommodate the electron domains around the central atom.

It is crucial to remember that $\pi$ bonds do not participate in hybridization. They are formed by the lateral overlap of unhybridized p-orbitals and do not affect the basic geometry of the molecule.

The Bent Nitrite Ion

Let's start with the nitrite ion, ${\text{NO}}_2^{-}$. Nitrogen, being in Group 15, has 5 valence electrons. In this ion, it forms one double bond and one single bond with the two oxygen atoms. The negative charge resides on the single-bonded oxygen atom.

After forming these bonds, nitrogen is left with one non-bonding pair of electrons, or a lone pair.

Now, let's calculate its steric number: - Number of $\sigma$ bonds = 2 (one from the single bond, one from the double bond) - Number of lone pairs = 1

A steric number of 3 corresponds to $s{p}^2$ hybridization. While the electron geometry is trigonal planar, the presence of the lone pair pushes the bonding pairs closer together, resulting in a bent molecular shape.

The Linear Nitronium Ion

Next up is the nitronium ion, ${\text{NO}}_2^{+}$. The positive charge indicates that nitrogen has lost one of its valence electrons, leaving it with only 4.

Nitrogen uses all 4 of these electrons to form two double bonds with the two oxygen atoms. Consequently, there are zero lone pairs left on the central nitrogen atom.

Let's find its steric number: - Number of $\sigma$ bonds = 2 (one from each double bond) - Number of lone pairs = 0

A steric number of 2 means the nitrogen atom is $sp$ hybridized. This hybridization leads to a perfectly linear geometry, with a bond angle of $180^{\circ }$.

The Tetrahedral Ammonium Ion

Finally, let's examine the ammonium ion, ${\text{NH}}_4^{+}$. Similar to the nitronium ion, the positive charge means nitrogen effectively has 4 valence electrons available for bonding.

It uses these 4 electrons to form four single $\sigma$ bonds with four hydrogen atoms. Once again, there are no lone pairs remaining on the nitrogen.

Calculating the steric number: - Number of $\sigma$ bonds = 4 - Number of lone pairs = 0

A steric number of 4 dictates an $s{p}^3$ hybridization. This results in a highly symmetrical tetrahedral geometry, with bond angles of approximately $109.5^{\circ }$.

By carefully drawing the Lewis structures and applying the steric number formula, we have successfully determined the hybridizations: - ${\text{NO}}_2^{-}$ is $s{p}^2$ hybridized. - ${\text{NO}}_2^{+}$ is $sp$ hybridized. - ${\text{NH}}_4^{+}$ is $s{p}^3$ hybridized.

This sequence perfectly matches option (d). The key takeaway here is to always account for formal charges and accurately count the lone pairs on the central atom. Happy learning!