Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The species in which the N-atom is in a state of $sp$ hybridisation is

The journey into the microscopic world of molecules is nothing short of fascinating. When we look at a chemical formula, we are merely seeing the ingredients. But how do these ingredients arrange themselves in three-dimensional space? This is where the elegant concept of hybridisation and the Valence Shell Electron Pair Repulsion (VSEPR) theory come into play.

In this problem, we are tasked with finding the nitrogen species that exhibits $sp$ hybridisation. To solve this, we must become molecular architects, drawing out the Lewis structures and calculating the steric number for each candidate. Let's dive deep into the thought process!

Steric Number

Before we analyze the options, we need a reliable tool. The Steric Number (SN) is our guiding light. It tells us how many "electron domains" or "parking spaces" are required around the central atom to accommodate all the bonded atoms and lone pairs.

The formula is beautifully simple:

Why do we ignore $\pi$ bonds? Because $\pi$ bonds are formed by the sideways overlap of unhybridized $p$-orbitals. They don't dictate the fundamental geometry of the molecule; they just add extra "glue" between atoms that are already connected by a $\sigma$ bond.

Once we have the steric number, the hybridisation is straightforward: SN = 2 $⟹$ $sp$ hybridisation (Linear geometry) SN = 3 $⟹$ $s{p}^2$ hybridisation (Trigonal planar geometry) SN = 4 $⟹$ $s{p}^3$ hybridisation* (Tetrahedral geometry)

Since we are hunting for an $sp$ hybridised nitrogen atom, our target must have a steric number of exactly 2. Let's put our candidates to the test.

1. The Nitrite Ion ($N{O}_2^{-}$) Imagine the nitrogen atom at the center. Nitrogen is in Group 15, meaning it brings 5 valence electrons to the table. The negative charge of the ion gives it one extra electron, making a total of 6 valence electrons to work with.

Nitrogen forms two bonds with the oxygen atoms (one single bond and one double bond, though resonance makes them equivalent). This uses up 3 of its electrons. Wait, let's count properly: Nitrogen forms a double bond with one oxygen (using 2 electrons) and a single bond with the other oxygen (using 1 electron). That leaves 2 electrons, which form exactly one lone pair.

So, we have: Number of $\sigma$ bonds = 2 Number of lone pairs = 1

A steric number of 3 means the nitrogen in $N{O}_2^{-}$ is $s{p}^2$ hybridised. It has a bent molecular geometry. This is not our answer.

2. The Nitrate Ion ($N{O}_3^{-}$) Now let's look at the nitrate ion. Nitrogen again starts with 5 valence electrons. It forms a double bond with one oxygen and single bonds with the other two oxygens (with one oxygen carrying a formal negative charge and nitrogen carrying a formal positive charge).

In this arrangement, nitrogen uses all 4 of its available valence orbitals to form bonds. It forms exactly 3 $\sigma$ bonds with the three oxygen atoms. Because all its valence electrons are involved in bonding, there are zero lone pairs left on the central nitrogen.

Number of $\sigma$ bonds = 3 Number of lone pairs = 0

Once again, a steric number of 3 points to $s{p}^2$ hybridisation. The molecule is perfectly trigonal planar. We must keep searching.

3. Nitrogen Dioxide ($N{O}_2$) This molecule is a classic anomaly—an odd-electron species. Nitrogen has 5 valence electrons. It forms two $\sigma$ bonds with the two oxygen atoms. After forming the necessary bonds (one double, one single), nitrogen is left with a single, unpaired electron.

Does this single electron count as a lone pair? Not quite, but it does occupy a hybrid orbital and exerts a repulsive force, albeit weaker than a full lone pair. For the sake of determining the basic geometry, we treat this odd electron as a domain, giving a steric number of approximately 3.

Number of $\sigma$ bonds = 2 Number of odd electrons = 1

Therefore, the nitrogen in $N{O}_2$ is also $s{p}^2$ hybridised, resulting in a bent shape. Still not what we are looking for!

4. The Nitronium Ion ($N{O}_2^{+}$) Finally, we arrive at $N{O}_2^{+}$. Let's carefully account for the electrons. Nitrogen normally has 5 valence electrons. However, the positive charge indicates that it has lost one electron. It now has exactly 4 valence electrons.

It uses these 4 electrons to form two double bonds, one with each oxygen atom. Let's calculate the steric number: Number of $\sigma$ bonds = 2 (one for each oxygen) Number of lone pairs = 0 (all 4 electrons are used in the double bonds)

Bingo! A steric number of exactly 2 means that the nitrogen atom must mix one $s$-orbital and one $p$-orbital to create two equivalent hybrid orbitals. This is the definition of $sp$ hybridisation. The two $sp$ hybrid orbitals arrange themselves as far apart as possible, resulting in a perfectly linear geometry with a bond angle of $180^{\circ }$.

We saw how lone pairs, odd electrons, and formal charges drastically alter the molecular geometry. The only species that perfectly satisfies the condition for $sp$ hybridisation is the nitronium ion, $N{O}_2^{+}$.

This problem beautifully illustrates that you don't need to memorize the shapes of hundreds of molecules. By mastering a few fundamental principles like Lewis structures and the steric number, you can deduce the geometry of almost any molecule you encounter!