Problem 51 - A Complete Resource Book in Chemistry for JEE Main

Question

What is the hybridization state of the central atom in the conjugate base of \(\mathrm{NH}_{4}^{+}\)ion? (a) sp (b) \(\mathrm{sp}^{3}\) (c) \(\mathrm{sp}^{3}\) (d) \(\mathrm{dsp}^{2}\)

Step-by-Step Solution

Answer

The hybridization state is \(\mathrm{sp}^3\).

Step 1: Identify the Central Atom and its Geometry in the Original Molecule

Ammonium ion, \(\mathrm{NH}_4^+\), has nitrogen (N) as the central atom. It forms a tetrahedral geometry with four hydrogen atoms, each forming a single bond, resulting in \(\mathrm{sp}^3\) hybridization.

Step 2: Determine the Conjugate Base

The conjugate base of \(\mathrm{NH}_4^+\) is formed by removing a proton (H\(^+\)) from it, resulting in \(\mathrm{NH}_3\) (ammonia).

Step 3: Analyze the Geometry and Hybridization of the Conjugate Base

In \(\mathrm{NH}_3\), nitrogen is again the central atom, which now has three hydrogen atoms bonded and one lone pair of electrons. This gives it a trigonal pyramidal shape. The hybridization involves 3 \(\sigma\) bonds and 1 lone pair, resulting in \(\mathrm{sp}^3\) hybridization.

Key Concepts

Conjugate Base

A conjugate base is formed when a proton (37; H\(^+\)) is removed from an acid. This term is often used in the context of acid-base chemistry. Understanding how a conjugate base operates can reinforce your knowledge about acid-base transformations and their consequences. For example, when an ammonium ion (\(\mathrm{NH}_4^+\)) releases a proton, it forms ammonia (\(\mathrm{NH}_3\)). The process of removing the proton alters the structure and the electronic configuration of the molecule, leading to a new species. Ammonia, as the conjugate base of the ammonium ion, showcases a key principle: removing a proton from an acid affects its acidity and its ability to further participate in chemical reactions.

Tetrahedral Geometry

Tetrahedral geometry is a term used to describe a molecular shape where a central atom is bonded to four substituents, positioned at the corners of a tetrahedron. This geometry is a result of the \(\mathrm{sp}^3\) hybridization, where one s orbital and three p orbitals mix to produce four equivalent \(\mathrm{sp}^3\) hybrid orbitals. These orbitals are directed in a manner that minimizes electron repulsion, leading to a bond angle of 109.5 degrees. In the ammonium ion (\(\mathrm{NH}_4^+\)), the central nitrogen atom is bonded equally to four hydrogen atoms, forming a perfect tetrahedral shape. Recognizing tetrahedral geometry is critical because it explains molecular structures' stability and reactivity, which are essential for understanding different chemical reactions and mechanisms.

Trigonal Pyramidal Shape

The trigonal pyramidal shape is seen in molecular geometries where a central atom is surrounded by three bonded atoms and one lone pair of electrons. This geometry results from \(\mathrm{sp}^3\) hybridization. Unlike tetrahedral molecules, the lone pair exerts additional electron repulsion, pushing the bonded atoms closer together and resulting in a bond angle slightly less than 109.5 degrees. A common example of this shape is found in ammonia (\(\mathrm{NH}_3\)). Here, the nitrogen atom, flanked by three hydrogen atoms and an unbonded electron pair, forms a triangular base with a pyramid-like structure on top. Recognizing this geometry helps to predict the molecule’s physical properties, such as polarity, and how it might interact with other molecules. Understanding the trigonal pyramidal shape is key to grasping various molecular interactions and chemical bonding principles.