Problem 183 - A Complete Resource Book in Chemistry for JEE Main

Question

Hybridization of the underlined atom changes in which of the following transitions? \(\quad\) [2002] (a) \(\mathrm{AIH}_{3}\) changes to \(\mathrm{AlH}_{4}^{-}\) (b) \(\mathrm{H}_{2} \mathrm{O}\) changes to \(\mathrm{H}_{3} \mathrm{O}^{+}\) (c) \(\mathrm{NH}_{3}\) changes to \(\mathrm{NH}_{4}^{+}\) (d) in all cases

Step-by-Step Solution

Answer

Option (a) has a change in hybridization from \( \text{sp}^2 \) to \( \text{sp}^3 \).

Step 1: Understand the Concept of Hybridization

Hybridization refers to the mixing of atomic orbitals to create new hybrid orbitals suitable for pairing of electrons to form chemical bonds. Different hybridizations involve different geometries and can be indicated as \( \text{sp}, \text{sp}^2, \text{sp}^3 \), etc.

Step 2: Analyze the Transition (a)

In \( \text{AIH}_{3} \), aluminum has three bonds with hydrogen and no lone pairs, corresponding to an \( \text{sp}^2 \) hybridization. In \( \text{AlH}_{4}^{-} \), the aluminum forms an additional bond, resulting in \( \text{sp}^3 \) hybridization due to the four equivalent bond pairs. Thus, the hybridization changes from \( \text{sp}^2 \) to \( \text{sp}^3 \).

Step 3: Analyze the Transition (b)

In \( \text{H}_{2} \text{O} \), the oxygen atom has two hydrogen bonds and two lone pairs, leading to an \( \text{sp}^3 \) hybridization. When \( \text{H}_{2} \text{O} \) becomes \( \text{H}_{3} \text{O}^{+} \), oxygen forms an additional bond with a hydrogen ion, but it maintains an \( \text{sp}^3 \) hybridization. Therefore, the hybridization does not change.

Step 4: Analyze the Transition (c)

In \( \text{NH}_3 \), nitrogen has three bonds with hydrogen and one lone pair, corresponding to an \( \text{sp}^3 \) hybridization. In \( \text{NH}_4^{+} \), the lone pair is used to form a bond with an additional hydrogen ion, still resulting in \( \text{sp}^3 \) hybridization. Thus, the hybridization remains \( \text{sp}^3 \), and there is no change.

Step 5: Determine the Overall Answer

Since we analyzed that only in \( \text{AIH}_3 \) to \( \text{AlH}_4^{-} \), the hybridization changes (from \( \text{sp}^2 \) to \( \text{sp}^3 \)), the answer is not 'in all cases'. The option where the hybridization of the underlined atom changes is option (a).

Key Concepts

Understanding Atomic Orbitals

Atomic orbitals are the regions around an atom's nucleus where electrons are most likely to be found. These orbitals have distinctive shapes and are named as s, p, d, and f.

• s orbitals are spherical and can hold up to 2 electrons.
• p orbitals are dumbbell-shaped and can hold up to 6 electrons across three different orientations.
• d and f orbitals have even more complex shapes and can hold more electrons.

The characteristics of these orbitals are fundamental to understanding chemical bonding and interactions between atoms. When atoms bond, atomic orbitals can overlap to form molecular orbitals, which are crucial in the formation of chemical bonds.

Insights on Chemical Bonds

Chemical bonds are the attractive forces that hold atoms together in molecules. These bonds form when atoms share or transfer electrons, leading to more stable electronic arrangements. The two primary types of chemical bonds are:

• Covalent Bonds: Atoms share electron pairs; common in organic compounds like water ( H_2O ).
• Ionic Bonds: The complete transfer of electrons from one atom to another, seen in compounds like sodium chloride (NaCl).

The strength and nature of these bonds can be influenced by the hybridization state of the orbitals involved. Hybridization affects the geometry and angles between bonds, playing a pivotal role in determining the molecule’s shape and properties.

Explaining sp3 Hybridization

The process of sp3 hybridization involves the combination of one s orbital and three p orbitals from the same atom to form four equivalent sp3 hybrid orbitals. This hybridization is characteristic of atoms that form four single bonds, such as carbon in methane ( CH_4 ) or nitrogen in ammonia ( NH_3 ).

• In sp3 hybridization, all four orbitals have equal energy levels and form a tetrahedral shape surrounding the atom.
• The bond angles in a perfect sp3 hybrid, such as methane, are approximately 109.5 degrees.
• This hybridization occurs because it allows the atom to form stronger, more directional sigma bonds, which contribute to the molecule's stability.

For instance, in the exercise, when AIH_3 changes to AlH_4^{-} , the hybridization shifts from sp2 to sp3, reflecting the creation of an additional bond for the aluminum atom.