Problem 20 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The correct order of hybridization of the central atom in the following species \(\mathrm{NH}_{3},\left[\mathrm{PtCl}_{4}\right]^{2-}, \mathrm{PCl}_{5}\) and \(\mathrm{BCl}_{3}\) is (a) \(d s p^{2}, d s p^{3}, s p^{2}, s p^{3}\) (b) \(s p^{3}, d s p^{2}, s p^{3} d, s p^{2}\) (c) \(d s p^{2}, s p^{2}, s p^{3}, d s p^{3}\) (d) \(d s p^{2}, s p^{3}, s p^{2}, d s p^{3}\)

Step-by-Step Solution

Answer

The correct order is (b) \(sp^3, dsp^2, sp^3d, sp^2\).

Step 1: Understand Hybridization

Hybridization involves the mixing of atomic orbitals to form new hybrid orbitals, which influence molecular geometry. The key is to consider the electron pairs (bonding and lone pairs) around the central atom to determine hybridization.

Step 2: Determine NH3 Hybridization

The ammonia molecule, NH3, consists of a nitrogen atom bonded to three hydrogen atoms, with one lone pair of electrons. This totals four regions of electron density, suggesting an \(sp^3\) hybridization.

Step 3: Determine [PtCl4]2- Hybridization

The \([\text{PtCl}_4]^{2-}\) ion includes a platinum atom surrounded by four chloride ions. Due to the d-orbital involvement in coordination compounds like these, the hybridization is \(dsp^2\), forming a square planar structure.

Step 4: Determine PCl5 Hybridization

The phosphorus in PCl5 is attached to five chlorine atoms, indicating five regions of electron density, which necessitates \(sp^3d\) hybridization, resulting in a trigonal bipyramidal geometry.

Step 5: Determine BCl3 Hybridization

In BCl3, the central boron atom is bonded to three chlorine atoms, with no lone pairs involved, suggesting \(sp^2\) hybridization, leading to a trigonal planar shape.

Step 6: Compare Hybridizations with Options

Now, order the hybridizations: NH3 (\(sp^3\)), \([\text{PtCl}_4]^{2-}\) (\(dsp^2\)), PCl5 (\(sp^3d\)), and BCl3 (\(sp^2\)). Match this order against the given options.

Key Concepts

Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. It is crucial in understanding how a molecule behaves, including its reactivity, color, phase of matter, and even biological activity. Geometry arises from the specific arrangement of atomic orbitals, which are determined by electron pair interactions.

Each type of hybridization influences the molecular geometry:

• For \(sp^3\) hybridization, as seen in \(\text{NH}_3\), the geometry is typically tetrahedral. However, due to one lone pair, ammonia's shape is trigonal pyramidal.
• In \(\text{PCl}_5\), \(sp^3d\) hybridization results in a trigonal bipyramidal shape.
• \([\text{PtCl}_4]^{2-}\) with \(dsp^2\) hybridization forms a square planar geometry.
• \(\text{BCl}_3\), characterized by \(sp^2\) hybridization, adopts a trigonal planar shape.

Understanding these geometries helps predict angles between bonds and potential molecular interactions.

Electron Pairs

Electron pairs, consisting of electrons occupying orbitals, are a fundamental component in determining molecular geometry and hybridization. They are classified as bonding pairs, which form bonds between atoms, and lone pairs, which do not bond but influence molecular shape.

The VSEPR (Valence Shell Electron Pair Repulsion) theory is used to predict molecular shapes based on electron pair repulsions. Here's how different hybridizations affect electron pair placement:

• In \(\text{NH}_3\), the lone pair repels the bonding pairs, decreasing bond angles slightly.
• \([\text{PtCl}_4]^{2-}\) has only bonding pairs around the central atom, creating a stable square planar shape without lone pair interference.
• \(\text{PCl}_5\) uses all five domains of electron density for bonding, yielding no lone pairs and a perfect trigonal bipyramidal shape.
• \(\text{BCl}_3\) also has only bonding pairs, contributing to its regular trigonal planar structure.

By considering electron pairs, students can better understand why molecules have specific shapes.

Hybrid Orbitals

Hybrid orbitals are the result of combining standard atomic orbitals (such as s, p, d) to form new, equivalent orbitals that can participate in bonding. This concept is essential in determining the structure and bonding properties of molecules.

Each type of hybridization involves a different combination of atomic orbitals:

• \(sp^3\) hybridization mixes one "s" and three "p" orbitals, forming four identical hybrid orbitals, as in \(\text{NH}_3\).
• \(sp^2\) hybridization combines one "s" and two "p" orbitals, creating three equivalent orbitals, seen in \(\text{BCl}_3\).
• \(dsp^2\) involves one "d," one "s," and two "p" orbitals, unique to coordination compounds like \([\text{PtCl}_4]^{2-}\).
• \(sp^3d\) incorporates one "s," three "p," and one "d" orbital, as needed for molecules like \(\text{PCl}_5\).

Mastering hybrid orbitals helps students envision how atoms come together and how electrons are arranged in molecules, impacting their chemical behavior.