Problem 127 - A Complete Resource Book in Chemistry for JEE Main

Question

Match the following: \(\begin{array}{ll}\text { List I } & \text { List II }\end{array}\) (Species) (Hybridization) 1\. \(\mathrm{BCI}_{3}\) (i) \(\mathrm{sp}^{3}\) 2\. \(\mathrm{NH}_{3}\) (ii) \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) 3\. \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) (iii) \(\mathrm{sp}^{2}\) 4\. \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (iv) \(\mathrm{sp}^{2} \mathrm{~d}\)

Step-by-Step Solution

Answer

1-iii, 2-i, 3-iv, 4-ii

Step 1: Determine Hybridization of BCI3 (Species 1)

The molecule \( \text{BCI}_3 \) is trigonal planar in shape. This geometry corresponds to \( \text{sp}^2 \) hybridization as it involves the mixing of one \( s \) orbital and two \( p \) orbitals to form three equivalent sp2 hybrid orbitals. Therefore, \( \text{BCI}_3 \) matches with \( (iii) \ \text{sp}^2 \) hybridization.

Step 2: Determine Hybridization of NH3 (Species 2)

The molecule \( \text{NH}_3 \) is pyramidal in shape with nitrogen forming three single bonds and one lone pair. This corresponds to an \( \text{sp}^3 \) hybridization, involving one \( s \) orbital and three \( p \) orbitals. Thus, \( \text{NH}_3 \) matches with \( (i) \ \text{sp}^3 \) hybridization.

Step 3: Determine Hybridization of [Cu(NH3)4]2+ (Species 3)

In the complex \( [\text{Cu(NH}_3)_4]^{2+} \), copper typically forms a square planar geometry in this complex, leading to \( \text{sp}^2 \text{d} \) hybridization, though often discussed in terms of \( \text{dsp}^2 \) due to involvement of d orbitals. But here it's primarily the concept of \( \text{sp}^2 \text{d} \) involved in creating a planar shape, matching \( (iv) \ \text{sp}^2 \text{d} \) hybridization.

Step 4: Determine Hybridization of [Fe(H2O)6]2+ (Species 4)

The complex \( [\text{Fe(H}_2\text{O})_6]^{2+} \) typically exhibits an octahedral geometry, which corresponds to \( \text{sp}^3\text{d}^2 \) hybridization. This involves mixing one \( s \), three \( p \), and two \( d \) orbitals. Hence, it matches with \( (ii) \ \text{sp}^3\text{d}^2 \) hybridization.

Key Concepts

Trigonal Planar

Trigonal planar geometry is observed in molecules where a central atom is bonded to three surrounding atoms. These atoms are evenly distributed around the central atom, creating a flat, triangular shape. The bond angles in a trigonal planar molecule are typically 120 degrees. This arrangement occurs due to the sp² hybridization of the central atom's orbitals.

Key features of trigonal planar molecules include:

• Flat, triangular geometry.
• Bond angles of approximately 120 degrees.
• Three atoms bonded to a central atom.

In the molecule boron trichloride (BCI₃), boron is the central atom, bonded to three chlorine atoms. The boron in BCI₃ undergoes sp² hybridization, using its s and two of its p orbitals to form three equivalent hybrid orbitals. This results in the trigonal planar shape, where the three chlorines are positioned symmetrically around boron.

Pyramidal Shape

A pyramidal shape is commonly seen in molecules with a central atom bonded to three other atoms and having one lone pair. This shape is distinct from the trigonal planar because of the lone pair which pushes the bonded atoms downwards, altering the geometric arrangement.

The nitrogen atom in ammonia (NH₃) is a classic example demonstrating sp³ hybridization. Here's what you need to know:

• Four hybrid orbitals are formed from one s and three p orbitals.
• Three of these orbitals form bonds with hydrogen atoms.
• The fourth orbital holds the lone pair.

The presence of the lone pair distorts the bond angles from the ideal tetrahedral angle of 109.5 degrees to about 107 degrees, forming a pyramidal shape rather than a planar one.

Square Planar

Square planar geometry is prominent in certain coordination complexes, where four ligands are arranged at the corners of a square around a central atom. This geometry is most often associated with transition metals.

In the complex ion \([ ext{Cu(NH}_3)_4]^{2+} \), copper exhibits a square planar geometry, a result of sp²d hybridization, sometimes alternatively described as dsp² hybridization due to the involvement of d orbitals. interesting characteristics include:

• Four ligands positioned at 90-degree angles, creating a square.
• Characteristics usually appear in complexes with d⁸ metal ions.

This geometry is particularly stable due to the minimization of electron repulsion between ligands, setting it apart from other potential shapes.

Octahedral Geometry

The octahedral geometry is characteristic of species where six ligands surround a central atom, positioned symmetrically at the corners of an octahedron. It is a common geometry for transition metal complexes, resulting from the mixing of one s, three p, and two d orbitals leading to sp³d² hybridization.

Consider the complex ion \([ ext{Fe(H}_2 ext{O})_6]^{2+} \):

• Iron in the center surrounded by six water molecules.
• Exhibits sp³d² hybridization.
• Creates equal bond angles of exactly 90 degrees between adjacent ligands.

The octahedral shape allows for optimal space filling and electron pair repulsion minimization, making it energetically favorable for many complexes.