Which of the following statement is false?

The repulsion among four hybrid orbitals is smallest if they point to the corners of a square.

The correct answer is option 3. The repulsion among four hybrid orbitals is smallest if they point to the corners of a square.

Let us go through each statement in explain why the third statement is false:

1. Octahedral molecular shape exists in \(d^2sp^3\) hybridization.

This statement is true. In \(d^2sp^3\) hybridization, one s orbital, three p orbitals, and two d orbitals mix to form six hybrid orbitals, which arrange themselves octahedrally. This hybridization is common in coordination compounds, such as \( [Fe(CN)_6]^{3-} \), where the central metal ion forms six bonds with ligands.

2. \(XeF_2\) involves \(sp^3d\) hybridization.

This statement is true. Xenon difluoride (\(XeF_2\)) has a linear structure with three lone pairs and two bond pairs around the central xenon atom. The five pairs of electrons (two bonding and three lone pairs) require \(sp^3d\) hybridization to arrange themselves in a trigonal bipyramidal electron geometry, with the lone pairs occupying the equatorial positions to minimize repulsion.

3. The repulsion among four hybrid orbitals is smallest if they point to the corners of a square.

This statement is false. The repulsion among four hybrid orbitals is smallest when they point to the corners of a tetrahedron. In a tetrahedral arrangement, the bond angles are approximately 109.5°, which allows for the optimal spatial separation of the electron pairs, minimizing electron-electron repulsion according to VSEPR (Valence Shell Electron Pair Repulsion) theory. If the four orbitals pointed to the corners of a square, the bond angles would be 90°, leading to higher repulsion because the electron pairs would be closer together.

4. Hybridization can take place only between orbitals of roughly the same energy.

This statement is true. Hybridization involves the mixing of atomic orbitals to form new hybrid orbitals. For effective hybridization, the atomic orbitals involved must have similar energies. This ensures that the resulting hybrid orbitals have comparable energy levels, allowing for the formation of stable bonds.

Detailed Explanation for the False Statement (3):

In molecular geometry, the arrangement of electron pairs around a central atom is determined by the principle of minimizing repulsion between them, as stated by VSEPR theory. For four electron pairs, the geometry that minimizes repulsion is tetrahedral, not square planar.

Tetrahedral Geometry: Four electron pairs around a central atom are most stable when arranged in a tetrahedral geometry, with bond angles of 109.5°. This arrangement allows each electron pair to be as far away from the others as possible, reducing repulsion to a minimum.

Square Planar Geometry: If four electron pairs were arranged in a square planar geometry, the bond angles would be 90°, resulting in increased repulsion compared to the tetrahedral arrangement. This higher repulsion is because the electron pairs are closer together.

To illustrate this with an example:

Methane (\(CH_4\)): The carbon atom in methane undergoes \(sp^3\) hybridization, forming four equivalent \(sp^3\) hybrid orbitals. These orbitals point to the corners of a tetrahedron, resulting in bond angles of 109.5°, which minimizes repulsion among the hydrogen atoms.

In summary, statement (3) is false because the tetrahedral arrangement, not the square planar arrangement, provides the lowest repulsion among four hybrid orbitals.