Question

Assuming pure 2s and 2p orbitals of carbon are used in forming CH4 molecule, which of the following statement is false?

• A. Three C-H bonds will be at right angle
• B. One C-H bond will be weaker than other three C-H bonds
• C. The shape of molecule will be tetrahedral
• D. The angle of C-H bond formed by s-s overlapping will be uncertain with respect to other three bonds.

Answer 1

Answer: (C)

The shape of molecule will be tetrahedral

Answer 2

To answer this question, we need to consider the hypothetical scenario where carbon uses its pure 2s and 2p orbitals to form bonds in methane (CH₄), instead of undergoing sp³ hybridization.

Carbon's electronic configuration is 1s² 2s² 2p². In its excited state, it would be 1s² 2s¹ 2pₓ¹ 2pᵧ¹ 2p₂¹. If pure orbitals are used:

1. One C-H bond would be formed by the overlap of carbon's 2s orbital and hydrogen's 1s orbital (C(2s)-H(1s) bond).
2. Three C-H bonds would be formed by the overlap of carbon's three 2p orbitals (2pₓ, 2pᵧ, 2p₂) and hydrogen's 1s orbitals (C(2p)-H(1s) bonds).

Let's evaluate each statement:

(a) Three C-H bonds will be at right angle

The three 2p orbitals (2pₓ, 2pᵧ, 2p₂) are mutually perpendicular to each other, meaning they are oriented at 90° angles. If these pure p orbitals form bonds with hydrogen's 1s orbitals, then the three C-H bonds formed by p-s overlap would indeed be at 90° to each other. So, this statement is TRUE.

(b) One C-H bond will be weaker than other three C-H bonds

The C(2s)-H(1s) bond and the C(2p)-H(1s) bonds are formed from different types of orbitals. Therefore, they will have different bond lengths and bond energies (strengths). Generally, bonds with higher s-character are shorter and stronger. The C(2s)-H(1s) bond has 100% s-character from carbon, while the C(2p)-H(1s) bonds have 0% s-character from carbon. Based on this, the C(2s)-H(1s) bond is expected to be stronger than the C(2p)-H(1s) bonds. If the 2s-1s bond is stronger, then the statement that "One C-H bond will be weaker than other three C-H bonds" is FALSE.

(c) The shape of molecule will be tetrahedral

For a molecule to have a tetrahedral shape, all four bonds must be equivalent, and the bond angles must be approximately 109.5°. This requires sp³ hybridization of the carbon atom. If pure 2s and 2p orbitals are used, three C-H bonds would be at 90° to each other (from p-s overlap), and the fourth C-H bond (from s-s overlap) would be different in nature and non-directional. This arrangement is clearly not tetrahedral. So, this statement is FALSE.

(d) The angle of C-H bond formed by s-s overlapping will be uncertain with respect to other three bonds.

The s orbital is spherically symmetric and does not have a specific direction. Therefore, the bond formed by C(2s)-H(1s) overlap would not have a fixed angular relationship with the directional C(2p)-H(1s) bonds. Its orientation relative to the other three bonds would not be geometrically constrained to a specific angle like 90° or 109.5°. So, this statement is TRUE.

Conclusion:

Both statements (b) and (c) are false. However, in multiple-choice questions, we typically look for the most definitively false statement. The tetrahedral geometry is a very specific arrangement that is fundamentally contradicted by the use of pure s and p orbitals (which would lead to 90° angles for p-bonds and a non-directional s-bond). The claim about relative bond strength in (b) is also false because the s-s bond is expected to be stronger, not weaker. In the context of introductory chemistry, the geometric consequences of non-hybridization (i.e., 90° angles for p-bonds and non-tetrahedral shape) are a more direct and universally accepted consequence than the precise quantitative comparison of s-s vs p-s bond strengths without further data. Therefore, (c) is a very clear and fundamental falsehood regarding the molecular geometry.

Given that typically only one option is correct/incorrect, and (c) is a direct consequence of using pure orbitals, making it definitively false.