Question

Arrange the following molecules in the correct order of their bond angles:

• A. A<C<D<B
• B. B<A<C<D
• C. C<B<A<D
• D. B<C<A<D

Hint:

Remember the structures and bond angles of these common allotropes: - White Phosphorus (\(P_4\)): A highly strained tetrahedron with 60° angles. - Sulfur (\(S_8\)): The stable form is a crown-shaped ring with ~108° angles. - Sulfur (\(S_6\)): A less stable chair-shaped ring with ~102° angles. The smaller the ring, the greater the angle strain and the smaller the bond angle for these non-planar rings.

Answer 1

The Correct Option is D

Let's determine the bond angles for each molecule. The question labels them A, B, C, D corresponding to S\(_8\), P\(_4\), S\(_6\), O\(_3\).
B. P\(_4\) (White Phosphorus): This molecule has a tetrahedral structure with a phosphorus atom at each vertex. The P-P-P bond angle within this strained tetrahedron is exactly 60°.
C. S\(_6\) (Cyclohexasulfur): This molecule exists in a "chair" conformation, similar to cyclohexane. The S-S-S bond angle in this ring structure is approximately 102.2°.
A. S\(_8\) (Cyclooctasulfur): This is the most common allotrope of sulfur. It has a puckered "crown" shape. The S-S-S bond angle in this eight-membered ring is approximately 107.8°, which is close to the tetrahedral angle, indicating less ring strain than in S\(_6\).
D. O\(_3\) (Ozone): As determined in a previous question (Q125), ozone is a bent molecule with a bond angle of about 116.8°. The central oxygen has 3 electron domains (a single bond, a double bond, a lone pair), leading to an angle slightly less than the ideal 120° of a trigonal planar arrangement.
Now, let's arrange these angles in increasing order:
P\(_4\) (60°)<S\(_6\) (102.2°)<S\(_8\) (107.8°)<O\(_3\) (116.8°)
In terms of the labels B, C, A, D, the order is:
B<C<A<D.
This matches option (D).