Question

If all bond angles in $AX_3$ molecule are the same, then which of the following conclusions is correct about $AX_3$?

• A. $AX_3$ must be polar.
• B. $AX_3$ must be planar.
• C. $AX_3$ must have at least 5 valence electrons.
• D. X must connect from central atom with either single bond or double bond.

Answer 1

Answer: (D)

X must connect from central atom with either single bond or double bond.

Answer 2

For an $AX_3$ molecule to have all identical bond angles, it must possess high symmetry. The two common VSEPR geometries that satisfy this condition are trigonal planar ($AX_3$ type, like $BF_3$ or $SO_3$) and trigonal pyramidal ($AX_3E$ type, like $NH_3$).

1. Polarity: Trigonal planar molecules like $BF_3$ are non-polar, so $AX_3$ is not necessarily polar. (Option A is incorrect).

2. Planarity: Trigonal pyramidal molecules like $NH_3$ are non-planar, so $AX_3$ is not necessarily planar. (Option B is incorrect).

3. Valence Electrons of A: In $BF_3$, the central atom Boron has 3 valence electrons. Thus, A does not necessarily have at least 5 valence electrons. (Option C is incorrect).

4. Bond Type: For all bond angles to be identical, the three A-X bonds must be equivalent in terms of bond order and electron domain repulsion. This implies that all bonds must be either single bonds (as in $BF_3$, $NH_3$) or double bonds (as in $SO_3$, or partial double bonds in $CO_3^{2-}$). If a triple bond were present, it would exert different repulsion, leading to unequal bond angles. Therefore, X must connect from the central atom with either a single bond or a double bond. (Option D is correct).