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Question

If all bond angles in $\ce{AX3}$ are the same, then which of the following are correct conclusions about $\ce{AX3}?$

(A) $\ce{AX3}$ must be polar.
(B) $\ce{AX3}$ must be planar.
(C) $\ce{AX3}$ must have at least 5 valence electrons.
(D) $\ce{X}$ must be connected to the central atom via single or double bonds.

My Attempt

This is clearly a case of $\mathrm{sp^2}$ hybridisation. So, I thought about molecules like $\ce{BF3}$. As the bond angle needs to be same, it must not be polar. $\mathrm{sp^2}$ has trigonal planar structure, so (B) must be correct.

But my book says that the answer is (D) only (this question was of single correct choice type).

Why are (B) and (C) not the answers and why is (D) the answer?

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    $\begingroup$ Bond angles in $\ce{NH3}$ are equal, but the molecular structure is not planar. That's a good place to start. $\endgroup$
    – Todd Minehardt
    Mar 13 '20 at 14:52
  • $\begingroup$ @ToddMinehardt Yes I see that rules out (B). But what do we mean by the option (C)? What does it mean when it says that the valence electrons of AX$_3$ is 5? Is it referring to the central atom or the whole molecule's number of valence electrons? $\endgroup$
    – Techie5879
    Mar 13 '20 at 14:56
  • $\begingroup$ @ToddMinehardt Also, isn't NH$_3$ polar?? $\endgroup$
    – Techie5879
    Mar 13 '20 at 15:02
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    $\begingroup$ Yes, it is polar. As for option (C) - it's unclear to me and I think it's poorly-worded - all you need to do to disprove it is come up with one example of a molecule that has neither 5 valence electrons on the central atom or in the whole molecule. Maybe $\ce{BH3}$? $\endgroup$
    – Todd Minehardt
    Mar 13 '20 at 15:12
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    $\begingroup$ Correct. And there you have disproved option (C), which leaves you with (D). This approach to problem-solving - find a counterexample to inform you yes/no - is generally useful for multiple choice questions such as this. $\endgroup$
    – Todd Minehardt
    Mar 13 '20 at 15:18
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(A) and (B) are answered well in the comments by Todd Minehardt and Techie5879. So, I'll summarise their points and clarify (C) to justify the answer (D).

(A) $\ce{AX3}$ must be polar.

Consider the case of $\ce{BH3}$ to disprove statement (A), since it is a non polar molecule with a trigonal planar structure (equal bond angles).

(B) $\ce{AX3}$ must be planar.

Consider the case of $\ce{NH3}$ to disprove statement (B), since it is has a trigonal pyramidal shape and isn't planar, but has equal bond angles.

(C) $\ce{AX3}$ must have at least 5 valence electrons.

Here, the valence electrons of the central atom is referred and not the total number of valence electrons in the molecule because, $\ce{AX3}$ has 3 bonds, thus, 6 electrons are involved in bonding at the very least. If 4 valence electrons are considered, that could equate to only 2 bonds, hence contradicting the existence of $\ce{AX3}$. With that mentioned, consider $\ce{BH3}$, it has 3 valence electrons, less than the 5 stated in the question. Hence (C) is disproved.

With (A), (B), and (C) disproved, (D) is the correct option by elimination.

(D) $\ce{X}$ must be connected to the central atom via single or double bonds.

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