The following question was asked in the Indian Olympiad Qualifier Chemistry Part I (IOQC) today:

For the given compound, %s character of phosphorus hybrid orbitals which contribute to various bonds are given in the table below. enter image description here The difference in %'s' character of various phosphorus bonds could be due to:

(A) The large size of bromine atom

(B) The large electronegativity difference between $\ce{P}$ and $\ce{O}$

(C) Increased overlap of $\sigma$-orbitals of terminal $\ce{P-O}$ bond

(D) Stronger covalent character of $\ce{P-O}$ in cyclic oxygen atoms

This is a multiple correct question, so any number of the options can be correct.

Due to the large size of bromine, the $\ce{P-Br}$ bond length will be longer than the $\ce{P=O}$ and $\ce{P-O}$ bonds; a longer bond length means lower s character, hence (A) should be correct according to me.

I am not sure how to approach the other options, any help is appreciated.


1 Answer 1


The most possible approach to answer the question is by Bent rule. Bromine being large, poor overlapping results in weak sigma bonding, and lower s character in the hybrid orbital of P. The oxygen of Exo or terminal P=O uses hybrid orbitals of more s-character as it overlaps to a greater extent. The oxygen of the cyclic ring therefore can not have a strong covalent sigma bond as Exo P=O. As there are two types of P-O bonds, B could not be the reason The answer is A and C

  • 1
    $\begingroup$ doi.org/10.1351/goldbook.BT07000 Bent's Rule does state the opposite of what you are saying. $\endgroup$ Mar 23, 2022 at 22:06
  • $\begingroup$ Bent rule applied for differentiation Exp P-O bond and P-O bond of the ring. Bent’s rule appears to be failed at Br atom due to its large size $\endgroup$ Mar 24, 2022 at 4:07
  • 1
    $\begingroup$ Either Bent's rule applies, or it doesn't. There are definitely other factors at play than just Bent's rule. It doesn't apply to the terminal oxygen bond; Bent's Rule would require less s-character not more. This approach to the question cannot be correct. Unfortunately. $\endgroup$ Mar 24, 2022 at 19:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.