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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.

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1 Answer 1

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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

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    $\begingroup$ 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
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    $\begingroup$ 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

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