Question

Do PH_3F_2 and XeS_2 exist, in accordance with Valence Bond Theory?

According to Drago's Rule, if the central atom of a compound is an element of the third period, and the attached atom has an electronegativity lesser than 2.5, then hybridisation does not occur, and bonding occurs solely in atomic orbitals.
In the case of a compound like PH_3F_2
, the fluorine atoms occupy the axial positions, while the hydrogen atoms occupy the equatorial positions. However, the electronegativity of fluorine is 4, hence the EN. difference with phosphorous is larger, making the axial bonds very strong, meanwhile the electronegativity of hydrogen is 2.1 (below 2.5), hence the equatorial bonds are very weak, and hybridisation shouldn't occur ie. PH_3F_2 shouldn't exist, according to Valence Bond Theory.
Searching for answers online just led me to this Quora post, contradicting the answer given, which I found inconclusive.
And I couldn't find any explanation for XeS_2 not existing, at all.
This doubt stemmed from the following JEE (competitive exam) question:
Identify the number of molecules which do not exist.
SBr_6, PH_5, XeF^−_3, PH_4F, PH_3F_2, OF_4, XeF^−_5, XeO_4, XeS_2
The answer given was: (7)
SBr_6, PH_5, XeF^−_3, PH_4F, PH_3F_2, OF_4,XeS_2
I understood the rationale behind all, except for PH_3F2 and XeS_2.

Answer 1

PH3F2 and XeS2 do not exist according to Valence Bond Theory due to the electronegativity differences and observed bond angles.

Step-by-step explanation:

According to Valence Bond Theory, PH3F2 and XeS2 should not exist. In the case of PH3F2, the electronegativity difference between phosphorus and fluorine is large, making the axial bonds very strong and the equatorial bonds very weak.

Therefore, hybridization should not occur and PH3F2 should not exist. Regarding XeS2, sulfur is a larger atom compared to oxygen, and the observed bond angle of sulfur compounds indicates less hybridization on sulfur than oxygen.

Therefore, XeS2 should not exist according to Valence Bond Theory.