Place the following in order of increasing F-A-F bond angle, where A represents the central atom in each molecule. PF3 OF2 PF4 + Group of answer choices PF3 < OF2 < PF4 + OF2 < PF3 < PF4 + PF4 + < OF2 < PF3 PF4 + < PF3 < OF2 OF2 < PF4 + < PF3 PF3 < PF4 + < OF2

PF4 + < PF3 < OF2

Explanation:

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In this case, the two non-shared pairs of electrons present on the OF2 occupy more space and causes the FOF bond angle to be the largest one (180°) due to its planar arrangement. Furthermore, PF3 has just 1 non-shared pair of electrons, therefore there is less space for the bonds to be trigonally arranged, it means, a smaller angle (120°). Finally, PF4 + shows a perfectly tetrahedral arrangement therefore it has the smallest bond angle (109.5°).

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PF3 < OF2 < PF4+

Explanation:

For PF3 angle is 96.3, for OF2 it's 103 (look through internet). Due to that PF4 + ion must be completely tetrahedral angle will be 109,5