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A molecule which has T.B.P. geometry, the axial bond length is always greater than that of equatorial bond length.

A molecule that has P.B.P. geometry, the axial orbital length of the central atom is greater than that of the equatorial orbital length.

In ${PCl}_{3} F_{2}$ molecule, the equatorial bond length is greater than that of the axial bond length

A molecule which has P.B.P geometry, equatorial orbital bond length of central atom is longer than that of axial orbital length.

answer is B, D.

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Detailed Solution

Since, In ${PCl}_{3} F_{2}$ molecule, the equatorial bond length is greater than the axial bond length. And also, in A molecule that has P.B.P geometry, equatorial orbital bond length of the central atom is longer than that of the axial orbital length.

So, correct option is B and D.

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