Trigonal bipyramidal geometry is shown by:
Show Hint
- \(\mathrm{XeOF_2}\)
- \(\mathrm{XeO_3F_2}\)
- \(\mathrm{FXeOSO_2F}\)
- \([\mathrm{XeF_8}]^{2-}\)
The Correct Option is B
Solution and Explanation
(1) \(\mathrm{XeOF_2}\): Xe=O uses 2 electrons and two Xe-F bonds use 2 electrons, so 4 electrons remain, giving 2 lone pairs. With 3 sigma bonds and 2 lone pairs there are 5 regions, sp\(^3\)d, but the shape is T-shaped, not a regular trigonal bipyramid.
(2) \(\mathrm{XeO_3F_2}\): 3 Xe=O and 2 Xe-F use 3 \(\times\) 2 + 2 = 8 electrons, so there are no lone pairs. 5 sigma bonds and 0 lone pairs give sp\(^3\)d hybridisation and a perfect trigonal bipyramidal geometry (3 oxygen atoms equatorial, 2 fluorine atoms axial).
(3) \(\mathrm{FXeOSO_2F}\): Xe has 2 sigma bonds (to F and to O) and 3 lone pairs, so the shape is linear.
(4) \([\mathrm{XeF_8}]^{2-}\): 8 bond pairs and 1 lone pair, sp\(^3\)d\(^3\), a square antiprismatic structure.
Only \(\mathrm{XeO_3F_2}\) is trigonal bipyramidal.
Answer: (2) \(\mathrm{XeO_3F_2}\)
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