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The correct order of F-M-F bond angle of compounds given in the options is (where M represents the central element)

  • A.\(\text{XeF}_{2} > \text{BF}_{3} > \text{OF}_{2} > [\text{NF}_{4}]^{+}\)
  • B.\(\text{XeF}_{2} > \text{BF}_{3} > [\text{NF}_{4}]^{+} > \text{OF}_{2}\)
  • C.\(\text{BF}_{3} > \text{XeF}_{2} > [\text{NF}_{4}]^{+} > \text{OF}_{2}\)
  • D.\(\text{XeF}_{2} > \text{OF}_{2} > \text{BF}_{3} > [\text{NF}_{4}]^{+}\)

Answer: B

The correct order is XeF2 > BF3 > [NF4]+ > OF2.

  • XeF2: three lone pairs occupy the equatorial positions of a trigonal bipyramid, so the molecule is linear, 180°.
  • BF3: three bond pairs and no lone pair, trigonal planar, 120°.
  • [NF4]+: four bond pairs, regular tetrahedron, 109.5°.
  • OF2: two bond pairs and two lone pairs, bent, about 103°, below the tetrahedral angle because lone pairs repel more strongly.

Putting BF3 ahead of XeF2 ignores the linear geometry, and orders placing OF2 above [NF4]+ or BF3 ignore the compression caused by its two lone pairs.

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