The F—Xe—F bond angle in XeF₄ is

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109°

103°

90°

60°

answer is C.

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Detailed Solution

In XeF₄ the F—Xe—F bond angle is $90 °$ . The Xe atom lies in the middle of the XeF₄ molecule, while the fluorine atoms are pointed in the direction of the square's four corners. It has a square pyramidal shape.

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