Determine the hybridisation and geometry of the central atom in ${\mathrm{X}\mathrm{e}\mathrm{F}}_2$ .

<p>Xenon has 8 valence electrons. In&nbsp;<span contenteditable="false"><math><msub><mrow><mrow><mi mathvariant="normal">X</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">F</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math>&nbsp;</span>, it forms two bonds with fluorine, leaving three lone pairs. The steric number =2 bonds +3 lone pairs <span contenteditable="false"><math><mo>=</mo><mn>5</mn></math>&nbsp;</span>.<br>As per the VSEPR Theory&nbsp; the electron? pair geometry due to SN = 5 will be trigonal bipyramidal. The hybridization type will be <span contenteditable="false"><math><msup><mrow><mrow><mi mathvariant="normal">s</mi><mi mathvariant="normal">p</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup><mtext></mtext><mi mathvariant="normal">d</mi></math>&nbsp;</span>hybridisation. However, the three lone pairs occupy equatorial positions, resulting in a linear molecular geometry.</p>

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