How $X e F_{2}$ and $X e F_{4}$ are prepared? What are their structure?

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Solution

$X e F_{2}$ and $X e F_{4}$ are obtained by a direct reaction between $X e$ and $F_{2}$ . The condition under which the reaction is carried out determines the product.
${X e}_{(g)} + {F_{2}}_{(g)} 673 K 1 bar - ------ \to {X e F_{2}}_{(s)}$
${X e}_{(g)} + 2 {F_{2}}_{(g)} 873 K 7 bar - ------ \to {X e F_{2}}_{(s)}$
Structure of $X e F_{2} \to$ The hybridisation of $X e F_{2}$ , is $s p^{3} d$ with two bond pair and three lone pair of electrons. Thus the structure is trigonal bypyramidal.
Structure of $X e F_{4} \to$ In case of $X e F_{4}$ , the hybridisation is $s p^{3} d^{2}$ with four bond pair and two lone pair of electrons. The two lone pairs will be opposite each other (to maintain symmetry), which makes for a square planar molecular geometry.

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