Question

Number of pi bonds in \[Xe{O_4}\]is:

Answer 1

Hint: The structure of xenon compound reveals the number of pi bonds present. The hybridization of xenon compounds determines the number of bonds present in between xenon and oxygen atoms.

Complete step by step answer: \[Xe{O_4}\] is the compound formed of xenon and fluorine atoms. Xenon is the central atom in \[Xe{O_4}\].
\[Xe{O_4}\] is synthesized by treating sodium or barium xenate with conc.\[{H_2}S{O_4}\]. Sodium and barium xenates have chemical formulas as \[N{a_4}Xe{O_6}\] and\[B{a_2}Xe{O_6}\] . The \[Xe{O_4}\] is purified using sublimation at\[195K\] .
$N{a_4}Xe{O_6} + 2{H_2}S{O_4} \to Xe{O_4} + 2N{a_2}S{O_4} + 2{H_2}O$
$B{a_2}Xe{O_6} + 2{H_2}S{O_4} \to Xe{O_4} + 2BaS{O_4} + 2{H_2}O$
In order to find the number of pi bonds in \[Xe{O_4}\], the number of structures has to be determined first. For this the VSEPR theory is used. Xenon is an element in the periodic table with atomic number \[54\] and electronic configuration is\[\left[ {Kr} \right]4{d^{10}}5{s^2}5{p^6}\] .
 

The number of valence electrons of Xenon in ground state is \[8\]. In the fourth excited, xenon atom has \[8\] unpaired electrons. One electron in s-orbital and three electrons in p-orbitals undergo \[s{p^3}\] hybridization.
As no lone pair of electrons is present the shape of \[Xe{O_4}\] is tetrahedral with the bond angle of\[{109^o}28'\]. The structure of the \[Xe{O_4}\] reveals that there are four sigma bonds and four pi bonds. The s and p orbitals are involved in sigma bonds and the d-orbitals are involved in pi bonds.The pi bonds are formed between the p-orbitals of oxygen and the d-orbitals of xenon. Thus they are also labeled as \[p\pi - d\pi \] bonds.

Note:
Xenon is a noble gas but it forms compounds in excited states. The outer 5d orbitals are used for pi bond formation.