
In ${\text{Xe}}{{\text{O}}_{\text{3}}}{{\text{F}}_{\text{2}}}$, the number of bond pair(s),$\pi $-bond(s) and lone pair(s) on ${\text{Xe}}$ atom respectively are.
A. $5,3,0$
B. $5,2,0$
C. $4,2,2$
D. $4,4,0$
Answer
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Hint: Xenon is a noble gas. It has a completely filled valence p shell and has 8 electrons in its valence shell. The valence shell electrons are excited to the empty d orbital and participate in bond formation.
Complete step by step solution:
We know that Xe has 8 electrons in its valence shell.
Since oxygen has a valency of 2, thus the three oxygen atoms form 1 sigma bond and 1 double bond each with the Xe atom.
Two fluorine atoms form 2 sigma bonds with Xe.
In total, we can say that xenon forms 5 sigma bonds and three pi bonds.
Since Xe will form three pi bonds, we remove three electrons from its outer electrons as they will take part in sideways overlap (to form pi bonds).
Thus 8 electrons are used up in bond formation, 2 in bonding with the two fluorine atoms and the remaining six are used in bond formation with the three oxygen atoms (since double bonds are formed). Thus, there are no lone pairs on the Xe atom.
Now, before bonding, the ${\text{5s, 5}}{{\text{p}}_{\text{x}}}{\text{, 5}}{{\text{p}}_{\text{y}}}{\text{,5}}{{\text{p}}_{\text{z}}}{\text{, 5}}{{\text{d}}_{{\text{xy}}}}$ orbitals hybridize to produce five ${\text{s}}{{\text{p}}^{\text{3}}}{\text{d}}$ hybrid orbitals of the same energy and same shape. All the 5 orbitals are singly occupied and overlap with p orbitals of the incoming oxygen and fluorine atoms to form five sigma bonds. The remaining three electrons, as we know, are involved in pi bond formation.
Thus, we can conclude by saying that in ${\text{Xe}}{{\text{O}}_{\text{3}}}{{\text{F}}_{\text{2}}}$, the number of bond pairs is 5, number of $\pi $ bonds is 3 and number of lone pairs is 0.
The correct answer is A.
Note: Xenon despite being a noble atom, forms compounds. This is because Xe atom is large in size and the valence electrons are loosely held by the nucleus. Highly electronegative elements like fluorine and oxygen attract the valence electrons of Xe forming compounds.
Complete step by step solution:
We know that Xe has 8 electrons in its valence shell.
Since oxygen has a valency of 2, thus the three oxygen atoms form 1 sigma bond and 1 double bond each with the Xe atom.
Two fluorine atoms form 2 sigma bonds with Xe.
In total, we can say that xenon forms 5 sigma bonds and three pi bonds.
Since Xe will form three pi bonds, we remove three electrons from its outer electrons as they will take part in sideways overlap (to form pi bonds).
Thus 8 electrons are used up in bond formation, 2 in bonding with the two fluorine atoms and the remaining six are used in bond formation with the three oxygen atoms (since double bonds are formed). Thus, there are no lone pairs on the Xe atom.
Now, before bonding, the ${\text{5s, 5}}{{\text{p}}_{\text{x}}}{\text{, 5}}{{\text{p}}_{\text{y}}}{\text{,5}}{{\text{p}}_{\text{z}}}{\text{, 5}}{{\text{d}}_{{\text{xy}}}}$ orbitals hybridize to produce five ${\text{s}}{{\text{p}}^{\text{3}}}{\text{d}}$ hybrid orbitals of the same energy and same shape. All the 5 orbitals are singly occupied and overlap with p orbitals of the incoming oxygen and fluorine atoms to form five sigma bonds. The remaining three electrons, as we know, are involved in pi bond formation.
Thus, we can conclude by saying that in ${\text{Xe}}{{\text{O}}_{\text{3}}}{{\text{F}}_{\text{2}}}$, the number of bond pairs is 5, number of $\pi $ bonds is 3 and number of lone pairs is 0.
The correct answer is A.
Note: Xenon despite being a noble atom, forms compounds. This is because Xe atom is large in size and the valence electrons are loosely held by the nucleus. Highly electronegative elements like fluorine and oxygen attract the valence electrons of Xe forming compounds.
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