
The molecule \[Xe{O_3}\] has
(A) three double bonded O-atoms
(B) Trigonal pyramidal geometry
(C) one lone pair and \[s{p^3}\] hybridisation
(D) All of these
Answer
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Hint: In order to find which of the following statements is correct for Xenon, we must first know about the atomic number and the electronic configuration of Xenon. On knowing this it will be easy to find the hybridisation, geometry and bonding present in Xenon.
Complete step by step answer:
In the Xenon trioxide the central atom present is Xenon. Xenon is a noble gas and its atomic number is 54. The electronic configuration for Xenon is given as \[Xe = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}5{s^2}4{d^{10}}5{p^6} = [Kr]4{d^{10}}5{s^2}5{p^6}\]
- Xenon being a noble gas it will be having octet electrons in the outermost shell.
The oxygen atom is having an atomic number 8 and the electronic configuration of Oxygen is given as
\[O = 1{s^2}2{s^2}2{p^4} = [He]2{s^2}2{p^4}\]
- The number of valence electrons present in the oxygen atom is 6. The six valence electrons of Xenon will combine with three oxygen atoms to form three Xenon-Oxygen double bonds. And there will be two electrons left, which will act as the lone pair.
- The number of bonding present in \[Xe{O_3}\] is 3. But the steric number for the central metal atom in \[Xe{O_3}\] is 4, this is due to lone pairs. Therefore, the hybridisation present in \[Xe{O_3}\] is \[s{p^3}\]. Usually when there is \[s{p^3}\] hybridisation the molecule will be having a tetrahedral geometry but due to the presence of the lone pair the geometry will be trigonal pyramidal geometry.
- From the above discussion, we can say that the Xenon trioxide will be having \[s{p^3}\] hybridisation, trigonal pyramidal geometry, three Xenon-oxygen double bonds and one lone pair. The correct option is option “D” .
Note: We have to always remember that the \[s{p^3}\] hybridisation will be having a tetrahedral geometry. But in this case, there is a lone pair present. So, due to lone pair repulsion the geometry becomes distorted and forms trigonal pyramidal geometry.
Complete step by step answer:
In the Xenon trioxide the central atom present is Xenon. Xenon is a noble gas and its atomic number is 54. The electronic configuration for Xenon is given as \[Xe = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}5{s^2}4{d^{10}}5{p^6} = [Kr]4{d^{10}}5{s^2}5{p^6}\]
- Xenon being a noble gas it will be having octet electrons in the outermost shell.
The oxygen atom is having an atomic number 8 and the electronic configuration of Oxygen is given as
\[O = 1{s^2}2{s^2}2{p^4} = [He]2{s^2}2{p^4}\]
- The number of valence electrons present in the oxygen atom is 6. The six valence electrons of Xenon will combine with three oxygen atoms to form three Xenon-Oxygen double bonds. And there will be two electrons left, which will act as the lone pair.
- The number of bonding present in \[Xe{O_3}\] is 3. But the steric number for the central metal atom in \[Xe{O_3}\] is 4, this is due to lone pairs. Therefore, the hybridisation present in \[Xe{O_3}\] is \[s{p^3}\]. Usually when there is \[s{p^3}\] hybridisation the molecule will be having a tetrahedral geometry but due to the presence of the lone pair the geometry will be trigonal pyramidal geometry.
- From the above discussion, we can say that the Xenon trioxide will be having \[s{p^3}\] hybridisation, trigonal pyramidal geometry, three Xenon-oxygen double bonds and one lone pair. The correct option is option “D” .
Note: We have to always remember that the \[s{p^3}\] hybridisation will be having a tetrahedral geometry. But in this case, there is a lone pair present. So, due to lone pair repulsion the geometry becomes distorted and forms trigonal pyramidal geometry.
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