Chapter 2, Problem 2F.14E

Interpretation Introduction

Interpretation:

Lewis structure, hybridization, and bond angle of ${\text{XeO}}_3$, ${\text{XeO}}_4$ and ${\text{XeO}}_6^{4-}$ has to be drawn. Also, molecule with largest bond length of $\text{Xe}-\text{O}$ bond has to be identified.

Concept Introduction:

Hybridization is calculated by the hybrid orbitals and to calculate hybrid orbitals we need to know the steric number that is given by,

  $\text{Steric number}=\left[\begin{array}{c}\left(\text{number of atoms bonded to the central atom}\right)+\\ \left(\text{number of lone pairs on the central atom}\right)\end{array}\right]$        (1)

The table that relates the steric number with hybridization is as follows:

$\begin{array}{cc}\text{Steric Number}& \text{Type of Hybridisation}\\ 2& \text{sp}\\ 3& {\text{sp}}^2\\ 4& {\text{sp}}^3\\ 5& {\text{sp}}^3\text{d}\\ 6& {\text{sp}}^3{\text{d}}^2\end{array}$

The table that relates steric number with geometry and bond angles is as follows:

$\begin{array}{cccc}\text{Steric number}& \text{Number of lone pairs}& \text{Molecular geometry}& \text{Bond angles}\\ 2& 0& \text{Linear}& 180°\\ 3& \begin{array}{c}0\\ 1\end{array}& \begin{array}{c}\text{Trigonal planar}\\ \text{Bent}\end{array}& 120°\\ 4& \begin{array}{c}0\\ 1\\ 2\end{array}& \begin{array}{c}\text{Tetrahedral}\\ \text{Trigonal pyramidal}\\ \text{Bent}\end{array}& 109.5°\\ 5& \begin{array}{c}0\\ 1\\ 2\\ 3\end{array}& \begin{array}{c}\text{Trigonal Bi-pyramidal}\\ \text{See-Saw}\\ \text{T-shaped}\\ \text{Linear}\end{array}& 90°\text{,120 }°,180°\\ 6& \begin{array}{c}0\\ 1\\ 2\end{array}& \begin{array}{c}\text{Octahedral}\\ \text{Square pyramidal}\\ \text{Square planar}\end{array}& 90°\text{,}180°\end{array}$

Answer to Problem 2F.14E

Hybridization of ${\text{XeO}}_3$ is ${\text{sp}}^3$ with bond angle $109.5°$.

Explanation of Solution

The molecule ${\text{XeO}}_3$ consists of one $\text{Xe}$ atom and three $\text{O}$ atoms.

The symbol for xenon is $\text{Xe}$; with atomic number 54. Its electronic configuration is $\left[\text{Kr}\right]4{\text{d}}^{10}{\text{5s}}^2{\text{5p}}^6$. Thus, it possesses 8 valence electrons.

The symbol for oxygen is $\text{O}$; with atomic number 8. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^4$. Thus, it possesses 6 valence electrons.

Thus total valence electrons are sum of the valence electrons for each atom in ${\text{XeO}}_3$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=8+6\left(3\right)\\ =26\left(\text{13pairs}\right)\end{array}$

The skeleton structure ${\text{XeO}}_3$ has three bonds that comprise 6 electrons. The electrons left to be allocated are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=26-6\\ =20\left(\text{10 pairs}\right)\end{array}$

To complete the octet of xenon it forms a double bond with three oxygen atom.

  $\begin{array}{c}\text{Total remaining electrons}=20-6\\ =14\left(7\text{ pairs}\right)\end{array}$

Hence, 14 electrons are allocated as 2 lone pairs on each oxygen atoms and 1 lone pair on xenon atom to complete their octet. The Lewis structure is as follows:

In the structure of ${\text{XeO}}_3$, xenon is bonded to 3 atoms and has 1 lone pair.

Substitute 3 for number of atoms bonded with central atom and 1 for lone pair on central atom in equation (1) to calculate steric number.

$\begin{array}{c}\text{Steric number}=3+1\\ =4\end{array}$

Since, the steric number of the molecule ${\text{XeO}}_3$ is 4 that is xenon is bonded to 3 atoms and has 1 lone pair, therefore, the hybridization of xenon $\text{Xe}$ in the molecule is ${\text{sp}}^3$ and acquires tetrahedral geometry to minimize the repulsion between bond pairs and bond angle formed is $109.5°$.

The molecule ${\text{XeO}}_4$ consists of one $\text{Xe}$ and four $\text{O}$ atoms.

The symbol for xenon is $\text{Xe}$; with atomic number 54. Its electronic configuration is $\left[\text{Kr}\right]4{\text{d}}^{10}{\text{5s}}^2{\text{5p}}^6$. Thus, it possesses 8 valence electrons.

The symbol for oxygen is $\text{O}$; with atomic number 8. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^4$. Thus, it possesses 6 valence electrons.

Thus total valence electrons are sum of the valence electrons for each atom in ${\text{XeO}}_4$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=8+6\left(4\right)\\ =32\left(\text{16 pairs}\right)\end{array}$

The skeleton structure ${\text{XeO}}_4$ has four bonds that comprise 8 electrons. The electrons left to be allocated are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=32-8\\ =24\left(\text{12 pairs}\right)\end{array}$

To complete the octet of xenon it forms a double bond with four oxygen atoms.

  $\begin{array}{c}\text{Total remaining electrons}=24-8\\ =16\left(8\text{ pairs}\right)\end{array}$

Hence, 16 electrons are allocated as 2 lone pairs on each oxygen atoms to complete its octet. The Lewis structure is as follows:

In the structure of ${\text{XeO}}_4$, xenon is bonded to 4 atoms and has 0 lone pair.

Substitute 4 for atoms bonded with central atom and 0 lone pair on central atom in equation (1) to calculate steric number.

$\begin{array}{c}\text{Steric number}=4+0\\ =4\end{array}$

Since, the steric number of the molecule ${\text{XeO}}_4$ is 4 that is xenon is bonded to 4 atoms and has 0 lone pair therefore the hybridization of xenon $\text{Xe}$ in the molecule is ${\text{sp}}^3$ and acquires tetrahedral geometry to minimize the repulsion between bond pairs and bond angle formed is $109.5°$.

The molecule ${\text{XeO}}_6^{4-}$ consists of one $\text{Xe}$ atom and six $\text{O}$ atoms.

The symbol for xenon is $\text{Xe}$; with atomic number 54. Its electronic configuration is $\left[\text{Kr}\right]4{\text{d}}^{10}{\text{5s}}^2{\text{5p}}^6$. Thus, it possesses 8 valence electrons.

The symbol for oxygen is $\text{O}$; with atomic number 8. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^4$. Thus, it possesses 6 valence electrons.

Four negative charges on molecule are added up in the total valence count.

Thus total valence electrons are sum of the valence electrons for each atom in ${\text{XeO}}_6^{4-}$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=8+6\left(6\right)+4\\ =48\left(\text{24 pairs}\right)\end{array}$

The skeleton structure ${\text{XeO}}_6^{4-}$ has six bonds that comprise 12 electrons. The electrons left to be allocated are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=48-12\\ =36\left(\text{18 pairs}\right)\end{array}$

To complete the octet of xenon it forms a double bond with two oxygen atoms.

  $\begin{array}{c}\text{Remaining electrons}=36-4\\ =32\left(\text{16 pairs}\right)\end{array}$

Hence, 32 electrons are allocated as 2 lone pairs on two doubly bonded oxygen atoms and 3 lone pairs on remaining oxygen atoms to complete their octet. The Lewis structure is as follows:

In the structure of ${\text{XeO}}_6^{4-}$, xenon is bonded to 6 atoms and has 0 lone pair.

Substitute 6 for number of atoms bonded with central atom and 0 for lone pair on central atom in equation (1) to calculate steric number.

$\begin{array}{c}\text{Steric number}=6+0\\ =6\end{array}$

Since the steric number of the molecule ${\text{XeO}}_6^{4-}$ is 6 that is xenon is bonded to 6 atoms and has 0 lone pair, therefore, the hybridization of xenon $\text{Xe}$ in the molecule is ${\text{sp}}^3{\text{d}}^2$ and acquires octahedral geometry to minimize the repulsions between bond pairs and bond angle formed is $90°$ and $180°$.

Bond order is the number of electrons involved between the two atoms in a molecule. Bond order is inversely related to bond length.

${\text{XeO}}_3$ molecule contains 3 $\text{Xe}-\text{O}$ double bonds; therefore bond order of $\text{Xe}-\text{O}$ bond is 2 as two electrons are involved between $\text{Xe}-\text{O}$ bonds.

In ${\text{XeO}}_4$ molecule, the bond order of $\text{Xe}-\text{O}$ bond is 2.

${\text{XeO}}_6^{4-}$ molecule contains 4 $\text{Xe}-\text{O}$ single bonds therefore its bond order is lesser as compared to other two molecules ${\text{XeO}}_3$ and ${\text{XeO}}_4$. Hence, ${\text{XeO}}_6^{4-}$ molecule has the longest $\text{Xe}-\text{O}$ bond distance.