Chapter 10, Problem 10.63P

(a)

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{PF}}_5$ is to be determined.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

Answer to Problem 10.63P

The molecular shape of ${\text{PF}}_5$ is trigonal bipyramidal.

Explanation of Solution

The total number of valence electrons of ${\text{PF}}_5$ is calculated as,

  $\text{Total valence electrons}\left(\text{TVE}\right)=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{Total number of P atom}\right)\\ \left(\text{Valence electrons of P}\right)\end{array}\right]\\ +\left[\begin{array}{l}\left(\text{Total number of F atom}\right)\\ \left(\text{Valence electrons of F}\right)\end{array}\right]\end{array}\right]$        (2)

Substitute 1 for the total number of $\text{P}$ the atom, 5 for valence electrons of $\text{P}$, 7 for the total valence electrons of $\text{F}$, and 5 for the value of the total number of $\text{F}$ atom in equation (1).

  $\begin{array}{c}\text{Total valence electrons}\left(\text{TVE}\right)=\left[\left(\text{1}\right)\left(\text{5}\right)+\left(5\right)\left(7\right)\right]\\ =40.\end{array}$

${\text{PF}}_5$ has 40 valence electrons. 10 electrons are used to form 5 single bonds with the five fluorine atoms and the remaining 30 electrons are placed on the five fluorine atoms as 3 lone pairs on each.

Thus, the Lewis structure of ${\text{PF}}_5$ is drawn as follows,

According to the Lewis structure of ${\text{PF}}_5$, there are five electron groups around the central atom $\text{P}$ with no lone pair. This gives a trigonal bipyramidal electron-group arrangement $\left({\text{AX}}_5\right)$ and gives trigonal bipyramidal molecular shape according to VSEPR.

Conclusion

The electron-group arrangement around the central atom in ${\text{PF}}_5$ is ${\text{AX}}_{\text{5}}$ and has a trigonal bipyramidal shape.

(b)

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{CCl}}_{\text{4}}$ is to be determined.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

Answer to Problem 10.63P

The molecular shape of ${\text{CCl}}_{\text{4}}$ is tetrahedral.

Explanation of Solution

The total number of valence electrons of ${\text{CCl}}_{\text{4}}$ is calculated as,

  $\text{Total valence electrons}\left(\text{TVE}\right)=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{Total number of C atom}\right)\\ \left(\text{Valence electrons of C}\right)\end{array}\right]\\ +\left[\begin{array}{l}\left(\text{Total number of Cl atom}\right)\\ \left(\text{Valence electrons of Cl}\right)\end{array}\right]\end{array}\right]$        (2)

Substitute 1 for the total number of $\text{C}$ the atom, 4 for valence electrons of $\text{C}$, 7 for the total valence electrons of $\text{Cl}$, 4 for the total number of $\text{Cl}$ the atom in equation (2).

  $\begin{array}{c}\text{Total valence electrons}\left(\text{TVE}\right)=\left[\left(1\right)\left(4\right)+\left(4\right)\left(7\right)\right]\\ =32.\end{array}$

The Lewis structure of ${\text{CCl}}_{\text{4}}$ is,

According to the Lewis structure of ${\text{CCl}}_{\text{4}}$, four electron groups are present around the central atom $\text{C}$ with no lone pair. This gives a tetrahedral electron-group arrangement $\left({\text{AX}}_4\right)$ and gives tetrahedral molecular shape according to VSEPR.

Conclusion

The electron-group arrangement around the central atom in ${\text{CCl}}_{\text{4}}$ is ${\text{AX}}_4$ and has a tetrahedral shape.

(c)

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is to be determined.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

Answer to Problem 10.63P

The molecular shape of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is trigonal pyramidal.

Explanation of Solution

The total number of valence electrons of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is calculated as,

  $\text{Total valence electrons}\left(\text{TVE}\right)=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{Total number of O atom}\right)\\ \left(\text{Valence electrons of O}\right)\end{array}\right]\\ +\left[\begin{array}{l}\left(\text{Total number of H atom}\right)\\ \left(\text{Valence electrons of H}\right)\end{array}\right]\\ -\left(\text{positive charge}\right)\end{array}\right]$        (3)

Substitute 1 for the total number of $\text{O}$ the atom, 6 for valence electrons of $\text{O}$, 1 for the total valence electrons of $\text{H}$, 3 for the total number of $\text{H}$ atom, and 1 for the positive charge in equation (3).

  $\begin{array}{c}\text{Total valence electrons}\left(\text{TVE}\right)=\left[\left(1\right)\left(6\right)+\left(3\right)\left(1\right)-1\right]\\ =8.\end{array}$

The total number of valence electrons in ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is 8. The Lewis structure of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is given as,

According to the Lewis structure of ${\text{H}}_{\text{3}}{\text{O}}^{+}$, four electron groups are present around the central atom $\text{C}$ as three bond pair and one lone pair. This gives a tetrahedral electron-group arrangement $\left({\text{AX}}_{\text{3}}\text{E}\right)$ and gives trigonal pyramidal ${\text{H}}_{\text{3}}{\text{O}}^{+}$ molecular shape according to VSEPR.

Conclusion

The electron-group arrangement around the central atom in ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is ${\text{AX}}_{\text{3}}\text{E}$ and has a trigonal pyramidal shape.

(d)

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{ICl}}_{\text{3}}$ is to be determined.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

Answer to Problem 10.63P

The molecular shape of ${\text{ICl}}_{\text{3}}$ is T-shaped.

Explanation of Solution

The total number of valence electrons of ${\text{ICl}}_{\text{3}}$ is calculated as,

  $\text{Total valence electrons}\left(\text{TVE}\right)=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{Total number of I atom}\right)\\ \left(\text{Valence electrons of I}\right)\end{array}\right]\\ +\left[\begin{array}{l}\left(\text{Total number of Cl atom}\right)\\ \left(\text{Valence electrons of Cl}\right)\end{array}\right]\end{array}\right]$        (4)

Substitute 1 for the total number of $\text{I}$ atom, 7 for valence electrons of $\text{I}$, 7 for the total valence electrons of $\text{Cl}$, and 3 for the total number of $\text{Cl}$ atom in equation (4).

  $\begin{array}{c}\text{Total valence electrons}\left(\text{TVE}\right)=\left[\left(\text{1}\right)\left(7\right)+\left(3\right)\left(7\right)\right]\\ =28.\end{array}$

With $\text{I}$ as central atom, first three bond pairs are placed between each $\text{I-Cl}$ bond then the remaining are placed as the lone pairs. Lewis structure of ${\text{ICl}}_{\text{3}}$ is,

According to the Lewis structure of ${\text{ICl}}_{\text{3}}$, there are 5 electron groups around the central atom $\text{I}$ of which three are bond pairs and 2 are lone pairs. This gives it an electron-group arrangement of the type ${\text{AX}}_3{\text{E}}_2$, molecular shape assigned would be T-shaped according to VSEPR.

Conclusion

The electron-group arrangement around the central atom in ${\text{ICl}}_{\text{3}}$ is ${\text{AX}}_3{\text{E}}_2$ and has a T-shape.

(e)

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{BeH}}_{\text{2}}$ is to be determined.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

Answer to Problem 10.63P

The molecular shape of ${\text{BeH}}_{\text{2}}$ is linear.

Explanation of Solution

The total number of valence electrons of ${\text{BeH}}_{\text{2}}$ is calculated as,

  $\text{Total valence electrons}\left(\text{TVE}\right)=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{Total number of Be atom}\right)\\ \left(\text{Valence electrons of Be }\right)\end{array}\right]\\ +\left[\begin{array}{l}\left(\text{Total number of H atom}\right)\\ \left(\text{Valence electrons of H }\right)\end{array}\right]\end{array}\right]$        (5)

Substitute 1 for the total number of $\text{Be}$ atom, 2 for valence electrons of $\text{Be}$, 1 for the total valence electrons of $\text{H}$, and 2 for the total number of $\text{H}$ atom in equation (5).

  $\begin{array}{c}\text{Total valence electrons}\left(\text{TVE}\right)=\left[\left(\text{1}\right)\left(2\right)+\left(2\right)\left(1\right)\right]\\ =4.\end{array}$

With $\text{Be}$ as central atom, first two bond pairs are placed between each $\text{Be-H}$ bond then the remaining are placed as the lone pairs. Lewis structure of ${\text{BeH}}_{\text{2}}$ is depicted as,

According to the Lewis structure of ${\text{BeH}}_{\text{2}}$, there are two electron groups around the central atom $\text{Be}$ and no lone pairs around $\text{Be}$. This gives a linear arrangement of the type ${\text{AX}}_2$, thus molecular shape assigned would be linear according to VSEPR.

Conclusion

The electron-group arrangement around the central atom in ${\text{BeH}}_{\text{2}}$ is ${\text{AX}}_2$ and has a linear shape.

(f)

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{PH}}_{\text{2}}{}^{-}$ is to be determined.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

Answer to Problem 10.63P

The molecular shape of ${\text{PH}}_{\text{2}}{}^{-}$ is bent.

Explanation of Solution

The total number of valence electrons of ${\text{PH}}_{\text{2}}{}^{-}$ is calculated as,

  $\text{Total valence electrons}\left(\text{TVE}\right)=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{Total number of P atom}\right)\\ \left(\text{Valence electrons of P}\right)\end{array}\right]+\\ \left[\begin{array}{l}\left(\text{Total number of H atom}\right)\\ \left(\text{Valence electrons of H }\right)\end{array}\right]+\\ \left(\text{negative charge}\right)\end{array}\right]$        (6)

Substitute 1 for the total number of $\text{P}$ atom, 5 for valence electrons of $\text{P}$, 1 for the total valence electrons of $\text{H}$, 2 for the total number of $\text{H}$ atom, and  1 for the negative charge in equation (6).

  $\begin{array}{c}\text{Total valence electrons}\left(\text{TVE}\right)=\left[\left(\text{1}\right)\left(5\right)+\left(2\right)\left(1\right)+1\right]\\ =8.\end{array}$

The Lewis structure of ${\text{PH}}_{\text{2}}{}^{-}$ is,

According to the Lewis structure of molecular shape of ${\text{PH}}_{\text{2}}{}^{-}$ would be T-shaped according to VSEPR. There are 4 electron groups around the central atom $\text{P}$ of which two are lone pairs and two bond pairs. This gives ${\text{AX}}_2{\text{E}}_2$ type arrangement, a bent V-shaped molecular shape according to VSEPR.

Conclusion

The electron-group arrangement around the central atom in ${\text{PH}}_{\text{2}}{}^{-}$ is ${\text{AX}}_2{\text{E}}_2$ and has a bent shape.

(g)

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{GeBr}}_{\text{4}}$ is to be determined.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

Answer to Problem 10.63P

The molecular shape of ${\text{GeBr}}_{\text{4}}$ is tetrahedral.

Explanation of Solution

The total number of valence electrons of ${\text{GeBr}}_{\text{4}}$ is calculated as,

  $\text{Total valence electrons}\left(\text{TVE}\right)=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{Total number of Ge atom}\right)\\ \left(\text{Valence electrons of Ge}\right)\end{array}\right]\\ +\left[\begin{array}{l}\left(\text{Total number of Br atom}\right)\\ \left(\text{Valence electrons of Br}\right)\end{array}\right]\end{array}\right]$        (7)

Substitute 1 for the total number of $\text{Ge}$ the atom, 4 for valence electrons of $\text{Ge}$, 7 for the total valence electrons of $\text{Br}$, 4 for the total number of $\text{Br}$ the atom, in equation (7).

  $\begin{array}{c}\text{Total valence electrons}\left(\text{TVE}\right)=\left[\left(1\right)\left(4\right)+\left(4\right)\left(7\right)\right]\\ =32.\end{array}$

The total number of valence electrons in ${\text{GeBr}}_{\text{4}}$ is 32. The Lewis structure of ${\text{GeBr}}_{\text{4}}$ is,

According to the Lewis structure of ${\text{GeBr}}_{\text{4}}$, four electron groups are present around the central atom $\text{Ge}$ with no lone pair. This gives a tetrahedral electron-group arrangement $\left({\text{AX}}_4\right)$ and gives tetrahedral molecular shape according to VSEPR.

Conclusion

The electron-group arrangement around the central atom in ${\text{GeBr}}_{\text{4}}$ is ${\text{AX}}_4$ and has a tetrahedral shape.

(h)

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{CH}}_{\text{3}}{}^{-}$ is to be determined.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

Answer to Problem 10.63P

The molecular shape of ${\text{CH}}_{\text{3}}{}^{-}$ is trigonal pyramidal.

Explanation of Solution

The total number of valence electrons of ${\text{CH}}_{\text{3}}{}^{-}$ is calculated as,

  $\text{Total valence electrons}\left(\text{TVE}\right)=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{Total number of C atom}\right)\\ \left(\text{Valence electrons of C}\right)\end{array}\right]\\ +\left[\begin{array}{l}\left(\text{Total number of H atom}\right)\\ \left(\text{Valence electrons of H}\right)\end{array}\right]\\ +\left(\text{negative charge}\right)\end{array}\right]$        (8)

Substitute 1 for the total number of $\text{C}$ atom, 4 for valence electrons of $\text{C}$, 1 for the total valence electrons of $\text{H}$, 3 for the total number of $\text{H}$ atom, and 1 for the negative charge in equation (8).

  $\begin{array}{c}\text{Total valence electrons}\left(\text{TVE}\right)=\left[\left(1\right)\left(4\right)+\left(3\right)\left(1\right)+1\right]\\ =8.\end{array}$

The total number of valence electrons in ${\text{CH}}_{\text{3}}{}^{-}$ is 8. These eight electrons are placed such that three electrons form bonding pair and the last resides on carbon as shown below:

According to the Lewis structure of ${\text{CH}}_{\text{3}}{}^{-}$, four electron groups are present around the central atom $\text{C}$ as three bond pairs and one lone pair. This gives a tetrahedral electron-group arrangement $\left({\text{AX}}_{\text{3}}\text{E}\right)$ and the electrons will best occupy the position as in a trigonal pyramidal shape according to VSEPR.

Conclusion

The electron-group arrangement around the central atom in ${\text{CH}}_{\text{3}}{}^{-}$ is ${\text{AX}}_{\text{3}}\text{E}$ and has a trigonal pyramidal shape.

(i)

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{BCl}}_{\text{3}}$ is to be determined.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

Answer to Problem 10.63P

The molecular shape of ${\text{BCl}}_{\text{3}}$ is trigonal planar.

Explanation of Solution

The total number of valence electrons of ${\text{BCl}}_{\text{3}}$ is calculated as,

  $\text{Total valence electrons}\left(\text{TVE}\right)=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{Total number of B atom}\right)\\ \left(\text{Valence electrons of B}\right)\end{array}\right]\\ +\left[\begin{array}{l}\left(\text{Total number of Cl atom}\right)\\ \left(\text{Valence electrons of Cl }\right)\end{array}\right]\end{array}\right]$        (9)

Substitute 1 for the total number of $\text{B}$ atom, 3 for valence electrons of $\text{B}$, 7 for the total valence electrons of $\text{Cl}$, and 3 for the total number of $\text{Cl}$ atom in equation (9).

  $\begin{array}{c}\text{Total valence electrons}\left(\text{TVE}\right)=\left[\left(\text{1}\right)\left(3\right)+\left(3\right)\left(7\right)\right]\\ =24.\end{array}$

These 24 electrons are placed such that three of these form bonding pairs and the remaining ones reside as lone pairs as shown below:

According to the Lewis structure of ${\text{BCl}}_{\text{3}}$, there are three electron groups around the central atom $\text{B}$. This gives a trigonal planar arrangement of the type ${\text{AX}}_3$, a trigonal planar molecular shape according to VSEPR.

Conclusion

The electron-group arrangement around the central atom in ${\text{BCl}}_{\text{3}}$ is ${\text{AX}}_3$ and has a trigonal planar shape.

(j)

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{BrF}}_4{}^{+}$ is to be determined.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

Answer to Problem 10.63P

The molecular shape of ${\text{BrF}}_4{}^{+}$ is seesaw.

Explanation of Solution

The total number of valence electrons of ${\text{BrF}}_4{}^{+}$ is calculated as,

  $\text{Total valence electrons}\left(\text{TVE}\right)=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{Total number of Br atom}\right)\\ \left(\text{Valence electrons of Br }\right)\end{array}\right]\\ +\left[\begin{array}{l}\left(\text{Total number of F atom}\right)\\ \left(\text{Valence electrons of F }\right)\end{array}\right]\\ -\left(\text{positive charge}\right)\end{array}\right]$        (10)

Substitute 1 for the total number of $\text{Br}$ atom, 7 for valence electrons of $\text{Br}$, 7 for the total valence electrons of $\text{F}$, 4 for the total number of $\text{F}$ atom, and 1 for the positive charge in equation (10).

  $\begin{array}{c}\text{Total valence electrons}\left(\text{TVE}\right)=\left[\left(\text{1}\right)\left(7\right)+\left(4\right)\left(7\right)-1\right]\\ =34.\end{array}$

The Lewis structure for ${\text{BrF}}_4{}^{+}$ is represented as follows:

According to the Lewis structure of ${\text{BrF}}_4{}^{+}$, there are 5 electron groups around the central atom of which one is lone pair and three are bond pairs. This gives an arrangement of type ${\text{AX}}_{\text{4}}\text{E}$ , and seesaw molecular shape according to VSEPR.

Conclusion

The electron-group arrangement around the central atom in ${\text{BrF}}_4{}^{+}$ is ${\text{AX}}_{\text{4}}\text{E}$ and has a seesaw shape.

(k)

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{XeO}}_{\text{3}}$ is to be determined.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

Answer to Problem 10.63P

The molecular shape of ${\text{XeO}}_{\text{3}}$ is trigonal pyramidal.

Explanation of Solution

The total number of valence electrons of ${\text{XeO}}_{\text{3}}$ is calculated as,

  $\text{Total valence electrons}\left(\text{TVE}\right)=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{Total number of Xe atom}\right)\\ \left(\text{Valence electrons of Xe}\right)\end{array}\right]\\ +\left[\begin{array}{l}\left(\text{Total number of O atom}\right)\\ \left(\text{Valence electrons of O}\right)\end{array}\right]\end{array}\right]$        (11)

Substitute 1 for the total number of $\text{Xe}$ the atom, 8 for valence electrons of $\text{Xe}$, 6 for the total valence electrons of $\text{O}$, 3 for the total number of $\text{O}$ the atom in equation (11).

  $\begin{array}{c}\text{Total valence electrons}\left(\text{TVE}\right)=\left[\left(1\right)\left(8\right)+\left(3\right)\left(6\right)\right]\\ =26.\end{array}$

The total number of valence electrons in ${\text{XeO}}_{\text{3}}$ is 26. The Lewis structure of ${\text{XeO}}_{\text{3}}$ is thus drawn as:

According to the Lewis structure of ${\text{XeO}}_{\text{3}}$, 4 electron groups are present as three bond pairs and 1 lone pair. This gives a tetrahedral electron-group arrangement $\left({\text{AX}}_{\text{3}}\text{E}\right)$ and gives trigonal pyramidal molecular shape according to VSEPR.

Conclusion

The electron-group arrangement around the central atom in ${\text{XeO}}_{\text{3}}$ is ${\text{AX}}_{\text{3}}\text{E}$ and has a trigonal pyramidal shape.

(l)

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{TeF}}_4$ is to be determined.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

Answer to Problem 10.63P

The molecular shape of ${\text{TeF}}_4$ is seesaw.

Explanation of Solution

The total number of valence electrons of ${\text{TeF}}_4$ is calculated as,

  $\text{Total valence electrons}\left(\text{TVE}\right)=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{Total number of Te atom}\right)\\ \left(\text{Valence electrons of Te }\right)\end{array}\right]\\ +\left[\begin{array}{l}\left(\text{Total number of F atom}\right)\\ \left(\text{Valence electrons of F }\right)\end{array}\right]\end{array}\right]$        (12)

Substitute 1 for the total number of $\text{Te}$ atom, 6 for valence electrons of $\text{Te}$, 7 for the total valence electrons of $\text{F}$, and 4 for the total number of $\text{F}$ atom in equation (12).

  $\begin{array}{c}\text{Total valence electrons}\left(\text{TVE}\right)=\left[\left(\text{1}\right)\left(6\right)+\left(4\right)\left(7\right)\right]\\ =34.\end{array}$

Analogous to ${\text{GeBr}}_{\text{4}}$ Lewis structure for ${\text{TeF}}_4$ may be drawn as follows:

According to the Lewis structure of ${\text{TeF}}_4$, there are 5 electron groups of which 4 are bond pairs and 1 is lone pair. This gives an arrangement of type ${\text{AX}}_{\text{4}}\text{E}$ , a seesaw molecular shape according to VSEPR.

Conclusion

The electron-group arrangement around the central atom in ${\text{TeF}}_4$ is ${\text{AX}}_{\text{4}}\text{E}$ and has a seesaw shape.