Chapter 10.1, Problem 10.5BFP

(a)

Interpretation Introduction

Interpretation:

Lewis structure with the lowest formal charges of ${\text{BeH}}_{\text{2}}$ is to be drawn.

Concept introduction:

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

Step 1: Find the central atom and place the other atoms around it. The atom in a compound which has the lowest group number or lowest electronegativity considered as the central atom.

Step 2: Calculate the total number of valence electrons.

Step 3: Connect the other atoms around the central atoms to the central atom with a single bond and lower the value of valence electrons by 2 of every single bond.

Step 4: Allocate the remaining electrons in pairs so that each atom can get 8 electrons.

Formula to calculate the formal charge of the atom is as follows:

  $\text{Formal}\text{charge}=\left(\begin{array}{l}\text{number}\text{of}\\ \text{valence}\\ \text{electrons}\end{array}\right)-\left(\left(\begin{array}{l}\text{number}\text{of}\\ \text{non-bonding}\\ \text{electrons}\end{array}\right)+\left(\frac{1}{2}\right)\left(\begin{array}{l}\text{number}\text{of}\\ \text{bonding}\\ \text{electrons}\end{array}\right)\right)$        (1)

Answer to Problem 10.5BFP

Lewis structure with lowest formal charges for ${\text{BeH}}_{\text{2}}$ is,

The formal charge on $\text{Be}$ is 0.

The formal charge on each $\text{H}$ is 0.

Explanation of Solution

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

  $\text{Total valence electrons}\left(\text{TVE}\right)=\left[\begin{array}{l}\left(\text{Total number of Be atom}\right)\\ \left(\text{Valence electrons of Be}\right)\\ +\left(\text{Total number of H atom}\right)\\ \left(\text{Valence electrons of H}\right)\end{array}\right]$        (1)

Substitute 3 for the total number of $\text{Be}$ atom, 2 for valence electrons of $\text{Be}$, and 2 for the number of $\text{H}$ atom, and 1 for valence electrons of $\text{H}$ in equation (1).

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

${\text{BeH}}_{\text{2}}$ has 4 valence electrons. Centre $\text{Be}$ atom form two single bonds with each $\text{H}$ atom. All electrons are used to form 2 single bonds. This leaves beryllium atom 4 electrons short to complete its octet. Thus, ${\text{BeH}}_{\text{2}}$ is an exception to the octet rule.

The Lewis structure of ${\text{BeH}}_{\text{2}}$ is as follows:

Substitute 2 for the number of valence electrons, 0 for the number of nonbonded electrons and 4 for the number of bonding electrons in equation (1) to calculate the formal charge on $\text{Be}$ atom.

  $\begin{array}{c}\text{Formal}\text{charge}=\left(2\right)-\left(\left(\text{0}\right)+\left(\frac{1}{2}\right)\left(4\right)\right)\\ =0\end{array}$

Substitute 1 for the number of valence electrons, 0 for the number of nonbonded electrons and 2 for the number of bonding electrons in equation (1) to calculate the formal charge on each hydrogen atom.

  $\begin{array}{c}\text{Formal}\text{charge}=\left(\text{7}\right)-\left(\left(6\right)+\left(\frac{1}{2}\right)\left(2\right)\right)\\ =0.\end{array}$

Conclusion

${\text{BeH}}_{\text{2}}$ has only one Lewis structure with the zero formal charges on $\text{Be}$ and $\text{H}$ atom.

(b)

Interpretation Introduction

Interpretation:

Lewis structure with the lowest formal charges of ${\text{I}}_3^{-}$ is to be drawn.

Concept introduction:

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

Step 1: Find the central atom and place the other atoms around it. The atom in a compound which has the lowest group number or lowest electronegativity considered as the central atom.

Step 2: Calculate the total number of valence electrons.

Step 3: Connect the other atoms around the central atoms to the central atom with a single bond and lower the value of valence electrons by 2 of every single bond.

Step 4: Allocate the remaining electrons in pairs so that each atom can get 8 electrons.

Formula to calculate the formal charge of the atom is as follows:

  $\text{Formal}\text{charge}=\left(\begin{array}{l}\text{number}\text{of}\\ \text{valence}\\ \text{electrons}\end{array}\right)-\left(\left(\begin{array}{l}\text{number}\text{of}\\ \text{non-bonding}\\ \text{electrons}\end{array}\right)+\left(\frac{1}{2}\right)\left(\begin{array}{l}\text{number}\text{of}\\ \text{bonding}\\ \text{electrons}\end{array}\right)\right)$        (1)

Answer to Problem 10.5BFP

The Lewis structure of ${\text{I}}_3^{-}$ is,

The formal charge on central $\text{I}$ is $-1$.

The formal charge on the terminal $\text{I}$ is 0.

Explanation of Solution

The total number of valence electrons of ${\text{I}}_3^{-}$ is calculated as,

  $\text{Total valence electrons}\left(\text{TVE}\right)=\left[\begin{array}{l}\left(\text{Total number of I atom}\right)\\ \left(\text{Valence electrons of I}\right)\\ +\left(\text{negative charge}\right)\end{array}\right]$        (1)

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

  $\begin{array}{c}\text{Total valence electrons}\left(\text{TVE}\right)=\left[\left(3\right)\left(7\right)+\left(1\right)\right]\\ =22\end{array}$

${\text{I}}_3^{-}$ has 22 valence electrons. One iodine serves as the central atom and forms two single bonds with other two $\text{I}$ atoms. Out of 22 electrons, 4 are used for the formation of 2 single bonds. Remaining 18 electrons are distributed as lone pairs of iodine atoms to complete the octet of each atom.

The Lewis structure of ${\text{I}}_3^{-}$ is as follows:

Substitute 7 for the number of valence electrons, 6 for the number of nonbonded electrons and 4 for the number of bonding electrons in equation (1) to calculate the formal charge on the central $\text{I}$ atom.

  $\begin{array}{c}\text{Formal}\text{charge}=\left(\text{7}\right)-\left(\left(\text{6}\right)+\left(\frac{1}{2}\right)\left(\text{4}\right)\right)\\ =-1\end{array}$

Substitute 7 for the number of valence electrons, 6 for the number of nonbonded electrons and 2 for the number of bonding electrons in equation (1) to calculate the formal charge on each terminal $\text{I}$ atom.

  $\begin{array}{c}\text{Formal}\text{charge}=\left(7\right)-\left(\left(6\right)+\left(\frac{1}{2}\right)\left(2\right)\right)\\ =0.\end{array}$

Conclusion

Only one Lewis structure is possible for ${\text{I}}_3^{-}$ and the central iodine has $-1$ formal charge.

(c)

Interpretation Introduction

Interpretation:

Lewis structure with the lowest formal charges of ${\text{XeO}}_3$ is to be drawn.

Concept introduction:

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

Step 1: Find the central atom and place the other atoms around it. The atom in a compound which has the lowest group number or lowest electronegativity considered as the central atom.

Step 2: Calculate the total number of valence electrons.

Step 3: Connect the other atoms around the central atoms to the central atom with a single bond and lower the value of valence electrons by 2 of every single bond.

Step 4: Allocate the remaining electrons in pairs so that each atom can get 8 electrons.

Formula to calculate the formal charge of the atom is as follows:

  $\text{Formal}\text{charge}=\left(\begin{array}{l}\text{number}\text{of}\\ \text{valence}\\ \text{electrons}\end{array}\right)-\left(\left(\begin{array}{l}\text{number}\text{of}\\ \text{non-bonding}\\ \text{electrons}\end{array}\right)+\left(\frac{1}{2}\right)\left(\begin{array}{l}\text{number}\text{of}\\ \text{bonding}\\ \text{electrons}\end{array}\right)\right)$        (1)

Answer to Problem 10.5BFP

Lewis structure of ${\text{XeO}}_3$ with lowest formal charge is,

The formal charge on $\text{Xe}$ is 0.

The formal charge on each $\text{O}$ is 0.

Explanation of Solution

The Lewis structures of ${\text{XeO}}_3$ are written as follows:

For structure I:

Substitute 8 for the number of valence electrons, 2 for the number of nonbonded electrons and 6 for the number of bonding electrons in equation (1) to calculate the formal charge on $\text{Xe}$ atom.

  $\begin{array}{c}\text{Formal}\text{charge}=\left(\text{8}\right)-\left(\left(2\right)+\left(\frac{1}{2}\right)\left(6\right)\right)\\ =+3\end{array}$

Substitute 6 for the number of valence electrons, 6 for the number of nonbonded electrons and 2 for the number of bonding electrons in equation (1) to calculate the formal charge on each single bonded oxygen atom.

  $\begin{array}{c}\text{Formal}\text{charge}=\left(\text{6}\right)-\left(\left(6\right)+\left(\frac{1}{2}\right)\left(2\right)\right)\\ =-1\end{array}$

For structure II:

Substitute 8 for the number of valence electrons, 2 for the number of nonbonded electrons and 8 for the number of bonding electrons in equation (1) to calculate the formal charge on $\text{Xe}$ atom.

  $\begin{array}{c}\text{Formal}\text{charge}=\left(\text{8}\right)-\left(\left(2\right)+\left(\frac{1}{2}\right)\left(8\right)\right)\\ =+2\end{array}$

Substitute 6 for the number of valence electrons, 6 for the number of nonbonded electrons and 2 for the number of bonding electrons in equation (1) to calculate the formal charge on each single bonded oxygen atom.

  $\begin{array}{c}\text{Formal}\text{charge}=\left(\text{6}\right)-\left(\left(6\right)+\left(\frac{1}{2}\right)\left(2\right)\right)\\ =-1\end{array}$

Substitute 6 for the number of valence electrons, 4 for the number of nonbonded electrons and 4 for the number of bonding electrons in equation (1) to calculate the formal charge on the double bonded oxygen atom.

  $\begin{array}{c}\text{Formal}\text{charge}=\left(\text{6}\right)-\left(\left(4\right)+\left(\frac{1}{2}\right)\left(4\right)\right)\\ =0\end{array}$

For structure III:

Substitute 8 for the number of valence electrons, 2 for the number of nonbonded electrons and 10 for the number of bonding electrons in equation (1) to calculate the formal charge on $\text{Xe}$ atom.

  $\begin{array}{c}\text{Formal}\text{charge}=\left(\text{8}\right)-\left(\left(2\right)+\left(\frac{1}{2}\right)\left(10\right)\right)\\ =+1\end{array}$

Substitute 6 for the number of valence electrons, 6 for the number of nonbonded electrons and 2 for the number of bonding electrons in equation (1) to calculate the formal charge on the single bonded oxygen atom.

  $\begin{array}{c}\text{Formal}\text{charge}=\left(\text{6}\right)-\left(\left(6\right)+\left(\frac{1}{2}\right)\left(2\right)\right)\\ =-1\end{array}$

Substitute 6 for the number of valence electrons, 4 for the number of nonbonded electrons and 4 for the number of bonding electrons in equation (1) to calculate the formal charge on each double bonded oxygen atom.

  $\begin{array}{c}\text{Formal}\text{charge}=\left(\text{6}\right)-\left(\left(4\right)+\left(\frac{1}{2}\right)\left(4\right)\right)\\ =0\end{array}$

For structure IV:

Substitute 8 for the number of valence electrons, 2 for the number of nonbonded electrons and 12 for the number of bonding electrons in equation (1) to calculate the formal charge on $\text{Xe}$ atom.

  $\begin{array}{c}\text{Formal}\text{charge}=\left(\text{8}\right)-\left(\left(2\right)+\left(\frac{1}{2}\right)\left(12\right)\right)\\ =0\end{array}$

Substitute 6 for the number of valence electrons, 4 for the number of nonbonded electrons and 4 for the number of bonding electrons in equation (1) to calculate the formal charge on each double bonded oxygen atom.

  $\begin{array}{c}\text{Formal}\text{charge}=\left(\text{6}\right)-\left(\left(4\right)+\left(\frac{1}{2}\right)\left(4\right)\right)\\ =0.\end{array}$

Conclusion

Four Lewis structures are possible for ${\text{XeO}}_3$. Among 4 structures, only one structure has zero formal charge on xenon atom.