Question: Calculate the formal charge on each second-row atom

a.

b.

c.

The formal charge is determined using the given formula:

$\mathbf{Formal}\mathbf{}\mathbf{Charge}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{valence}\mathbf{}{\mathbf{e}}^{\mathbf{-}}\mathbf{s}\mathbf{}\mathbf{of}\mathbf{}\mathbf{an}\mathbf{}\mathbf{atom}\mathbf{)}\mathbf{}\mathbf{-}\mathbf{(}\mathbf{lone}\mathbf{}\mathbf{pair}\mathbf{}{\mathbf{e}}^{\mathbf{-}}\mathbf{}\mathbf{of}\mathbf{}\mathbf{an}\mathbf{}\mathbf{atom}\mathbf{)}$$\mathbf{-}\left(\frac{1}{2}\times \mathrm{bonding}\mathrm{pair}\mathrm{of}{e}^{-}s\right)$

a. The formal charge of N (second-row element) is calculated as follows:

• The total count of valence electrons of nitrogen is five.
• The number of lone pair of electrons is 0.

• The total count of bonding pair electrons is eight as all the electrons of nitrogen are used.

$\begin{array}{rcl}\mathrm{Formal}\mathrm{charge}\mathrm{of}\mathrm{N}& =& 5-0-\left(\frac{1}{2}\times 8\right)\\ & =& 5-4\\ & =& +1\end{array}$

Therefore, the formal charge of N is +1.

b. The formal charge of C (methyl carbon) is calculated as follows:

• The total number of valence electrons of C is four.
• The number of lone pair of electrons is 0.
• The total number of bonding pair electrons is eight as all the electrons of N are used.

$\begin{array}{rcl}\mathrm{Formal}\mathrm{charge}\mathrm{of}\mathrm{C}& =& 4-2-\left(\frac{1}{2}\times 6\right)\\ & =& 4-5\\ & =& -1\end{array}$

Therefore, the formal charge of C is -1.

c. The formal charge of O (extreme left) is calculated as follows:

• The total valence electrons of O are six.
• The number of lone pairs of electrons is four.
• The total number of bonding pair electrons is four as all the electrons of O are used.

role="math" localid="1648189065351" $\begin{array}{rcl}\mathrm{Formal}\mathrm{charge}\mathrm{of}\mathrm{O}& =& 6-4-\left(\frac{1}{2}\times 4\right)\\ & =& 6-6\\ & =& 0\end{array}$

Therefore, the formal charge of O is 0.

c. The formal charge of O (middle) is calculated as follows:

• The total valence electrons of O are six.
• The number of lone pairs of electrons is two.

The total number of bonding pair electrons is six as all the electrons of O are used

$\begin{array}{rcl}\mathrm{Formal}\mathrm{charge}\mathrm{of}\mathrm{O}& =& 6-4-\left(\frac{1}{2}\times 4\right)\\ & =& 6-6\\ & =& 0\end{array}$

Therefore, the formal charge of O is +1.

d. The formal charge of O (extreme right) is calculated as follows:

• The total valence electrons of O are six.
• The number of lone pairs of electrons is six.
• The total number of bonding pair electrons is two as all the electrons of O are used.

$\begin{array}{rcl}\mathrm{Formal}\mathrm{charge}\mathrm{of}\mathrm{O}& =& 6-6-\left(\frac{1}{2}\times 2\right)\\ & =& 6-7\\ & =& -1\end{array}$

Therefore, the formal charge of O is -1.