Question

Determine the formal charges on all the atoms in the following Lewis diagrams

$H-\ddot{N}=\ddot{0}$and $H-\ddot{N}=\ddot{0}$

Which one would best represent bonding in the$HNO$molecule?

Short answer

The first option that is$H-\ddot{N}=\ddot{0}$more favorable because the formal charges have lower absolute values.

Step-by-step solution

Determine the formal charges of all atoms

Each atom in a molecule can be assigned a formal charge in the molecule:

Formal charge of $X=$group of $X^{-}$(lone electrons on$X+\frac{1}{2}$electrons in bond to $X$)

Now, in $HNO:$

$H-$-in the 1st group, lone electrons, bond of electrons

Then, the formal charge of$H=1-\left(0+\frac{1}{2}·2\right)=0$

Formal charge of $N=5-\left(2+\frac{1}{2}\times 2\times 3\right)=5-5=0$

And, the formal charge of

$O=6-\left(4+\frac{1}{2}\times 2\times 2\right)=6-6=0$

So, in $HON:$

$H-\ddot{0}=\ddot{N}$

$H-$in the 1st group, lone electrons, bond of electrons

Then, the formal charge of $H=1-\left(0+\frac{1}{2}·2\right)=0$

Formal charge of $O=6-\left(2+\frac{1}{2}\times 2\times 3\right)=6-5=1$

And, the formal charge of $0=5-\left(4+\frac{1}{2}\times 2\times 2\right)=5-6=-1$

Since in the first option the formal charges are all whereas in the second option, oxygen has formal charge of $-1$and nitrogen of$+1.$

Therefore, structures where atoms have lower formal charges are preferred.

Hence, the first option$H-\ddot{N}=\ddot{0}$is more favorable because the formal charges have lower absolute values.