Question

The formal charges of ${{{N}}_{\left( {{1}} \right)}}$, ${{{N}}_{\left( {{2}} \right)}}$ and ${{O}}$ atoms in ${{:}}{{{\ddot N}}_{\left( {{1}} \right)}} = {{{N}}_{\left( {{2}} \right)}} = {{\ddot O:}}$: are respectively.
A) $ + 1, - 1,0$
B) $ - 1, + 1,0$
C) $ + 1, + 1,0$
D) $ - 1, - 1,0$

Answer 1

Hint: To solve this we must know that the formal charge is the charge assigned to an atom in a molecule assuming that the bonded atoms equally share the electrons regardless of their electronegativities. Determine the formal charge using the formula to calculate the formal charge.

Formula Used:${{\text{Formal charge}}} = {{\text{Valence electrons in free atom}}} - {{\text{Non - bonding electrons}}} - \dfrac{1}{2}{{\text{Bonding electrons}}}$

Complete step-by-step answer:
We are given a structure as follows:
${{:}}{{{\ddot N}}_{\left( {{1}} \right)}} = {{{N}}_{\left( {{2}} \right)}} = {{\ddot O:}}$
We have to calculate the formal charges of ${{{N}}_{\left( {{1}} \right)}}$, ${{{N}}_{\left( {{2}} \right)}}$ and ${{O}}$ atoms.
The formula to calculate the formal charge is as follows:
${\text{Formal charge}} = {{\text {Valence electrons in free atom}}} - {{\text {Non - bonding electrons}}} - \dfrac{1}{2}{{\text {Bonding electrons}}}$
The non-bonding electrons are the lone pairs of electrons that are shown on the atoms.
Now, calculate the formal charge of ${{{N}}_{\left( {{1}} \right)}}$ atom as follows:
We know that nitrogen has 5 valence electrons. From the given structure, non-bonding electrons of ${{{N}}_{\left( {{1}} \right)}}$ are 4 and bonding electrons are 4. Thus,
${{\text{Formal charge of} }}{{{N}}_{\left( {{1}} \right)}} = {{5}} - {{4}} - \dfrac{1}{2}\times 4$
${{\text{Formal charge of }}}{{{N}}_{\left( {{1}} \right)}} = {{5}} - {{4}} - 2$
${{\text{Formal charge of} }}{{{N}}_{\left( {{1}} \right)}} = - 1$
Thus, the formal charge of ${{{N}}_{\left( {{1}} \right)}}$ atom is $ - 1$.

Now, calculate the formal charge of ${{{N}}_{\left( {{2}} \right)}}$ atom as follows:
We know that nitrogen has 5 valence electrons. From the given structure, non-bonding electrons of ${{{N}}_{\left( {{2}} \right)}}$ are 0 and bonding electrons are 8. Thus,
${{\text{Formal charge of} }}{{{N}}_{\left( {{2}} \right)}} = {{5}} - {{0}} - \dfrac{1}{2}\times 8 $
${{\text{Formal charge of} }}{{{N}}_{\left( {{2}} \right)}} = {{5}} - {{0}} - 4$
${{\text{Formal charge of} }}{{{N}}_{\left( {{2}} \right)}} = + 1$

Thus, the formal charge of ${{{N}}_{\left( {{2}} \right)}}$ atom is $ + 1$.
Now, calculate the formal charge of ${{O}}$ atom as follows:
We know that oxygen has 6 valence electrons. From the given structure, non-bonding electrons of ${{O}}$ are 4 and bonding electrons are 4. Thus,
${{\text{Formal charge of O}}} = 6 - 4 - \dfrac{1}{2}\times 4$
${{\text{Formal charge of O}}} = {{6}} - {{4}} - 2$
${{\text {Formal charge of O}}} = 0$
Thus, the formal charge of ${{O}}$ atom is 0.
Thus, the formal charges of ${{{N}}_{\left( {{1}} \right)}}$, ${{{N}}_{\left( {{2}} \right)}}$ and ${{O}}$ atoms in ${{:}}{{{\ddot N}}_{\left( {{1}} \right)}} = {{{N}}_{\left( {{2}} \right)}} = {{\ddot O:}}$: are respectively $ - 1, + 1,0$.

Thus, the correct option is (B) $ - 1, + 1,0$.

Note: The formal charge keeps a track of all the electrons for every atom in a molecule. Thus, formal charge helps in predicting reactivity of a molecule. Formal charge helps in determining the lowest energy Lewis structures.