To determine the number of resonating structures for ${\mathrm{NO}}_2$, we first draw its Lewis structure.

The total number of valence electrons in ${\mathrm{NO}}_2$ is: Nitrogen (Group 15) = 5 valence electrons; Oxygen (Group 16) = 6 valence electrons. So, $5+2\times 6=17$ valence electrons.

Since ${\mathrm{NO}}_2$ has an odd number of valence electrons, it is a free radical. The odd electron typically resides on the central atom (Nitrogen) or a less electronegative atom.

Step 1: Draw the skeletal structure.
N is the central atom: O—N—O

Step 2: Distribute valence electrons.
Place 2 electrons in each single bond: $17-4=13$ electrons remaining.
Distribute remaining electrons as lone pairs to satisfy octets (or duets for H) for terminal atoms first, then central atom.
Each oxygen needs 6 more electrons to complete its octet. So, $2\times 6=12$ electrons used for lone pairs on oxygens. $13-12=1$ electron remaining.
This one electron goes to the central nitrogen atom.

Step 3: Form multiple bonds if necessary to satisfy octets.
Nitrogen has 2 (from bonds) + 1 (lone electron) = 3 electrons. It needs 5 more electrons to complete its octet. One oxygen has an octet, the other has an octet. To satisfy nitrogen's octet, one of the oxygen atoms must form a double bond with nitrogen.

Step 4: Draw the resonating structures.
There are two possible ways to form a double bond:

Structure 1:
$\mathrm{O}=\mathrm{N}-\ddot{\mathrm{O}}:$
Here, one oxygen forms a double bond with nitrogen, and the other forms a single bond. The odd electron is on the nitrogen atom. The formal charges are: O (double bond) = 0, N = +1, O (single bond) = -1.

Structure 2:
$:\ddot{\mathrm{O}}-\mathrm{N}=\mathrm{O}$
This is an equivalent structure where the double bond is with the other oxygen atom. The odd electron is still on the nitrogen atom. The formal charges are: O (single bond) = -1, N = +1, O (double bond) = 0.

These two structures are resonance contributors, meaning the actual structure is a hybrid of these two. Therefore, there are two resonating structures for ${\mathrm{NO}}_2$.

Option Analysis:

• A) Six: Incorrect. There are only two equivalent ways to place the double bond and the odd electron.
• B) Four: Incorrect. This number is too high for ${\mathrm{NO}}_2$.
• C) Five: Incorrect. This number is too high for ${\mathrm{NO}}_2$.
• D) Two: Correct. As derived from the Lewis structures, there are two equivalent resonance forms.

Correct Answer: (D)