1. Determine the shape of each species using VSEPR theory

We check the central atom, number of bond pairs, and lone pairs.

2. $\mathrm{N_3^-}$ (azide ion)

Structure is resonance stabilized and has the general form $\mathrm{N-N-N}$. The central nitrogen is $sp$ hybridized with two regions of electron density.

So, shape is: $\text{linear}$

So, not bent.

3. $\mathrm{NO_2^-}$ (nitrite ion)

Central atom = N.

Valence electrons: $5 + 2(6) + 1 = 18$

Around central N, there are 3 electron domains:

• 2 bond pairs
• 1 lone pair

Electron geometry: trigonal planar
Molecular shape: $\text{bent}$

So, bent.

4. $\mathrm{I_3^-}$ (triiodide ion)

Central atom = I.

The central iodine has 5 electron pairs around it:

• 2 bond pairs
• 3 lone pairs

This is $AX_2E_3$. In trigonal bipyramidal electron arrangement, the 3 lone pairs occupy equatorial positions. Thus the two bonded atoms occupy axial positions.

So, shape is: $\text{linear}$

So, not bent.

5. $\mathrm{O_3}$ (ozone)

Central atom = O.

Central oxygen has 3 electron domains:

• 2 bond pairs
• 1 lone pair

Thus molecular shape is: $\text{bent}$

So, bent.

6. $\mathrm{SO_2}$

Central atom = S.

Sulfur has 3 electron domains around it:

• 2 bonding regions
• 1 lone pair

Hence shape is: $\text{bent}$

So, bent.

7. Count bent molecules/ions

Bent species are:

• $\mathrm{NO_2^-}$
• $\mathrm{O_3}$
• $\mathrm{SO_2}$

Total number: $3$

8. Comparison with stored answer

Derived answer = $3$
Stored correct answer = $3$

They match.