What is the shape of the `IBr_2^-` ion ?

Options

A. Linear
B. Bent shape with bond angle of about `90^@`
C. Bent shape with bond angle of about `109^@`
D. Bent shape with bond angle of about `120^@`

Answer

To determine the shape of the \( \text{IBr}_2^- \) ion, we can follow these steps:

### Step 1: Determine the Valence Electrons
Iodine (I) is in group 17 of the periodic table and has 7 valence electrons. Each bromine (Br) atom, also in group 17, has 7 valence electrons. Since there are two bromine atoms, they contribute a total of \( 2 \times 7 = 14 \) electrons. The negative charge on the ion indicates that there is one additional electron. Therefore, the total number of valence electrons is:
\[ 7 \, (\text{I}) + 14 \, (\text{2 Br}) + 1 \, (\text{negative charge}) = 22 \, \text{valence electrons} \]

### Step 2: Draw the Lewis Structure
In the Lewis structure, iodine will be the central atom, bonded to two bromine atoms. We will place the remaining electrons as lone pairs around the iodine:
- Connect Iodine to each Bromine with a single bond (2 electrons).
- This leaves us with \( 22 - 2 = 20 \) electrons.
- Distribute the remaining electrons as lone pairs around the iodine and bromine atoms.

### Step 3: Count the Steric Number
The steric number is calculated as follows:
\[ \text{Steric Number} = \text{Number of sigma bonds} + \text{Number of lone pairs} \]
In this case:
- There are 2 sigma bonds (one for each I-Br bond).
- There is 1 lone pair on the iodine.

Thus, the steric number is:
\[ \text{Steric Number} = 2 + 1 = 3 \]

### Step 4: Determine Hybridization
The hybridization corresponding to a steric number of 3 is \( \text{sp}^2 \).

### Step 5: Determine the Geometry
With a steric number of 3, the electron geometry is trigonal planar. However, the presence of the lone pair will affect the molecular shape.

### Step 6: Determine the Molecular Shape
According to VSEPR theory, the lone pair will occupy one of the positions in the trigonal planar arrangement. This will push the two bromine atoms closer together, resulting in a bent shape. However, since we have a steric number of 5 (considering the lone pair), the actual shape will be linear because the lone pair occupies the equatorial position in a trigonal bipyramidal arrangement.

### Conclusion
The shape of the \( \text{IBr}_2^- \) ion is linear.

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