What is the geometry of the `IB r_(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^(@)`

Explanation

To determine the geometry 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 has 7 valence electrons.
- Since there are two bromine atoms, they contribute a total of \( 7 \times 2 = 14 \) electrons.
- The negative charge on the ion adds one additional electron.
- Therefore, the total number of valence electrons is:
\[ 7 \, (\text{from I}) + 14 \, (\text{from 2 Br}) + 1 \, (\text{negative charge}) = 22 \, \text{valence electrons} \]

### Step 2: Draw the Lewis Structure
- Place iodine in the center and connect it to the two bromine atoms with single bonds.
- Each bond uses 2 electrons, so 4 electrons are used for the two bonds.
- This leaves us with \( 22 - 4 = 18 \) electrons.
- Distribute the remaining electrons to satisfy the octet rule. Each bromine will get 6 electrons (3 lone pairs), and iodine will have 2 lone pairs (4 electrons) remaining.
- The Lewis structure looks like this:
```
Br
|
I - Br
```

### Step 3: Calculate the Steric Number
- The steric number is calculated by adding the number of sigma bonds and lone pairs around the central atom (iodine).
- There are 2 sigma bonds (one for each Br) and 2 lone pairs on iodine.
- Thus, the steric number is:
\[ \text{Steric Number} = 2 \, (\text{sigma bonds}) + 2 \, (\text{lone pairs}) = 4 \]

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

### Step 5: Determine the Geometry
- For \( \text{sp}^3 \) hybridization with 2 lone pairs, the arrangement of electron pairs will be tetrahedral.
- However, because lone pairs occupy more space and repel more strongly than bonding pairs, they will occupy the equatorial positions in a trigonal bipyramidal arrangement to minimize repulsion.
- This results in a linear arrangement of the two bromine atoms, with the bond angle being 180 degrees.

### Conclusion
- Therefore, the geometry of the \( \text{IBr}_2^{-} \) ion is **linear**.

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