The expected shape of `Br_3^-` ion is

Options

A. linear
B. trigonal planar
C. pyramidal
D. angular

Explanation

To determine the expected shape of the `Br_3^-` ion, we will follow these steps:

### Step 1: Determine the Valence Electrons
Bromine (Br) is in group 17 of the periodic table and has 7 valence electrons. Since there are three bromine atoms in the ion, we will calculate the total valence electrons contributed by bromine.

- Total valence electrons from 3 Br = 3 × 7 = 21 electrons

### Step 2: Account for the Charge
The `Br_3^-` ion has a negative charge, which means we need to add one additional electron to the total count.

- Total valence electrons = 21 + 1 = 22 electrons

### Step 3: Calculate the Steric Number
The steric number can be calculated using the formula:
\[ \text{Steric Number} = \text{(Valence Electrons of Central Atom)} + \text{(Monovalent Atoms Attached)} + \text{(Anionic Charge)} \]

In this case:
- Central atom: 1 Br (7 valence electrons)
- Monovalent atoms attached: 2 Br
- Anionic charge: +1

So, the steric number calculation would be:
\[ \text{Steric Number} = 7 + 2 + 1 = 10 \]

### Step 4: Determine Hybridization
The steric number of 5 indicates the hybridization of the central atom. For a steric number of 5, the hybridization is \( sp^3d \).

### Step 5: Identify Lone Pairs
In the `Br_3^-` ion, the central bromine atom will have 3 bonding pairs (with the two other bromine atoms) and 2 lone pairs of electrons.

### Step 6: Determine the Molecular Geometry
According to VSEPR theory, when there are 5 regions of electron density (3 bonding pairs and 2 lone pairs), the molecular shape is determined by the positions of the bonding pairs. The lone pairs will occupy equatorial positions to minimize repulsion, leading to a trigonal bipyramidal arrangement.

However, since we are only considering the positions of the bonded atoms for the shape, the molecular shape will be linear.

### Conclusion
The expected shape of the `Br_3^-` ion is **linear**.