A substance which is not paramagnetic is .

Options

A. `Cr(CIO_(4))_(3)`
B. `KMnO_(4)`
C. `TiCI_(3)`
D. `VOBr_(2)`

Explanation

To determine which substance is not paramagnetic among the given options, we will analyze the oxidation states and electron configurations of each compound.

### Step-by-Step Solution:

1. **Analyzing CrClO4 (Chromium Chlorate)**:
- The charge of ClO4 is -1, and since there are three ClO4 groups, the total negative charge is -3.
- To balance this, chromium must have an oxidation state of +3 (since the overall charge of the compound is 0).
- The electron configuration for chromium in the +3 state is:
- Ground state: [Ar] 4s² 3d⁵
- After losing 3 electrons (2 from 4s and 1 from 3d), the configuration becomes [Ar] 3d³.
- In this configuration, there are 3 unpaired electrons, making it paramagnetic.

2. **Analyzing KMnO4 (Potassium Permanganate)**:
- The oxidation state of manganese (Mn) can be calculated as follows:
- The charge of O is -2, and there are 4 O atoms, contributing -8.
- The overall charge of the ion is -1, thus Mn must be +7 to balance this.
- The electron configuration for manganese in the +7 state is:
- Ground state: [Ar] 4s² 3d⁵
- After losing 7 electrons (2 from 4s and 5 from 3d), the configuration becomes [Ar].
- This configuration has all paired electrons, making it **not paramagnetic**.

3. **Analyzing TiCl3 (Titanium Trichloride)**:
- The oxidation state of titanium (Ti) can be calculated:
- Each Cl contributes -1, and there are 3 Cl atoms, contributing -3.
- Thus, Ti must be +3 to balance the charge.
- The electron configuration for titanium in the +3 state is:
- Ground state: [Ar] 4s² 3d²
- After losing 3 electrons (2 from 4s and 1 from 3d), the configuration becomes [Ar] 3d¹.
- This configuration has 1 unpaired electron, making it paramagnetic.

4. **Analyzing VBr2 (Vanadium Bromide)**:
- The oxidation state of vanadium (V) can be calculated:
- Each Br contributes -1, and there are 2 Br atoms, contributing -2.
- Thus, V must be +4 to balance the charge.
- The electron configuration for vanadium in the +4 state is:
- Ground state: [Ar] 4s² 3d³
- After losing 4 electrons (2 from 4s and 2 from 3d), the configuration becomes [Ar] 3d¹.
- This configuration has 1 unpaired electron, making it paramagnetic.

### Conclusion:
The only substance that is **not paramagnetic** is **KMnO4**, as it has all paired electrons in its electron configuration.