Paramagnetism results from unpaired electrons in a molecule. We will determine the number of paramagnetic molecules by analyzing each molecule's electronic configuration:
- O\(_2\) (Oxygen): Oxygen (16 electrons) has the molecular orbital configuration: \[ (\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^1 (\pi_{2p_y})^1 (\pi_{2p_x}^*)^0 (\pi_{2p_y}^*)^0 \]. This configuration indicates 2 unpaired electrons, making O\(_2\) paramagnetic.
- N\(_2\) (Nitrogen): Nitrogen (14 electrons) has the molecular orbital configuration: \[ (\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 \]. All electrons in N\(_2\) are paired, classifying it as diamagnetic.
- F\(_2\) (Fluorine): Fluorine (18 electrons) has the electronic configuration: \[ (\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi_{2p_x}^*)^1 (\pi_{2p_y}^*)^1 \]. This configuration shows 2 unpaired electrons, making F\(_2\) paramagnetic.
- B\(_2\) (Boron): Boron (10 electrons) has the electronic configuration: \[ (\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\pi_{2p_x})^1 (\pi_{2p_y})^1 \]. This configuration reveals 2 unpaired electrons, making B\(_2\) paramagnetic.
- Cl\(_2\) (Chlorine): Chlorine (18 electrons) has the electronic configuration: \[ (\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi_{2p_x}^*)^1 (\pi_{2p_y}^*)^1 \]. This configuration shows 2 unpaired electrons, making Cl\(_2\) paramagnetic. Consequently, the paramagnetic molecules are O\(_2\), F\(_2\), and B\(_2\). The total count of paramagnetic molecules is 3.