Answer
The atoms expected to be paramagnetic are lithium (\( \mathrm{Li} \)) with one unpaired electron, nitrogen (\( \mathrm{N} \)) with three unpaired electrons, and nickel (\( \mathrm{Ni} \)) with two unpaired electrons.
Work Step by Step
To determine which atoms would be expected to be paramagnetic and the number of unpaired electrons in each atom, we need to examine their ground-state electron configurations.
1. Lithium (\( \mathrm{Li} \)): The electron configuration of lithium is 1s² 2s¹. It has one unpaired electron in the 2s orbital, making it paramagnetic with one unpaired electron.
2. Nitrogen (\( \mathrm{N} \)): The electron configuration of nitrogen is 1s² 2s² 2p³. It has three unpaired electrons in the 2p orbitals, making it paramagnetic with three unpaired electrons.
3. Nickel (\( \mathrm{Ni} \)): The electron configuration of nickel is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁸. It has two unpaired electrons in the 3d orbitals, making it paramagnetic with two unpaired electrons.
4. Tellurium (\( \mathrm{Te} \)): The electron configuration of tellurium is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁴. It has no unpaired electrons, making it diamagnetic (not paramagnetic).
5. Barium (\( \mathrm{Ba} \)): The electron configuration of barium is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s². It has no unpaired electrons, making it diamagnetic (not paramagnetic).
6. Mercury (\( \mathrm{Hg} \)): The electron configuration of mercury is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶. It has no unpaired electrons, making it diamagnetic (not paramagnetic).
Therefore, the atoms expected to be paramagnetic are lithium (\( \mathrm{Li} \)) with one unpaired electron, nitrogen (\( \mathrm{N} \)) with three unpaired electrons, and nickel (\( \mathrm{Ni} \)) with two unpaired electrons.
