Questions: How many unpaired electrons are present in the ground state Bi atom?

Solution Steps

Bismuth (Bi) has an atomic number of 83, which means it has 83 electrons in its neutral state. The electron configuration of Bi can be determined by filling the orbitals in order of increasing energy:

1. \(1s^2\)
2. \(2s^2\)
3. \(2p^6\)
4. \(3s^2\)
5. \(3p^6\)
6. \(4s^2\)
7. \(3d^{10}\)
8. \(4p^6\)
9. \(5s^2\)
10. \(4d^{10}\)
11. \(5p^6\)
12. \(6s^2\)
13. \(4f^{14}\)
14. \(5d^{10}\)
15. \(6p^3\)

Thus, the electron configuration of Bi is:
\[ \text{Bi: } [\text{Xe}] \, 4f^{14} \, 5d^{10} \, 6s^2 \, 6p^3 \]

The unpaired electrons are found in the outermost orbitals. In the case of Bi, the outermost electrons are in the \(6p\) orbital, which has the configuration \(6p^3\).

The \(p\) subshell can hold a maximum of 6 electrons, distributed over 3 orbitals. In the \(6p^3\) configuration, each of the three \(p\) orbitals will have one electron, resulting in three unpaired electrons.

Final Answer