Question

Write all 4 quantum numbers for an unpaired electron in a silver atom.

Answer 1

Electronic configuration of silver is = [Kr] 4d¹⁰5s¹

1) n = 5 The unpaired electron is in the 5th shell.

2) l = 0 the value of the angular momentum quantum number is given by the subshell in which the electron is located.

Electron is in the s subshell

l = 0 the s subshell
l = 1 the p subshell
l = 2 the d subshell
l = 3 the f subshell

Therefore, you can say that silver's valence electron will have l = 0.

3) m_l = 0 the magnetic quantum number.

In this case, the s subshell can only hold one orbital, the s-orbital. This means that the value of ml, which is determined by the value of l, can only be m_l = 0.

4) m_s=+1/2 the spin quantum number is assigned by keeping in mind that each orbital can only hold two electrons, one having spin-up and one having spin-down.

In this case, the single electron located in the s-orbital is considered to have spin-up.

Therefore, you can say that the quantum number set that describes the highest-energy electron in an atom of silver will be

n = 5, l = 0, m_l = 0, m_s = +12

This set describes an electron located in the fifth energy shell, in the s subshell, in the 5s orbital, that has spin-up.