Question

practice: electron configurations
part 1: write out the electron configurations for the following atoms.

1. be:
2. f:
3. p:
4. cu:
5. hg:

part 2: write out the electron configurations in noble gas notation for the following atoms.

1. b:
2. ca:
3. cl:
4. mn:
5. pb:

Explanation:

Step1: Determine the atomic number

• For Be (Beryllium), atomic number \(Z = 4\).
• For F (Fluorine), \(Z=9\).
• For P (Phosphorus), \(Z = 15\).
• For Cu (Copper), \(Z=29\).
• For Hg (Mercury), \(Z = 80\).
• For B (Boron), \(Z=5\).
• For Ca (Calcium), \(Z = 20\).
• For Cl (Chlorine), \(Z=17\).
• For Mn (Manganese), \(Z=25\).
• For Pb (Lead), \(Z = 82\).

Step2: Write the electron - configuration using Aufbau principle, Pauli exclusion principle and Hund's rule

• Be:
• Fill orbitals starting from the lowest energy. \(1s\) can hold 2 electrons and \(2s\) can hold 2 electrons.
• Electron - configuration: \(1s^{2}2s^{2}\)
• F:
• \(1s^{2}2s^{2}2p^{5}\) (after filling \(1s\) and \(2s\), fill \(2p\) orbitals. \(2p\) has 3 sub - orbitals and we put 5 electrons following Hund's rule)
• P:
• \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}\) (fill \(1s\), \(2s\), \(2p\), then \(3s\) and \(3p\))
• Cu:
• Due to the stability of half - filled and fully - filled orbitals. The expected configuration \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{9}\) is not correct. The actual configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10}\) (electron from \(4s\) moves to \(3d\) to make \(3d\) fully - filled)
• Hg:
• \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^{10}\)
• B (Noble - gas notation):
• The noble gas before B is He (\(1s^{2}\)). So, \([He]2s^{2}2p^{1}\)
• Ca (Noble - gas notation):
• The noble gas before Ca is Ar (\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}\)). So, \([Ar]4s^{2}\)
• Cl (Noble - gas notation):
• The noble gas before Cl is Ne (\(1s^{2}2s^{2}2p^{6}\)). So, \([Ne]3s^{2}3p^{5}\)
• Mn (Noble - gas notation):
• The noble gas before Mn is Ar. So, \([Ar]4s^{2}3d^{5}\)
• Pb (Noble - gas notation):
• The noble gas before Pb is Xe (\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}\)). So, \([Xe]6s^{2}4f^{14}5d^{10}6p^{2}\)

Answer:

1. \(1s^{2}2s^{2}\)
2. \(1s^{2}2s^{2}2p^{5}\)
3. \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}\)
4. \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10}\)
5. \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^{10}\)
6. \([He]2s^{2}2p^{1}\)
7. \([Ar]4s^{2}\)
8. \([Ne]3s^{2}3p^{5}\)
9. \([Ar]4s^{2}3d^{5}\)
10. \([Xe]6s^{2}4f^{14}5d^{10}6p^{2}\)