Question

i. configuration writing practice
a. write a ground state electron configuration for each neutral atom. ground state
means that all of the lowest possible energy levels (up to the proper number of
electrons for the element) are filled.

1. na: _
2. pb: _
3. sr: _
4. u: _
5. n: _
6. ag: _
7. ti: _
8. ce: _
9. cl: _
10. hg: _

Explanation:

Step1: Determine the atomic number of each element

• For Na (Sodium), atomic number \(Z = 11\)
• For Pb (Lead), \(Z=82\)
• For Sr (Strontium), \(Z = 38\)
• For U (Uranium), \(Z=92\)
• For N (Nitrogen), \(Z = 7\)
• For Ag (Silver), \(Z=47\)
• For Ti (Titanium), \(Z = 22\)
• For Ce (Cerium), \(Z=58\)
• For Cl (Chlorine), \(Z = 17\)
• For Hg (Mercury), \(Z=80\)

Step2: Use the Aufbau principle, Pauli exclusion principle and Hund's rule to write electron configurations

• Na:
• \(1s^{2}2s^{2}2p^{6}3s^{1}\)
• Pb:
• \([Xe]4f^{14}5d^{10}6s^{2}6p^{2}\)
• Sr:
• \([Kr]5s^{2}\)
• U:
• \([Rn]5f^{3}6d^{1}7s^{2}\)
• N:
• \(1s^{2}2s^{2}2p^{3}\)
• Ag:
• \([Kr]4d^{10}5s^{1}\)
• Ti:
• \([Ar]3d^{2}4s^{2}\)
• Ce:
• \([Xe]4f^{1}5d^{1}6s^{2}\)
• Cl:
• \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}\)
• Hg:
• \([Xe]4f^{14}5d^{10}6s^{2}\)

Answer:

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