Question

match the following as either be (beryllium) or f (fluorine)
valence electrons = 2
electron configuration is
1s² 2s² 2p⁵
total electrons of 9
electron configuration is
1s² 2s²
valence electrons = 7
total electrons of 4
be (beryllium)
f (fluorine)

Explanation:

Step1: Recall atomic number and electron configuration rules

• Beryllium (Be) has an atomic number of 4. So, total electrons = 4. Its electron configuration is \(1s^{2}2s^{2}\). The valence electrons (in the outermost shell, here \(n = 2\)) are 2.
• Fluorine (F) has an atomic number of 9. So, total electrons = 9. Its electron configuration is \(1s^{2}2s^{2}2p^{5}\). The valence electrons (in \(n=2\) shell: \(2s^{2}2p^{5}\)) are \(2 + 5=7\).

Answer:

• Valence electrons = 2: Be (Beryllium)
• Electron configuration is \(1s^{2}2s^{2}2p^{5}\): F (Fluorine)
• Total electrons of 9: F (Fluorine)
• Electron configuration is \(1s^{2}2s^{2}\): Be (Beryllium)
• Valence electrons = 7: F (Fluorine)
• Total electrons of 4: Be (Beryllium)