Question

1. identify the element with the electron configuration of $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$
2. identify the element with the electron configuration. $ne3s^{2}3p^{4}$
3. a barium atom (gains/loses) electrons when it forms a barium ion. what is the symbol for a barium ion?
4. a fluorine atom (gains/loses) electrons when it forms a fluorine ion. what is the symbol for a fluorine ion?

Explanation:

34.

Count the total number of electrons. For \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}\), sum the exponents: \(2 + 2+6 + 2+4=16\). The element with atomic number 16 is sulfur (\(S\)).

\([Ne]\) has 10 electrons. For \([Ne]3s^{2}3p^{4}\), add the electrons from \(3s^{2}3p^{4}\) (\(2 + 4=6\)). Total electrons \(=10 + 6=16\). The element with atomic number 16 is sulfur (\(S\)).

Barium (\(Ba\)) is in Group 2. Metals in Group 2 lose 2 electrons. The ion symbol is \(Ba^{2+}\).

Answer:

Sulfur (\(S\))

35.