Question

which element from group 17 will form an ion with the electron configuration 1s²2s²2p⁶3s²3p⁶ upon gaining one electron?
iodine (i)
chlorine (cl)
bromine (br)
fluorine (f)

in a lewis dot diagram, the formation of a sulfide ion (s²⁻) from a sulfur atom would be represented by:
sulfur with 6 dots losing 2 dots.
sulfur with 6 dots gaining 1 dot.
sulfur with 6 dots gaining 2 dots, enclosed in brackets with a 2⁻ charge.
sulfur with 8 dots losing 2 dots.

Explanation:

Step1: Analyze electron - configuration for Group 17 element

Group 17 elements have 7 valence electrons. When they gain one electron, they achieve a noble - gas electron configuration. The electron configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$ is that of argon (a noble gas). Chlorine ($Cl$) has an atomic number of 17 with electron configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}$. When it gains one electron, it has the configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$. Iodine, bromine, and fluorine do not have this resulting electron - configuration upon gaining one electron.

Step2: Analyze Lewis - dot diagram for sulfide ion formation

A sulfur atom has 6 valence electrons (represented by 6 dots in a Lewis - dot diagram). To form a sulfide ion ($S^{2 - }$), it gains 2 electrons. In the Lewis - dot diagram, this is represented as sulfur with 6 dots gaining 2 dots, and the resulting structure is enclosed in brackets with a 2 - charge to indicate the ion.

Answer:

1. B. Chlorine (Cl)
2. C. Sulfur with 6 dots gaining 2 dots, enclosed in brackets with a 2 - charge.