Question

What is the formula of the ion formed when phosphorus achieves a noble-gas electron configuration?

Answer 1

Answer:

d. p3-

Explanation:

Phosphorus has fifteen electrons in its orbitals and five electrons in its outermost shell. An atom is said to have noble gas configuration when it has eight electron in its outermost shell.

In order to have eight electrons in its outermost shell, phosphorus must accept three electrons and form an ion of the sort P^3-. This ion now has eighteen electrons (argon configuration) and eight electrons in the outermost shell.

Answer 2

Answer:

D. P^3-

Explanation:

Phosphorus has an electronic configuration of [Ne] 3s² 3p³ that is, 15 electrons in its shells.

At noble gas configuration, Phosphorus obtains a stable octet state of [Ne] 3s² 3p^6 (Argon electronic configuration) is 18 electrons by accepting 3 electrons.

Phophorus in this state becomes P^3-

Electronic configuration

P = 1s2 2s2 2p6 3s2 3p3

= [Ne] 3s2 3p3

P^3- = Ar = 1s2 2s2 2p6 3s2 3p6

= [Ne] 3s23p6