Beryllium has an atomic number of #2#. What is its electron configuration? Will it lose #2# electrons or gain #6# to form an ion?

1 Answer
Truong-Son N.
Jan 23, 2016

It wants the configuration that helium has, which is two valence electrons. To be more specific, it wants a DUET, not an octet.

Beryllium (#"Be"#) has the atomic number #4# (not #2#), so its electron configuration is #1s^2 2s^2#.

#"He"# has the atomic number #2#, so its electron configuration is #1s^2#.

Since it is much easier to lose two electrons than to gain six, the #"He"#-like electron configuration (#1s^2#) is much more favored than the #"Ne"#-like configuration (#1s^2 2s^2 2p^6#).

Related questions
See all questions in Octet Rule
Impact of this question
10831 views around the world
You can reuse this answer
Creative Commons License