3.9 (sixth edition)

Moderators: Chem_Mod, Chem_Admin

Posts: 61
Joined: Fri Sep 28, 2018 12:18 am

3.9 (sixth edition)

Postby melodyzaki2E » Wed Oct 24, 2018 7:12 pm

Which M2 ions (where M is a metal) are predicted to have the following ground-state electron configurations: (a) [Ar]3d7; (b) [Ar]3d6; (c) [Kr]4d4; (d) [Kr]4d3? Where do you start when you approach this problem?

Jovian Cheung 1K
Posts: 48
Joined: Fri Sep 28, 2018 12:16 am

Re: 3.9 (sixth edition)

Postby Jovian Cheung 1K » Wed Oct 24, 2018 10:41 pm

Note that for all these metal ions, the two 4s2 electrons have been removed to form the M2+ ions.
Thus if you put the two 4s electrons back into the configuration, it would be the configuration of the original metal M.
E.g. for [Ar]3d6, the original configuration of the metal atom would be [Ar]3d64s2. You can then check with the periodic table (count 6 elements from left to right in the 3d block) to deduce that this is the configuration of Fe.
Hope this helped! :o)

Return to “Sigma & Pi Bonds”

Who is online

Users browsing this forum: No registered users and 1 guest