Write electron configurations for the following ions, and determine which have noble-gas configurations.?

Question

(Type your answer using the format [Ar] 4s2 3d10 4p2 for [Ar]4s23d104p2.)

a) P3+
b) As3-
c) Ba2+
d) Tl+
e) Sc3+
f) Mn3+

Answer 1

Find the element in the periodic table. Identify the immediately preceding noble gas.

The difference between the atomic number of the element, and that of the preceding noble gas, is the number of electrons in the ATOM that you would have to find homes for.

In your case, you are dealing with cations, so take away the charge to find the number of electrons remaining.

Then find homes for the electrons, using the lowest energy orbitals first. (But remember the special rule for cations, given below)

If there were no electrons remaining at this stage, then the cation must have the same configuration as the noble gas.

If you get a question like this regarding anions, you do it in just the same way, but this time, of course, the charge represents added electrons.

IMPORTANT: Special rule for cations; in all cations (and actually in all chemical compounds) 3d is lower in energy than 4s, 4d than 5s etc, and 4f is lower than 5d and 6s. (Textbooks often say that the rule is that you remove 4s before you remove 3d. This comes to the same thing in the end, but doesn't tell you why it happens.)

Answer 2

a) P3+ 1s2 2s2 2p6 3s2 or [Ne] 3s2 (b) As3- 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 (noble gas) or [Kr] (c) Ba2+ 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 (noble gas) or [Kr] (d) Tl+ 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 or [Xe] 6s2 4f14 5d10 (e) Sc3+ 1s2 2s2 2p6 3s2 3p6 (noble gas) or [Ar] (f) Mn3+ 1s2 2s2 2p6 3s2 3p6s 3d4 or [Ar] 3d4