Electron configuration of indium cation

Question

Just had a lecture there where the lecturer was doing revision on electronic configuration.

He stated that the electronic configuration of $$\ce{In^3+=[Kr] 5s^2 3p^1}$$

He didn't include the $\ce{4d^10}$ orbital. He said it doesn't matter and if we write the full E.C out it's incorrect. It basically goes against everything we studied in 1st year general chemistry. Also I believe he didn't take into account the fact that $\ce{In^3+}$ has $46 \, e^-$ not $49 \, e^-$.

Should the electronic configuration not be $$\ce{[Kr] 5s^0 4d^10 3p^0}= \ce{[Kr] 4d^10}$$

Can someone shed some light on whether he is right.

Oscar Lanzi · Accepted Answer · 2017-10-31T19:15:23.093

5

Yup, it's $[\ce{Kr}]4d^{10}$. Post-transition ($p$-block) elements rarely if ever ionize or even covalently bond their $d$ subshell electrons. Although we use the bracketed core designation only for noble gas cores, by the time we get to Group 12 or 13 the $d$ electrons are also effectively part of the core.

edited Oct 31 '17 at 19:15

answered Oct 31 '17 at 12:58

Oscar Lanzi

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for example $\ce{Se^2-}$ would be $\ce{[Kr]}$? – Patrick Moloney Oct 31 '17 at 13:01
1

You could do that, although usually when writing electron configurations we go to the next lower noble gas, thus $[\ce{Ar}]3d^{10}4s^24p^6$. – Oscar Lanzi Oct 31 '17 at 13:04

Electron configuration of indium cation

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