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According to a Blog https://ericscerri.blogspot.com/2012/06/trouble-with-using-aufbau-to-find.html written by Dr.Eric Scerri,

On moving from the $\ce{Sc^3+}$ ion to that of $\ce{Sc^2+}$ it is plain to see that the additional electron enters a 3d orbital and not a 4s orbital as the sloppy scheme dictates. Similarly on moving from this ion to the $\ce{Sc^1+}$ ion the additional electron enters a 4s orbital as it does in finally arriving at neutral scandium atom or Sc.

According to it, the last electron enters the 4s orbital. Why is Scandium a d-block element, then?

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    $\begingroup$ Please do use MathJax where it is appropriate (e.g. to typeset chemical formulae using $\ce{...}$) and don't use it where it is inappropriate (e.g. to place emphasis on text by simply wrapping it in $...$). Please consider reading FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange?. $\endgroup$
    – orthocresol
    Jul 6, 2021 at 8:03
  • $\begingroup$ Is the configuration of Sc [Ar] 4s2 4p1? $\endgroup$
    – Poutnik
    Jul 6, 2021 at 8:43
  • $\begingroup$ @Poutnik,it is indeed,[Ar]4s2 3d1,and as per the blog,the last electron enters 4s,so why it isn't a s block element? $\endgroup$
    – Dheeraj Gujrathi
    Jul 6, 2021 at 10:26
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    $\begingroup$ Why do you think element blocks are named according to where the last formally added electron enters? $\endgroup$
    – Poutnik
    Jul 6, 2021 at 11:34
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    $\begingroup$ Seems to be a duplicate (in effect) of Blocks on the Periodic Table. $\endgroup$
    – StephenG - Help Ukraine
    Jul 6, 2021 at 22:33

1 Answer 1

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I got the answer after a short surfing,The answer to this question is pretty simple,the blocks are made ,by taking the basis of classification as the Differentiating electrons,wherever the differentiating electron exists in a subshell,that is the block of that Element. Differentiating electrons simply mean that the electron which differentiates the element's Electronic configuration from the preceeding element's Electronic Configuration, for example Though the last electron enters in 4s subshell of Scandium atom, it is a 3d electron which makes the difference between the atom and the previous one.Since Ca is $[Ar] 4s^2$ and Sc is $[Ar]3d^1 4s^2$

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