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I was preparing for an exam and I saw one question which is difficult to understand:

The electronic configuration $\mathrm{(1s)^2(2s)^2(2p)^6}$ does not belong to:

a) any element
b) anion
c) cation
d) a transition metal in its ground state
e) noble gas in its ground state

According to me its (e) because of helium since it electronic configuration is $\mathrm{(1s)^2}$.

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  • $\begingroup$ What does that writing mean? Where does it lead you, on the periodic table? Tip: why are s, p and d blocks on the periodic table called this way, and how are they organized? $\endgroup$ – The_Vinz Jun 17 at 5:00
  • $\begingroup$ on the periodic table ; the existence of this electronic configration $\endgroup$ – Feemo Fellow Jun 17 at 5:02
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    $\begingroup$ Just to make it clear, the notation for electron configuration (by the way, yours is in the first paragraph of this Wikipedia article) should be reviewed as a whole, e.g. $\mathrm{(1s)^2(2s)^2(2p)^6},$ not part by part. Otherwise I don't understand why you keep on insisting on $\mathrm{(1s)^2}.$ $\endgroup$ – andselisk Jun 17 at 5:15
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    $\begingroup$ If that's your answer, I strongly suggest reading carefully the relative chapter on your book, since you lack the fundamental understanding required for giving a proper answer $\endgroup$ – The_Vinz Jun 17 at 5:15
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    $\begingroup$ ok sir i will try it again, thank sir for your co operation $\endgroup$ – Feemo Fellow Jun 17 at 5:18

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