The last shell has 2 electrons and the second last has eighteen electrons. Byt we know that second last shell cannot have more than 8 electrons. Then how is this possible. Please explain in simple manner.
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1$\begingroup$ How do you know that the second half-shell cannot have more than $8$ electrons ? Is this rule valid ? Never seen before.. $\endgroup$– MauriceJan 17 at 15:14
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$\begingroup$ It depends if by a shell is meant orbitals grouped by similar energy or the same main quantum number. It is 18 for the former but 8 for the latter. E.g. for Rb with 5s1 electron, there are 8 electrons with main quantum number 4 but 18 if 3d electrons are involved energetically. For mercury, energetically, there come also 14 4f electrons. $\endgroup$– PoutnikJan 17 at 15:21
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2 Answers
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The second last shell can have more than 8 electrons if it is M shell or above. According to the 2n2 rule (n= principal quantum number), O shell can theoretically can have upto 100 electrons.
Mercurys 5s, 5p, 5d subshells are full having 2,6 and 10 electrons respectively and there seems to be no problem.
In reality, there is no rule preventing the second last shell having more than 8 electron. Look at Zinc! It also has 18 electron in its M shell.
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It is actually the last shell that typically has no more than eight electrons in a typical neutral atomic ground state. The second to last shell can have up to eighteen in the ground state.
This is dictated by the manner in which shells and subshells are filled in typical ground state configurations of neutral atoms. Take the $n=3$ shell. It can hold a total of $2×3^2=18$ electrons, but only eight are there in argon and then when we move to potassium and calcium, the $4s$ orbital fills before we get any electrons into the $3d$ subshell. Then the last shell becomes $n=4$ instead of $n=3$. Only after it becomes the second to last shell does $n=3$ gain ten more electrons (filling the $3d$ subshell) as we go from scandium to zinc, and then it gets to eighteen.
