Question
How many electrons can be accommodated in a $M$ and $N-$shell?
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Answer
Following the order: $KLMN, N$ is fourth shell of an atom. The number of electrons in a shell depends on the element, its oxidation state and its electronic state. Considering a electrically neutral atom in the ground state, the elements in $H$ -Ar range does have no electron in the $N$ shell ( $K , L$ and M shells are being filled). $K$ is the first element where $N$ shell begins to be filled ([Ar]4s ${ }^1$ configuration) followed by $Ca \left([ Ar ] 4 s^2\right)$. The next elements are transition metals (Sc$Zn$ ) which, owing to the filling of 3d orbitals, does not have its $N$ shell changed (except by a couple of exceptions). After $Zn$ , in $Ga - Kr$ range, $N$ shell is filled with $3 r d$ to $8$ th electrons which occupy $4$ p orbitals. The next two elements, Rb and Sr still have $8$ electron in $N$ shell because $5$ s orbitals are filled. Then $4$ d orbitals are filled across $Y - Cd$ transition metals, growing $N$ shell from $9$ to $18$ electrons. In the next elements ( $In - Xe$ ) $5$ p orbitals are filled and $N$ shell remains unchanged, the same happening to Cs and Ba where $6$ s orbitals are filled. The $N$ shell are filled again in the lanthanoid series ($La-Yb$) where $4 f$ orbitals are occupied. In the last element of this series (Yb) $N$ shell is full, with $32$ electrons, and no further electron can be added to this shell.
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