If the n^{th} specific period of the periodic | vedclass

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(B) The number of elements in a period is given by the number of electrons that can be filled in the orbitals corresponding to that period. For a give

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$\frac{n}{2}$

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(B) The number of elements in a period is given by the number of electrons that can be filled in the orbitals corresponding to that period. For a given period $n$,the orbitals filled are $ns$,$(n-2)f$,$(n-1)d$,and $np$. The total number of electrons is $2 + 2(2l+1) + 2(2l+1) + 6 = \frac{(n+2)^2}{2}$. Alternatively,for a period $n$,the maximum azimuthal quantum number $l$ is related to the number of orbitals. Given the formula $\frac{(n+2)^2}{2}$,we equate this to the capacity of orbitals. For the $n^{th}$ period,the maximum value of the azimuthal quantum number $l$ is $\frac{n}{2}$ if $n$ is even,or $\frac{n+1}{2}$ if $n$ is odd. However,checking the options against the standard derivation for this specific theoretical model,the correct expression is $\frac{n}{2}$.

List-$A$	List-$B$
$(1)$ Principal quantum number	$(A)$ Shape of the orbital
$(2)$ Azimuthal quantum number	$(B)$ Spin of the electron
$(3)$ Magnetic quantum number	$(C)$ Orientation of the orbital in space
$(4)$ Spin quantum number	$(D)$ Size of the orbital

If the $n^{th}$ specific period of the periodic table contains $\frac{(n+2)^2}{2}$ elements,then the maximum value of the azimuthal quantum number associated with that period is:

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