How many elements are possible for $$1st$$ period of periodic table if azimuthal quantum number can have integral values from $$0$$, $$1$$,$$2$$... (n$$ $$+$$ $$1)?

Question

How many elements are possible for $$1st$$ period of periodic table if azimuthal quantum number can have integral values from $$0$$, $$1$$,$$2$$... (n$$ $$+$$ $$1)?

Infinity Learn · Accepted Answer

$$n=1$$                              $$n=2l=0$$,$$1$$,$$2$$                         $$l=0$$,$$1$$,$$2$$,$$3s$$,p,d                               s,p,d,$$f1s21p6$$⏟Ist period $$2s21d10$$…    According to question value of azimuthal quantum number is $$0$$ to (n+1)Therefore$$, for n $$=$$ $$1$$, l $$=$$ $$0$$, $$1$$, $$2$$                    for n $$=$$ $$2$$ , l $$=$$ $$0$$, $$1$$, $$2$$, $$3Now$$ applying $$(n+l) rule we get subshellfilling order like this :$$1s$$,$$1p$$,$$2s$$,...After $$1p$$, the second shell starts filling only the elements having electrons present in $$1s$$ and $$1p$$ will belong to $$1st$$ period.Total Number of such element possible $$is2$$× $$(capacity$$ of $$1s$$ $$subshell)+6$$ $$(capacity$$ $$of1p$$ $$subshell)which$$ is equal to '$$8$$' elements.

How many elements are possible for 1st period of periodic table if azimuthal quantum number can have integral values from 0, 1,2... (n + 1)?

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