• question_answer The quantum number of four electrons are given below: (I) $n=4$, $l=2$, ${{m}_{1}}=1,$${{m}_{s}}=-\frac{1}{2}$ (II) $n=3$, $l=2$, ${{m}_{1}}=1,$ ${{m}_{s}}=+\frac{1}{2}$ (III) $n=4$, $l=1$, ${{m}_{1}}=0,$${{m}_{s}}=+\frac{1}{2}$ (IV) $n=3$, $l=1$, ${{m}_{1}}=1,$ ${{m}_{s}}=-\frac{1}{2}$ The Correct order of their increasing energies will be: A) $I<III<II<IV$   B) $IV<II<III<I$ C) $I<II<III<IV$   D) $IV<III<II<I$

 Higher-the value of$(n+l)$ Higher will be energy of orbital. If $(n+l)$ are equal, then higher the value n higher will be energy I. $n+l=6$ $n=4$ II. $n+l=5$ $n=3$ III. $n+l=5$ $n=4$ \[IV