Which of the following sets of quantum number | vedclass

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(D) The energy of an orbital is determined by the $(n + l)$ rule.For option $A$: $n = 4, l = 0$,so $(n + l) = 4 + 0 = 4$.For option $B$: $n = 3, l = 0

Vedclass · Accepted Answer

$n = 3, l = 2, m = 1, s = +\frac{1}{2}$

Vedclass · Answer

(D) The energy of an orbital is determined by the $(n + l)$ rule.For option $A$: $n = 4, l = 0$,so $(n + l) = 4 + 0 = 4$.For option $B$: $n = 3, l = 0$,so $(n + l) = 3 + 0 = 3$.For option $C$: $n = 3, l = 1$,so $(n + l) = 3 + 1 = 4$.For option $D$: $n = 3, l = 2$,so $(n + l) = 3 + 2 = 5$.Comparing the $(n + l)$ values,option $D$ has the highest value of $5$,which corresponds to the $3d$ orbital. Therefore,it has the highest energy.

Which of the following sets of quantum numbers belongs to the highest energy?

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The electrons identified by quantum numbers $n$ and $l$ :
$A. n=4, l=1$ $B. n=4, l=0$
$C. n=3, l=2$ $D. n=3, l=1$
can be placed in order of increasing energy as :

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