The electrons identified by quantum numbers n | vedclass

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(A) To determine the energy order,we use the $(n+l)$ rule:$A. n=4, l=1 \implies n+l = 4+1 = 5$ ($4p$ orbital)$B. n=4, l=0 \implies n+l = 4+0 = 4$ ($4s

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$D < B < C < A$

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(A) To determine the energy order,we use the $(n+l)$ rule:$A. n=4, l=1 \implies n+l = 4+1 = 5$ ($4p$ orbital)$B. n=4, l=0 \implies n+l = 4+0 = 4$ ($4s$ orbital)$C. n=3, l=2 \implies n+l = 3+2 = 5$ ($3d$ orbital)$D. n=3, l=1 \implies n+l = 3+1 = 4$ ($3p$ orbital)According to the $(n+l)$ rule,orbitals with lower $(n+l)$ values have lower energy.If $(n+l)$ values are equal,the orbital with the lower $n$ value has lower energy.Comparing the values:$D (n+l=4, n=3) Thus,the order of increasing energy is $D < B < C < A$.

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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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 :