The quantum numbers n and l for four electron | vedclass

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(B) According to the $(n+l)$ rule,the energy of an orbital increases as the value of $(n+l)$ increases. If two orbitals have the same $(n+l)$ value,th

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$(4) < (2) < (3) < (1)$

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(B) According to the $(n+l)$ rule,the energy of an orbital increases as the value of $(n+l)$ increases. If two orbitals have the same $(n+l)$ value,the one with the lower $n$ value has lower energy.Calculating $(n+l)$ for each:$(1) \, n=4, l=1 \implies n+l = 5$$(2) \, n=4, l=0 \implies n+l = 4$$(3) \, n=3, l=2 \implies n+l = 5$$(4) \, n=3, l=1 \implies n+l = 4$Comparing the values:For $(n+l) = 4$: $(4)$ has $n=3$ and $(2)$ has $n=4$. Thus,$(4) For $(n+l) = 5$: $(3)$ has $n=3$ and $(1)$ has $n=4$. Thus,$(3) Combining these,the increasing order of energy is $(4) < (2) < (3) < (1)$.

The quantum numbers $n$ and $l$ for four electrons are given as: $(1) \, n = 4, l = 1; \, (2) \, n = 4, l = 0; \, (3) \, n = 3, l = 2; \, (4) \, n = 3, l = 1$. Arrange them in increasing order of their energy.

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