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

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(A) 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 	ext{ (4p orbital)}$$(2) n = 4, l = 0 \implies n+l = 4 	ext{ (4s orbital)}$$(3) n = 3, l = 1 \implies n+l = 4 	ext{ (3p orbital)}$$(4) n = 3, l = 2 \implies n+l = 5 	ext{ (3d orbital)}$Comparing the values:For $(3)$ and $(2)$,both have $(n+l) = 4$. Since $(3)$ has $n=3$ and $(2)$ has $n=4$,$(3) For $(4)$ and $(1)$,both have $(n+l) = 5$. Since $(4)$ has $n=3$ and $(1)$ has $n=4$,$(4) Combining these,the increasing order of energy is $(3) < (2) < (4) < (1)$.

Arrange the following orbitals in increasing order of energy based on the given values of principal quantum number $(n)$ and azimuthal quantum number $(l)$:
$(1) n = 4, l = 1$
$(2) n = 4, l = 0$
$(3) n = 3, l = 1$
$(4) n = 3, l = 2$

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Arrange the following orbitals in increasing order of energy based on the given values of principal quantum number $(n)$ and azimuthal quantum number $(l)$:$(1) n = 4, l = 1$$(2) n = 4, l = 0$$(3) n = 3, l = 1$$(4) n = 3, l = 2$