The quantum numbers of six electrons are given below. Arrange them in order of increasing energy. If any of these combinations have the same energy,list them:
$(1)$ $n = 4, l = 2, m_l = 2, m_s = -1/2$
$(2)$ $n = 3, l = 2, m_l = 1, m_s = +1/2$
$(3)$ $n = 4, l = 1, m_l = 0, m_s = +1/2$
$(4)$ $n = 3, l = 2, m_l = -2, m_s = -1/2$
$(5)$ $n = 3, l = 1, m_l = -1, m_s = +1/2$
$(6)$ $n = 4, l = 1, m_l = 0, m_s = +1/2$

(A) The energy of an orbital is determined by the $(n + l)$ rule. If $(n + l)$ values are equal,the orbital with the lower $n$ value has lower energy.
$(1)$ $n=4, l=2 \rightarrow n+l = 6$ $(4d)$
$(2)$ $n=3, l=2 \rightarrow n+l = 5$ $(3d)$
$(3)$ $n=4, l=1 \rightarrow n+l = 5$ $(4p)$
$(4)$ $n=3, l=2 \rightarrow n+l = 5$ $(3d)$
$(5)$ $n=3, l=1 \rightarrow n+l = 4$ $(3p)$
$(6)$ $n=4, l=1 \rightarrow n+l = 5$ $(4p)$
Comparing $(n+l)$ values:
$(5)$ $3p$ $(n+l=4)$
$(2)$ and $(4)$ $3d$ $(n+l=5)$
$(3)$ and $(6)$ $4p$ $(n+l=5)$
$(1)$ $4d$ $(n+l=6)$
Since $(2)$ and $(4)$ have the same $(n+l)$ and $n$,they have the same energy. Similarly,$(3)$ and $(6)$ have the same energy.
Increasing order of energy: $(5) < (2) = (4) < (3) = (6) < (1)$.