The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination$(s)$ has/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. Lower $(n+l)$ value implies lower energy. If $(n+l)$ values are equal,the orbital with lower $n$ has lower energy.
$(1).$ $n=4, l=2 \implies n+l = 6$ $(4d)$
$(2).$ $n=3, l=2 \implies n+l = 5$ $(3d)$
$(3).$ $n=4, l=1 \implies n+l = 5$ $(4p)$
$(4).$ $n=3, l=2 \implies n+l = 5$ $(3d)$
$(5).$ $n=3, l=1 \implies n+l = 4$ $(3p)$
$(6).$ $n=4, l=1 \implies n+l = 5$ $(4p)$
Comparing $(n+l)$ values: $(5) < (2) = (4) = (3) = (6) < (1)$.
For orbitals with the same $(n+l)$ value,the one with lower $n$ has lower energy. Thus,$3d$ $(n=3)$ has lower energy than $4p$ $(n=4)$.
Therefore,the increasing order of energy is: $(5) < (2) = (4) < (3) = (6) < (1)$.