The orbital with maximum energy is

Two electrons $A$ and $B$ in an atom have the following set of quantum numbers:
$A: n=3, l=2, m_l=-2, m_s=+1/2$
$B: n=3, l=0, m_l=0, m_s=+1/2$
What is true for $A$ and $B$?

Write information about the four quantum numbers.

If the principal quantum number $n=6,$ the correct sequence of filling of electrons will be

In a multi-electron system,the energy of an electron of $2p_y$ orbital is :-

The arrangement of orbitals on the basis of energy is based upon their $(n + l)$ value. Lower the value of $(n + l)$,lower is the energy. For orbitals having same values of $(n + l)$,the orbital with lower value of $n$ will have lower energy.
$(I)$ Based upon the above information,arrange the following orbitals in the increasing order of energy.
$(a)$ $1s, 2s, 3s, 2p$
$(b)$ $4s, 3s, 3p, 4d$
$(c)$ $5p, 4d, 5d, 4f, 6s$
$(d)$ $5f, 6d, 7s, 7p$
$(II)$ Based upon the above information,solve the questions given below:
$(a)$ Which of the following orbitals has the lowest energy? $4d, 4f, 5s, 5p$
$(b)$ Which of the following orbitals has the highest energy? $5p, 5d, 5f, 6s, 6p$