If $Br^{-}$ configuration is $[Ar]\, 3d^{10}\,4s^2\,4p^6$,then $Br^{+2}$ configuration will be identical to which element?

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? $5p, 5d, 5f, 6s, 6p$

Which one of the following statements is incorrect?

For an electron with $n = 3$,there is only one radial node. The orbital angular momentum of the electron will be

What would be the electronic configuration of $Cs (Z=55)$ in each case?
$(a)$ If there were three possibilities of the electron spin.
$(b)$ If the quantum number,$l$,could have the value $n$,and if all the rules governing electron configuration were otherwise valid.

The number of $d$-electrons in $Fe^{2+}$ (atomic number of $Fe = 26$) is not equal to that of the