The electronic configuration of the element with atomic number $27$ is:

Which of the following principles/rules limits the maximum number of electrons in an orbital to two?

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

The maximum number of electrons that can have principal quantum number,$n=3$,and spin quantum number,$m_s=-1/2$,is

The number of valence electrons in an atom with electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^3$ is

Which electronic level would allow the hydrogen atom to absorb a photon but not to emit a photon?