The orbital with maximum energy is

An element has the electronic configuration $1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^5, 4s^1$. It is a

Which of the following electronic configurations is closer to the nucleus: $4s^2 3d^{10}$ or $4s^0 3d^{10}$?

Assertion :- In $H$-atom,the energy of $3d$-level is smaller than $4s$-level.
Reason :- In multi-electron species,an orbital with lower value of $(n+\ell)$ has energy smaller than the orbital with larger value of $(n+\ell)$.

Identify which of the following sets of quantum numbers are not possible and explain why.
$(a) n = 0, l = 0, m_l = 0, m_s = +1/2$
$(b) n = 1, l = 0, m_l = 0, m_s = -1/2$
$(c) n = 1, l = 1, m_l = 0, m_s = +1/2$
$(d) n = 2, l = 1, m_l = 0, m_s = -1/2$
$(e) n = 3, l = 3, m_l = 3, m_s = +1/2$
$(f) n = 3, l = 1, m_l = 0, m_s = +1/2$

The atomic number of the element (in ground state) having the maximum number of unpaired $3p$ electrons is: