If the nitrogen atom had electronic configuration $1s^7$,it would have energy lower than that of the normal ground state configuration $1s^2, 2s^2, 2p^3$ because the electrons would be closer to the nucleus. Yet,$1s^7$ is not observed because it violates:

The number of correct statements from the following:
$A.$ For $1s$ orbital,the probability density is maximum at the nucleus.
$B.$ For $2s$ orbital,the probability density first increases to maximum and then decreases sharply to zero.
$C.$ Boundary surface diagrams of the orbitals enclose a region of $100\%$ probability of finding the electron.
$D.$ $p$ and $d$-orbitals have $1$ and $2$ angular nodes respectively.
$E.$ Probability density of $p$-orbital is zero at the nucleus.

The atomic numbers of elements $X, Y$ and $Z$ are $19, 21$ and $25$ respectively. The number of electrons present in the $M$-shell of these elements follow the order:

What is the difference between the ground state and the excited state of a hydrogen atom?

Arrange the orbitals of $H$ atom in the increasing order of their energy:
$3p_x, 2s, 4d_{xy}, 3s, 4p_z, 3p_y, 4s$

$A$ $3p$ orbital has