The maximum number of electrons which each sub-shell can occupy is

Match List-$I$ with List-$II$ :

List-$I$ (Quantum number)	List-$II$ (Orbital)
$a. n=2, \ell=1$	$i. 2s$
$b. n=3, \ell=2$	$ii. 3s$
$c. n=3, \ell=0$	$iii. 2p$
$d. n=2, \ell=0$	$iv. 3d$

Choose the correct answer from the options given below :

For $s-$orbitals,since $\psi$ (orbital) is independent of angles,the probability $(\psi^2)$ is

The maximum number of electrons that can have the set of quantum numbers $n=4$,$m_l=0$,and $m_s=\frac{1}{2}$ is

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 $(n+\ell)$.

The electronic configuration of the ferric ion $(Fe^{3+})$ is: