The shape of $\mathrm{p}-$orbital is dumbbell like.

The correct decreasing order of energy, for the orbitals having, following set of quantum numbers:

(A) $\mathrm{n}=3,1=0, \mathrm{m}=0$

(B) $\mathrm{n}=4,\mathrm{l}=0, \mathrm{m}=0$

(C) $\mathrm{n}=3,\mathrm{l}=1, \mathrm{m}=0$

(D) $\mathrm{n}=3,1=2, \mathrm{m}=1$

Given below are the quantum numbers for $4$ electrons.

A. $\mathrm{n}=3,\mathrm{l}=2,{\mathrm{m}}_1=1,{\mathrm{m}}_{\mathrm{s}}=+\frac{1}{2}$

B. $\mathrm{n}=4,\mathrm{l}=1,{\mathrm{m}}_1=0,{\mathrm{m}}_{\mathrm{s}}=+\frac{1}{2}$

C. $\mathrm{n}=4,\mathrm{l}=2,{\mathrm{m}}_1=-2,{\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}$

D. $\mathrm{n}=3,\mathrm{l}=1,{\mathrm{m}}_1=-1,{\mathrm{m}}_{\mathrm{s}}=+\frac{1}{2}$

The correct order of increasing energy is

The number of given orbitals which have electron density along the axis is

${\mathrm{p}}_{\mathrm{x}},{\mathrm{p}}_{\mathrm{y}},{\mathrm{p}}_{\mathrm{z}},{\mathrm{d}}_{\mathrm{xy}},{\mathrm{d}}_{\mathrm{yz}},{\mathrm{d}}_{\mathrm{xz}},{\mathrm{d}}_{\mathrm{z}}^2,{\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}$

Consider the following pairs of electrons:

(A) (a) $\mathrm{n}=3, \mathrm{l}=1, {\mathrm{m}}_{\mathrm{l}}=1,{\mathrm{m}}_{\mathrm{s}}=+\frac{1}{2}$
(b) $\mathrm{n}=3, \mathrm{l}=2,{\mathrm{m}}_{\mathrm{l}}=1,{\mathrm{m}}_{\mathrm{s}}=+\frac{1}{2}$
(B) (a) $\mathrm{n}=3, \mathrm{l}=2,{\mathrm{m}}_{\mathrm{l}}=-2,{\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}$
(b) $\mathrm{n}=3, \mathrm{l}=2,{\mathrm{m}}_{\mathrm{l}}=-1,{\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}$
(C) (a) $\mathrm{n}=4, \mathrm{l}=2,{\mathrm{m}}_{\mathrm{l}}=2,{\mathrm{m}}_{\mathrm{s}}=+\frac{1}{2}$
(b) $\mathrm{n}=3,\mathrm{l}=2,{\mathrm{m}}_{\mathrm{l}}=2,{\mathrm{m}}_{\mathrm{s}}=+\frac{1}{2}$

The pairs of electrons present in degenerate orbitals is/are:

How are a $1\mathrm{s}$ orbital and a $2\mathrm{s}$ orbital in an atom similar? How do they differ? 

How are a $2p_x$ orbital and a $2p_y$ orbital in an atom similar? How do they differ?