When $$n =2$$  and $$l =1,$$  then subshell is $$2p.$$  The number of orbitals in $$p$$ - subshell
$$\eqalign{ & = \left( {2l + 1} \right) \cr & = \left( {2 \times 1 + 1} \right) \cr & = 3 \cr} $$
Total (maximum) number of electrons
$$\eqalign{ & = 2 \times {\text{number of orbitals}} \cr & = 2 \times 3 \cr & = 6 \cr} $$
( as each orbital contains 2 electrons )

(A) $${Z_{eff}}$$  for an electron in a $$2s$$  orbital is greater than that in a $$2p$$  orbital ( $$s$$  orbital is more tightly bound to the nucleus than $$p$$  orbital ).
(B) Energy of $$2s < 2p,$$  lower the value of $$n + l,$$  lower is the energy.
(C) $${Z_{eff}}$$  for an electron in $$1s$$  orbital is greater than $${Z_{eff}}$$  for an electron in a $$2s$$  orbital.
(D) The two electrons present in any orbital have spin quantum numbers with opposite sign $$\left( {{\text{i}}{\text{.e}}.,\,{m_s} = + \frac{1}{2}\,\,{\text{and}}\,\, - \frac{1}{2}} \right).$$

Electronic configuration of $$Cr$$ atom ( $$z$$ = 24)
$$\eqalign{ & = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^5},4{s^1} \cr & \operatorname{When} \,\ell = 1,\,p - {\text{subshell}}, \cr & {\text{Numbers}}\,{\text{of}}\,{\text{electrons}} = 12 \cr & {\text{when}}\,\ell = 2,\,d{\text{ - subshell}}, \cr & {\text{Numbers}}\,{\text{of}}\,{\text{electrons}} = 5 \cr} $$

TIPS/Formulae :
$$\eqalign{ & {\text{Total nodes}} = n - l \cr & {\text{No}}{\text{. of radial nodes}} = n - l - 1 \cr & {\text{No}}{\text{. of angular nodes}} = l \cr & {\text{For}}\,3p\,\,{\text{sub - shell,}}\,\,n = 3,\,\,l = 1 \cr & \therefore \,\,\,{\text{No}}{\text{. of radial nodes}} = n - l - 1 = 3 - 1 - 1 = 1 \cr & \therefore \,\,\,{\text{No}}{\text{. of angular nodes}} = l = 1 \cr} $$

The wave nature of an electron is proved by Davisson and Germer experiment. In this experiment the scattering pattern of an electron is similar to that of $$X$$ - rays.

$$3{d^6}$$  configuration means $$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^6}$$

Radius of n^th Bohr orbit in H-atom
$$ = 0.53\,{n^2}\mathop {\text{A}}\limits^{\text{o}} $$
Radius of 2^nd Bohr orbit $$ = 0.53 \times {\left( 2 \right)^2}$$
$$ = 2.12\mathop {\text{A}}\limits^{\text{o}} $$

$$\eqalign{ & {\text{For hydrogen atom}}\,\,Z = 1 \cr & \therefore \,\,{\text{Ionisation energy,}}\,{E_H} = \frac{{2{\pi ^2}m{e^4}}}{{{n^2}{h^2}}}\,\,...{\text{(i)}} \cr & {\text{For }}H{e^ + }{\text{ }}ion,\left( {H{e^ + } = 1{s^1}} \right) \cr & {\text{so, }}(H{e^ + } = H)\,{\text{ionisation energy,}} \cr & {{\text{E}}_{H{e^ + }}} = \frac{{2{\pi ^2}m{e^4}{Z^2}}}{{{n^2}{h^2}}}\,\,...{\text{(ii)}} \cr & Eq{\text{ (i)}}/Eq{\text{ (ii)}},{\text{ we get}} \cr & {E_{H{e^ + }}} = {E_H} \times {Z^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 13.6 \times 4 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 54.4\,\,eV \cr} $$

$$\eqalign{ & {\text{Radius of an orbit,}} \cr & {r_n} = \frac{{{n^2}{h^2}}}{{4{\pi ^2}m{e^2}Z}} \cr & \,\,\,\,\,\, = \frac{{0.529{n^2}}}{Z}\mathop {\text{A}}\limits^{\text{o}} \cr & {\text{For }}H{\text{ - }}atom,Z{\text{ }} = 1 \cr & {\text{If}}\,\,\,{r_1} = r\,\,\left( {{\text{according to question}}\,\,{r_1} = r} \right) \cr & \therefore {r_n} = \frac{{r \times {n^2}}}{1} \cr & \,\,\,\,\,\,\,\,\,\,\, = r{n^2} \cr} $$

Electron can undergo transition from higher state to all lower states by loss of energy.