Atomic Structure MCQ Questions & Answers in Physical Chemistry | Chemistry

Bohr model can explain spectrum of atoms/ions containing one electron only.

When azimuthal quantum number is 3
$$\eqalign{ & m = \left( {2l + 1} \right) \cr & \,\,\,l = 3 \cr & m = \left( {2 \times 3 + 1} \right) \cr & \,\,\,\,\,\, = 7\,{\text{orbitals}} \cr} $$
then total values of $$m = \left( {2 \times 3 + 1} \right) = 7\,{\text{orbitals}}{\text{.}}$$      We know that, one orbital contains two electrons. Hence, total number of electrons $$ = 7 \times 2 = 14.$$
Alternative
Total number of electrons
  $$ = 4l + 2$$
$$\eqalign{ & = 4 \times 3 + 2 \cr & = 12 + 2 \cr & = 14\,{\text{elctrons}} \cr} $$

$$\eqalign{ & {\text{According to de Broglie's equation,}} \cr & \lambda = \frac{h}{{mv}} = \frac{{6.626 \times {{10}^{ - 34}} \times 3600}}{{0.2 \times 5}}; \cr & \lambda = 2.385 \times {10^{ - 30}}\,m \cr} $$

$$\eqalign{ & \left( {\text{i}} \right)n = 3,l = 2 \Rightarrow 3d \cr & \left( {{\text{ii}}} \right)n = 4,l = 3 \Rightarrow 4f \cr} $$

Total no. of orbitals in $${n^{th}}$$  shell $$ = {n^2}$$
∴ Total no. of orbitals in 3^rd shell $$\left( {n = 3} \right) = {3^2} = 9$$

The electronic configuration of Rubidium $$(Rb=37)$$  is
$$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}4{s^2}4{p^6}5{s^1}$$
Since last electron enters in $$5s$$  orbital
$${\text{Hence}}\,\,n = 5,\,\,l = 0,\,\,m = 0,\,\,s = \pm \frac{1}{2}$$

$$\eqalign{ & {\text{Frequency}}\left( \nu \right) \cr & = \frac{c}{\lambda } = \frac{{3 \times {{10}^8}m\,{s^{ - 1}}}}{{8 \times {{10}^{15}}{s^{ - 1}}}} \cr & = 0.375 \times {10^{ - 7}}m \cr & = 3.75 \times {10^1}nm \approx 4 \times {10^1}nm \cr} $$

$${\text{Radius of}}\,\,{n^{th}}\,\,{\text{orbit}}$$     $$ = \frac{{0.529 \times {n^2}}}{Z}\mathop {\text{A}}\limits^{\text{o}} = \frac{{52.9 \times {n^2}}}{Z}pm$$
$${\text{Radius}}\left( {{r_2}} \right) = 1.3225\,nm$$     $$ = 1322.5\,pm = \frac{{52.9\,n_2^2}}{Z}$$
$${\text{Radius}}\left( {{r_1}} \right) = 211.6\,pm = \frac{{52.9\,n_1^2}}{Z}$$
$$\eqalign{ & \therefore \,\,\,\frac{{{r_2}}}{{{r_1}}} = \frac{{1322.5}}{{211.6}} = \frac{{n_2^2}}{{n_1^2}} \cr & \Rightarrow 6.25 = \frac{{n_2^2}}{{n_1^2}} = {\left( {\frac{{{n_2}}}{{{n_1}}}} \right)^2} \cr & \Rightarrow \frac{{{n_2}}}{{{n_1}}} = \sqrt {6.25} = 2.5 \cr} $$
$$\therefore \,\,\,{n_1} = 2,{n_2} = 5$$     thus, the transition is from 5^th orbit to 2^nd orbit. It belongs to Balmer series.
$$\eqalign{ & \bar \upsilon = R\left( {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right) \cr & \Rightarrow \bar \upsilon = 1.097 \times {10^7}\left( {\frac{1}{{{2^2}}} - \frac{1}{{{5^2}}}} \right) \cr & = 1.097 \times {10^7}\left( {\frac{1}{4} - \frac{1}{{25}}} \right) \cr & = 1.097 \times {10^7} \times \frac{{21}}{{100}} \cr} $$
$$\lambda = \frac{1}{{\bar \upsilon }} = \frac{{100}}{{1.097 \times 21 \times {{10}^7}}}m$$       $$ = 4.34 \times {10^{ - 7}}\,m = 434 \times {10^{ - 9}}\,m$$
$$\eqalign{ & \lambda = 434\,nm \cr & {\text{Thus, it lies in the visible region}}{\text{.}} \cr} $$

$$\eqalign{ & _{15}P:\left[ {Ne} \right]3{s^2}3{p^3} \cr & _{24}Cr:\left[ {Ar} \right]3{d^5}4{s^1} \cr & _{26}Fe:\left[ {Ar} \right]4{s^2}3{d^6} \cr & _{14}Si:\left[ {Ne} \right]3{s^2}3{p^2} \cr} $$

$$^{39}{K^ + }$$  and $$^{40}{K^ + }$$  contains same number of electrons so they are isoelectronic. They have same atomic number but different mass numbers so they are also isotopes.