Atomic Structure MCQ Questions & Answers in Physical Chemistry | Chemistry

Firstly the electrons are filled in increasing order of energy and then rearrange the subshells in increasing order as
$$_{26}Fe = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^6},4{s^2}$$

$$G{d_{64}} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}},$$      $$4{s^2}4{p^6}4{d^{10}}4{f^7},5{s^2}5{p^6}5{d^1},6{s^2}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {Xe} \right]4{f^7},5{d^1},6{s^2}$$

For $$1s$$  orbital, the probability density is maximum at the nucleus and it decreases sharply as we move away from it. On the other hand, for $$2s$$  orbital, the probability density first decreases sharply to zero and again starts increasing.

$${H_\alpha }$$  line of Balmer series means first line of Balmer series.
$$\eqalign{ & {n_1} = 2,\,{n_2} = 3 \cr & \bar v = \frac{1}{{{\lambda _\alpha }}} = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) = \frac{{5R}}{{36}} \cr & \therefore \,\,{\lambda _\alpha } = \frac{{36}}{{5R}} = X \cr} $$
$${H_\beta }$$  line of Balmer series means, second line of Balmer series, $${n_1} = 2,\,{n_2} = 4.$$
$$\eqalign{ & \bar v = \frac{1}{{{\lambda _\beta }}} \cr & = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{4^2}}}} \right) \cr & = \frac{{3R}}{{16}} \cr & \therefore {\lambda _\beta } = \frac{{16}}{{3R}} = X \cr & {\text{when}}\,\,\frac{{36}}{{3R}} = X \cr & {\text{Then}}\,\,\frac{{16}}{{3R}} \cr & = \frac{{X \times 5R \times 16}}{{36 \times 3R}} \cr & = \frac{{80X}}{{108}}\mathop {\text{A}}\limits^{\text{o}} \cr} $$

Calculating number of electrons
\[\begin{gathered} 1.\,\,\,\left. \begin{gathered} BO_3^{3 - } \to 5 + 8 \times 3 + 3 = 32 \hfill \\ CO_3^{2 - } \to 6 + 8 \times 3 + 2 = 32 \hfill \\ NO_3^ - \to 7 + 8 \times 3 + 1 = 32 \hfill \\ \end{gathered} \right\}\,\,{\text{iso - electronic }}\,{\text{species}} \hfill \\ \hfill \\ 2.\,\,\,\,\left. \begin{gathered} SO_3^{2 - } \to 16 + 8 \times 3 + 2 = 42 \hfill \\ CO_3^{2 - } \to 32 \hfill \\ NO_3^ - \to 32 \hfill \\ \end{gathered} \right\}\,\,{\text{not iso - electronic}}\,{\text{ species}} \hfill \\ \hfill \\ {\text{3}}{\text{.}}\,\,\,\,\left. \begin{gathered} C{N^{2 - }} \to 6 + 7 + 1 = 14 \hfill \\ {N_2} \to 7 \times 2 = 14 \hfill \\ C_2^{2 - } \to 6 \times 2 + 2 = 14 \hfill \\ \end{gathered} \right\}\,\,{\text{iso - electronic}}\,{\text{ species}} \hfill \\ \hfill \\ 4.\,\,\,\,\left. \begin{gathered} PO_4^{3 - } \to 15 + 8 \times 4 + 3 = 50 \hfill \\ SO_4^{2 - } \to 16 \times 4 + 8 + 2 = 50 \hfill \\ CLO_4^ - \to 17 + 8 \times 4 + 1 = 50 \hfill \\ \end{gathered} \right\}\,\,{\text{iso - electronic }}\,{\text{species}} \hfill \\ \end{gathered} \]
Hence the species in option (B) are not isoelectronic.

$$Cr\left( {24} \right):3{d^5}4{s^1},C{r^{3 + }}:3{d^3}$$
$$Fe\left( {26} \right):3{d^6}4{s^2},F{e^{3 + }}:3{d^5};$$      $$Mn\left( {25} \right):3{d^5}4{s^2},M{n^{2 + }}:3{d^5}$$
$$Co\left( {27} \right):3{d^7}4{s^2},C{o^{3 + }}:3{d^6};$$      $$Sc\left( {21} \right):3{d^1}4{s^2},S{c^{3 + }}:3{d^0}$$
Thus, $$F{e^{3 + }}$$  and $$M{n^{2 + }}$$  have same electronic configuration.

$$\eqalign{ & \lambda = \frac{h}{{mv}} \cr & = \frac{{6.63 \times {{10}^{ - 34}}}}{{1.67 \times {{10}^{ - 27}} \times 1 \times {{10}^3}}} \cr & = 3.97 \times {10^{ - 10}}{\text{meter}} \cr & = 0.397\,{\text{nanometer}} \cr} $$

Electrons in an atom occupy the extra nuclear region.

$$\eqalign{ & {n_1} + {n_2} = 4;{n_2} - {n_1} = 2; \cr & \therefore \,\,{n_1} = 1,\,{n_2} = 3 \cr & \bar \nu = R{\left( 2 \right)^2}\left[ {\frac{1}{{{1^2}}} - \frac{1}{{{3^2}}}} \right] \cr & = \frac{{32R}}{9} \cr} $$

$$\eqalign{ & {n_1} + {n_2} = 4\,\,\,.....\left( {\text{i}} \right) \cr & {n_2} - {n_1} = 2\,\,\,.....\left( {{\text{ii}}} \right) \cr & {n_1} = 1,{n_2} = 3 \cr & \bar \upsilon = {R_H}{Z^2}\left( {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right) \cr & = {R_H} \times {3^2} \times \left[ {\frac{1}{{{1^1}}} - \frac{1}{{{3^2}}}} \right] \cr & = 8\,{R_H} \cr} $$