Atomic Structure MCQ Questions & Answers in Physical Chemistry | Chemistry

Energy of absorbed photon = Sum of the energies of emitted photon
$$\eqalign{ & \frac{{hc}}{\lambda } = \frac{{hc}}{{{\lambda _1}}} + \frac{{hc}}{{{\lambda _2}}}\,\,or\,\,\frac{1}{\lambda } = \frac{1}{{{\lambda _1}}} + \frac{1}{{{\lambda _2}}} \cr & \frac{1}{{355 \times {{10}^{ - 9}}}} = \frac{1}{{680 \times {{10}^{ - 9}}}} + \frac{1}{{{\lambda _2}}} \cr & \frac{1}{{{\lambda _2}}} = \frac{1}{{355 \times {{10}^{ - 9}}}} - \frac{1}{{680 \times {{10}^{ - 9}}}} \cr & = 1.346 \times {10^6} \cr & {\text{or}}\,\,\,\,{\lambda _2} = \frac{1}{{1.346 \times {{10}^6}}} \cr & = 743 \times {10^{ - 9}}m \cr & = 743nm \cr} $$

$$\eqalign{ & p = \frac{h}{\lambda }{\text{(de - Broglie equation)}} \cr & \lambda = \frac{h}{{mv}}\,\,\left( {\because \,\,p = mv} \right) \cr & h = 6.625 \times {10^{ - 34}} \cr & \,\,\,\,\, \approx 6.63 \times {10^{ - 34}}kg/s \cr & \lambda = \frac{{6.63 \times {{10}^{ - 34}}kg\,{m^2}/s}}{{{{10}^{ - 3}}kg \times 100m/s}} \cr & \,\,\,\,\, = 6.63 \times {10^{ - 33}}m \cr} $$

$$\eqalign{ & {\text{For }}L{i^{2 + }}{\text{ ion}} \cr & E = - 13.6 \times \frac{{{Z^2}}}{{{n^2}}} = - 13.6 \times \frac{{{{\left( 3 \right)}^2}}}{{{{\left( 2 \right)}^2}}} \cr & = \frac{{ - 13.6 \times 9}}{4} = - 30.6\,eV \cr} $$

$${\text{Uncertainty in the speed of ball}}$$       $$ = \frac{{100 \times 10}}{{100}} = 10\,m/s$$
$$\Delta x.m\Delta v = \frac{h}{{4\pi }}$$
$$\eqalign{ & {\text{Uncertainty in the position,}} \cr & \Delta x = \frac{h}{{4\pi m\Delta v}} \cr & \,\,\,\,\,\,\,\, = \frac{{6.626 \times {{10}^{ - 34}}}}{{4 \times 3.14 \times 10 \times {{10}^{ - 3}} \times 10}} \cr & \,\,\,\,\,\,\,\, = 5.27 \times {10^{ - 34}}\,m \cr} $$

Hydrogen has three isotopes:
Protium $$\,\left( {{}_1{H^1}} \right),$$  deuterium $$\left( {{}_1{H^2}} \right)$$  and tritium $$\left( {{}_1{H^3}} \right).$$

The energy of an orbital is given by $$\left( {n + l} \right)$$  in (D) and (C). $$\left( {n + l} \right)$$  value is $$(3+2) = 5$$   hence they will have same energy, since there n values are also same.

When $$l$$ = 3, as $${m_l} \simeq \left( {2l + 1} \right) = 7$$

The outermost shell configuration of the element shows 8 electrons hence it is a noble gas.

According to Aufbau principle, the orbital of lower energy $$(2s)$$  should be fully filled before the filling of orbital of higher energy starts.

$$\left( {\text{A}} \right)\,\,{N^{3 - }} = 7 + 3 = 10{e^ - },$$      $${O^ - } \to 8 + 2 = 10{e^ - }$$
$${F^ - } = 9 + 1 = 10{e^ - },$$     $${S^{ - - }} \to 16 + 2 = 18{e^ - }$$     $${\text{(not}}\,\,{\text{iso}}\,\,{\text{electronic)}}$$
$$\left( {\text{B}} \right)\,\,L{i^ + } = 3 + 1 = 4{e^ - },$$     $$N{a^ + } = 11 - 1 = 10{e^ - },$$
$$M{g^{ + + }} = 12 - 2 = 10{e^ - }$$
$$C{a^{ + + }} = 20 - 2 = 18{e^ - }$$     $${\text{(not}}\,\,{\text{iso}}\,\,{\text{electronic)}}$$
$$\left( {\text{C}} \right)\,\,{K^ + } = 19 - 1 = 18{e^ - },$$      $$C{\ell ^ - } = 17 + 1 = 18{e^ - },$$
$$C{a^{ + + }} = 20 - 2 = 18e,$$     $$S{c^{3 + }} = 21 - 3 = 18{e^ - }$$     $${\text{(iso}}\,\,{\text{electronic)}}$$
$$\left( {\text{D}} \right)\,\,B{a^{ + + }}56 - 2 = 54e,$$     $$S{r^{ + + }}38 - 2 = 36{e^ - }$$
$${K^ + } = 9 - 1 = 18{e^ - },$$     $$C{a^{ + + }} = 20 - 2 = 18{e^ - }$$     $${\text{(not iso}}\,\,{\text{electronic)}}$$