Atomic Structure MCQ Questions & Answers in Physical Chemistry | Chemistry

TIPS/Formulae : According to de-Broglie’s equation
$$\eqalign{ & \lambda = \frac{h}{p} = \frac{h}{{mv}} \cr & {\text{Given,}}\,h{\text{ = 6}}{\text{.6}} \times {\text{1}}{{\text{0}}^{ - 34}}\,Js,\,\,m = 200 \times {10^{ - 3}}kg \cr & v = \frac{5}{{60 \times 60}}m/s \cr & \lambda = \frac{{6.6 \times {{10}^{ - 34}}}}{{200 \times {{10}^{ - 3}} \times \frac{5}{{\left( {60 \times 60} \right)}}}} = 2.38 \times {10^{ - 10}}m \cr} $$

In boundary surface diagram (1) the four lobes lie between $$y$$  and $$z$$ - axis $$\left( {{d_{yz}}} \right)$$  whereas, in boundary surface diagram (2) the four lobes lie on the $$x$$  and $$y$$ - axis $$\left( {{d_{{x^2} - {y^2}}}} \right).$$

$$v = 2.188 \times {10^6}\,Z/n$$

$$\eqalign{ & E = h\upsilon = \frac{{hc}}{\lambda } \cr & = \frac{{\left( {6.626 \times {{10}^{ - 34}}J\,\,s} \right)\left( {3 \times {{10}^8}m\,\,{s^{ - 1}}} \right)}}{{0.5 \times {{10}^{ - 10}}m}} \cr & = 3.98 \times {10^{ - 15}}J \cr} $$

$$\eqalign{ & {\text{Ionisation energy of}}\,H = 2.18 \times {10^{ - 18}}J\,ato{m^{ - 1}} \cr & \therefore \,\,{E_1}\left( {{\text{Energy of Ist orbit of }}H{\text{ - }}atom} \right) \cr & = - 2.18 \times {10^{ - 18}}J\,ato{m^{ - 1}} \cr & \therefore \,\,{E_n} = \frac{{ - 2.18 \times {{10}^{ - 18}}}}{{{n^2}}}J\,ato{m^{ - 1}} \cr & Z = 1\,{\text{for}}\,H{\text{ - }}atom \cr & \Delta E = {E_4} - {E_1} \cr & = \frac{{ - 2.18 \times {{10}^{ - 18}}}}{{{4^2}}} - \frac{{ - 2.18 \times {{10}^{ - 18}}}}{{{1^2}}} \cr & = - 2.18 \times {10^{ - 18}} \times \left[ {\frac{1}{{{4^2}}} - \frac{1}{{{1^2}}}} \right] \cr & \Delta E = - 2.18 \times {10^{ - 18}} \times - \frac{{15}}{{16}} \cr & \,\,\,\,\,\,\,\,\,\, = + \,2.0437 \times {10^{ - 18\,}}J\,ato{m^{ - 1}} \cr & \therefore \,\,\nu = \frac{{\Delta E}}{h} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{2.0437 \times {{10}^{ - 18}}J\,ato{m^{ - 1}}}}{{6.625 \times {{10}^{ - 34}}J\,s}} \cr & \,\,\,\,\,\,\,\,\,\, = 3.084 \times {10^{15}}{s^{ - 1}}ato{m^{ - 1}} \cr} $$

$$\eqalign{ & \because \,\,{\text{Total energy }}\left( {{E_n}} \right) = KE + PE \cr & {\text{In first excited state}} \cr & = \frac{1}{2}m{v^2} + \left[ { - \frac{{Z{e^2}}}{r}} \right] \cr & = + \frac{1}{2}\frac{{Z{e^2}}}{r} - \frac{{Z{e^2}}}{r} \cr & {\text{Energy of first excited state is}}\,\,3.4\,eV \cr & - 3.4\,eV = - \frac{1}{2}\frac{{Z{e^2}}}{r} \cr & \therefore \,\,\,\,\,KE = \frac{1}{2}\frac{{Z{e^2}}}{r} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = + 3.4\,eV \cr} $$

$$\eqalign{ & E = h\nu \cr & = 6.63 \times {10^{ - 34}} \times 2.47 \times {10^{15}} \cr & = 1.640 \times {10^{ - 18}}J \cr} $$

$$_{10}^{20}Ne$$  contains 10 protons and 10 neutrons
$$\therefore \,\,{M_1} = 10\,{m_p} + 10{m_n}$$
$$_{20}^{40}Ca$$  contains 20 protons and 20 neutrons
$$\eqalign{ & \therefore \,\,{M_2} = 20\,{m_p} + 20\,{m_n} \cr & \therefore \,\,{M_2} = 2{M_1} \cr} $$

Average atomic mass of $$Fe$$
$$\eqalign{ & = \frac{{\left( {54 \times 5} \right) + \left( {56 \times 90} \right) + \left( {57 \times 5} \right)}}{{100}} \cr & = 55.95 \cr} $$

Aufbau principle is violated in option (C) as electrons switch to $$p$$ orbitals without completely filling the $$s$$ orbital which has comparatively lower energy.