Atomic Structure MCQ Questions & Answers in Physical Chemistry | Chemistry

To write the electronic configuration of an atom, it is better if we remember the atomic number of noble gases and the orbitals follow the noble gas. The atomic number of $$Ca$$  is 20 and its nearest noble gas is argon $$(Ar = 18).$$
Hence, the electronic configuration of $$Ca = \left[ {Ar} \right]4{s^2}.$$

$$\eqalign{ & {\left( {K.E.} \right)_1} = h{\upsilon _1} - h{\upsilon _0} \cr & {\left( {K.E.} \right)_2} = h{\upsilon _2} - h{\upsilon _0} \cr & {\text{As}}\,\,\,{\left( {K.E.} \right)_1} = 2 \times {\left( {K.E.} \right)_2} \cr & \therefore \,\,\left( {h{\upsilon _1} - h{\upsilon _0}} \right) = 2\left( {h{\upsilon _2} - h{\upsilon _0}} \right) \cr} $$
$${\text{or}}$$  $${\upsilon _0} = 2{\upsilon _2} - {\upsilon _1} = 2 \times \left( {2 \times {{10}^{16}}} \right) - \left( {3.2 \times {{10}^{16}}} \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.8 \times {10^{16}}\,Hz\,\,\,{\text{or}}\,\,\,8 \times {10^{15}}\,Hz$$

$$\eqalign{ & {\text{Power of the bulb}} = 100\,watt = 100\,J\,{s^{ - 1}} \cr & {\text{Energy of one photon}}\,\,E = h\upsilon = \frac{{hc}}{\lambda } \cr & = \frac{{6.626 \times {{10}^{ - 34}}Js \times 3 \times {{10}^8}m\,{s^{ - 1}}}}{{400 \times {{10}^{ - 9}}\,m}} \cr & = 4.969 \times {10^{ - 19}}\,J \cr & {\text{Number of photons emitted}} \cr & = \frac{{100\,J\,{s^{ - 1}}}}{{4.969 \times {{10}^{ - 19}}J}} \cr & = 2.012 \times {10^{20}}{s^{ - 1}} \cr} $$

The energy of second Bohr orbit of hydrogen atom $$\left( {{E_2}} \right)$$  is $$ - 328\,kJ\,mo{l^{ - 1}}$$
$$\eqalign{ & \,\,\,\,\,\,\,{E_n} = - \frac{{1312}}{{{n^2}}}kJ\,mo{l^{ - 1}} \cr & \therefore \,\,{E_2} = - \frac{{1312}}{{{2^2}}}kJ\,mo{l^{ - 1}} \cr & {\text{If}}\,\,n = 4 \cr & \therefore \,\,{E_4} = - \frac{{1312}}{{{4^2}}}kJ\,mo{l^{ - 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 82\,kJ\,mo{l^{ - 1}} \cr} $$

The radius of nucleus is of the order of $$1.5 \times {10^{ - 13}}$$   to $$6.5 \times {10^{ - 13}}cm$$    or 1.5 to 6.5 Fermi $$\left( {1\,{\text{Fermi}} = {{10}^{ - 13}}cm} \right)$$

$$\eqalign{ & \Delta E = 2.178 \times {10^{ - 18}}\left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) = \frac{{hc}}{\lambda } \cr & 2.17 \times {10^{ - 18}} \times \frac{3}{4} = \frac{{hc}}{\lambda } = \frac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{\lambda } \cr & \lambda = \frac{{6.62 \times \times {{10}^{ - 34}}}}{{2.17 \times {{10}^{ - 18}}}}\frac{{3 \times {{10}^8} \times 4}}{{ \times 3}} \cr & = 1.214 \times {10^{ - 7}}m \cr} $$

The fifth $$d$$ - orbital $$\left( {{d_{{z^2}}}} \right)$$  has a doughnut shaped electron cloud in the centre whereas others have clover leaf shape.

$$\eqalign{ & {\text{de Broglie wavelength}}\,\,\lambda = \frac{h}{{mv}} \cr & \frac{{{\lambda _1}}}{{{\lambda _2}}} = \frac{{{m_2}{v_2}}}{{{m_1}{v_1}}};\,\frac{1}{4} = \frac{1}{9} \times \frac{{{v_2}}}{{{v_1}}} \cr & \frac{{{v_2}}}{{{v_1}}} = \frac{9}{4};\frac{{{v_1}}}{{{v_2}}} = \frac{4}{9} \cr & KE = \frac{1}{2}m{v^2} \cr & \frac{{K{E_1}}}{{K{E_2}}} = \frac{{{m_1}}}{{{m_2}}} \times \frac{{v_1^2}}{{v_2^2}} \cr & = \frac{9}{1} \times {\left( {\frac{4}{9}} \right)^2} \cr & = \frac{{16}}{9} \cr} $$

According, to Pauli’s exclusion principle “no two electrons in an atom can have the same values of all the four quantum numbers.”
In $$1{s^2}$$
$$\eqalign{ & {\text{for }}I{\text{ electron}}\,\,n = 1,\,l = 0,m = 0,s = + \frac{1}{2} \cr & {\text{for }}II{\text{ electron}}\,\,n = 1,\,l = 0,m = 0,s = - \frac{1}{2} \cr} $$
It means if the values of $$n, l,$$  and $$m$$ are same, then the value of spin quantum number must be different, i.e. $$ + \frac{1}{2}$$  and $$ - \frac{1}{2}.$$

Shell with $$n = 3$$   has $$3s,3p\left( {{p_x},{p_y},{p_z}} \right)$$    and $$3d$$ $$\left( {{d_{xy}},{d_{xz}},{d_{yz}},{d_{{x^2} - {y^2}}},{\text{and}}\,\,{d_{{z^2}}}} \right)$$      orbitals.
$$s$$  has no nodal plane.
Each of $${{p_x},{p_y},{p_z}}$$  has one nodal plane, which means a total of three nodal planes.
$${d_{{z^3}}}$$  has no nodal plane.
Each of $${{d_{xy}},{d_{xz}},{d_{yz}},{d_{{x^2} - {y^2}}}}$$    has two nodal planes, which means a total of eight nodal planes.
Hence, for $$n = 3,$$  a total of 11 nodal planes are there.