Atomic Structure MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & {E_{{1_{H{e^ + }}}}} = {E_{{1_H}}} \times {Z^2} \cr & \therefore \,\, - 871.6 \times {10^{ - 20}} = {E_{{1_H}}} \times 4 \cr & \therefore \,\,{E_{{1_H}}} = - 217.9 \times {10^{ - 20}}J \cr} $$

$$\eqalign{ & n{\text{ = 3,}}\,l = 0\,{\text{means 3s - orbital and}}\,n + 1 = 3 \cr & n{\text{ = 3,}}\,l = 1\,{\text{means 3p - orbital }}n + 1 = 4 \cr & n{\text{ = 3,}}\,l = 2\,{\text{means 3d - orbital }}n + 1 = 5 \cr & n{\text{ = 4,}}\,l = 0\,{\text{means 4s - orbital }}n + 1 = 4 \cr} $$
Increasing order of energy among these orbitals is
$$3s< 3p< 4s< 3d$$
∴ $$3d$$ has highest energy.

$$\eqalign{ & {\text{Kinetic energy of any particle}} = \frac{3}{2}KT \cr & {\text{Also}}\,K.E. = \frac{1}{2}m{v^2} \cr & \frac{1}{2}m{v^2} = \frac{3}{2}KT \Rightarrow {v^2} = \frac{{3KT}}{m} \cr & v = \sqrt {\frac{{3KT}}{m}} \cr & {\text{De - broglie wavelength}} \cr & = \lambda = \frac{h}{{mv}} = \frac{h}{{m\sqrt {\frac{{3KT}}{m}} }} \cr & \lambda = \frac{h}{{\sqrt {3KTm} }}\lambda \propto \frac{1}{{\sqrt m }} \cr} $$
$${\text{Mass of electron < }}$$    $${\text{mass of neutron}}\,\lambda {\text{(electron) }}$$     $${\text{ > }}\lambda {\text{(neutron)}}$$

The ground state electronic configuration, of $$C{u^ + }\,ion$$   is $$1{s^2},2{s^2},2{p^6},3{s^2},3{p^6},3{d^{10}}$$
$$n = 3$$  ( number of shells ) ; Number of subshells occupied $$= 6,$$  Number of filled orbitals $$= 14 ;$$  There is no unpaired electron.

$${p_x}$$ orbital being dumbell shaped, have number of nodal planes $$ = 1,$$ in $$yz$$  plane.

$$\eqalign{ & {\text{Potential energy = work done}} \cr & = \int_\infty ^r { - \frac{{Z{e^2}dr}}{{{r^2}}}} \cr & = - \frac{{Z{e^2}}}{r} \cr} $$

The orbital of the electron having $$n = 3,l = 1\,$$   and $$m = - 1$$   is $$3{p_z}\left( {{\text{as}}\,n{l_m}} \right)$$   and an orbital can have a maximum number of two electrons with opposite spins.
∴ $$3{p_z}$$ orbital contains only two electrons or only 2 electrons are associated with $$n = 3,l = 1\,,m = - 1.$$

Spin angular momentum of the electron, a vector quantity can have two orientations relative to the chosen axis. These two orientations take the values of $$ + \frac{1}{2}$$  or $$ - \frac{1}{2}$$  and are called two spin states of the electrons.

Atomic number 22 has the following configuration : $$_{22}X = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^2}$$

J.J. Thomson proposed that an atom possesses a spherical shape in which the positive charge is uniformly distributed. This model was able to explain the overall neutrality of the atom.