Atomic Structure MCQ Questions & Answers in Physical Chemistry | Chemistry

The electronic configuration of $$Cu,Th,Cs$$   and $$K$$ are
$$\eqalign{ & {\text{Cu}}\left( {29} \right) = 2,8,18,1 \cr & C{u^ + } = 2,8,18 \cr & Th\left( {90} \right) = 2,8,18,32,18,10,2 \cr & T{h^{4 + }} = 2,8,18,32,18,8 \cr & Cs\left( {55} \right) = 2,8,18,18,8,1 \cr & C{s^ + } = 2,8,18,18,8 \cr & K\left( {19} \right) = 2,8,8,1 \cr & {K^ + } = 2,8,8 \cr} $$
Thus, $$C{u^ + }$$  has 18 electrons in the outermost shell.

$$\eqalign{ & {\text{Mass of an electron}} = 9.1096 \times {10^{ - 31}}\,kg \cr & 1\,g\,\,{\text{or}}\,\,{10^{ - 3}}\,kg = \frac{1}{{9.1096 \times {{10}^{ - 31}}}} \times {10^{ - 3}} \cr & = 1.098 \times {10^{27}}\,{\text{electrons}} \cr} $$

The configuration does not follow Hund's rule of maximum multiplicity.

Total number of subshells $$ = \left( {2l + 1} \right)$$
∴ Maximum number of electrons in the subshell
$$\eqalign{ & = 2\left( {2l + 1} \right) \cr & = 4l + 2 \cr} $$

$$\eqalign{ & \lambda = \frac{h}{{mv}} \cr & h = 6.6 \times {10^{ - 34}}J{\text{ - }}s \cr & m = 1000\,kg \cr & v = 36\,km/hr = \frac{{36 \times {{10}^3}}}{{60 \times 60}}m/\sec = 10\,m/\sec \cr & \therefore \,\,\lambda = \frac{{6.6 \times {{10}^{ - 34}}}}{{{{10}^3} \times 10}} = 6.6 \times {10^{ - 38}}m \cr} $$

$${\text{For }}4p{\text{ electron}}\,n = 4,l = 1,$$      $$m = - 1,0 + 1\,{\text{and}}\,s = + \frac{1}{2}\,{\text{or}}\, - \frac{1}{2}$$

$${E_1} = \frac{{hc}}{{{\lambda _1}}},{E_2} = \frac{{hc}}{{{\lambda _2}}}$$
$$\frac{{{E_1}}}{{{E_2}}} = \frac{{hc}}{{{\lambda _1}}} \times \frac{{{\lambda _2}}}{{hc}}$$     $${\text{or}}\,\,\,\frac{{{E_1}}}{{{E_2}}} = \frac{{{\lambda _2}}}{{{\lambda _1}}} = \frac{{400}}{{800}};\frac{{{E_1}}}{{{E_2}}} = \frac{1}{2}$$       $${\text{or}}\,\,\,{E_2} = 2{E_1}$$

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

$$3s,3p,4s,3d,4p,5s,4d,5p$$       is the correct order of orbitals with increasing energy.

According to Rutherford's experiment. "The central part consisting of whole of the positive charge and most of the mass, called nucleus, is extremely small in size compared to the size of the atom."