Atomic Structure MCQ Questions & Answers in Physical Chemistry | Chemistry

Sommerfeld modified Bohr's theory. According to him electrons move in elliptical orbits in addition to circular orbits.

Electronic configuration of $$F{e^{2 + }}\,{\text{is}}\,\,\left[ {Ar} \right]3{d^6}4{s^0}.$$
∴ Number of electrons = 6
$$Mg - 1{s^2}2{s^2}2{p^6}3{s^2}\left( {6s\,{\text{electrons}}} \right)$$
It matches with the $$6d$$ electrons of $$F{e^{2 + }}$$
$$Cl - 1{s^2}2{s^2}2{p^6}3{s^2}3{p^5}\left( {11p\,\,{\text{electrons}}} \right)$$
It does not match with the $$6d$$ electrons of $$F{e^{2 + }}.$$
$$Fe - \left[ {Ar} \right]3{d^6}4{s^2}\left( {6d\,{\text{electrons}}} \right)$$
It matches with the $$6d$$ electrons of $$F{e^{2 + }}.$$
$$Ne - 1{s^2}2{s^2}2{p^6}\left( {6p\,{\text{electrons}}\,} \right)$$
It matches with the $$6d$$ electrons of $$F{e^{2 + }}.$$
Hence, $$Cl$$ has $$11p$$  electrons which does not matches in number with $$6d$$ electrons of $$F{e^{2 + }}.$$

The value of $$n =3$$  and $$l = 1$$  suggests that it is a $$3p$$  - orbital while the value of $${m_l} = 0$$  [ magnetic quantum number ] shows that the given $$3 p$$  - orbital is $$3{p_z}$$ in nature.

Hence, the maximum number of orbitals identified by the given quantum number is only $$1,$$ i.e. $$3{p_z}.$$

$$l = 4;$$
number of degenerate orbitals $$ = 2l + 1 = 9;$$
maximum total spins $$ = 9 \times \frac{1}{2}$$
maximum multiplicity $$ = 2s + 1$$
$$ = 2 \times \frac{9}{2} + 1 = 10$$
minimum total spins $$ = \frac{1}{2}$$
minimum multiplicity $$ = 2 \times \frac{1}{2} + 1 = 2$$

Revolving electrons possess energy.

Angular momentum of electron in $$d$$ - orbital is
$$\eqalign{ & = \sqrt {l\left( {l + 1} \right)} \frac{h}{{2\pi }};\,{\text{for }}d{\text{ - orbital, }}l = 2 \cr & = \sqrt {2\left( {2 + 1} \right)} \,\,\,\,\,\left( {\because \,\,\rlap{--} h = \frac{h}{{2\pi }}} \right) \cr & h = \sqrt 6 \,\,h \cr} $$

$$\eqalign{ & {\lambda _p} = \frac{h}{{\sqrt {2eV{m_p}} }}; \cr & {\lambda _{H{e^{2 + }}}} = \frac{h}{{\sqrt {2 \times 2eV{m_{H{e^{2 + }}}}} }} \cr & = \frac{h}{{\sqrt {2 \times 2eV \times 4{m_p}} }} \cr & \therefore \,\,\frac{{{\lambda _p}}}{{{\lambda _{H{e^{2 + }}}}}} = 2\sqrt 2 \cr} $$

NOTE: Energy is emitted when electron falls from higher energy level to lower energy level and energy is absorbed when electron moves from lower level to higher level.
$$1s$$ is the lowest energy level of electron in an atom.
∴ An electron in $$1s$$  level of hydrogen can absorb energy but cannot emit energy.

Isobars are the atoms with same mass number ( i.e., sum of protons and neutrons ) but different atomic number ( i.e., no. of protons ).

In $$'X'$$  one electron can exchange energy with 5 orbitals while in $$'Y'$$  energy exchange is between 3 orbitals hence half-filled orbitals are more stable.