Not more than two electrons can be present in same atomic orbital. This is Pauli's exclusion principle.

$$ns$$  orbital has $$\left( {n - 1} \right)$$  nodes.

According to Heisenberg’s uncertainty principle
$$\eqalign{ & \Delta x \times \Delta v = \frac{h}{{4\pi m}} \cr & {\text{Here, }}\Delta x = {\text{uncertainty in position}} \cr & \Delta v = {\text{uncertainty in velocity}} \cr & h = {\text{Planck's constant}}\,\,\,\left( {6.626 \times {{10}^{ - 27}}Js} \right) \cr & m = {\text{mass of electron }}\left( {9.1 \times {{10}^{ - 28}}kg} \right) \cr & {\text{Here,}}\,\,\Delta v = 0.001\% \,\,{\text{of}}\,\,3 \times {10^4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.001}}{{100}} \times 3 \times {10^4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.3\,cm/s \cr & \therefore \,\,\Delta x = \frac{h}{{4\pi m\,\Delta v}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{6.626 \times {{10}^{ - 27}}}}{{4 \times 3.14 \times 9.1 \times {{10}^{ - 28}} \times 0.3}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.93\,cm \cr} $$

Species having same number of electrons are isoelectronic calculating the number of electrons in each species given here, we get.
$$\eqalign{ & C{N^ - }\left( {6 + 7 + 1 = 14} \right);\,{N_2}\left( {7 + 7 = 14} \right); \cr & O_2^{2 - }\left( {8 + 8 + 2 = 18} \right);\,C_2^{2 - }\left( {6 + 6 + 2 = 14} \right); \cr & O_2^ - \left( {8 + 8 + 1 = 17} \right);\,N{O^ + }\left( {7 + 8 - 1 = 14} \right) \cr & CO\left( {6 + 8 = 14} \right);\,NO\left( {7 + 8 = 15} \right) \cr} $$
From the above calculation we find that all the species listed in choice (B) have 14 electrons each so it is the correct answer.

Two electrons occupying the same orbital has equal spin but the directions of their spin are opposite. Hence, spin quantum number, $$s,$$ ( represented $$ + \frac{1}{2}\,and - \frac{1}{2}$$   ) distinguishes them.

Radial probability function curve for $$1s$$ is (D).
$${\text{Here}}\,P\,{\text{is}}\,\,4\pi {r^2}{R^2}.$$

Energy of $$1\,mole$$  of photons,
$$\eqalign{ & E = {N_0} \times h\,\upsilon \cr & \,\,\,\,\,\, = \frac{{{N_0} \times h \times c}}{\lambda } \cr & \,\,\,\,\,\, = \frac{{6.023 \times {{10}^{23}} \times 6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{253.7 \times {{10}^{ - 9}}}} \cr} $$
Energy converted into $$KE = \left( {472.2 - 430.53} \right)kJ$$
$$\% $$ of energy converted into
$$\eqalign{ & KE = \frac{{\left( {472.2 - 430.53} \right)}}{{472.2}} \times 100 \cr & \,\,\,\,\,\,\,\,\,\, = 8.82\% \cr} $$

The expression for orbital angular momentum is
Angular momentum $$ = \sqrt {l\left( {l + 1} \right)} \left( {\frac{h}{{2\pi }}} \right)$$
$$\eqalign{ & {\text{For }}d{\text{ orbital,}}\,l = 2. \cr & {\text{Hence,}}\,L = \sqrt {2\left( {2 + 1} \right)} \left( {\frac{h}{{2\pi }}} \right) = \sqrt 6 \left( {\frac{h}{{2\pi }}} \right) \cr} $$

TIPS/Formulae : The element having highest tendency to accept the electron will be most electronegative element.
Configuration $$n{s^2},\,n{p^5}$$  means it requires only one electron to attain nearest noble gas configuration. So, it will be most electronegative element among given choices.

$$\eqalign{ & E = \frac{{hc}}{\lambda }\,\,\,{\text{or}}\,\,\,\lambda = \frac{{hc}}{E} \cr & \lambda = \frac{{6.626 \times {{10}^{ - 34}}J\,\,s \times 3 \times {{10}^8}m\,\,{s^{ - 1}}}}{{3.03 \times {{10}^{ - 19}}J}} \cr & \,\,\,\,\, = 6.56 \times {10^{ - 7}}m\,\,\,{\text{or}}\,\,\,656\,nm \cr} $$