Answer: Option A. -> $\text{N < C < S < P}$
:
A
In a group, the size of an atom increases as one proceeds from top to bottom.
This is due to the successive addition of shells.
In a period, the size of an atom decreases from left to right.
This is because the effective nuclear charge increases from left to right in the same period, thereby bringing the outermost shell closer to the nucleus.
So, the correct order of atomic size are as follows:
N < C < S < P

Answer: Option B. -> $B{e}^{-}$
:
B

Here we are going to compare these with each other and find out which is the least stable. But these species might not exist due to inherent instability. 
Electronic configuration of $L{i}^{-}$ - ($2$) fully-filled, so relatively more stable. 

$B{e}^{-}$ - ($2,1$) less stable than $Be$ ($2$);

$B^{-}$ -  ($2,2$) relatively stable;
$C^{-}$ -  ($2,3$)  relatively stable.

Answer: Option C. -> Vander Waals radius
:
C

As inert gasses do not combine to form molecules, they have Vander Waals radius.

Answer: Option B. -> 69% ${Mg}^{+}$ + 31 % ${Mg}^{+}$
:
B

Energy absorbed for converting $M{g}_{\left(g\right)}$ $\to$ ${M{g}_{\left(g\right)}}^{+}$ = 750 kJ

Energy left unconsumed = 1200 = 750 = 450 kJ.

This energy will be required to convert ${M{g}_{\left(g\right)}}^{+}$ to ${M{g}_{\left(g\right)}}^{+2}$

Thus % of ${M{g}_{\left(g\right)}}^{+2}$ = $\frac{450}{1450}$ $\times$ 100 = 31%

% of ${M{g}_{\left(g\right)}}^{+}$ = 100 - 31 = 69%

Answer: Option D. -> s - electron
:
D

For a given shell, the penetration power of a subshell decreases from s to f, s > p > d > f.
Higher the penetration power, higher is the energy required to remove an electron.

Answer: Option D. -> (i), (ii), (iii)
:
D

Down the group ionization potential (ionization enthalpy) decreases with increase in size.
Across a period, ionization potential decreases and then increases at the noble gasses. 

Elements of group 2(IIA) have I.E more than that of group 13(IIA) as they have completely filled 's' orbital. Be > B

Elements of group 15(VA) have more than that of group 16(VIA) as they have half filled 'P' subshell. N > O

Inert gas elements have the highest I.P in a given period due to a stable electronic configuration.

Answer: Option B. -> Si
:
B

Among the Ionization Potentials given there is a huge jump from $I{P}_4$ to $I{P}_5$, which shows that element attains stable inert gas configuration after losing 4 electrons.
It should be an element from ${14}^{th}$ group (IVA). Among the given elements, the answer will be Si. 

Answer: Option C. -> 38.3, 35.1
:
C

Second ionization potential is the energy required to remove an electron from a uni-positively charged ion.
Let's look at the electronic configurations. 

$O$ $\to$ $2,6$  
$O^{+}$ $\to$ $2,5$

$F$ $\to$ $2,7$ 
$F^{+}$  $\to$ $2,6$

 $I.P_1$
$O$ $\to$ $O^{+}$   $F$ $\to$ $F^{+}$
 $I.P_2$

$O$ $\to$  $O^{2+}$  $F^{+}$ $\to$$F^{2+}$

 

$I.P_1$  of   $F>O$   as fluorine has smaller atomic size as compared to O.

$I.P_2$  of    $O>F$ as $O^{+}$ as it is harder to ionise the O+ compared to F+.

Answer: Option C. -> Type of bonding in the crystal lattice
:
C
Electron is removed from an isolated neutral gaseous atom and so type of bonding in the crystalline lattice is not associated with ionization enthalpy.

Answer: Option D. -> Na and Mg
:
D
The ionization energy increases normally from sodium to magnesium as Mg has a higher effective nuclear charge as compared to Na. 
In the case of Be and B, and N and O, Be is more stable because of its electronic configuration.