Answer: Option B. -> Be−
:
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 stablethan $Be$ ($2$);
$B^{-}$ - ($2,2$) relatively stable;
$C^{-}$ - ($2,3$) relatively stable.

Answer: Option C. -> Vander Waals radius
:
C
As inert gasses do not combineto form molecules, they have VanderWaals radius.

Answer: Option B. -> 520 J, 495 J
:
B
Number of moles of Li = $\frac{7\times {10}^{-3}}{7}$ = ${10}^{-3}$
Number of moles of Na= $\frac{23\times {10}^{-3}}{23}$ = ${10}^{-3}$
The amount of energy required for 1 mole of Li = 520 kJ.
For${10}^{-3}$ of Li = 250 $\times$${10}^{-3}$ = 520 J
The amount of energy required for 1 mole of Na = 495 kJ.
For${10}^{-3}$ of Na = 495 $\times$${10}^{-3}$ = 495 J

Answer: Option B. -> Be > Mg > Ca
:
B
Down the group ionization enthalpy decreases with an increase in atomic size.
Along theperiod it increases due to an increase in effective nuclear charge and decrease in size.
K > Na > Li ---> same group, going upwards ---> so decreasing
Be > Mg > Ca ----> same groupbut going downwards, so it should increase.
B > C > N ---> same period, going left to right ---> so increasing
Ge > Si > C ----> same group, going upwards. ---> so decreasing

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 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 A. -> An element which has high electronegativity always has high electron gain enthalpy.
:
A
Elements with high electronegativity usually have high electron gain enthalpies, but this does not happen all the time.
If you take the example of nitrogen, it's electronegativity is one of the highest.
But if you look at its electron gain enthalpy, you actually need to supply energy to add the electron because nitrogen has a half-filled configuration which is stable.

Answer: Option B. -> CH4
:
B
As we go left to right in a period, as the non-metallic character increases and the acidic character of hydrides increases.

Answer: Option A. -> Energy is released in (I) and absorbed in (II)
:
A
(I)$K^{+}\left(g\right)\stackrel{+e^{-}}{⟶}K\left(g\right)⟶$ Negative electron gain enthalpy
(II)$K\left(g\right)\stackrel{+e^{-}}{⟶}{K}^{-}\left(g\right)⟶$ Positive electron gain enthalpy
In first process energy is released while in the second energy is absorbed.

Answer: Option A. -> s and p-block
:
A
s and p-block elements expect inert gases are called representative elements as their valence electrons represent their group number