(1)

From left to right across a period, number of protons as well as number of electrons increase. Increase in number of protons results in stronger attraction of outermost electron which contributes towards increase of ionization energy. Increase in number of electrons increases inter-electronic repulsion, which contributes towards decrease of ionization energy. Effect of increase in number of protons outweighs the effect of increase in number of electrons. Thus ionization energy increases from left to right across a period. (If nothing is mentioned then trend is taken as left to right across a period).

(4)

${\mathrm{IE}}_1 \mathrm{of} \mathrm{Carbon}-1086{\mathrm{kJmol}}^{-1}$

${\mathrm{IE}}_1 \mathrm{of} \mathrm{Al}-577{\mathrm{kJmol}}^{-1}$

${\mathrm{IE}}_1 \mathrm{of} \mathrm{Si}-786{\mathrm{kJmol}}^{-1}$

${\mathrm{IE}}_1 \mathrm{of} \mathrm{N}-1402{\mathrm{kJmol}}^{-1}$

(1)

Ionization enthalpy increases from left to right across a period. Thus in second period element expected order of ionization enthalpy is:

$\mathrm{Li}\left(2{\mathrm{s}}^1\right)<\mathrm{Be}\left(2{\mathrm{s}}^2\right)<\mathrm{B}\left(2{\mathrm{s}}^{2}2{\mathrm{p}}^1\right)<\mathrm{C}\left(2{\mathrm{s}}^{2}2{\mathrm{p}}^2\right)<\mathrm{N}\left(2{\mathrm{s}}^{2}2{\mathrm{p}}^3\right)<\mathrm{O}\left(2{\mathrm{s}}^{2}2{\mathrm{p}}^4\right)<\mathrm{F}\left(2{\mathrm{s}}^{2}2{\mathrm{p}}^5\right)<\mathrm{Ne}\left(2{\mathrm{s}}^{2}2{\mathrm{p}}^6\right)$

But removal of electron from stable fully filled subshell of Be and half flled subshell of N is difficult. Thus Be and N have ionization energy more than expected value. Hence actual order of ionization energy is:

$\mathrm{Li}\left(2{\mathrm{s}}^1\right)<\mathrm{B}\left(2{\mathrm{s}}^{2}2{\mathrm{p}}^1\right)<\mathrm{Be}\left(2{\mathrm{s}}^2\right)<\mathrm{C}\left(2{\mathrm{s}}^{2}2{\mathrm{p}}^2\right)<\mathrm{O}\left(2{\mathrm{s}}^{2}2{\mathrm{p}}^4\right)<\mathrm{N}\left(2{\mathrm{s}}^{2}2{\mathrm{p}}^3\right)<\mathrm{F}\left(2{\mathrm{s}}^{2}2{\mathrm{p}}^5\right)<\mathrm{Ne}\left(2{\mathrm{s}}^{2}2{\mathrm{p}}^6\right)$

                  (E)                    (C)                                       (A)                    (B)                    (D)

(2)

Ionization enthalpy decreases down the group. In group 13, $\mathrm{Ga}$ and $\mathrm{Tl}$ have more ionization enthalpy than expected value because of poor shielding of $d$ and $f$ electrons respectively. Actual order of first ionization enthalpy for group 13 elements is: $\mathrm{B}>\mathrm{Tl}>\mathrm{Ga}>\mathrm{Al}>\mathrm{In}$

(2)

There is huge jump on moving from $I{E}_3$ to $I{E}_4$. This means removing fourth electron is very difficult. In other words, removing three electrons are relatively easy. i.e. the element has 3 valence electrons. So it is a group 13 element.

(2)

More negative is the electron gain enthalpy value, more is non metallic character in the element and hence more is electronegativity. More is electronegativity of the element (A, B, C or D), higher will be the electronegativity difference between the element and ‘E’. More is electronegativity difference, more is ionic character. As electron gain enthalpy becomes less negative in the order: B > A > C > D, ionic character decreases in the order: EB > EA > EC > ED.

(4)

Ionization energy of ${\mathrm{Mn}}^{2+}\left({}_{18}[\mathrm{Ar}\right]3d^5)$ is more than that of ${\mathrm{Fe}}^{2+}\left({}_{18}[\mathrm{Ar}\right]3d^6)$, as ${\mathrm{Mn}}^{2+}$ has stable half-filled configuration.

(1)

Statement I: Ionization energy increases from left to right in a periodic table, hence group 14 elements have higher ionization energy than group 13 elements in respective periods.

Statement II: Group 14 elements have higher boiling points than group 13 elements of respective periods.

(4)

Order of first ionization enthalpy of group 13 elements is: B > Tl > Ga > Al > In.

(3)

${}_{87}\mathrm{Fr}$ has lowest energy among given elements.