Periodic Table and Periodicity

In general, bond enthalpy or bond dissociation energy of halogen molecules $\left(X_2\right)$ decreases down the group from $\mathrm{Cl}_2$ to $\mathrm{I}_2$.

$\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{F}_2>\mathrm{I}_2$

The decrease in bond enthalpy from $\mathrm{Cl}_2$ to $\mathrm{I}_2$ is due to increase in atomic size that increases the distance between two atoms. Hence, it becomes easier to break them apart. The exceptionally low bond dissociation energy of fluorine molecule is due to very small size of fluorine atom. This brings the non - bonding electrons on fluorine nearer to each other, resulting in a much greater lone pair-lone pair repulsion, which weakens the covalent bond and lowers its dissociation energy. Hence, option (d) is correct.

Since, $\mathrm{Zn}$ and $\mathrm{Cd}$ both have completely filled outermost shells, i.e. have no unpaired electrons. Thus, they can not form covalent and metallic bonds strongly and show boiling point less than $500^{\circ} \mathrm{C}\left[\mathrm{Zn}\left(146^{\circ} \mathrm{C}\right)\right.$ and $\left.\mathrm{Cd}\left(321^{\circ} \mathrm{C}\right)\right]$

Hence, option (b) is the correct.

Due to lanthanoid contraction poor screening offered by $[n-1] d$-electrons, size of first series of $d$-block elements initially decreases and finally increases in last.

Thus, (Ni) has least while $\mathrm{Zn}$ has maximum radius among the given options.

Hence, option (d) is the correct.

Electronegativity decreases down the group and increases on moving left to right in a period. Thus, Li is more electronegative than Na and F is more electronegative than Cl and Br. Br is less electronegative than Cl because of its larger size. Also, lesser be the electronegative of metal element and more is the electronegativity of halogen atom, more be the electronegativity difference of a given species.

Thus, Na and F show the maximum electronegative difference and option(b) is the correct.

The correct increasing order of basic strength is

$\mathrm{NH}_3>\mathrm{PH}_3>\mathrm{AsH}_3>\mathrm{SbH}_3$

$\mathrm{NH}_3$ is the most basic because of its small size, the electron density of electron pair is concentrated over small region. As the size increases, the electron density gets diffused over a large surface area. Hence, the ability to donate the electron pair (basicity) decreases. The other three statements given are correct.

$\mathrm{C}^{4-}, \mathrm{N}^{3-}$ and $\mathrm{O}^{2-}$ are isoelectronic species.

The ionic radius of isoelectronic species decreases with increase in nuclear charge.

Hence, the order of ionic radius is

species : $\mathrm{C}^{4-}>\mathrm{N}^{3-}>\mathrm{O}^{2-}$

Ionic radii $(\mathop A\limits^o): 2.60>1.71>1.40$

Assertion and reason both are correct and reason is the correct explanation of the assertion. Electron gain enthalpy becomes less negative as the size of an atom increases down the group. This is because within a group screening effect increases on going downward and the added electron would be farther away from the nucleus.