Consider the isoelectronic species, ${\mathrm{Na}}^{+},<mspace\; width="1em"></mspace>{\mathrm{Mg}}^{2+},<mspace\; width="1em"></mspace>{\mathrm{F}}^{-},<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{O}}^{2-}$. The correct order of increasing length of their radii is :

1. ${\mathrm{F}}^{-}<{\mathrm{O}}^{2-}<{\mathrm{Mg}}^{2+}<{\mathrm{Na}}^{+}$

2. ${\mathrm{Mg}}^{2+}<{\mathrm{Na}}^{+}<{\mathrm{F}}^{-}<{\mathrm{O}}^{2-}$

3. ${\mathrm{O}}^{2-}<{\mathrm{F}}^{-}<{\mathrm{Na}}^{+}<{\mathrm{Mg}}^{2+}$

4. ${\mathrm{O}}^{2-}<{\mathrm{F}}^{-}<{\mathrm{Mg}}^{2+}<{\mathrm{Na}}^{+}$

The order of screening effect of electrons of s, p, d and f orbitals of a given shell of an atom on electrons in its outer shell is :

1. s > p > d > f

2. f > d > p > s

3. p < d < f < s

4. f > p > s > d

The first ionisation enthalpies of Na, Mg, Al, and Si are in the order of:

1. Na < Al < Mg < Si

2. Na > Mg > Al > Si

3. Na < Mg < Al < Si

4. Na > Mg > Al < Si

Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy) is:

The period number in the long form of the periodic table is equal to : 

The elements in which electrons are progressively filled in 4f-orbitals are called:

The correct order of size of the given species is: 

The formation of oxide ion O^2-(g), from oxygen atom, requires first an exothermic and then an endothermic step as shown below

$\mathrm{O}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}\to {\mathrm{O}}^{-}\left(\mathrm{g}\right);<mspace\; width="1em"></mspace>∆{\mathrm{H}}^{⊝}=-141<mspace\; width="1em"></mspace>\mathrm{kJ}<mspace\; width="1em"></mspace>{\mathrm{mol}}^{-1}$
${\mathrm{O}}^{-}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}\to {\mathrm{O}}^2\left(\mathrm{g}\right);<mspace\; width="1em"></mspace>∆{\mathrm{H}}^{⊝}=+780<mspace\; width="1em"></mspace>\mathrm{kJ}<mspace\; width="1em"></mspace>{\mathrm{mol}}^{-1}$

Thus, the process of formation of O^2- in the gas phase is unfavourable even though O^2- is isoelectronic with neon. It is due to the fact that:

The outermost electronic configuration of the last element of the p-block in the 6th period is represented by:

1. $7{\mathrm{s}}^{2}7{\mathrm{p}}^6$

2. $5{\mathrm{f}}^{14}6{\mathrm{d}}^{10}7{\mathrm{s}}^{2}7{\mathrm{p}}^0$

3. $4{\mathrm{f}}^{14}5{\mathrm{d}}^{10}6{\mathrm{s}}^{2}6{\mathrm{p}}^6$

4. $4{\mathrm{f}}^{14}5{\mathrm{d}}^{10}6{\mathrm{s}}^{2}6{\mathrm{p}}^4$