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

The correct order of size of the given species is: 

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

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

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

Boron has lesser ionization enthalpy than Beryllium, because:

Match the element in Column I with that in Column II.

The formula for the oxide formed by Eka-aluminum is:

1. ${\mathrm{EO}}_4$

2. ${\mathrm{E}}_2{\mathrm{O}}_3$

3. ${\mathrm{E}}_3{\mathrm{O}}_2$

4. EO

The increasing order of non-metallic character of the elements N, P, O, and S is: 

1. P < S < N < O

2. O < S < N < P 

3. O < N < S < P 

4. N < S < O < P