Question

Arrange the atoms in Exercise 103 in order of increasing first ionization energy.

Short answer

(a) Te< Se< S (d) Rb< Na< Be

(b) K< Ni< Br (e) Sr < Se< Ne

(c) Ba< Si< F (f) Fe< P < O

Step-by-step solution

Determine the general trend of the ionization energy in the periodic table

1st ionization energy increases on moving left to right in a period and decreases down the group.

Check that the elements in (a) belong to a group

Te, S, and Se belong to the chalcogen group (Group 16). Ionization energy decreases down the group.

So, the correct order of increasing ionization energy is Te< Se< S.

For elements in (b), check that they belong to the same period.

K (19) = [Ar]$4{\mathrm{s}}^1$, Ni (28) =[Ar]$3{\mathrm{d}}^{8}4{\mathrm{s}}^2$, Br (35) =[Ar]$4{\mathrm{s}}^{2}3{\mathrm{d}}^{10}4{\mathrm{p}}^5$

All of these belong to$4^{\mathrm{th}}$ period. Ionization energy increases in a period.

So, the correct order is K< Ni< Br.

For the rest of the elements, check their electronic configuration and position in the periodic table.

(c) F (9) = [He] $2{\mathrm{s}}^{2}2{\mathrm{p}}^5$= Group 17 & period 2

Si (14) = [Ne]$3{\mathrm{s}}^{2}3{\mathrm{p}}^2$= Group 14 & period 3

Ba (56) = [ Xe] $6{\mathrm{s}}^2$= Group 2 & period 6

The correct order of increasing ionization energy will be Ba< Si< F.

(d) Be (4) = [He]$2{\mathrm{s}}^2$ = Group 2 and period 2

Na (11) = [Ne] $3{\mathrm{s}}^1$= Group 1 and period 3

Rb (37) = [Kr]$5{\mathrm{s}}^1$ = Group 1 and periods 5

Ionization energy increases in a period, so group 2 members will have larger ionization energy than group 1. Also, ionization energy decreases down the group.

So, the correct order is Rb < Na <Be.

(e) Ne (10) = [He]$2{\mathrm{s}}^{2}2{\mathrm{p}}^6$ = Group 18 and period 2

Se (34) = [Ar] $3{\mathrm{d}}^{10}4{\mathrm{s}}^{2}4{\mathrm{p}}^4$= Group 16 and period 4

Sr (38) = [Kr] $5{\mathrm{s}}^2$= Group 2 and periods 5

Ne belongs to the last group, so its ionization energy will be the highest. Sr is larger than Se, so its ionization energy will be less.

So, the correct order will be Sr< Se< Ne.

(f) O (8) = [He]$2{\mathrm{s}}^{2}2{\mathrm{p}}^4$= Group 16 and period 2

P (15) = [Ne]$3{\mathrm{s}}^{2}3{\mathrm{p}}^3$= Group 15 and period 3

Fe (26) = [Ar]$3{\mathrm{d}}^{6}4{\mathrm{s}}^2$= Group 8 and period 4

$4^{\mathrm{th}}$period Fe will have a larger size than P & O, so its ionization energy will be smaller. Similarly, O will have larger first ionization energy than P.

So, the correct order will be Fe < P < O.