Consider the following orders –
(1) $F_{2} > C l_{2} > B r_{2} > I_{2} :$ boiling point
(2) $F_{2} > C l_{2} > B r_{2} > I_{2} :$ oxidizing nature
(3) $F_{2} > C l_{2} > B r_{2} > I_{2} :$ EN
(4) $F_{2} > C l_{2} > B r_{2} > I_{2} :$ BDE
(5) $F_{2} > C l_{2} > B r_{2} > I_{2} :$ EA
(6) $F_{2} > C l_{2} > B r_{2} > I_{2} :$ Reactivity
(7) $H O C l > H C l O_{2} > H C l O_{3} > H C l O_{4} :$ Acidic nature
(8) $H O C l > H C l O_{2} > H C l O_{3} > H C l O_{4} :$ Oxidizing nature
Then calculate $(x^{2} + y^{2})$ when $x$ is correct order and $y$ is incorrect order.

34.0

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Solution

(1) The boiling point increases down the group because of the van der waals forces.
$F_{2} < C l_{2} < B r_{2} < I_{2}$
This statement is incorrect.

(2) Members of the halogen family act as strong oxidizing agents on account of their electron accepting tendency both in the molecular as well as well as atomic form.
$F_{2} > C l_{2} > B r_{2} > I_{2}$
This statement is correct.

(3) Electronegativity decreases down the group as the no. of valence electrons in an atom increase down the group.
$F_{2} > C l_{2} > B r_{2} > I_{2}$
This statement is correct.

(4) Bond dissociation energy of halogen family decreases down the group as the sie of atom increases.
$C l_{2} > B r_{2} > F_{2} > I_{2}$
Fluorine is an exception because of inter electronic repulsion is present in small atom fluorine.
This statement is incorrect.

(5) Since the atomic size increases down the group, electron affinity generally decreases. An electron will not be as attached to the nucleus, resulting in a low electron affinity.
$C l_{2} > F_{2} > B r_{2} > I_{2}$
This statement is incorrect.

(6) The reactivity of the halogens decreases down the group.
$F_{2} > C l_{2} > B r_{2} > I_{2}$
This statement is correct.

(7) Oxygen is more electronegative than chlorine with an increases in the no. of O atoms attached to Cl, more electrons are pulled away from O—H bond and weaker becomes the O—H bond.
$H C l O_{4} > H C l O_{3} > H C l O_{2} > H C l O$
This statement is incorrect.

(8) Higher the oxidation state of chlorine stronger will be the oxidising power.
$H C l O > H C l O_{2} > H C l O_{3} > H C l O_{4}$
This statement is incorrect.
Here $x = 3, y = 5$
The value of $(x^{2} + y^{2}) = (3^{2} + 5^{2}) = 34$

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