(a) Complete the following chemical equations :

(i) ${P}_{4}+NaOH+{H}_{2}O\to$

(ii) $Xe{F}_{4}+{O}_{2}{F}_{2}\to$

(b) How would you account for the following situations ?

(i) The acidic strength of these compounds increases in the following order:

$P{H}_{3}>{H}_{2}S>HCl$

(ii) The oxidising power of oxoacids of chlorine follows the order :

$HCI{O}_{4}>HCI{O}_{3}>HCI{O}_{2}>HCIO$

(iii) In vapour state sulphur exhibits paramagnetic behaviour.

Concept Videos :-

#2 | 15th Group: Properties: II
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Concept Questions :-

group 15 ,preparation and properties

(a) (i)

(ii)

(b) (i) As the electronegativity increases in the same period from left to right so their electronegativity are in the increasing order, P<S<Cl.

In the same way the acid strength is also in the increasing order

i.e. $P{H}_{3}<{H}_{2}S

(ii) $HCI{O}_{4}>HCI{O}_{3}>HCl{O}_{2}>HClO$

o.s. = +4     +3         +2          +1

Acidic strength of oxoacids of the same halogen increases with increase in oxidation number of the halogen because of the relative stability of the anions left after removal of proton. Thus as the number of oxygen atoms in the anion increases, the dispersal of the negative charge through $p\pi -p\pi$ back bonding also increases and hence stability and acidic strength increases.

(iii) In vapour state sulphur partly exists as ${S}_{2}$ molecule which has two unpaired electrons in the antibonding $\pi$ orbitals and hence exhibits paramagnetism.

Difficulty Level:

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