p-Block Elements

In case of Lassaigne’s test of halogens, it is necessary to remove NaCN and Na 2 S from the sodium extract, if nitrogen and sulphur are present.

This is done by boiling the sodium extract with conc.

HNO 3 .

NaCN + HNO 3 $\to$ NaNO 3 + HCN $\uparrow$ Na 2 S + 2HNO 3 $\to$ 2NaNO 3 + H 2 S $\uparrow$

Pyrosilicate contains two units of SiO 4 4– joined along a corner containing oxygen atom.

The correct order of increasing bond angles is ClO

< Cl 2 O < ClO 2

In BF 3 , p-p overlap between B and F is maximum due to identical size and energy of p-orbitals, so electron deficiency in boron of BF 3 is neutralized partially to the maximum extent by back donation.

Also, the tendency to back donate decreases from F to I.

So the order will be: BBr 3 > BCl 3 > BF 3

Among the given molecules, only diborane(B 2 H 6 ) is electron deficient i.e., it does not complete octet.

Thus, it acts as a Lewis acid.

NH 3 and H 2 O being electron rich species behave as Lewis base.

The oxidation state can be calculated as : H 4 P 2 O 5 : +4 + 2x + 5(– 2) = 0 $\Rightarrow$ 2x – 6 = 0 $\Rightarrow$ x = +3 H 4 P 2 O 6 : +4 + 2x + 6(– 2) = 0 $\Rightarrow$ 2x – 8 = 0 $\Rightarrow$ x = +4 H 4 P 2 O 7 : +4 + 2x + 7(– 2) = 0 $\Rightarrow$ 2x – 10 = 0 $\Rightarrow$ 2x = 10 $\Rightarrow$ x = +5

Hydrolysis of (CH 3 ) 2 SiCl 2 followed by condensation polymerization.

The given elements belong to 13th group.

The elements mainly exhibit +3 and +1 oxidation states.

As we know, the stability of lower oxidation state i.e., +1 state, increases on moving down the group due to inert pair effect.

The, stability follows the order : Al < Ga < In < Tl

Standard reduction potential of halogens are positive and decrease from fluorine to iodine.

So, F 2 is the strongest oxidizing agent.

O 3 molecule can be represented as