p-Block Elements

The $H-X$ bond strength decreases from $HF$ to $HI.$ i.e. $HF > HCl > HBr > HI.$ Thus $HF$ is most stable while $HI$ is least stable. This is evident from their decompositions reaction: $HF$ and $HCl$ are stable upto $1473K,HBr$ decreases slightly and $HI$ dissociates considerably $713K.$ The decreasing stability of the hydrogen halide is also reflected in the values of dissociation energy of the $H-X$ bond

$\matrix{ {H - F} & {H - Cl} & {H - Br} & {H - I} \cr {135kcal\,mo{l^{ - 1}}} & {103kcal\,mo{l^{ - 1}}} & {87kcal\,mo{l^{ - 1}}} & {71kcal\,mo{l^{ - 1}}} \cr } $

The solution of aluminium chloride in water is acidic due to hydrolysis.

$AlC{l_3} + 3{H_2}O\buildrel \, \over \longrightarrow Al{\left( {OH} \right)_3} + 3HCl.$

On heating it till dryness $Al{\left( {OH} \right)_3}$ is converted into $A{l_2}{O_3}$

$Al{\left( {OH} \right)_3}\buildrel \Delta \over \longrightarrow A{l_2}{O_3} + 3{H_2}O$

$A{l_2}C{l_6} + 12{H_2}O\,\,\rightleftharpoons\,\,2{\left[ {Al{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }} + 6C{l^ - }$

The fluorine has low dissociation energy of $F$-$F$ bond and reaction of atomic fluorine is exothermic in nature.

The maximum valency of beryllium is $+2$ white that of aluminium is $+3.$

Grey tin $\,\,\,$ $\rightleftharpoons$ $\,\,\,$ white tin

Grey tin is brittle and crumbles down to powder in very cold climate

The conversion of grey tin to white tin is accompanied by increase in volume. This is known as tin plaque or tin disease.

Helium is heavier than hydrogen although it is non-inflammable

$4HCl + \mathop {{O_2}}\limits_{air} \to \mathop {2C{l_2} + 2{H_2}O}\limits_{cloud\,\,\,of\,\,\,white\,\,\,fumes} $

Glass is a translucent or transparent amorphous super-cooled solid solution or we can say super cooled liquid of silicates and borats having a general formula ${R_2}O.\,MO.\,6Si{O_2}.$ where $R=Na$ or $K$ and $M$ $=Ca, Ba, Zn$ or $Pb.$

In graphite, carbon is $s{p^2}$ hybridized. Each carbon is thus linked to three other carbon atoms forming hexagonal rings. Since only three electrons of each carbon are used in making hexagonal ring, fourth electron of each carbon is free to move. This makes graphite a good conductors of heat and electricity. Further graphite has a two dimensional sheet like structure. These various sheets are held together by weak van der Waal's force of attraction. due to these weak forces of attraction, one layer can slip over the other. This makes graphite soft and a good lubricating agent.

When phosphine gas (PH₃) is mixed with chlorine gas (Cl₂), the reaction between them depends on the conditions and the proportions of the reactants. Let's consider the options :

1. Option A : PCl₃ and HCl are formed and the mixture warms up.

   
   
   • When phosphine gas reacts with chlorine gas, one of the possible products is phosphorus trichloride (PCl₃) along with hydrogen chloride (HCl). This reaction is exothermic, meaning it releases heat and causes the mixture to warm up.



2. Option B : PCl₅ and HCl are formed and the mixture cools down.

   
   
   • Phosphorus pentachloride (PCl₅) can be formed if an excess of chlorine is used, but this reaction is also exothermic. The formation of PCl₅ is less likely with PH₃ as it requires a higher proportion of chlorine and controlled conditions.



3. Option C : PH₃·Cl₂ is formed with warming up.

   
   
   • There is no known stable compound like PH₃·Cl₂ formed directly from the reaction of phosphine and chlorine gas. This option does not represent a typical chemical reaction product.



4. Option D : The mixture only cools down.

   
   
   • Chemical reactions between phosphine and chlorine are exothermic, so this statement is incorrect.

Based on this analysis, the most likely and correct reaction is described in Option A : PCl₃ and HCl are formed, and the mixture warms up. This reaction is feasible and aligns with the known chemistry of phosphine and chlorine gases.

Among the given compounds, the $N{H_3}$ is most basic. Hence has highest proton affinity.

In the formation of $Xe{F_2},\,s{p^3}d\,\,$ hybridisation occurs which gives the molecule a trigonal bipyramidal structure.



In the formation of $Xe{F_4},\,s{p^3}{d^2}$ hybridization occurs which gives the molecule an octahedral structure.



In the formation of $Xe{F_6},\,s{p^3}{d^3}$ hybridization occurs which gives the molecule a pentagonal bipyramidal structure.

${}_7N = 1{s^2}2{s^2}\,2{p^3};\,$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,{}_{15}P = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^3}$

NOTE : In phosphorous the $3d$-orbitals are available. Hence phosphorous can from pentahalides also but nitrogen can not form pentahalide due to absence of $d$-orbitals.

When H₂S is passed through Hg₂S we get a mixture of mercurous sulphide and mercury(HgS + Hg₂S).

$CaOC{l_2} - $or it can also be written as $Ca\mathop {\left( {OCl} \right)}\limits_{{x_1}} \mathop {Cl}\limits_{{x_2}} $

hence oxidation no of $Cl$ in $OC{l^ - }$ is

$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2 + {x_2} = - 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x^2} = 2 - 1 = + 1 \cr} $

now oxidation no. of another $Cl$ is $-1$ as it is present as $C{l^ - }.$