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Chapter 7- The p-Block Elements Interview Questions Answers

Question 1 :

Why are pentahalides morecovalent than trihalides?

Answer 1 :

In pentahalides, the oxidationstate is +5 and in trihalides, the oxidation state is +3. Since the metal ionwith a high charge has more polarizing power, pentahalides are more covalentthan trihalides.

Question 2 :

Why is BiH3 the strongest reducing agent amongst allthe hydrides of Group 15elements?

Answer 2 :

As we move down a group, theatomic size increases and the stability of the hydrides of group 15 elementsdecreases. Since the stability of hydrides decreases on moving from NH3 to BiH3, the reducing character of the hydridesincreases on moving from NH3 to BiH3.

Question 3 :

Why is N2 less reactive at room temperature?

Answer 3 :

The two N atoms in N2 are bonded to each other by very strongtriple covalent bonds. The bond dissociation energy of this bond is very high.As a result, N2 isless reactive at room temperature.

Question 4 :

Mention the conditions requiredto maximise the yield of ammonia.

Answer 4 :

Ammonia is prepared using theHaber’s process. The yield of ammonia can be maximized under the followingconditions:

(i) High pressure ( 200atm)

(ii) A temperature of 700 K

(iii) Use of a catalyst such as iron oxide mixed with smallamounts of K2Oand Al2O3

Question 5 :

How does ammonia react with asolution of Cu2+?

Answer 5 :

NH3 acts as a Lewis base. It donates itselectron pair and forms a linkage with metal ion.

Question 6 :

What is the covalence of nitrogenin N2O5?

Answer 6 :

From the structure of N2O5, it is evident that the covalence of nitrogenis 4.

Question 7 : Bond angle inis higher than that in PH3. Why?

Answer 7 : In PH3, P is sp3 hybridized.Three orbitals are involved in bonding with three hydrogen atoms and the fourthone contains a lone pair. As lone pair-bond pair repulsion is stronger thanbond pair-bond pair repulsion, the tetrahedral shape associated with sp3 bonding is changed to pyramidal. PH3 combines with a proton to form  in which the lone pair is absent. Due to theabsence of lone pair in, there is no lone pair-bond pair repulsion. Hence,the bond angle in is higher than the bond angle inPH3.


Question 8 :

What happens when whitephosphorus is heated with concentrated NaOH solutionin an inert atmosphere of CO2?

Answer 8 :

White phosphorous dissolves inboiling NaOH solution (in a CO2 atmosphere) to give phosphine, PH3.

Question 9 :

What happens when PCl5 is heated?

Answer 9 :

All the bonds that are present inPCl5 arenot similar. It has three equatorial and two axial bonds. The equatorial bondsare stronger than the axial ones. Therefore, when PCl5 is heated strongly, it decomposes to formPCl3.

Question 10 :

Write a balanced equation for thehydrolytic reaction of PCl5 inheavy water.

Answer 10 :



Chapter 7- The p-Block Elements Contributors


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