Important Questions for Class 11 Chapter 11 - The p-Block Elements

Important questions based on NCERT syllabus for Chapter 11 - The p-Block Elements:

Question-1: How can you explain higher stability of BCl3 as compared to TlCl3?

Solution: BCl3 is quite stable. Because there is absence of d- and f-electrons in boron three valence electrons (2s2 2px1) are there for bonding with chlorine atom. In Tl the valence s-electron (6s2) are experiencing maximum inert pair effect. Thus, only 6p1 electron is available for bonding. Therefore, BCl3 is stable but TlCl3 is comparatively unstable.

Question-2: What are electron deficient compounds? Are BCl3 and SiCl4 electron deficient species? Explain.

Solution: Electron deficient species are those in which the central atom in their molecule has the tendency to accept one or more electron pairs. They are also known as Lewis acid. BCl3 and SiCl4 both are electron deficient species.
Since, in BCl3, B atom has only six electrons. Therefore, it is an electron deficient compound.
In SiCl4 the central atom has 8 electrons but it can expand its covalency beyond 4 due to the presence of d-orbitals.
Thus, SiCl4 should also be considered as electron deficient species.

Question-3: Explain the difference in properties of diamond and graphite on the basis of their structures.

Solution: Since diamond exists as a three dimensional network solid, it is the hardest substance known with high density and high melting point.
Whereas in graphite, any two successive layers are held together by weak forces of attraction. This makes graphite soft.
In graphite, carbon atom is sp2 hybridized whereas in diamond, carbon atom is sp3 hybridized.
Unlike diamond, graphite is good conductor of heat and electricity.