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Chapter 11 The p Block Elements Solutions

Question - 1 : -
Discuss the pattern of variation in the oxidation states of
(i) B to Tl (ii) C to Pb.

Answer - 1 : -

(i) B to Tl
Common oxidation states are +1 and +3. The stability of +3 oxidation state decreases from B to Tl. +1 oxidation state increases from B to Tl.
(ii) C to Pb
The common oxidation states are +4 and +2. Stability of +4 oxidation state decreases from C to Pb.
Details can be seen from the text part.

Question - 2 : - How can you explain higher stability of BCl3 as compared to TlCl3?

Answer - 2 : -

BCl3 is quite stable. Because there is absence ofd- and f-electrons in boron three valence electrons (2s2 2px1) are there for bonding with chlorine atom. In Tlthe 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 - 3 : -
Why does BF3 behave as Lewis acid ?

Answer - 3 : -

BF3 behaves as a Lewis acid because centralboron atom has only six electrons (three pairs) after sharing with theelectrons of the F atoms. It is an electron deficient compound and, therefore,behaves as a Lewis acid.
Boron Halides. The halides of boron are covalent in naturebecause the central boron atom, as stated earlier, is very small in size andcannot part with the valence electrons to give a trivalent B3+ Thus, BF3, BCl3, BBr3 and BI3 are all covalent in nature.
In these halides, the central B atom is sp2 hybridised with a vacant 2p orbital (1 .v2 2s1 2pxl2pyi2pz°). The hybridised orbitals are directedtowards the three corners of an equilateral triangle and are involved incovalent bond formation with the half filled p-orbitals of the halogen atoms (ns2p5).

The empty 2p orbital lies perpendicular to thehybridised orbitals. Since it is empty, it can accept an electron pair fromelectron donor species (Lewis bases) resulting in co-ordinate or dative bond formation. Withthree shared pairs of electrons on the central boron atom, boron halidesare Lewis acids i.e., electron deficient molecules and take part in Lewis acid– base reactions. In the compounds thus formed, boron atom undergoes a changein state of hybridisation from sp2 to sp3. These are tetrahedral in nature.

Question - 4 : - Consider the compounds BCl3 and CCl4, How will they behave towards water ?

Answer - 4 : - In BCl3 (Batom is sp2 hybridised), the B atomhas incomplete octet and unhybridised 2p-orbital which can take up electronpair from the H2Omolecule to form addition product.

In this way, a Cl atom has been replaced by OH groupon reacting with water. Similarly, the other two Cl atoms will also be replacedby the OH groups as follows :
This shows that boron trichloride has undergone hydrolysis. Butthis is not possible with carbon tetrachloride (CCl4). The carbon atom has a complete octet and there is noscope of forming addition product with H2O molecules. As a result, carbon tetrachloride does notget hydrolysed. When added to water, it even does not mix and forms a separateoily layer.

Question - 5 : -
Is boric acid a proton acid ? Explain.

Answer - 5 : -

Boric acid is not a proton acid. It is a Lewis acid and acceptselectron pair from hydroxyl ion of H2Omolecule.
B(OH)3 + 2HOH→[B(OH)4]– + H3O+

 

Question - 6 : -
Explain what happens when boric acid is heated ?

Answer - 6 : -

Preparation of Boric Acid. Boric acid can be prepared by the following methods.
(a) From borax. A hot and concentrated solution containing borax is boiled with hydrochloric or sulphuric acid. The solution upon concentration and cooling gives crystals of boric acid.
(b) From Colemanite. Sulphur dioxide gas is passed through the hot concentrated solution of mineral colemanite made in water. The solution upon concentration followed by cooling gives crystals of boric acid. Calcium bisulphite remains in solution as it is highly soluble in water.

(c) From boron compounds by hydrolysis. Certain boron compounds upon boiling withwater (upon hydrolysis) give boric acid.
BCl3 + 3H2O→ H3BO3 + 3HCl
BN + 3H2O → H3BO3 + NH3

Question - 7 : -

Describe the shapes of BF3 and BH4–. Assign thehybridisation of boron in these species.

Answer - 7 : - In BF3, boron is SP2 hybridized.
∴ shape of BF3 = planar.
In [BH4]–, boron is sp3 hybridized, thus the shape istetrahedral.

Question - 8 : - Write reactions to justify amphoteric nature of aluminium.

Answer - 8 : -

Aluminium reacts with acidas well as base. This shows amphoteric nature of aluminium.
2Al(s) + 6HCl(dil.) ——> 2AlCl3(aq) + 3H2(g)
2Al(s) + 2NaOH(aq) + 6H2O(l) ———-> 2Na+ [Al(OH)4]–(aq) + 3H2(g)

Question - 9 : - What are electron deficient compounds? Are BCl3 and SiCl4 electron deficientspecies? Explain.

Answer - 9 : -

Electron deficient species are those inwhich the central atom in their molecule has the tendency to accept one or moreelectron 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 isan electron deficient compound.
In SiCl4 the central atom has 8 electrons but it canexpand its covalency beyond 4 due to the presence of d-orbitals.
Thus, SiCl4 should also be considered aselectron-deficient species.

Question - 10 : -

Write the resonance structure of CO32- andHCO3– .

Answer - 10 : -

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