Chemistry Class 11 NCERT Solutions: Chapter 11 the p Block Elements Part 1

Q: 1. Discuss the pattern of variation in the oxidation states of

(i) Equation

(ii) C to Pb

Answer:

The electric configuration of group Equation elements is Equation . Therefore, the most common oxidation state exhibited by them should be Equation . However, it is only boron and aluminium which practically show the +3-oxidation state. The remaining elements, i.e., Ga, In, Tl, show both the Equation and +3 oxidation states. On moving down the group, the +1 state becomes more stable. For example, Equation is more stable than Tl Equation This is because of the inert pair effect. The two electrons present in the s-shell are strongly attracted by the nucleus and do not participate in bonding. This inert pair effect becomes more and more prominent on moving down the group. Hence, Equation is unstable, Equation is fairly stable, and Equation is very stable.

Element and Oxidation State

Group 13 Element

Oxidation State

B

+3

Al

+3

Ga, In, Tl

+1, +3

The stability of the Equation oxidation state decreases on moving down the group

(ii) C to Pb

The electronic configuration of group 14 elements is Equation . Therefore, the most common oxidation state becomes more and more common on moving down the group. C and Si mostly show the Equation state. On moving down the group, the higher oxidation state becomes less stable. This is because of the inert pair effect. Thus, although Ge, Sn, and Pb show both the Equation and +4 states, the stability of the lower oxidation state increases and that of the higher oxidation state decreases on moving down the group.

Elements and Oxidation State

Group 14 Element

Oxidation State

C

+4

Si

+4

Ge, Sn, Pb

+2, +4

Equation

Q: 2. How can you explain higher stability of Equation as compared to Equation ?

Answer:

Boron and thallium belong to group 13 of the periodic table. In this group, the Equation oxidation state becomes more stable on moving down the group. Equation is more stable than Equation because the Equation oxidation state of B is more stable than the Equation oxidation state of Equation . In Equation the Equation state is highly oxidising and it reverts back to the more stable Equation state.

Q: 3. Why does boron trifluoride behave as a Lewis acid?

Answer:

The electric configuration of boron is Equation . It has three electrons in its valence shell. Thus, it can form only three covalent bonds. This means that there are only six electrons around boron and its octet remains incomplete. When one atom of boron combines with three fluorine atoms, its octet remains incomplete. Hence, boron trifluoride remains electron – deficient and acts as a Lewis acid.

Q: 4. Consider the compounds, Equation and Equation .How will they behave with water? Justify.

Answer

Being a Lewis acid Equation readily undergoes hydrolysis. Equation acid is formed as a result. Equation

Equation completely resists hydrolysis. Carbon does not have any vacant orbital. Hence, it cannot accept electrons from water to form an intermediate. When Equation and water are mixed, they form separate layers.

Equation

Q: 5. Is boric acid a protic acid? Explain.

Answer:

Boric acid is not a protic acid. It is a weak monobasic acid, behaving as a Lewis acid

Equation

It behaves as an acid by accepting a pair of electrons from Equation ion.

Q: 6. Explain what happens when boric acid is heated.

Answer:

On heating orthoboric acid Equation at Equation or above, it changes to metaboric acid Equation . On further heating, this yields boric oxide Equation

Equation

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