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INTRODUCTION - Chapter 11 P Block Elements class 11 ncert solutions Chemistry - SaraNextGen [2024]


 

Question 11.1:

Discuss the pattern of variation in the oxidation states of

(i) B to Tl and (ii) C to Pb.

Answer:

(i) B to Tl

The electric configuration of group 13 elements is ns2 np1. Therefore, the most common oxidation state exhibited by them should be +3. 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 +1 and +3 oxidation states. On moving down the group, the +1 state becomes more stable. For example, Tl (+1) is more stable than Tl (+3). 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, Ga (+1) is unstable, In (+1) is fairly stable, and Tl (+1) is very stable.

 

Group 13 element

Oxidation state

B

+3

Al

+3

Ga, In, Tl

+1, +3

(ii) C to PbThe stability of the +3 oxidation state decreases on moving down the group.

The electronic configuration of group 14 elements is ns2 np2. Therefore, the most common oxidation state exhibited by them should be +4. However, the +2 oxidation state becomes more and more common on moving down the group. C and Si mostly show the +4 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 +2 and + 4 states, the stability of the lower oxidation state increases and that of the higher oxidation state decreases on moving down the group.

 

 

Group 14 element

Oxidation state

C

+4

Si

+4

Ge, Sn, Pb

+2, +4

Question 11.2:https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/204/3939/NCERT%20Solution_6-11-08_Sonali_11_Chemistry_11_38_GSX_LVN_html_m269bce01.gif

How can you explain higher stability of BClas compared to TlCl3?

Answer:

Boron and thallium belong to group 13 of the periodic table. In this group, the +1 oxidation state becomes more stable on moving down the group. BCl3 is more stable than TlCl3 because the +3 oxidation state of B is more stable than the +3 oxidation state of Tl. In Tl, the +3 state is highly oxidising and it reverts back to the more stable +1 state.

Question 11.3:

Why does boron trifluoride behave as a Lewis acid?

Answer:

The electric configuration of boron is ns2 np1. 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.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/204/3942/NCERT%20Solution_6-11-08_Sonali_11_Chemistry_11_38_GSX_LVN_html_572fbb33.jpg

Question 11.4:

Consider the compounds, BCland CCl4. How will they behave with water? Justify.

Answer:

Being a Lewis acid, BClreadily undergoes hydrolysis. Boric acid is formed as a result.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/204/3943/NCERT%20Solution_6-11-08_Sonali_11_Chemistry_11_38_GSX_LVN_html_m3d088c26.gif

CCl4 completely resists hydrolysis. Carbon does not have any vacant orbital. Hence, it cannot accept electrons from water to form an intermediate. When CCl4 and water are mixed, they form separate layers.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/204/3943/NCERT%20Solution_6-11-08_Sonali_11_Chemistry_11_38_GSX_LVN_html_m4b1c83a5.gif

Question 11.5:

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.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/204/3944/NCERT%20Solution_6-11-08_Sonali_11_Chemistry_11_38_GSX_LVN_html_179334a2.gif

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

Question 11.6:

Explain what happens when boric acid is heated.

Answer:

On heating orthoboric acid (H3BO3) at 370 K or above, it changes to metaboric acid (HBO2). On further heating, this yields boric oxide B2O3.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/204/3946/NCERT%20Solution_6-11-08_Sonali_11_Chemistry_11_38_GSX_LVN_html_19832338.gif

Question 11.7:

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

Answer:

(i) BF3

As a result of its small size and high electronegativity, boron tends to form monomeric covalent halides. These halides have a planar triangular geometry. This triangular shape is formed by the overlap of three sp2 hybridised orbitals of boron with the sp orbitals of three halogen atoms. Boron is sp2 hybridised in BF3.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/204/3947/NCERT%20Solution_6-11-08_Sonali_11_Chemistry_11_38_GSX_LVN_html_74329b87.jpg

(ii) BH4–

Boron-hydride ion (BH4) is formed by the sp3 hybridisation of boron orbitals. Therefore, it is tetrahedral in structure.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/204/3947/NCERT%20Solution_6-11-08_Sonali_11_Chemistry_11_38_GSX_LVN_html_m365e1edb.jpg

Question 11.8:

Write reactions to justify amphoteric nature of aluminium.

Answer:

A substance is called amphoteric if it displays characteristics of both acids and bases. Aluminium dissolves in both acids and bases, showing amphoteric behaviour.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/204/3948/NCERT%20Solution_6-11-08_Sonali_11_Chemistry_11_38_GSX_LVN_html_6132cead.gif

Also Read : Page-No-324:-Chapter-11-P-Block-Elements-class-11-ncert-solutions-Chemistry

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