Chemistry NCERT Exemplar Solutions Class 11th Chapter Eleven

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(i) When moving down the periodic table within a group, ionization enthalpy generally decreases due to increased atomic radius. However, in the case of gallium (Ga) and aluminum (Al), gallium experiences a higher effective nuclear charge due to less shielding from its inner electrons, resulting in a higher ionization enthalpy compared to aluminum.

(ii) As boron is smaller in size and the sum of its first three ionization enthalpies i.e. ΔH₁+ΔH₂+ΔH₃ is very large so boron does not allow to lose its all three valence electrons and exist as +3 ion rather it shares electrons to make covalent bonds.

(iii) As compared to boron, aluminium is larger in size and has vacant d-orbital through which it can expand its coordination number from +4 to +6. Can easily acquire the sp³d² hybridisation State. On the other hand, boron is smaller in size with no d-orbitals and can expand its coordination number maximum up to +4 so boron can maximum the form BF₄⁻ but not BF₆²⁻.

(iv) Lead is extra stable in its +2 oxidation state than in +4 oxidation state due to inert pair effect.

(v) Due to the inert Pair effect, lead is more stable in its +2 oxidation state than in +4. So if it is present in +4 oxidation state in any compound it can readily reduce itself by gaining two electrons. Thus acting as an oxidising agent.

Reaction: Pb⁴⁺+2e⁻→Pb²⁺

While for Sn2+, +4 oxidation state is extra stable than +2 oxidation state so in any compound where acid is present in +2 oxidation state it loses two more electron and oxidises itself to +4 oxidation state. Thus acting as a reducing agent.

Reaction: Sn²⁺→Sn⁴⁺+2e⁻

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The following reaction takes place in the test tube

1. i) When Aluminium is treated with dilute hydrochloric acid.

Al + HCl  -à AlCl₃

ii) When Aluminium is treated with dilute sodium hydroxide solution.

Al  +  NaOH -àNaAlO₂ + H₂

In both the cases, the Hydrogen atoms are evolved. So, when a burning match stick is brought near the test tube we hear a pop sound.

Aluminium does not react with concentrated nitric acid. A "layer" of "Aluminium Oxide" is formed, when aluminium reacts with the nitric acid because "nitric acid" is an oxidising agent. Due to the "layer" of “Aluminium oxide”, nitric acid cannot react with the “inner aluminium metal.

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(A) Carbon has small size and large electronegativity, it forms strong n−pπ bonding with two oxygen atoms forming a separate CO₂ molecule.

While in SiO₂ silicon is larger in size with comparatively less electronegativity than carbon it shows no tendency to form n−pπ bonding rather forms Single covalent bond with oxygen. Thus, SiO₂ possess 3D network-like structure in which each Silicon is bonded to 4 oxygen atoms.

(B) Carbon is smaller in size and lacks d-orbitals hence can have a maximum coordination number of four and sp³ hybridisation only.

Whereas Silicon has d-orbital, it can amplify its hybridisation up to sp³d² (CN=6) and can complete its Octet. Therefore, Silicon forms SiF₆²⁻.

whereas the corresponding flugre-compound of carbon is not known.

This is a Short Answers Type Questions as classified in NCERT Exemplar

(A) sarbon in CCl₄ does not have a vacant d-orbital to accommodate the electrons from OH of water molecules. Also CCl₄ is nonpolar covalent compounds whereas H₂O is polar. So, no strong interaction occurs between them. Hence CCl₄ is miscible in water.

Whereas in SiCl₄, silicon has bigger size than carbon and have d-orbitals for accommodation of electrons donated by OH of water in the process of hydroxylation. This leads to a strong interaction and silicon acid Is formed as a product. SiCl₄ is completely miscible in water.

(B) As we move from carbon to silicon, the size increases and electronegativity decreases. This results in the decrease in M-M bond energy. Hence the catenation property decreases.

Bond energy is the amount of energy required to fragment the atoms combined in a molecule are born into separate atoms. Carbon has maximum bond energy as compared to silicon. Therefore, shows maximum catenation property.