Ncert Solutions Chemistry Class 11th

The necessary conditions for a molecule to be aromatic are:

• It should have a single cyclic cloud of delocalised n-electrons above and below the plane of the molecule.
• It should be planar. This is because complete delocalization of n-electrons is possible only if the ring is planar to allow cyclic overlap of p-orbitals.
• It should contain Huckel number of electrons, i.e., (4n + 2) n-electrons where n = 0, 1, 2, 3 etc.
  A molecule which does not satisfy any one or more of the above conditions is said to be non-aromatic.

Benzene is a resonance hybrid of two canonical forms. In the resonance hybrid, all the six pi electrons are completely delocalized. This results in resonance stabilization.

The structures of cis- and trans-isomer of hex-2-ene are:

The boiling point of a molecule depends upon dipole-dipole interactions. Since cis-isomer has higher dipole moment, therefore, it has higher boiling point.

3.40. Within a period, the oxidising character increases from left to right. Therefore, among F, O and N, oxidising power decreases in the order: F > O > N. However, within a group, oxidising power decreases from top to bottom. Thus, F is a stronger oxidising agent than Cl. Further because O is more electronegative than Cl, therefore, O is a stronger oxidising agent than Cl. Thus, overall decreasing order of oxidising power is: F > O > Cl > N, i.e., option (b) is correct.

A combustion reaction is a reaction in which a substance reacts with oxygen gas, there is a formation of carbon dioxide, water with the evolution of light and heat.

(i) 2C₄H₁₀ (g) +13 O₂ (g)?8CO₂ (g)+10H₂O (g) + Heat

(ii) 2C₅H₁₀ (g) +15 O₂ (g)?10CO₂ (g)+10H₂O (g) + Heat

(iii) 2C₆H₁₀ (g) +17 O₂ (g)?12CO₂ (g)+10H₂O (g) + Heat

The ozonolysis of 4-Ethylhex-3-ene gives propanal and pentan-3-one.
The structural formula of the alkene (4-Ethylhex-3-ene) is as shown.

(i) Write the structures of propanal and pentan-3-ene with their oxygen atoms facing each other, we have,

 

(ii) Remove oxygen atoms and join the two fragments by a double bond, the structure of the alkene is

3.39. In a period, the non-metallic character increases from left to right. Thus, among B, C, N and F, non-metallic character decreases in the order: F > N > C > B. However, within a group, non-metallic character decreases from top to bottom. Thus, C is more non-metallic than Si. Therefore, the correct sequence of decreasing non-metallic character is: F > N > C > B > Si, i.e., option (c) is correct.

(i) An aldehyde with molar mass of 44 u is ethanal, CH₃CH=0

(ii) Write two moles of ethanal side by side with their oxygen atoms pointing towards each other.

 

(iii) Remove the oxygen atoms and join them by a double bond, the structure of alkene 'A' is

As required, but-2-ene has three C—C, eight C—H a? bonds and one C—C? -bond.

3.38. In a period, metallic character decreases as we move from left to right. Therefore, metallic character of K, Mg and Al decreases in the order: K > Mg > Al. However, within a group, the metallic character, increases from top to bottom. Thus, Al is more metallic than B. Therefore, the correct sequence of decreasing metallic character is: K > Mg > Al > B, i.e., option (d) is correct

Step 1. Write the structure of the products side by side with their oxygen atoms pointing towards each other.

Step 2. Remove the oxygen atoms and join the two ends by a double bond, the structure of the alkene 'A' is