Ncert Solutions Chemistry Class 12th

In the presence of sulphuric acid (H₂SO₄), KI produces HI as follows : -

2KI + H₂SO₄ → 2KHSO₄ + 2KI

Since H₂SO₄ is an oxidising agent, it oxidises HI (produced in the reaction to I₂)

2HI + H₂SO_{4 →} I₂ + SO₂ +;

As a result, the reaction between an alcohol and HI to produce alkyl iodide cannot occur. Therefore, sulphuric acid is not used during the reaction of alcohols with KI. Instead, a non-oxidising acid such as H3PO4 is used in the reaction to get the desired product.

A few things you can remember here, while solving NCERT Solutions for Haloalkanes And Haloarenes.

• Sulphuric acid is a powerful oxidising agent. We know this because the sulphur atom in sulphuric acid is in its highest possible oxidation state (+6). It can only be reduced. And it can readily accept electrons from other substances.

• It oxidises the necessary intermediate, hydrogen iodide (HI), to iodine (I2). We know this because iodide ions are strong reducing agents, and they can simply give up electrons. So when sulphuric acid accepts electrons, it can convert the iodide to elemental iodine.

• The alcohol cannot be converted to alkyl iodide. The reason for this is that the side reaction consumes the iodide ions necessary to act as a nucleophile and replace the alcohol's -OH group. This is the desired alkyl iodide.

• A non-oxidising acid like phosphoric acid (H3PO4) is used instead to produce the desired alkyl iodide. Phosphoric acid is a weaker oxidising agent than sulphuric acid and does not react with the iodide ions. That allows the intended reaction with the alcohol to proceed.  
  
  
  

7.31

In interhalogen compound ICl and I₂ the atoms are bonded by covalent bonds

Reactivity refers to the rate at which a chemical species will undergo reaction in time. For reaction to take place the compound needs to be broken into separate elements first, then the individual elements react with other elements to form new compounds. So any compound which can easily break into their individual elements can react faster.

The covalent bonds between dissimilar atoms I and Cl atoms in ICl are weaker than between similar atoms in I₂. Therefore the bond between ICL will break easily so the I and Cl atom will be easily available to form another compound as compared to the strong bond between I

Therefore bond dissociation enthalpy of all bond is less than that of I₂ Therefore ICl is more reactive than I₂.

10.1 From the name of the compound it is clear that the parent ring is pentane and chloro and methyl groups are attached in the straight chain at 2nd and 5th position respectively. Hence, the structure is as follows:

From the name of the compound given it is clear that the parent group is hexane with 2 attachments namely chloro and ethyl groups at 1 and 4 positions respectively. Hence, the structure is as follows: -

From the name of the compound given it is clear that the parent group is heptane with tertiary butyl and iodine groups attached at 4 and 3 positions respectively. Hence, the structure is as follows: -

From the name of the compound given it is clear that the parent group is an alkene with the unsaturation present in between 2^nd and 3^rd carbon and 2 bromine atoms attached at 1^st and 4^th carbon. Hence, the structure is as follows : -

From the name of the compound given it is clear that the parent group is benzene with bromine group attached at 1^st carbon, secondary butyl group attached at 4^th carbon and a methyl group attached at 2^nd carbon. Hence, the structure is a follows : -

8.48 To calculate magnetic moment of the complex species, we use the spin formula:

μ =√n(n+2) BM

1. [K₄ [Mn(CN)₆]

⇒μ = 2.2 BM (given)

We can see from the above calculation that the given value(2.2) is close to n=1. It means that it has only one unpaired electron Also in this complex Mn is in +2 oxidation state,i.e., as Mn²⁺.Thus when CN^- ligands approach Mn²⁺ ion, The electrons in 3d do not pair up.

The atomic number of Manganese (Mn) is Z = 25

The electronic configuration of ₂₅Mn= [Ar] 3d⁵ 4s²

And, the electronic configuration of Mn²⁺=[Ar] 3d⁵

Thus CN^- is a strong ligand.

The hybridization involved is d²sp³ forming inner orbital octahedral complex

2. [Fe(H2O)₆]^2+.

⇒ μ = 5.3 BM (given)

We can see from the above calculation that the given value(5.3) is close to n = 4. It means that it has four unpaired electrons. In this complex, Fe is in +2 state, i.e., as Fe²⁺. This means that the electrons in 3d do not pair up when the ligands, H₂O molecules approach.

The atomic number of Iron(Fe)is Z = 26

The electronic configuration of ₂₆Fe= [Ar] 3d⁶ 4s²

And, the electronic configuration of Fe²⁺=[Ar] 3d⁶

Thus H₂O is a weak ligand. To accommodate the electrons donated by six H₂O molecules, the hybridization will be sp³d². Hence it will be an outer orbital octahedral complex.

3. K₂ [MnCl₄]

⇒μ = 5.9 BM (given)

We can see from the above calculation that the given value(5.9) is close to n=5. It means that it has five unpaired electrons

In this complex, Mn is in +2 state, i.e., as Mn²⁺. Hence, we can say that Cl- is a weak ligand and does not cause the pairing of electrons.

The atomic number of Manganese (Mn) is Z = 25

The electronic configuration of ₂₅Mn= [Ar] 3d⁵ 4s²

And, the electronic configuration of Mn²⁺=[Ar] 3d⁵

Thus, the hybridization involved will be sp³ and the complex will be tetrahedral complex

Being Lewis bases (those who donate electrons) the ligands with less electronegativity will be stronger. Therefore, in general halogen or oxygen donors (F^-, Cl^-,Br^-, H₂O) are weak field ligands and the ones in which carbon or nitrogen atom is the donor (eg-CN^-,CO,NH₃) are strong field ligands.