Ncert Solutions Chemistry Class 12th

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alok kumar singh

Contributor-Level 10

5.30 

The catalytic reaction that depends upon the pore structures of the catalysts and the size of reactant and product molecules is called shape selective catalysis. Zeolites are good shape- selective catalysts.

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Payal Gupta

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8.10 The 5f orbitals (in actinoids) have a poorer shielding effect than 4f orbitals (in lanthanoids). Thus, the effective nuclear charge experienced by electrons in outer shells in case of actinoids is much more that experienced by lanthanoids.

As the effective nuclear charge experienced is high, the electrons are attracted with much force, hence the size of the atom decreases. Hence, actinoid contraction is greater from element to element than lanthanoid contraction.

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alok kumar singh

Contributor-Level 10

5.29 

Catalysis by zeolites is dependent on shape. Because zeolites are shape-selective catalysts. They are alumino silicates which are microporous in nature. It has Honeycomb structure. That makes them shape selective. In zeolites, some si atoms are replaced by Al to form Al-O-Si network.

Reactants are very sensitive to the pore size of zeolites. Zeolites are used in petrochemical industry. Ex- ZSM-5 used to convert alcohol into gasoline.

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Vishal Baghel

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(i) CH3CH2CH2NH2 + HCl → CH3CH2CH2N+H3Cl-

The final product is (N-propyl ammonium chloride.)

(ii) (C2H5)3N + HCl → (C2H5)3N+HCl-

The final product is (Tri ethyl ammonium chloride)

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alok kumar singh

Contributor-Level 10

5.28

Activity - It is the ability of the catalyst to accelerate the Reaction. It mostly depends upon the Chemisorption strength.

Selectivity - It is an ability to direct reaction to yield of a particular product i.e., One catalyst cannot be a catalyst for other reactions.

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Payal Gupta

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8.9 The stability in aqueous condition depends on the hydration energy of the ions when they bond to the water molecules. And, the hydration energy is the amount of heat released as an ionic substance is dissolved and its constituent ions are hydrated or surrounded by water molecules.

Now, in Cu2+ and Cu+ ion, Cu2+ has a greater charge density than the Cu+ ion and so forms much stronger bonds releasing more energy. Therefore, in an aqueous medium, Cu2+ ion is more stable than Cu+ ion. This is because the energy required to remove one electron from Cu+ to Cu2+, is compensated by the high hydration energy of Cu2+.

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Vishal Baghel

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1- Alkyl group contribute inductive effect which increases the basic strength of

NH32H5NH2< (C2H5)2NH.

Then C6H5NH2 is having –I effect that reduces strength. And C6H5CH2NH2 increases the basic strength but not as much as C2H5 group.

Hence final order will be C6H5NH2<3

2- By taking into consideration –R effect and steric hindrance of groups we can arrange them in the order

C6H5NH2< C2H5NH2< (C2H5)3N< (C2H5)2NH.

Because (C2H5)3N has a lot of steric hindrances that reduces the basic strength.

3- In C6H5NH2, N is directly attached to the ring that causes delocalization of electrons of the benzene ring. Whereas in case of C6H5CH2NH2 it is not directly connected to benze

...more

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Vishal Baghel

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Benzene into aniline

When Benzene is treated with HNO3/H2SO4 it forms nitrobenzene. When Nitrobenzene reduced with Sn/HCL it forms Aniline because Sn/HCl is a reducing Mixture.

When Benzene is reacted with nitrating mixture it forms nitrobenzene. When it Reduced H2/Pd in ethanol or Sn/HCl, it forms Aniline. When Aniline reacts 2 times with CH3Cl It forms N, N- dimethylaniline.

When 1,4-dichlorobutane reacts with NaCN it forms Di cyanide compound, After Hydrogenation it forms the Hexane 1,6-Diamine.

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alok kumar singh

Contributor-Level 10

5.27

Oxidation of sulphur dioxide into sulphur trioxide in the presence of

(3) Oxidation of ammonia into nitric acid in presence of platinum gauze in Ostwald's process

→ 4NH3 (g)+ 5O2 (g) Pt (s) 4NO +6H2O

(4) Hydrogenation of vegetable oils in presence of Nickel as catalyst

Vegetable oil (l)+H2 (g) Ni → Vegetable ghee (s)

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Payal Gupta

Contributor-Level 10

8.8 Given, Z=atomic number = 27 = [Ar]3d7 4s2

⇒ M2+ = [Ar]3d7

Hence, 3 unpaired electrons are present.

The spin only magnetic moment μ= √n(n+2)

Where n is the number of unpaired electrons.

Hence, μ =√(3*(3+2))

μ = √15 BM or μ = 3.87 BM

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