The ability of cation to distort the electron cloud of anion is called polarising power

THAT IS OLIGO SACCHARIDES.

Oligo saccharides are the carbohydrates on hydrolysis give two to ten units of monosaccharides

Ans. Dichlorodiphenyldichloroethylene (DDE) is a chemical compound formed by the loss of hydrogen chloride from DDT, of which it is one of the more common breakdown products.

Because by losing two electrons it attains exactly half filled outermost d orbital

According to the formula oxidation state of  N in  HNO₃ is  +5 ,as calculated below.

Let  the oxidation state of N be  x , then applying the rules to calculate O.N. , we get

+1  +(x)   + (- 2 X 3 )  =  0

{tex}\therefore{/tex}   x   =  (  +6  - 1 )

=  +5 

We know that , N is restricted to a maximum covalency of  4 , since only  four (1s  and 3p  )  orbitals are available for bonding , and also due to the absence of a d-orbital in the atom. However , there are cases where   oxidation state of N in compounds like  viz .  N₂ O₅ ,  HNO₃    is  +5 . This is due to resonating structures  of  these compounds .  Their resonating structures well support and justify the oxidation number of N as  +5  in the compound  as follows - 

Resonating structure of HNO₃ results into formation of a coordinate bond , where N donates both the electrons to oxygen atom. 

and the double bond is formed with one of the oxygen atoms  giving  a charge as  +2 to N

A single bond with the other oxygen atom gaining a negative charge over it  gives  a charge of  +2  to  N

&  OH^- group  gives  a charge to N as  =  +1

Adding up all the charges provided to N in HNO₃  ,

Oxidation  No.  of  N  in  HNO₃   =   2+2+1  =  5 

similarly the oxidation state of N  in  N₂ O₅ can  also be  made out through its structure as given above. 

It is defined as the no. of  moles of solute present in 1 L of solution

Study :=}

Catalyst is a substance which can change the speed of the chemical reaction without itself undergoing any change in its mass and chemical composition. and the phenomenon is known as catalysis.

In case of p-dichlorobenzene both the chlorine atoms pull the electron cloud towards themseleves woth equal magnitude and in the opposite directiona and hence its dipole moment turns out to be zero while in the case of p dihydroxybenzene O atom is sp³ hybridized and lone pair is not in the plane of the molecule so electron pull is same in magnitude but in different directions hence its dipole moment does not cancel out

Azo dye test

Tollin's test

Fehling test

Ans. 30 Important Name Reactions are Following :

1. Sandmeyer Reaction
2. Gatterman Reaction
3. Balz-Schiemann Reaction
4. Finkelstein Reaction
5. Swartz Reaction
6. Wurtz Reaction
7. Fitting Reaction
8. Wurtz - Fittig Reaction
9. Kolbe's Reaction
10. Reimer- Tiemann Reaction
11. Rosenmund Reduction
12. Gattermann - Koch Reaction
13. Stephen Reaction or Stephen Reduction
14. Clemmensen Reduction
15. Wolff - Kishner Reduction
16. Haloform Reaction (Iodoform Reaction)
17. Aldol Condensation
18. Cannizzaro Reaction
19. Friedel - Crafts Reaction
20. Grignard Synthesis
21. Esterification Reaction or Fischer Esterification
22. Williamson Synthesis
23. Diazotisation Reaction
24. Etard Reaction
25. Hell - Volhard Zelinsky Reaction
26. Decarboxylation Reaction
27. Hofmann Bromide Reaction
28. Gabriel Phthalimide Synthesis
29. Coupling Reaction
30. Carbylamine Reaction

Ans. The radius ratio for AgBr is intermediate. Thus, it shows both frenkel and schottky defects.

The major defect in AgBr is the Frenkel defect. It has a rock salt structure i.e. CCP lattice of Br with atoms of Ag occupying all octahedral holes. Ag moves from octahedral to tetrahedral sites causing only cations to precipitate.

Schottky defect arise due to missing of ions from their lattice point and frenkel arise when the missing ions occupy interstitial sites.

in AgBr, Ag⁺ ion is small in size  and when removed from lattice point they can occupy interstitial site  and therefore show both frenkel and schottky defect. SCHOTTKY Defect in AgBr is exhibited due to precipitation of both Cations and Anions.

You can download from cbse guide itself

All these hydrides are covalent in nature and have pyramidal structure. These involve sp³ hybridization of the central atom and one of the tetrahedral positions is occupied by a lone pair.Due to the presence of lone pair, the bond angle in NH₃ is less than the normal tetrahedral angle. It has been found to be 107 degrees. Down the group the bond angle decreases as:

In all these hydrides, four electron pairs, three bond pairs and one lone pair surround the central atom. Now, as we move down the group from N to Bi, the size of the atom goes on increasing and its electro negativity decreases. Consequently, the position of bond pair shifts more and more away from the central atom in moving from NH₃ to BiH₃. For e.g., the bond pair in NH₃ is close to N in N-H bond than the bond pair in P-H bond in PH₃. As a result, the force of repulsion between the bonded pair of electrons in NH₃ is more than in PH₃. In general, the force of repulsion between bonded pairs of electrons decreases as we move from NH₃ to BiH₃ and therefore, the bond angle also decreases in the same order.

Ans. Nitrogen does not have D orbitals whereas Phosphorus on the other hand can expand itself into the d-orbital and thus can share all 5 of it's outer shell electrons.Also Due to small size of Nitrogen, there is a lot of inter-electronic repulsion. So the N-N single bond is very weak.

Make a list of all the questions from these chapters and learn them by heart.

Revise the answers everyday and try to write them too. 

Practice these questions thoroughly. 

All the best. 

Ans. 

Ans. 

Aniline is convered to phenyl isocyanide by its reaction with KOH and chloroform i.e. CHCl₃

C₆H₅-NH₂ + KOH + CHCI₃----------> C₆H₅-NC

Ans. Aniline reacts with NaNO₂/HCl  to form benzenediazonium chloride Which on Reduction with Cu forms Benzene.

Then Benzene reacts with Methyl Chloride in presence of anhydrous AlCl₃ Form Toluene.

Toluene in the presence of UV reacts with chlorine atom forms Benzyl Chloride. 

It would be better if you solve these questions directly from NCERT. 

These chapters have most of the direct questions in paper. And you can refer to the sample papers too. 

Learn the most important and most frequent questions from these chapters by heart. 

All the best! 

Ans. 1) Acetaldehyde to acetone

2) Acetone to Tertabutyl alcohol

3) Ethyl alcohol to acetone