Subject

Chemistry

Class

CBSE Class 12

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 Multiple Choice QuestionsShort Answer Type

1. Distinguish between ‘rate expression’ and ‘rate constant’ of a reaction.

The rate expression can be defined as the stoichiometric coefficients of reactants and products. An expression in which the rate of reaction is given in terms of the molar concentration of the reactants, with each term raised to some power, which may or may not is the stoichiometric coefficient of the reacting species in a balanced chemical equation. 

The rate constant can be defined as the rate of the reaction when the concentration of each of the reactant is taken as unity. 

 Example: 2NO(g)+O2(g)--- 2NO2(g)

The rate expression for the above reaction can be written as follows:

 Rate = k [NO]2 [O2] (Experimentally determined)

Now, if the concentration of NO and O2 is taken to be unity, then the rate constant is found to be equal to the rate of the reaction.

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2.

State reasons for each of the following: 

SF6 is kinetically an inert substance. 


The kinetic inertness of SF6 can be explained on the basis of its structure. 

The six fluoride (F-) atoms protect the sulphur atom from attack by the regents to such an extent that even thermodynamically most favorable reactions like hydrolysis do not occur. 

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3. Draw the structure of XeF2 molecule. 

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4.

State reasons for each of the following: 

(i) All the P-Cl bonds in PCl5 molecule are not equivalent. 

(ii) Sulphur has a greater tendency for catenation than oxygen.


(i) In gaseous and liquid state, PCl5 has a trigonal bipyramidal structure. In this structure, the two axial P-Cl bonds are longer and less stable than the three equatorial P-Cl bonds. This is because of the greater bond pair - bond pair repulsion in the axial bonds. Hence, all the bonds in PCl5 are not equivalent. 

(ii) Because of stronger S-S bonds as compared to O-O bonds, sulphur has a greater tendency for catenation than oxygen.

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5.

Define electrophoresis ? 


1. The migration of charged colloidal particles or molecules through a solution under the influence of an applied electric field usually provided by immersed electrodes. Also called ionophoresis, phoresis.

2. A method of separating substances, especially proteins, and analyzing molecular structure based on the rate of movement of each component in a colloidal suspension while under the influence of an electric field.

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6.

The chemistry of corrosion of iron is essentially an electrochemical phenomenon. Explain the reactions occurring during the corrosion of iron in the atmosphere.


The process of corrosion is a redox reaction and involves simultaneous oxidation and reduction reactions. It can, therefore, be referred to as an electrochemical reaction. 

In the process of corrosion, due to the presence of air and moisture, oxidation takes place at a particular spot of an object made of iron. That spot behaves as the anode. The reaction at the anode is can be written as follows. 

 Anode reaction:Fe space left parenthesis straight s right parenthesis space rightwards arrow space Fe to the power of 2 plus end exponent space left parenthesis aq right parenthesis space plus 2 straight e to the power of minus
Electrons released at the anodic spot move through the metallic object and go to another spot of the object. There, in the presence of H+ ions, the electrons reduce molecular oxygen. This spot behaves as the cathode. These H+ ions come either from H2CO3, which are formed due to the dissolution of carbon dioxide from the air into water or from the dissolution of other acidic oxides from the atmosphere in water. 

The reaction corresponding at the cathode is written as follows. 

 

Cathode reaction:
 straight O subscript 2 left parenthesis straight g right parenthesis space plus 4 straight H to the power of plus space plus 4 straight e to the power of minus space rightwards arrow 2 straight H subscript 2 straight O left parenthesis straight l right parenthesis
Thus comma space the space overall space reaction space is space colon space

2 Fe left parenthesis straight s right parenthesis space plus space straight O subscript 2 left parenthesis straight g right parenthesis space plus 4 straight H to the power of plus left parenthesis aq right parenthesis space rightwards arrow 2 Fe to the power of 2 plus end exponent space left parenthesis aq right parenthesis space plus 2 straight H subscript 2 straight O left parenthesis straight l right parenthesis

 Also, ferrous ions are further oxidised by atmospheric oxygen to ferric ions. 

These ferric ions combine with moisture, present in the surroundings, to form a hydrated ferric oxide (Fe2O3x H2O) i.e., rust. 

 

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7.

State reasons for each of the following: 

The N-O bond in  is shorter than the N-O bond in 


The shorter N - O bond in  is due to the existence of resonance in. The resonating structures can be drawn as follows.

 


                               

 

Due to resonance in, the two bonds are equivalent. This leads to a decrease in bond length.

Thus, the N - O bond length in resembles a double bond.

Now, the resonating structures for can be drawn as: 

 

 

 

As seen from the above resonating structures of NO subscript 3 superscript minus, the three oxygen atoms are sharing two single bonds and one double bond. So, the real N - O bond length resembles a single bond closely.  

In NO subscript 2 superscript minus  the lone pair is delocalized between the 2 oxygen groups. Bond order equal to 1+1/2=3/2.

In NO subscript 3 superscript minus lone pair shared between three oxygen atom hence bond order =1+1/3=4/3.

Greater the bond order shorter the bond. Hence bond length of NO subscript 2 superscript minus less than that of NO subscript 3 superscript minus

 

This explains the existence of shorter bond length of the N - O bond NO subscript 2 superscript minus in  than in NO subscript 3 superscript minus .

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8. Express the relation between conductivity and molar conductivity of a solution held in a cell.

The molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length, the conductivity of  the solution is given by the following relation.

 

 straight capital lambda subscript straight m equals straight k straight A over straight l
Now comma space straight l equals 1 space and space straight A equals space straight V space left parenthesis volume space containing space 1 space mole space of space the space electrolyte. right parenthesis
Therefore comma space straight capital lambda subscript straight m space equals kV

 

 

 

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9.

Determine the values of equilibrium constant (KC) and  incrementG° for the following reaction: 

Ni left parenthesis straight s right parenthesis space plus 2 Ag to the power of plus left parenthesis aq right parenthesis space rightwards arrow Ni to the power of 2 plus end exponent space left parenthesis aq right parenthesis space plus 2 Ag space left parenthesis straight s right parenthesis comma space straight E to the power of 0 equals 1.05 straight V space left parenthesis IF space equals 96500 space mol to the power of negative 1 end exponent right parenthesis space


Ni left parenthesis straight s right parenthesis space plus 2 Ag to the power of plus left parenthesis aq right parenthesis space rightwards arrow Ni to the power of 2 plus end exponent space left parenthesis aq right parenthesis space plus 2 Ag space left parenthesis straight s right parenthesis comma space straight E to the power of 0 equals 1.05 straight V space left parenthesis IF space equals 96500 space mol to the power of negative 1 end exponent right parenthesis space
space space space
space space space space space space space straight E subscript Ni to the power of 2 plus end exponent over Ni end subscript superscript 0 space equals negative 0.25 space straight V
space space space space
space space space space space space straight E subscript Ag to the power of plus over Ag end subscript superscript 0 space equals 0.80 straight V

The space galvanic space cell space of space the space given space cell space reaction space is space depicted space as colon
Now comma space the space standard space cell space potential space is
straight E subscript cell superscript 0 space equals straight E subscript straight R superscript 0 space minus straight E subscript straight L superscript 0

equals space 0.80 minus left parenthesis negative 0.25 right parenthesis
equals 1.05 straight V
increment subscript straight r straight G to the power of 0 space equals negative nFE subscript cell superscript 0

In space the space given space equation comma space straight n space equals 2
straight F space equals 96500 space straight C space mol to the power of negative 1 end exponent
straight E subscript cell superscript 0 space equals space 1.05 space straight V space
Then comma
increment subscript straight r straight G to the power of 0 space equals space minus 2 space straight x space 96500 space straight C space mol to the power of negative 1 end exponent space straight x space 1.05 space straight V space
space space space space space space space space space space equals space minus 202650 space CV space mol to the power of negative 1 end exponent
space space space space space space space space space space space equals 202650 space straight J space mol to the power of negative 1 end exponent
space space space space space space space space space space space equals space minus 202.65 space KJ space mol to the power of negative 1 end exponent
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10.

‘Crystalline solids are anisotropic in nature’. What does this statement mean?


All crystalline solids are not anisotropic. Those crystalline solids which are anisotropic have their atoms arranged and spaced in a different manner in three different planes (X, Y and Z). Therefore, the physical properties of crystalline solids such as electrical resistance or refractive index show different values when measured along different directions in the same crystals.

Example NaCl, Quartz, Ice, HCl, Iron, etc.

 

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