session ending examination class:xi sub: chemistry …

SESSION ENDING EXAMINATIONCLASS:XI SUB: CHEMISTRY

TIME: 3 Hours M.MARKS:70BLUE PRINT

S.NO CHAPTER 1 MARKS 2 MARKS 3 MARKS 5 MARKS TOTAL

1 Some basic concepts of

chemistry

3×1 3

2 Structure of atom 1×1 2×1 3×1 6

3 Classification of Elements

and Periodicity in

properties

1×1 3×1 4

4 Chemical Bonding and

Molecular Structure

2×1 3×1 5

5 States of Matter: gas and

Liquids

1×1 3×1 4

6 Thermodynamics 1×1 5×1 6

7 Equilibrium 2×1 5×1 7

8 Redox Reaction 3×1 3

9 Hydrogen 3×1 3

10 S-Block Elements 3×1 4* 3+4*

11 P-Block Elements 2×1 3×1 5

12 Organic Chemistry-Basic

principles and tech.

1×1 2×1 3×1 6

13 Hydro-carbon 3×1 5×1 8

14 Environmental Chemistry 3×1 3

Total 5 5 12 3+1* 70

“Value Education with Training”

DAV CENTENARY PUBLIC SCHOOL, PASCHIM ENCLAVE, NEW DELHI - 87


TIME: 3 Hours M.MARKS:70GENERAL INSTRUCTIONS:

(i) All questions are compulsory.

(ii) Marks for each question are indicated against it.

(iii) Questions no.1to 5 is very short answer questions and carries 1 marks each.

(iv) Questions no.6to 10 is short answer questions and carries 2 marks each.

(v) Questions no. 11to22 are also short answer question and carry 3 marks each.

(vi) Questions no.23 carry 4 marks and Q.No.25 to 26 are long answer questions of 5

marks each.

(vii) Use long Tables, if necessary. Use of calculators is not allowed.

Q1. . Which gas law is shown by the following graph?

PV

P

Q2. . what is inductive effect ?.

Q3. Write the electronic configuration of Cr+ ion. [Atomic number of Cr = 24]

Q4. Predict the sign of S for the reaction:

2NaHCO3 (s) Na2CO3 (s) + CO2 (g) + H2O (g)

Q5. . Write the IUPAC name and symbol for the element with atomic number 118.

Q6. . Account for the following:

a) NH3 molecules has more dipole moment that NF3 although N – F bond is more polar than

N – H bond.

b) H2O is a liquid while H2S is a gas.

Q7. what are electrophiles and nucleophiles ? explain with examples.



Q8. which of the following are lewis acids ?

AlCl3, H2O, BF3, H+ ,NH3,and NH4+.

Q9. . What is the state of hybridization of carbon in

(a) CO32- (b) diamond

Q10. . The mass of an electron is 9.1 × 10-31 kg. if its kinetic energy (K.E.) is 3.0× 10-25 J, calculate its

wavelength. (h = 6.626 × 10-34 J s)

Q11. a) State Pauli’s exclusion principle.

b) Account for the following:

i) Bohr’s orbits are called stationary states.

ii) A moving cricket ball cannot have wave character.

Q12. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction:

CaCO3 (s) + 2HCl (aq) CaCl2 (aq) + CO2 (g) + H2O (l)

What mass of CaCO3 is required to react completely with 25 ml of 0.75 M HCl?

[Molar mass/g mol-1 : Ca = 40 u, C = 12 u, O = 16 u, H = 1 u, Cl = 35.5 u]

Q13. Account for the following:

a) Ionization enthalpy of nitrogen (Z = 7) is more than that of oxygen (Z = 8).

b) Noble gases have large positive electron gain enthalpies.

c) An anion is always bigger than its parent atom.

Q14. . a) Write the molecular orbital configuration of 2+. Calculate its bond order and predict its

magnetic behaviour.

b) Using VSEPR model, predict the geometry of SF6 molecule.

Q15. (a) Write van de Waals equation for ‘n’ moles of a real gas.

b) An open beaker at 27 0C is heated to 477 0C. what fraction of air would have been

expelled out?

Q16. . Explain the following terms:

a) Acid rain

b) Green chemistry

c) Ozone layer depletion

Q17. What happens when?

(A). sodium metal is dropped in water?

(b). sodium metal dissolves in liquid ammonia?

(c). Sodium metal is heated in free supply of air?



OR

Comment on the following observations:

a) The mobility’s of the alkali metal ions in aqueous solution are:

Li+ < Na+ < K+ < Rb+ < Cs+

b) Beryllium and magnesium do not give colour to flame whereas other alkaline earth metals

do so.

c) Alkaline earth metals cannot be obtained by chemical reduction methods.

Q18. Explain the following with suitable reasons:

a) Diamond is covalent, yet it has high melting point.

b) CO2 is a gas but SiO2 is a solid.

c) B to TI, +1 oxidation state becomes more stable.

Q19. ( a) Arrange the following in order of property mentioned against each:

i) (CH3)3C+, CH3CH2CH+, CH3CH2CH2CH2+ [Increasing stability]

CH3

ii) HCOOH, CH3COOH, CH3CH2COOH [Increasing acidic strength]

b) Write the name of isomerism among the following compounds:

CH3 – O – CH2CH2CH3 and C2H5 – O – C2H5

Q20. How is hydrogen peroxide prepared industrially? How does it concentrated? Explain why it is stored in coloured wax-lined glass or plastic bottles?

Q21. Balance the following ionic reaction with the help of oxidation number method: (Acidic

medium) N2H4 (l) + Cl 3 (aq) NO (g) + (aq)

Q20. Write IUPAC names of the products obtained by the ozonolysis of the following compounds:

(i) 2-ethylbut-1-ene

(ii) 1-phenylbut-1-ene ; Explain the reactions.

Q23. Amitabh was getting his house white-washed. His friend Suresh came to see his house. He found that quick lime was being soaked in water and decanted solution was being taken for white-washing by the worker. Suresh found that the lime solution that was used in the process was not as white as the walls after the white-wash. He asked Amitabh, “How is it that the walls look so sparkingly white while lime water that is used, does not.”

(a) How can you explain the difference in the whiteness of the lime water and the walls?

(b) How does lime water react with chlorine gas? Name the product that is formed.

Q24. (a) Define solubility product. Write solubility product expression for Zr3(PO4)4.

(b) Calculate the pH of 0.01 M CH3COOH solution. [Ka(CH3COOH) = 1.74 × 10-5]

(c) Explain why NaCl is precipitated when HCl (g) is passed through the saturated solution of

NaCl.



OR

(a) equal volumes of 0.002 MN solutions of Na(IO3) and Cu(CLO3)2 are mixed together. Will it

lead to the precipitation of Cu(LO3)3 = 7.4 × 10 8]

(b). define Le-Chatelier’s principle. What is the effect of:

(i). addition of H2

(ii). Removal of CO

On the equilibrium : 2H2 (g) + CO (g) CH3OH?

Q25. (a) State first law of thermodynamics. Heat (q) and work done (W) individually are not state functions but their sum is always a state function. Explain why? (b) Calculate the standard enthalpy change ( rH0) and standard internal energy change ( ru0)for the following reaction at 300 K:

OF2 (g) + H2O (g) O2 (g) + 2HF (g)

Standard enthalpy of formation ( rH0 ) of various species are given as below : ( rH0)/Kj mol-1 :

OF2 (g) =23.0 H2O (g)=-241.8 , HF (g) =268.6 J K -1, R=8.314 JK -1 mol. -1

OR

(a) Predict the sign of the entropy change for each of the following changes;

(i)

(ii)

(b) For a reaction at 25 0C , 2A (g) + B (g) 2D (g)

U 0 = - 10.5kJ and S 0 = -44.1 JK -1. Calculate G 0 for the reaction and predict whether the

Reaction is spontaneous or not at 25 0C.

Q26. (a) Explain the following reactions with suitable examples:

(i) Wurtz reaction.

(ii) Friedel-craft, alkylation reaction.

(b) An alkene ‘A’ on ozonolysis gives a mixture of ethanol and pentane-3-one. Write structure and IUPAC name of ‘A’. ( c) give one chemical test to distinguish between ethane and ethyne.

OR

(a) Write suitable reason for the following:

(i) C – C bond length in benzene ring is 139 pm which is in between C – C single bond 154 pm and C = C double bond 133 pm.(ii) Trans-2-butene has higher melting point than cis-isomer.

(b) Give a chemical test to distinguish between but-1-yne and but-2-yne.

(c ) How will you carry out following conversions:

(i) Ethene to benzene

(ii) 1-bromopropane to 2-bromopropane.




TIME: 3 Hours M.MARKS:70MARKING SCHEME

Ans.No. Brief Answers Marks

1 The graph represents Boyle’s law. 1

2. polarization of a -bond caused by the polarization of adjacent . -bond is

called inductive effect.

1

3. 1s,22s22p6 3s23p63d5. 1

4. Reactant NaHCO3 is a solid and it has low entropy. Among the products, there

are one solid and two gases. Therefore the entropy increases.

1

5. Ununoctium(Uuo) 1

6. a) This is because in the case of NH3, the orbital dipole due to lone pair is in the

same direction as the resultant dipole moment of the N – H bonds, whereas in

NF3, the orbital dipole is in the direction opposite to the resultant dipole

moment of the three N – F bonds. The orbital dipole because of the lone pair

decreases the effect of the resultant N – F bond moments which result in low

dipole moment o b) There is hydrogen bonding in H2O while there is no such

bonding in H2S. Due to associated H2O molecules, the boiling point of H2O is

raised.

1

1

7. 1. Electrophiles are electron loving reagents. They are electron deficient and

so, act as lewis acid and seek electrons.

Examples : H+, H3O+,NO2+(from HNO3), SO2OH+,NO+,Br+,AlCl3,BF3 are neutral

species which act as electrophiles.

2. Nucleophiles are nucleus loving reagents. They are electron rich and so, act

as a lewis base and donate electrons.

Example: Cl-,Br-.

1

1

8. (I) AlCl3: lone pair deficient.

(I) BF3: lone pair deficient.

(II) H+ : electron deficient.

(III) NH4+: electron deficient.

1/2x4

9. (a) carbon atom in CO32- is sp2 hybridized.

(b) carbon in diamond is sp3

1

1



10.K.E = 1

2

2 or v = 2× . .

Substituting the values, we have =2×3.0×10 25 2 2

9.1×10 31

= 812 m s-1

=6.626×10 34

9.1×10 31×812

= 8967× 10-10 m

= 896.7 nm

½

½

½

½

11. a) Pauli’s exclusion principle state: No two electrons in an atom can have the

same set of four quantum numbers.

It can also be stated as: only two electrons may exist in the same orbital and

these electrons must have opposite spin.

b) i) According to Bohr, electrons in an atom move around the nucleus in

orbits of fixed radius and energy. Therefore Bohr’s orbits are called stationary

states.

ii) Objects like cricket ball have a large mass (compared to an electron or a

photon). The wavelength associated with such moving objects is so short that

their wave properties cannot be detected .

1

1

1

12. 1000 ml of 1.0 M HCl = 1 mol of HCl

25 ml of 0.75 M HCl = 1

1000× 25 × 0.75mole of HCl

= 0.01875 mole of HCl

CaCO3 (s) + 2HCl (aq) CaCl2 (aq) + CO2 (g) + H2O (l)

2 moles of HCl react with = 100 g of CaCO3

0.01875 Mole of HCl react with = 1002× 0.01875 = 0.9375 g.

½

½

½

½

1

13. a) Nitrogen has the electronic configuration 1s22s22 12 12 1 , which is stable

half-filled configuration in contrast with the configuration of O

1s22s22 22 12 1 , which is not as stable. Therefore ionization enthalpy of N is

more than that of O.

b) Noble gases have a stable electronic configuration. These elements do not

accept the electron easily. Therefore they have large positive electron gain

1

1

1



enthalpies.

c) An anion is formed when a neutral atom gains one or more electrons.

Atom + electron/s Anion

The electrons that are added to the atom suffer repulsion from the electrons

already present. Moreover, the added electrons generally occupy the next

orbital, which is further away from the nucleus. Therefore the anion is always

bigger than its parent atom.

14. a) The M.O. configuration of 2+:

( 1s)2( *1s)2( 2s)2( *2s)2( 2pz)2( 2pz)2( 2 2 2 2)( 2 1 =

2 1)

Bond order = 2

= 10 6

2= 2

Magnetic behavior : paramagnetic because of the presence of two unpaired

electrons.

b) SF6 has octahedral geometry according to VSEPR theory.

1

½

½

1

15. a) Van de Waals equation for n moles of real gas is:

+2

2(V – nb) = nRT

b) Let the initial and final volumes be V1 and V2.

Applying Charles law: 2

1= 2

1

Substituting the values, we have 2

1=

477+273

27+273=

750

300

Or 2

1= 2.5

Subtracting 1 from both sides, we get

2

1

1 = 2.5 12 1

1

= 1.5

Or fraction of air expelled out = 1.5.

½

½

½

1

½

16. a) Acid rain: it refers to the ways in which acid from the atmosphere is

deposited on the earth’s surface. Oxides of nitrogen and sulphur which are

acidic in nature can be blown by wind along with solid particles in the

atmosphere and finally settle down either on the ground as dry deposition, or

1



in water, fog and snow as wet deposition.

b) Green chemistry: Green Chemistry is a way of thinking and is about utilizing

the existing knowledge and principles of Chemistry and other sciences to

reduce the adverse impact on environment. Utilization of existing knowledge

base for reducing the chemical hazards along with the developmental

activities is the foundation of green chemistry.

c) Ozone layer depletion: the upper stratosphere consists of ozone which

protects us from the harmful effects of harmful UV radiations coming from the

sun. it is important to maintain the ozone shield. In recent years there have

been reports of the depletion of this protective ozone layer because of the

presence of certain chemicals in the stratosphere. The main reason for the

depletion of ozone layer is the release of chlorofluorocarbons compounds

(CFCs), also known as freons. These chemicals are used in refrigerators and air-

conditioners. Once these are released into the atmosphere. Chlorine radicals

produced from CFCs cause the breakdown of ozone.

1

1

17. (a). Sodium metal is dropped in water

2Na + 2H2 2NaOH + H2

Hydrogen gas is released.

(b). sodium metal dissolves in liquid ammonia

Na + (x + y) NH [Na(NH3)x]+ + [e(NH3)y]-

A deep blue solution is formed which is conducting in nature.

(c). sodium metal is heated in free supply of air

2Na + O2 2 O2

Or

a) Hydration enthalpies of alkali metal ions decrease with increase in ionic

sizes. The higher the hydration energy, the lower will be the ionic mobility. Cs+

with the lowest hydration enthalpy has the maximum mobility and Li+ the

lowest ionic mobility.

b) in flame, the electrons are excited to higher energy levels. When they drop

back to the ground state, the difference of energy is emitted in the form of

visible light. Electrons in Be and Mg are too strongly found to get excited by

the flame. Hence, these elements do not impart colour to the flame.

c) alkaline earth metals have negative value of reduction potential in solution.

Hence they cannot be obtained by chemical reduction method.

1

1

1

1

1

1



18. a) Diamond has a rigid three dimensional network of carbon atoms, each

carbon is tetrahedrally attached to four carbons around it. Because of this,

diamond has high melting point.

b) In CO2 molecule, carbon atom undergoes sp hybridization. Two sp

hybridized orbitals overlap with the two p-orbital’s of oxygen to form two

sigma bonds. Other two electrons of carbon form bonding with

oxygen atoms. SiO2 is a covalent three dimensional network solid in which

each silicon atom is tetrahedrally bonded to four oxygen atoms. Network

compounds have generally high melting and boiling points. Therefore, CO2 is a

gas but SiO2 is a solid.

c) as we move from B to TI, due to poor shielding effect of d and f-orbitals, the

ns2 electrons of the metals are held more and more tightly by the increasing

positive charge of the nucleus. Therefore only p1 electron can take part in

bonding. Thus +1 oxidation state becomes more stable. This is called inert pair

effect

1

1

1

19. a.

i. CH3CH2CH2CH2+ < CH3CH2CH+ < (CH3)3C+

CH3

ii. CH3CH2COOH < CH3COOH < HCOOH

b. Metamerism

1

1

1

20. In industry, H2O2 is prepared by the auto oxidation of 2-alkylanthraquinols

2-ethylanthraquinol H2O + oxidized product

1% H2O2 is formed. It is extracted with water and concentrated to ~ 30% (by

mass) by distilled under reduced pressure. It is further concentrated to ~ 85%

by careful distillation under reduced pressure. The remaining water can be

frozen out to obtain pure H2O2.

H2O2 decomposes slowly on exposure to light

2H2O2 (l) 2H2O (l) + O2 (g)

To avoid this, it is stored in coloured wax-lined glass or plastic bottles.

1

1

1

21. 22 H4 (l) + Cl+5

3 N+2 O (g) + Cl- (aq)

In this, N2H4 acts as a reducing agent, whereas Cl 3 acts as an oxidizing agent.

Total increase in oxidation number of N = 2× 4 = 8

½

½



And total decrease in oxidation number of Cl = 1 × 6 = 6

To balance increase/decrease in oxidation number, multiplying N2H4 by 3

and Cl 3 by 4, we get

3N2H4 (l) + 4Cl 3 (aq) NO (g) + (aq)

To balance N and Cl atoms, multiplying NO by 6 and Cl—by 4, we get

3N2H4 (l) + 4Cl 3 (aq) 6NO (g) + 4 (aq)

Balancing O atoms by adding 6H2O to R.H.S., we get

3N2H4 (l) + 4Cl 3 (aq) 6NO (g) + 4 (aq) + 6H2O (l)

In this, H atoms get automatically balanced.

½

½

½

½

22. (i) (CH3CH2)2 – C = CH2 (CH3CH2)2 –C=O +HCHO

(ii)Zn/H2O Pentan-3-one & methanal

(i) O3

(ii) CH3CH2-CH=C6H5 CH3CH2CHO +C6H5CHO

(ii) Zn/H2O Propanal & Benjaldihyde

11\2

11\2

23. (a) Coating of lime water on the walls reacts with CO2 present in the

atmosphere to give calcium carbonate which has a bright white colour. The

reaction is explained as under:

Ca(OH)2 + CO2 CaCO3 + water

(b) On passing Cl2 gas through lime water, we obtain bleaching powder.

Ca(OH)2 + 2Cl2 CaCl2 + Ca(OCl)2 + 2H2O

2

2

24. (a). solubility product of a substance may be defined as the product of the

concentration of the ions raised to the number of respective ions in a

saturated solution.

For Zr3(PO4)4 , the solubility product expression may be given by

Ksp = [Zr4+]3 [PO3-]4

(b). pH of 0.01 M CH3COOH solution

Use the

Substracting the values in the above equation, we have

= =1.74×10 5

0.01= 1.74 × 10 3 = 17.4 × 10 4

= 4.17 × 10 2 = 0.0417

2

2



[H+] = [H3O+] = C = 0.01 × 0.0417

= 0.000417

pH = - log (0.000417)

= 3.38

(c). on passing HCL (g) through a saturated solution of NaCL, the ionic product

[Na+] [Cl-] becomes more than the solubility product. Therefore, precipitation

of NaCl takes place.

Or

NaIO3+ + I 3

Cu(ClO3)2 Cu 2+ + 2Cl 3

Concentration of Cu2+ ions in the solution will be 0.002 M and that of I 3 ions

will be 0.002 in the solution before mixing. After mixing, the volume will

become double; therefore, the concentration will be reduced to half. Thus the

concentration will be

[Cu2+] = 0.0022

= 0.001

[I 3 ] = 0.0022

= 0.001

Ionic product,

[Cu+][IO3]2 = (0.001)(0.001)2

= 1× 10-3 × 1 × 10-6

= 1 × 10-9

As the ionic product is less than solubility product, precipitation of Cu(IO3)2 will

not take place.

(b) le- chatelier’s principle states that change in any factor such as

temperature, pressure, concentration etc., will cause the equilibrium to shift in

such a direction so as to reduce or counteract the effect of the change.

(i) Addition of H2 will shift the equilibrium in the forward direction.

(ii) Removal of CO will shift the equilibrium in the backward direction.

1

½

½

1

1

1

1

25. (a) First law of thermodynamics: The energy of an isolated system is constant.

q And w can depending on how the change is carry out. But the sum of q and

1



W i.e., U will depend only upon the initial and final state .That is why the sum

i.e. U is is a state function .

(b) rH0fH (products ) - fH (reactants)

= [0 +2× (-268.6)] - [23.0 + (-241.8)]

= (-573.2)-(23.0-241.8)

=-573.2 + 218.8

= -354.4 kJ = -354400 J

U = rH - -354400-[1×8.314×300]

= -354400-2494

=-351906 J or 351.9 kJ

OR

a) (i) Hg (l) Hg (g)

Disorder increase in going from liquid state to gaseous state.

Therefore S = -ve

(ii) Steam Water

Disorder decrease in moving from gaseous state to liquid state.

Therefore S = +ve

ng = 2 – 3 = -1

Ho = Uo + ngRT

= -10.5 kJ + (-1)(8.314 × 10-3 × 298 kJ)

= (-10.5 – 2.48) kJ

= -12.98 kJ

According to gibbs Helmholtz equation

Go = Ho - T o

= -12.98 kJ – 298 (-44.1 × 10-3 kJ)

= -12.98 kJ + 13.14 kJ

= 0.16 kJ

As Go is +ve, the reaction would not occur spontaneously.

½

½

1

1

1

½

1

½

½

1

½

½

½

26. (i) Wurtz reaction

CH3I + 2Na + ICH3 CH3 – CH3 + 2NaI

When an alkyl halide is treated with sodium metal in dry ether, bigger alkane is

formed.

1



ii) Friedal-crafts alkylation reaction

Cl + CH3 Cl + Cl CH 3

CH 3

(b) Alkene CH3CHO + CH3CH2COCH2CH3

Alkene reacts with ozone and gives ozonide products. This ozonide on

hydrolysis will give the products as under. Thus A has the structure

CH3 – CH2 – C = CH – CH3

CH 2

CH 3

3-Ethylpent-2-ene

( c) Distinction between ethane the ethyne:

Ethyne gives a white precipitate with ammoniacal silver nitrate (Tollen’s

reagent).

CH CH AgC CAg

Silver diacetylide

Ethane does not give this test.

Or

(a) (i) this is because of the resonance structure of benzene.

Thus, all the C – H bonds in benzene are equivalent and they have the same

bond length (139 pm) which is in between the C – C single bond length and C –

C double bond length.

(ii) trans-2-butene has a higher melting point than cis-isomer.

H CH3 CH3 CH3

C = C C = C

CH3 H H H

1

1

1

1

1

1


Cl + CH ClAnhyd. AlCl 3


Molecules of trans isomer can fit more tightly into one another giving a

compact lattice as compared to cis-isomer. Thus, the force of attraction

between molecules in trans-isomer will be more and hence, trans-isomer will

melt at a higher temperature.

(b) Distinction between but-1-yne and but-2-yne

CH3CH2C CH CH3C C – CH3

But-1-yne but-2-yne

But-1-yne has an acidic hydrogen (terminal hydrogen). It will give Tollen’s

reagent test and Fehling solution test. But-2-yne has no acidic hydrogen and

hence it will not give these tests.

CH3CH2C CH CH3CH2C CAg CH3CH2C CCu

White ppt. Red ppt.

( C ) (i) Ethene to benzene

CH2 = CH2 CH2ClCH2Cl CH2 = CHCl CH CH C6H6

(ii) 1-Bromopropane to 2-Bromopropane

CH3CH2CH2Br CH3CH = CH2 CH3 – CH – CH3

Br

1-Bromopropane propane 2-Bromopropane

1

1

1

**********************************************************





S.No Chapters V.S.A S.A S.A

Long

Answers

Total

Marks

1 Some basic concepts of chemistry 1 1 0 3

2 Structure of Atom - 1 1 - 5

3

Classification of Elements and Periodicity in

property 1 1 4

4 Chemical Bonding and Molecular Structure 1 - - 1 6

5 States of Matter 1 - 1 - 4

6 Thermodynamics 1 2*2 - - 5

7 Equilibrium - - 2*3 - 6

8 Redox Reactions - - 1 - 3

9 Hydrogen - - 1 - 3

10 S-block Elements - 1 1 - 5

11 P-Block Elements - - 1 1 8

12

Organic Chemistry: Some basic Principles And

Techniques - 2 6

13 Hydrocarbons - - 1 1 8

14 Environmental Chemistry - - 1(VBQ) - 4

TOTAL 5 5 12+1 3 70










(vi) Questions no.23 carry 4 marks and Q.No.25 to 26 are long answer questions of 5 marks

each.


1. What does a mole represent?

2. Which electrons take part in bond formation?

3. Locate the position of element with atomic number 33.

4. Define Charles law?

5. What is the entropy change when a liquid vaporizes?

6. Mention the conditions under which the reaction is spontaneous?

7. How many grams of NaOH should be dissolved to make 100 ml of 0.15 M NaOH solution?

8. (a.) Why Rutherford’s model was discarded?

(b.) Write the electronic configuration of S2-

9. The reaction of NH2

be -742.7 Kj/mol of cyanamide at 298 K. calculate enthalpy change for the reaction

NH2CN(s)+3/2 O2 (g) N2 (g) + CO2 (g) + H2O(l)

OR

Calculate the enthalpy change for the reaction:-

H2 (g) +Cl2 (g) 2HCl (g)

Given that bond energies H-H, Cl-Cl and H-Cl bonds are 433,244 and 431Kj/mol respectively

10 . Why is Li2CO3 decomposed at a lower temperature whereas Na2CO3 is decomposed at a higher temperature?



11. Write the equilibrium constant for the reaction :

(a) Fe3+

(aq) + 30H- (aq) Fe(OH)3 s

(b) Cu(NO3)2 (s) 2CuO (s) + 4NO2(g) + 02 (g)

ii) What is the conjugate acid of NH3?

12. Account for the following.-

(i) Na and K impart colour to the flame but Mg does not?

(ii) Lithium is the best reducing agent in aqueous solution?

(iii) Which out of the two- lithium or sodium forms nitrides?

13. (a) What is inert pair effect? Give one example.

(b) Why is CO2 gas whereas SiO2 is solid?

14. (a) Which of the following carboanion is more stable?

(i) (CH3)3CCH2 CH3CHCH 2CH3 CH3CH2CH2

(b) Draw the resonating structures for CO32-.

c) Among carbocations which is the most stable?

15. Explain the following

a) CCl4 does not give white precipitate with silver nitrate solution on heating. Why?

b) Why is O2NCH2O- is expected to be more stable than CH3CH2O-

16.Complete the following

(a)CH3Br + Mg dry ether

(b) 2CH3CH2I + Na dry ether

c) CH3COONa + NaOH +CaO

17. The kinetic energy of an electron is 4.55x10-25 J. the mass of electron 9.1x10-31 Kg. Calculate

velocity, momentum and the wavelength of the electron?

18. Give reasons:



a) Size of cation is always smaller than the parent atom.

b) Ionization enthalpy of nitrogen is greater than that of oxygen.

c) Electron gain enthalpies of beryllium and magnesium are positive.

19. What will be the pressure exerted by mixture of 3.2 g of methane and 4.4 g of carbon dioxide

contained in a 9 dm3 flask of 27o C.

20.The pH of 0.1 M solution of HCNO is 2.34. Calculate the ionization constant for the acid and its

degree of ionization in the solution.

21. (a) Identify the substance oxidized and reduced:

N2H4 + 2H202 N2 + 4H2O

(b) Calculate the oxidization No. of Carbon in CH3CH2OH

(c) Balance the following:

MnO4–+Fe2+ Mn2++Fe3 (Acidic medium)

OR

(a) Balance the following equation Fe2++Cr2O72- 3++ Cr3+

(b) What is salt bridge? Write its application also.

22.(a) Give two properties of Water which are due to Hydrogen bonding.

(b) Explain the Structure of H2O2.

23.Ramu had reared fishes ia local pond in the fields of his father.He was very fond of them.His

father had sprayed insecticides in the field to protect the paddy crop. However after a month

Ramu found that the percentage of fish in the pond had sharply declined.There was a thick green

covering of filamentous material and the water was putrifying.Ramu persuaded his father to turn

the direction of the irrigated water away from the pond?

(a) What may be the possible reasons for Ramu’s observation?

b) What do you mean by BOD

(c)What are the values depicted from this text about Ramu and his father?.

24. Give reason:

(a) Concentrated HNO3 can be transported in aluminum containers.

(b) Graphite is used as lubricant.

(c) A mixture if a dil. NaOH and aluminum piece is used to open holes.

(d) Carbon shows catenation but silicon does not.

(e) Tin (II) is a reducing agent but Pb(II) is not.



OR

A certain salt(X) gives the following results

(a) Its aqueous solution is alkaline to litmus.

(b) It swells up to glassy material (Y) on strong heating.

(c) When concentrated H2SO4 is added to hot solution of X, white crystals of an acid Z

separates out.

(d) Write equations for ii and iii. Also identify X, Y, Z.

25.(a) i) How many sigma and Pi bonds are present in CH2=CH—

ii) Why H2+ is more stable than H2 ?

iii) Why is B2 molecule paramagnetic?

(b) Discuss orbital structures of following molecules on the basis of hybridization

i) BH3

ii) BeCl2

OR

a) on the basis of VSEPR theory explain shapes of followingi) NH3

ii) H2Ob) Which of the two NH3 or NF3 has higher dipole moment?

c) Why N2 is more stable than O2?

d) Why is ethyl alcohol miscible with water?

26. a) Why terminal alkynes are acidic in nature?

b) Write IUPAC name of following

CH3—C—CH2—COOH

O

c) Write the following conversionsi) Propene to 2,3 -Dimethyl butane

ii) Ethene to Ethyne

iii) Benzene to Benzene sulphonic acid

OR

I. What effect does branching of an alkane chain has on its boiling point?

II. Define Ozonolysis Reaction?

III. Define cracking?

IV. Why benzene is extra ordinary stable though it contains three double bonds?

V. Why Nitro-benzene doesn’t undergo Friedel-Craft alkylation?



SESSION ENDING EXAMINATION

CLASS:XI SUB: CHEMISTRY


Q No1.2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

Value Points6.022 x 1023

Valance electrons present in the outermost shell take part in the bond formationPeriod – 4 and group -13

At constant pressure, volume of a given mass of a gas increases or decreases by 1/273 of it’s volume at 00 C for every 10 rise or fall in temperature.

Positive or increases.-

–ve the reaction is spontaneousMolarity = WB/MB x 1000/vol. of sol. in ml.

MB = 23+16+1 = 40 WB = 0.15 x 40 x 100/1000 = .6g

(a) Because it couldn’t explain the stability of atom. (b) 1s2 2s2 2p6 3s2 3p6

ng = 2-3/2 = ½

Correct substitution of values,Correct answer -741.5 kJ/ mol OR rH=£B.E of H-H - £B.E of product =[B.E of H-H – B.E of cl-cl] =[2 x B.E of H-Cl bond]Correct substitutionCorrect Answer= -185KJ.

Lithium is less electropositive than sodiumLithium is smaller in size , thus lattice enthalpy of Li2CO3 is lessKc = 1 .

[Fe][OH]

Kc = [NO2(g)]4[O2(g)]Conjugate acid is NH4

+

Due to small size if Na and K, less energy required for the excitation of e-, whereas in Mg atom, due to small size, large amount of energy is required which is not possible in Bunsen flame.Due to most negative reduction potential.

The reluctance of valence s-electrons to participate in bond formalin.

Marks

11

1/2,1/2

1

1

11

1+1

1+1

11

11

1+1

1+1+1



13.(i)

(ii)

14.(a)(b)

c)

15(a)(b)

16(i)(ii)(iii)17

E.g.:- In gp 13 stability of +1 oxidation state increases down the group.

gas-dimensional network structure.

Correct resonating structuresTertiary carbocation due to inductive effect and hyperconjugation

Because CCl4 is covalent and cannot give chlorine ion in solution.Because of electron withdrawing nature of NO2

CH3MgBrCH3CH2CH2CH3 + 2NaICH4

K.E= ½ mv2

x 4.55 x 10-25)/(9. 1 x 10-31) 103 m/sMomentum, =mv =9.1 x 10-31 x 103

= 9.1 x 10-28 kg m/s

= 6.629 x 10-34 .

9.1 x 10-31 x 103

=7.28 x 10-7 mBecause of less number of protons in cation than in parent atom, the effective nuclear charge is more in cation than in parent atom.Because nitrogen has exactly half filled electronic configuration i.e 1s22s22p3

Because they have stable configuration

P = nRT V

PCH4 = (WCH4.RT)/CH4 ..V)

PCH4 = 3.2 x 0.0821 x 300 = 0.55 atm 16 9

PCO2 = 4.4 x 0.0821 x 300 = 0.27 atm44 9

Total pressure = PCH4+PCO2

=0.55+0.27=0.82 atm

1+1+1

1+1

1

1

1

1

11/2+11/2

111

1/2

1/2

½½

½½


CH3CH2CH2


18 (1)

(2)

(3)

19.

(a)

(b)

20.(a)(b)

21.(a)

(b)

(c)

21.

+ + CNO-

pH=2.34=-log[H+][H+]=Antilog3.86= 4.57 x 10-3 M[CNO-]=[H+]=4.57 x 10-3 MKa= (4.57 x 10-3 )( 4.57 x 10-3 ) 0.1 =2.09 x 10-4

a-4)0.1)}

= 0.0457Substance oxidized N2H4

Substance reduced H2O2

C2xH6

+1O-2

2x+6-2=0 x=-2Balancing can be done by any method

8H++MnO4-+5Fe2+ 2++4H2O+5Fe3+

OR Fe2++Cr2O7

2-+H+ 3++Cr3++H2O Oxidation no. inc by Fe2++Cr2

6+O72-+H+ 3++Cr3++H20

| 6Fe2++Cr2O7

2-+14H+ 3++2Cr3++7H2O

Correct answer, maintains the internal circuit.

(i) It exist as liquid at room temperature(ii) Its boiling point is high H2O2 has a non planar structureThe O-H bond are on different planes thus the structure of H2O2 is like open book.

1

1

1

1

1/2

1

11/2

1

1

1

½½

1/2

11/2

½½



22.(1)

(2)

23.a)b)

c)

24.(a)

(b)

(c)

(d)

(e)

Algal bloom and eutrophicationThe amount of oxygen consumed by micro organism in decomposing organic waste of sewage water.

Ramu has knowledge of environmental chemistry and is ecofriendly.The father is ignorant and self obsessed.

Because HNO3 reacts with aluminum to form a thin film of aluminum oxide on its surface which protects it from further reaction

2Al(s) + 6HNO3 2O3(s) + 6NO2(g) + 3H2O(l)Graphite has layered structure which are held by weak Vander Waals forces thus graphite cleanes easily between the layers. Therefore, it is very soft and slippery that’s why it is used as lubricant.

NaOH reacts with Al to evolve H2 gas Thus the pressure of the gas produced can be used for clogged drains.

2Al(s) + 2NaOH(aq.) + 2H2 2(aq.) + 3H2(g)Because carbon is small in size and form C- -

Due to inert pair effect Pb2+ is more stable than Pb4+ whereas Sn4+ is more stable than Sn2+ thus Sn2+ is a good reducing agent and Pb2+ is not.

OR2B4O7

2 + B2O3

3BO3

Na2B4O7.10H2 2B4O7 2 +B2O3

Glassy bead

Na2B4O7.10H2O + H2SO4 3BO3 + Na2SO4 + 5H2O

(ii) Both the ions have the same bond order (O.S) but they differ in their configuration

H2+ 1

H2- 2 1

Since H2- ion has an electron in the * bonding molecular orbit it is therefore,

less stable.(iii) B2

2 2 2x]1

y]1

Since B2 has two unpaired electron thus it is paramagnetic

(i) correct structure, Hybridization(ii) correct structure, Hybridization OR

1

1

1,1

1

1+1+2

1

1

1

1

1



(b)

25.(a)

(b)

26.a)

b)

(a) correct shapes(b) NH3 due to more +ve dipole moment(c) N2 has bond order 3 whereas has O2 Thus due to more BO of N2,it is

more stable(d) Because of hydrogen bonding

Due to more s-character of the C-H bond of 1-alkyne 3- Oxobutanoic acid

HBr(i) CH3-CH=CH2 3-CH-CH3 Dry ether CH3-CH-CH-CH3

| Na | | Br CH 3

CH3

Br2/Ccl4(ii)CH2=CH2- -CH2-CH2- CH2

ORi) As branching increases, the surface area of alkene approaches that of a sphere. Since a sphere has minimum surface area, therefore, Vander walls forces of attraction are minimum and hence the boiling point of the alkene decreases with branching.ii) When an alkene is treated with ozone at low temperature ozonoid is obtained which when heated with Zn dust and water gives aldehydes and ketones depending upon the nature of alkene. Reaction is called ozonolysis.

iii) The thermal decomposition of higher hydrocarbons into lower hydrocarbons in presence or absence of a catalyst is called cracking.iv) Due to resonance.v)The NO2 groups strongly deactivates the benzene ring for the attack of an electrophile.

1*3=3

2

1

1

1

1+1

1+111

1

1

1

1

1





S.No. Unit VSA(1Mark)

SAI(2Mark)

SAII(3Mark)

VB (4Mark)

LA (5Mark)

Total

Unit I Some Basic concept of chemistry

2(1) 3(1) 5(2)

Unit II Structure of atom 1(1) 2(1) 3(1) 6(3)Unit III Classification of elements

& periodicity in properties1(1) 3(1) 4(2)

Unit IV Chemical Bonding & Molecular Structure

1(1) 5(1) 6(2)

Unit V States Of Matter 1(1) 3(1) 4(2)Unit VI Thermodynamics 2(1) 3(1) 5(2)Unit VII Equlibrium 1(1) 2(1) 3(1) 6(3)Unit VIII Redox Reaction 3(1) 3(1)Unit IX Hydrogen 3(1) 3(1)Unit X s-Block elements 5(1) 5(1)Unit XI p-Block elements 2(1) 3(1) 5(2)Unit XII Organic chemistry-Some

Basic Principles & Techniques

6(2) 6(2)

Unit XIII Hydrocarbon 3(1) 5(1) 8(2)Unit XIV Environmental Chemistry 4(1) 4(1)Total 5(5) 10(5) 36(12) 4(1) 15(1) 70(26)










(vi) Questions no.23 carry 4 marks and Q.No.25 to 26 are long answer questions of 5 marks

each.


Q.1 Calculate the total no. of angular nodes and radial nodes in 3p orbital (1)

Q.2 Which of the following have high lattice energy CsF, CsCl, CsBr, CsI.and Why?(1)

Q.3 Which has high electron affinity-F or Cl. (1)

Q.4 Write the conjugate base for HF & H2SO4. (1)

Q.5 Writes Vander Waal’s equation. (1)

Q.6 An element with massno.81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.(2)

Q.7 Write the expression for the equilibrium constant Kc for each of the following reaction:

(i) Fe+3(aq)+3OH- (aq) ---- Fe(OH)3 (s)

(ii) I2 (s)+5F2(g) ---- 2IF5(g) (2)

Q.8How many significant figures are present in the following calculations:

(i) 5 X 5.364 (ii)0.0125+0.7676+0.0215 (2)

Q.9 Give reasons:

(i) Atomic radius of Ga is less than that of Al.

(ii)BF3 is not easily hydrolysed by water.



Q.10 fH0 of HClif bond energy of H-H bond is 437kJmol-1Cl-Cl bond is 244kJmol-1 and H-Cl is 433 kJmol-1.

Q.11 An organic compound containing carbon, Hydrogen, Oxygen gave the percentage composition as C=40.687% H=5.085% The vapour density of the compound is 59. Calculate the molecular formula of the compound.

Q.12 (i) Define these: Aufbau Principle Hund’s Rule of maximum multiplicity

(ii) Write the electronic configuration of Cu+

Q.13 A student forgot to add the reaction mixture to the round bottomed flask at 270c but instead he placed the flask on the flame.After a lapse of time he realised his mistake and using the pyrometer he found that temperature of flask was 4770c. What fraction of air would have been expelled out? (3)

Q.14 Account for these:

(i) size of Na+ is lesser than Na.

(ii) Nitrogen has high ionisation energy than Oxygen.

(iii)Sixth period of the periodic table should have 32 elements.(3)

Q.15 (i) Balance the following equation by ion electron method:

Cr2O7-2(aq)+ SO2(g)--- Cr+3(aq)+SO4

-2(aq)( IN ACIDIC MEDIUM)

(II) Assign oxidation no. of Mn& N in KMnO4 And NO3- (3)

Q.16

(i) For the reaction 2Cl (g)---

r0

r will be zero at equilibrium?

(ii) why the entopy of perfectly crystalline substance is taken as zero? (3)

Q.17 Write the following naming reactions

(i) Wurtz Reaction (ii) Friedal-Craft Alkylation (iii) Peroxide effect (3)

Q.18

(i) Calculate the solubility of A2X3 in pure water, assuming that neither reacts with water.Ksp=1.1X10-23

(II) What will be the pH of 1M Na2SO4 sol? (3)

Q.19 (i) Write one method to remove the permanent hardness of water.

(ii) Is demineralised water fit for drinking purposes ?How can it be made useful?



(iii)Write one laboratory method of preparation of hydrogen gas(3)

Q.20 0.3780g of organic chloro compound gave 0.5740g of silver chloride in carius estimation .Calculate the percentage of chlorine present in the compound. (3)

Q.21 Write balanced equation :

(i) BF3+LiH --

(ii) NaH +B2H6--

(iii) B(OH)3 + H2O-- (3)

Q.22

(i) Write IUPAC Name of following compound: OHC-CH2-CH2-CHO

(ii) Explain InTolune CH3 group is O &P directing in nature.

(iii) Why HCOOH is more acidic than CH3COOH? (3)

Q.23 In slum areas in winter season people often burn coke angithis to heat up their rooms & then sleep their whole night.Sometimes cases of unconsciousness &death are reported from these houses,not because of their houses catches fire but because of the poisonous gases accumulated in their room.Even in urban areas, in winter where people keep their electric heaters on the whole night,incidents occur due to overheating of the electric heater or short circuiting.

After reading the paragraph answer the following questions:

(i) What values are expressed in above paragraph?(ii) How is poisoning caused by coke angithi ?(iii) what steps should be taken immediately as a part of treatment? (4)

Q.24

(i) Draw energy level diagram of O2. Calculate its bond order& write its magnetic property.

(ii) On the basis of VSEPR Theory Explain the Geometory Of; ClF3& CH4

Q.25 Give reasons for each:

(i) BeO is insoluble but BeSO4 Is soluble in water.

(ii)Be & Mg Do not impart colour to the flame .

(iii)Gypsum is added in the final stages of prepration of cement.

(iv)Alkali metals give deep blue color when dissolved in liquid ammonia.

(v)Li Forms oxide Na forms peroxide & K forms superoxide (5)

OR



(i) Write the balanced equation for the reaction of:Na2O2 and water Na2O and CO2

(ii) Write the various steps involved in the Solvay process for the prepration of Na2CO3

Q.26 (i) Arrange Benzene, n-Hexane &ethyne in decreasing order of acidic behaviour. Give reason.

(ii)Propanal& pentane-3-one are the ozonolysis products of an alkene .What is the structural formula of the alkene?

(iii) What happened on heating of phenol with Zn dust. Write reaction.(iv) Convert propyneino acetone.

OR

(i)Explain Markovnikov,sRule.write its mechanism.

(ii) Convert These:

Benzene into acetophenone.

Ethene into Benzene.





S.No Answer Value Point1 angular nodel=1

Radial node=n-l-1=3-1-1=11/21/2

2 CsF. Small size of anion ½+1/23 Cl 14 F- HSO4

- ½+1/25 (p+an2/V2) (v-nb)=nRT 16 A=p+n=81

P=x n= X+31.7% of x=1.317xX=35 symbol Br35

81

1/21/21

7 Kc=1/[Fe+3].[OH-]3

Kc=[IF5]2/[F2]511

8 (i)=4 (ii)=4 1+1=29 i) poor shielding of d orbital in Ga

(ii)Strong B-F bond due to back bonding 1

110 (i) Bond energy is the amount of energy released when bond is

formed between t wo isolated atoms in ground state. (ii) 1/2 H2 (g) + 1/2Cl2 fH0 = ?

fH0 = -

fH0 = {1/2 x 437 + 1/2 x 244} -{433} = - 92.5kJmol-1

1

1

11 writing correct sep.calculating empirical formula (C2H3O2)correct molecular formula (C4H6O4)

111

12 .(i) Aufbau Principle: the orbitals are filled in increasing order of energy.Hund’s Rule: electron pairing did not take place until each orbital is first singly filled with parallel spin(ii) 3d10

1+1=21

13 consider Volume =V At 270cV1=V V2=? T1=300K T2=750KV1/T1=V2/T2 (1)V2=2.5VVolume Expelled=2.5V-V=1.5 V Fraction of air expelled=1.5v/2.5V=0.6

1

1

114 (i) High nuclear charge in Na+ after removal of electron,

No of shell reduces.(ii) Completely half filled configuration in N , symmetrical

distribution leading to stability.(iii) n=6 involve filling of4f,5d,6s&6p orbital hence

32elements only

1+1+1=3



15 Correct balanced equation Cr2O7

-2(aq)+3 SO2(g) +2H+--- 2Cr+3(aq)+3SO4-2(aq) +H2O

O.S. Of Mn=+7 N= +5

2

½+1/216 - -ve

r=0 at equilibrium (iii) due to orderly arrangement of atoms.

111

17 (i) Wurtz synthesis: Higher alkanes are prepared by heating an alkylhalide (RX) with sodium metal in dry ether solution.

R-X +2Na + XR dry ether R-R + 2NaXCH3Br + 2Na + BrCH3dry ether CH3CH3 + 2NaBr(ii)

(iii)CH3-CH=CH2 +HBr--Peroxide------ CH3CH2CH2Br

1

1

1

18 (i) KSP=(2S)3. (3s)3=1.1 x10-23

S=1.0X 10-5mol/l (ii) pH =7 (neutral soi. Of strong acid & strong base)

111

19 (i)Treatment with washing soda (sodium carbonate):MCl2 + Na2CO3--- MCO3+2NaCl Or any Method

(ii) No, By adding ions(iii) Generally prepared by the reaction of granulated zinc

with dilute hydrochloric acid as underZn(s)+2HCl --- ZnCl2 +H2 (g)

1+1+1=3

20 Write formula Value substitution Write Ans% of chlorine = 35.5*mass of AgCl formed * 100/143.5 * mass of the substance = 35.5*0.5740*100/143.5*0.3780

= 19.6 %

111

21 B2H6+ LiFNa+[BH4]-

[B(OH)4]-+ H3O+

111

22 .(i) butane1,4dial(ii) due to Hyperconjugation.(III) HCOOH is more acidic +I effect in CH3COOH

111

23 (i)Awareness regarding the harmful effect of coke , Careless attitude ,ignorance regarding harmful effect of coke(ii)coke burns to produce CO , which combine with Hb to form Carboxyhaemoglobin& function of hb to carry Oxygen to different parts get inhibited.(iii)Any Two steps

Carry the patient into fresh air immediately.Artificial respiration, if not possible to breathe properly.

1

1

2

24 Energy level diagram 2



Bond order =2 Paramagnetic (ii) ClF3- T shape CH4- Tetrahedral

½½2

25 Correct Reason (i) BeO is covalent & BaSO4 is ionic(ii) Small size &high ionization energy(iii) To slow down the process of setting of cement&

impart greater strength.(iv) Due to the formation of ammoniated electron(v) Smaller cation can stablise smaller anion & larger

cation can stablise larger anion.ORNaOH +H2O2

Na2CO3

Correct steps in Solvay Process In Solvay process ,CO2 is passed through brine saturated with ammonia .when NaHCO3

being sparingly soluble gets precipitated.NaCl+ NH3 + CO2 - NaHCO3 + NH4Cl (i) Sodium bicarbonate thus formed is filtered, dried and then heated when sodium carbonate is

Obtained. 2 NaHCO3 ------> Na2CO3 + CO2 + H2O CO2 needed for the reaction shown in eq.(i) is prepared by heating calcium carbonate and the quick lime,CaO thus formed is dissolved in water to form slaked lime, Ca(OH)2.

NH3 needed for the purpose is prepared by heating NH4Cl

1+1+1+1+1=5

113



26 (i) ethyne> benzene> HexaneDue to greater S-Character. (ii)3-Ethylhex-3-ene

(iii)(iv)

ORDuring addition across unsymmetrical compound the negative part go toward that side which contain less no. of Hydrogen atom.CH3-CH=CH2 +HBr-- CH3 CH(Br) CH3

Conversion:

(1+1=2)

111

***************************





S.no

Name of the chapter Weightage

of the

chapter

VSA

-1

SA-

2

SA-

3

SA-4

Value

based

SA-

5

Total=70

1 Some basic concepts in Chemistry

5 1(1) 2(2) 3(5)

2 Structure of atom 6 1(2) 2(3) 3(8)

3 Classification of elements and periodicity in properties

4 1(1) 1(3) 2(4)

4 Chemical bonding and molecular structure

7 1(1) 2(3) 3(7)

5 State of matter 4 1(1) 1(3) 2(4)

6 Thermodynamics 4 1(1) 1(3) 2(4)

7 Equilibrium 7 1(2) 1(5) 2(7)

8 Redox reaction 3 1(3) 1(3)

9 Hydrogen 3 1(3) 1(3)

10 s-Block elements 5 1(2) 1(3) 2(5)

11 p-Block element 5 1(5) 1(5)

12 Organic-Some basic principles and techniques

6 2(3) 2(6)

13 Hydrocarbons 7 1(2) 1(5) 2(7)

14 Environmental Chemistry 4 1(4)

15 Total 70 5(1)=5

5(2)=10

12(3)=36

1(4)=4

3(5)=15

30(70)





(viii) All questions are compulsory.

(ix) Marks for each question are indicated against it.

(x) Questions no.1to 5 is very short answer questions and carries 1 marks each.

(xi) Questions no.6to 10 is short answer questions and carries 2 marks each.

(xii) Questions no. 11to22 are also short answer question and carry 3 marks each.

(xiii) Questions no.23 carry 4 marks and Q.No.25 to 26 are long answer questions of 5 marks

each.

(xiv) Use long Tables, if necessary. Use of calculators is not allowed.

Q1. Define limiting reagent.

Q2. Write IUPAC name and the symbol for the element with atomic number 120.

Q3. Write the Vander Waals equation for ‘n’ mole of a gas.

Q4. Predict the sign of entropy change for the reaction:2O3 2(g)

Q5. What is the state of hybridization of ‘S’ in SF4 molecule?

Q6. Calculate the number of molecule and number of atoms of each type in 5.3g of Na2CO3.(Atomic mass of sodium =23g ,carbon =12g and Oxygen = 16g)

Q7.Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040(Assume the density of water as one).

Q8. Calculate the Ph of solution obtain when 4g of sodium hydroxide has been dissolved in to 500 ml of the solution.

ORDerive the relationship between Kp and Kc.

Q9. Account for the following K2CO3 cannot be prepared by Solvey ammonium processAlkali metals are not found in nature

Q10 An alkene ‘A’ upon ozonolysis gives methnal and propan 2-one .What should be the structure of the alkene.

Q11. Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.



Q12. (a) How many electrons in an atom may have the following quantum numbers?n = 4, ms = – ½ (ii) n = 3, l = 0 (b) What is the maximum number of emission lines when the excited electron of a Hatom in n = 6 drops to the ground state?

Q13 Consider the following species : Al3+, F- , Mg2+ , Ne ,Na1+ O2- and N3-

(i) What is common in them?(ii) Arrange them in the order of increasing ionic radii.(iii) What are these called as.

Q14 What is meant by the term bond order ? Calculate the bond order of : O2, O2+ O2

- and O22-

species and arrange these species according to their decreasing order of stability.

Q15. What is meant by hybridization .Explain sp and sp2 hybridization with an example?

Q16. For a reaction .0 =-10.5 kJmol-1 0 = - 44.10 Jmol-1 0 for the reaction

and predict whether the reaction may occur spontaneously.

Q17. 2.9 g of a gas at 950 C occupied the same volume as 0.184 g of hydrogen at 17 0 C at the same pressure. What is the molar mass of the gas?

Q18. (a) What is the oxidation number of ‘S’ in HSO4- ion?

(b)Balance the following ionic equation by ion electronic method.P4(S) + OH-

3 + H2PO2-

Q19. (a) Write a chemical equation to justify that hydrogen peroxide function as oxidizing. (b) Explain why hydrogen peroxide is stored in wax lined glass or plastic bottles (c ) Ice float on the surfaced of the water.

Q20 Explain the following terms(i) Beryllium and magnesium do not give colour to the flame whereas other alkaline

earth metals do so , Why?(ii) When alkali metals dissolved in dilute liquid ammonia, the solution acquire blue

colour, why?(iii) Alkali and alkaline earth metals cannot be obtained by chemical reduction method,

why?

Q21. Write the IUPAC name of following organic compound.(i) CH3CH2COCH2OCH3

(ii) CH2=CHCH (CH3) CH=CH2

Write the structure of the compound whose IUPAC name is 4-Hydroxy phenol.

Q22. (a) Find the total number of sigma and pi bonds in the given molecule.CH3CH (CH3) COCH2NO2

(b) What are carbocations . Arrange the primary secondary and tertiary carbocations according to their decreasing order of satiability.



Q23. A trainee pilot was flying his plane in stratosphere . His senior advised him not to fly the aero plane in the stratosphere.Answer the following the following questions:In your opinion , why the senior pilot advised his trainee not to fly the aeroplane in the stratosphere ?Write the possible chemical reaction affecting the stratosphere Mention the possible value associated with your reply.

Q24. (a) What will be the Conjugate acids for the following Bronsted bases,(i) NH2

- ,NH3,HCOO- and H2O(b) What is Le chatlier principle? Under what conditions would we get maximum yield of ammonia in the Haber’s process? OR

Q. (a) What are buffer solution .Explain with an example.

(b) A sample of HI(g) is placed in a flask at a pressure of 0.2 atm . At equilibrium, the partial pressure of HI (g) is 0.04 atm. What is Kp for the given equilibrium?

Q25.(i) Is boric acid a protic acid? Explain. (ii) Suggest a reason why the B-F bond in BF3 is shorter than B-F bond in BF4

-.(Iii) Diamond is used as abrasive Draw the structure of diborane .

OR

Q (a) When a metal (X) is treated with sodium hydroxide , a white precipitate (A) is obtained , which is soluble in excess of sodium hydroxide to give soluble complex (B) . Compound (A) is soluble in dil HCl to form compound (C) . The compound (A ) when heated strongly gives (D) which is used to extract metal . Identify X ,(A) ,(B) ,(C) and (D) . Write suitable equation to support your identity. (b) Explain Why Conc. HNO3 cannot be transported in aluminium container.( c) Explain Why Conc. Aluminium utensils are not kept in water overnight .

Q26. (a)The addition of HBr to propene yields 2-Bromo-propane, while in the presence of benzoyl peroxide the same reaction yields 1-Bromopropane. Explain and give the mechanism. (b) Explain Wurtzs reaction with an example .Why this method is not suitable for preparing odd number of carbons alkanes. 3+2

OR (a) Draw cis and trans isomers of But-2ene. Which isomer have higher boiling point and why?(b) Explain the following terms with an example.(i) Friedel craft reaction (ii) Kolbe’s electrolysis (iii) Kharash effect




TIME: 3 Hours M.MARKS:70Ans1. The reactant which get completely consumed 1 mark

Ans2. Ubn unbinilium ½ +1/2

Ans3. Correction equation 1 marks

Ans4. Entropy increases 1 marks

Ans5. SP3 1 Marks

Ans 6. 3.01 x1022 , 6.01 x1022 ,9.03 x1022 2 marksAns7. Correct formula ½ mark Correct Calculation- 1 M Correct Calculation - 1/2 M

Ans8. Correct calculation , answer 13 – 2 marks

Ans9. (i) Because KHCO3 is soluble in water . (ii) These are reactive metals 1+1

Ans10. 2-Methyl But-1-ene 2

Ans11. Correct formula ,,correct calculation and correct answer 1+ 1 +1

Ans12. (i) 16 electrons, 2 electrons 2(ii) 15 lines 2+1

Ans13. (i) All have 10 electrons (ii) Correct order Iii Iso -electronic species ½+1+1/2

Ans14. Definition Bond order O2 = 2 O2

+= 2.5 O2-1 = 1.5 O2

2- =1.0Correct order 1+1+1

Ans15. Correct definition , Correct explanation with example 1+2marks

= 12.978Kj Mol-1

G=0.164Kj Mol-1

Ans17. Apply formula PV=w/MRTM= 40 g mol-1 3marks

Ans 18. O.N = +6 Correct balancing 1+2

Ans19. Write one equation in support of your answer 1+1Due to its oxidation in air in presence of sunlight 1



Ans20. (i) Due to high ionization enthalpy of the Be and Mg(ii) Due to ammoniated electronIii Very reactive matal 1+1+1 marks

Ans21. Correct IUPAC name 1+1+1

Ans22. Correct number of sigma and Pi bonds Correct explanation and correct order 1+1+1Ans23. Chloro fuloro carbon Freon gas released by aero plane reacts with stratosphere

Correct equation Concern for environmental pollution. 1+1+1+1

Ans24. (i) NH3 NH4+ , HCOOH and H3O+

High pressure , low temperature ,use of excess N2 And H2 2+3

Ans25 (i) It is a Lewis acid(ii) SP2 hybridisation in BF3Iii it is hard

Iv Correct shape 1+1+1+2

OR

X= Al metalA=Al(OH)3

B=NaAlO2

C=AlCl3D= Al2O3

Ans26. (i) Peroxide effect 1+1+1(ii) Correct wurtz reaction 2

OR

Correct drawing of Cis and trans isomers 2 MarksCorrect explanation of each effect 1+1+1

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session ending examination class:xi sub: chemistry …

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