cbse class 11 chemistry sample paper 2013 (5) · 2020-05-24 · 1 xi - chemistry question paper –...

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XI - CHEMISTRY

QUESTION PAPER – 5

Max. Time: Three Hours Max. Marks: 70

General Instructions:

1. All questions are compulsory.

2. Question no. 1 to 8 are very short answer questions and carry 1 mark each.

3. Question no. 9 to 18 are short answer questions and carry 2 marks each.

4. Question no. 19 to 27 are also short answer questions and carry 3 marks

each.

5. Question no. 28 to 30 are long answer questions and carry 5 marks each

6. Use log tables if necessary, use of calculators is not allowed.

1. What happens when sodium metal is heated in free supply of air? 2. Given:

Determine ∆H Ө for the reaction 2Al2O3 (s) → 4Al (s) + 3O2 (g).

3. Explain why BeH 2 molecule has zero dipole moment although the Be- H

bonds are polar? 4. Predict the shape of the PH 3 molecule according to VSEPR theory.

5. Which isotope of hydrogen is radioactive?

6. Write the correct IUPAC name of the compound given below:

7. How many mono substituted derivatives of naphthalene are possible?

8. Name any two gases responsible for greenhouse effect.

9. Arrange the following ions in order of increasing ionic radius:K+, P3-, S2-,

Cl-. Give reason.

10.The successive ionization energies of a certain element are I1 = 589.5

kJ/mol, I2 = 1145 kJ/mol, I3 = 4900 kJ/mol, I4 = 6500 kJ/mol, I5 = 8100

kJ/mol. This pattern of ionization energies suggests that the unknown

element is: a) K b) Si c) Ca d) As. Explain.

11.A sample of gas occupies 3.00 L at 760 torr. Calculate the volume the gas

will occupy if the pressure is changed to 1.45 atm and the temperature

remains constant.

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12.A mixture of hydrogen and oxygen at 1 bar pressure contains 20 % by

mass of hydrogen .Calculate the partial pressure of hydrogen.

13.Complete the following reactions:

14.Explain :

i) Alkali metals are soft and can be cut with help of a knife.

ii) Potassium is more reactive than sodium.

15.

i) How would you distinguish between BeSO4and BaSO4?

ii) Which is thermally most stable alkaline earth metal carbonate among

MgCO3, CaCO3, SrCO3, BaCO3? Why?

16.Why SiCl4 can be easily hydrolysed but CCl4 cannot be hydrolysed easily?

Explain with reaction.

17.Arrange benzene, hexane and ethyne in decreasing order of acidic

behaviour .Also give reasons for this behaviour.

18.State the difference between classical smog and photochemical smog.

19.50 kg of N2 (g) & 10.0 kg H2 (g) of are mixed to produce NH3 (g) identify

the limiting reagent. Also, calculate the amount of NH 3 formed.

20.

i) Calculate the wavelength in nm, of visible light having a frequency

of 4.37 x 10-14/s.

ii) What are frequency and wavelength of a photon emitted during a

transition from n = 6 to n = 1 state in the hydrogen atom.

21.

i) Explain why the following electronic configuration is not

possible:

ii) Write electric configurations of Cu & Cu2+.

22.

i) Draw the resonating structures of carbon dioxide molecule.

ii) Why is NF3 trigonal pyramidal while BF3 is trigonal planar, though

both are tetra atomic molecules?

iii) State the hybridization of carbon atoms in the fig below numbered




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1 & 2:

23.Consider the reaction:

Indicate the direction in which the equilibrium will shift when:

(i) Concentration of SO2 is increased.

(ii) Concentration of SO3 is increased.

(iii) Temperature is increased.

24.

i) Calculate the concentration of hydroxyl ion in 0.1 M solution of

ammonium hydroxide having Kb = 1.8 x 10-5.

ii) Ksp valueof two sparingly soluble salts Ni (OH)2 and AgCN are 2 x 10-15

and 6.0 x 10-17 respectively. Which salt is more soluble?

25.Write the balanced equation by half-reaction method:

(in acidic medium)

26.

i) Draw the structural isomers of pentane.

ii) Give the IUPAC names of the following compounds.

a)

b)

c)

27.

i) 0.2475 g of an organic compound gave on combustion 0.4950 g of

carbon dioxide and 0.2025 g of water. Calculate the percentage of




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C and H in it.

ii) What will happen during Lassaigne's test for nitrogen if the

compound also contains sulphur?

28.

i) In a process, 701 J of heat is absorbed by a system and 394 J of

work is done by the system. What is the change in internal energy

for the process?

ii) The equilibrium constant for the reaction is 10. Calculate the value

of ∆Gθ. Given R = 8.0 J/mol, T = 300 K.

Or

Calculate the lattice energy for the change: Li+(g) + Cl-(g) � LiCl(s)

Given that:

,

,

29. i) When happens when borax solution is acidified. Write the chemical

reactions for the reaction.

ii) Explain why BF3 exists whereas BH3 does not?

iii) SiO2 is solid but CO2 is a gas at room temperature.

Or

When a metal X is treated with NaOH a white precipitate (A) is

obtained, which is soluble in excess of NaOH to give soluble

complex (B). Compound (A) is soluble in dilute HCl to form

compound (C).The compound (A) when heated strongly gives D

which is used to extract metal. Identify (X), (A), (B), (C) & D. Write

suitable equations to support their identities.

30. Complete the following reactions:

i)




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ii)

iii)

iv)

v)

Or

a) Convert benzene to

b) Convert to




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ANSWERS - QUESTION PAPER – 5

1. On heating sodium in free supply of air, it forms sodium peroxide.

2 Na(s) + O2 (Air) → Na2O2

2. = 2 x (+ (1,670)) kJ/mol = + 3,340 kJ/ mol.

3. BeH2 is a linear molecule with H-Be-H bond angle as 180°. Although the

Be-H bonds are polar, the bond polarities cancel each other and the net

dipole moment is zero.

4. Trigonal pyramidal.

5. Tritium.

6. 2, 3-Dimethylpentane.

7. 3.

8. Carbon dioxide and methane.

9.

All the ions are isoelectronic with 18 electrons. If the number of electrons

is the same, as the number of protons increase, the nuclear charge

increases and hence the outermost electrons will experience a greater

force of attraction towards the nucleus. This results in the decrease in

ionic radii. Since the nuclear charge decreases K+ to P3-, from the ionic

radii increases from K+ to P3-.

10.The unknown element is Ca. Here the third ionization energy is very high

which suggest that the removal of the third electron is difficult. The

electronic configuration of calcium is [Ar] 4s2 first two electrons can be

removed without much difficulty. But the removal of third electron from

the stable electronic configuration of argon is difficult. Hence, the third

ionization energy is high.

11.Based on Boyle’s law, P1V1 = P2V2 (at constant temperature)

760 torr/760 torr/atm x 3L = 1.45 atm x V2 L

V2 = 2.07 L.

12.Since 20 % by mass of hydrogen is present in the mixture, mass of

hydrogen in the mixture is 20 g in 100 g of the mixture present.

Remaining mass in the mixture is of oxygen = 80 g

No.of moles of H = 20/2 g/mol = 10 mol




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No. of moles of O = 80/32 g/mol = 2.5 mol

Mole fraction of hydrogen

Partial pressure of hydrogen

Therefore, the partial pressure of hydrogen is 0.8 bar.

13.

14.

i. Alkali metals have large atomic size with only one valence electron.

Thus, they have weak metallic bonding between the atoms of the

metal Because of weak metallic bonding; alkali metals are soft and

can be cut with a knife.

ii. Reactivity of metals depends on ionization enthalpy. Smaller is the

ionization enthalpy, greater is the reactivity. Potassium has a larger

atomic size than sodium. Thus, the ionization enthalpy of potassium is

less than sodium. Hence, potassium is more reactive than sodium.

15.

i. Barium and Beryllium sulphate can be distinguished by solubility test.

Beryllium sulphate is soluble in water and barium sulphate is insoluble

in water.

ii. Barium carbonate is thermally most stable alkaline earth metal

carbonate because; its ion being larger in size is more stabilized by

larger carbonate ion through the formation of stable lattice.

16. In SiCl4, Si atom has empty d-orbitals in its valence shell. These empty




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d orbitals of Si can accept lone pair of electrons from water molecule.

Eventually this results in hydrolysis forming silicon tetrachloride and

silicon hydroxide.

Carbon atom on the other hand does not have any vacant d-orbitals in its

valence shell. Hence, it cannot accept the electron pair from water

molecule. Thus, CCl 4 does not hydrolyse.

17.The decreasing order of acidic behaviour is: Ethyne > benzene > n-

pentane. The C-H bond in ethyne, benzene and n-pentane are formed

overlap. Now, greater the percentage s character, greater is the

electronegativity. The C- H bond in ethyne, benzene and n – pentane is

formed by sp – s, sp2 – s, sp3 – s overlap. Now, greater the percentage s

character, greater is the electronegativity. Therefore, sp-hybridised

carbon in ethyne is more electronegative than sp2 hybridised carbon of

benzene which in turn is more electronegative than sp3 hybridised carbon

of n – pentane. Thus the polarity of C – H bond is in the order: Ethyne >

Benzene > Pentane.

18.

Classical smog Photochemical smog

Occurs in cool humid climate. Occurs in warm, dry and sunny

climate.

It is a mixture of smoke, fog and

sulphur dioxide.

Components of photochemical smog

result from the action of sunlight on

unsaturated

hydrocarbons & oxides of nitrogen

produced by automobiles &

factories.




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19.

------------------- (1)

According to eqn (1),

1 mole of nitrogen gas reacts with 3 moles of hydrogen gas

5.0 x 103 moles of hydrogen will give = 2/3 x 5 x 103 moles of NH3

= 3.3 x 103 moles of NH3

Mass of NH3 produced = 3.3 x 103 x 17 g of NH3 = 56.1 kg

20. i)




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∆E = 2.18 x 10-18 ( - )

Since ∆E is negative energy, the frequency of photon is given by

21.

i. For n = 1,

Value of l = n – 1 = 1 – 1= 0

For each value of l,

Value of ml = - 1, 0, +1

Therefore,

For n= 1, l = 0,

ml =0. So the value of ml =1 is not possible.

ii. The electronic configuration of Cu = 1s22s22p63s23p63d104s1. The

electronic configuration of Cu2+ = 1s22s22p63s23p63d9

22.

i. Resonating structures of carbon dioxide molecule




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ii. In NF3, N atom involves sp3 hybridization and one position is

occupied by a lone pair. Therefore the molecule is trigonal pyramidal.

But in BF3, B involves sp2 hybridization having trigonal planar

geometry. Thus NF3 is trigonal pyramidal while BF3 is trigonal planar,

even though both are tetra atomic molecules.

iii.

23.

i. If the concentration of SO2 is increased the equilibrium will shift in the

forward direction to consume the reactant SO2.

ii. If the concentration of SO3 is increased the equilibrium will shift in the

backward direction to consume the product SO3.

iii. If the temperature is increased, the equilibrium will shift in the

backward direction as the increase in temperature will be

compensated by absorbing heat.

24.

i.

Kb = [NH4+] [OH-]/[NH4OH]

ii.




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Since solubility of Ni(OH)2 is more than AgCN, Ni(OH)2 is more soluble

than AgCN.

25.

Step 1 : Write the oxidation no.s and separate the reaction into oxidation

half and reduction half reactions

Step 2: The half reactions are:

Step 3: Balance the Ox. No. by adding electrons

Step 4: There are no oxygen atoms. So, balance the hydrogen atoms.




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Since the reaction takes place in acidic medium, the balancing of

hydrogen atoms is done by adding the appropriate number of hydrogen

ions to the deficient side.

Step 5: Add the two reactions to get the balanced redox reaction.

26.

i. There are three structural isomers of pentane:

ii.

a) 2, 2, 3-Trimethylhexane

b) Ethylcyclopentane

c) 3-Ethylhexane

27.




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(ii) Blood red colouration due to Fe(CNS)will be produced.

28.

i. Heat absorbed by the system (q)= + 701 J

Work done by the system (w) = - 394 J

Change in the internal energy (∆U) = q + w = + 701 + ( - 394)

= + 307 J

ii. ∆Gθ = - 2.303 RT log K

Or

Substituting the values, we get

∆Hθ = - 12.978 kJ

Substituting the values for ∆Gθ = ∆Hθ - T∆Sθ , we get

∆Gθ = - 2816.2 J.

Since the value of ∆Gθ is negative, the reaction is spontaneous.

29.

i) Borax solution on acidification forms boric acid.




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ii) BF3 is trigonal planar molecule. Due to pπ - pπ back bonding lone

pair of electrons of F is back donated to B atom. This delocalization

reduces the deficiency of electrons of boron thereby increasing the

stability of BF3 molecule.

The mechanism is as follows:

Due to absence of lone pair of electrons on H atom this compensation

does not occur in BH3. In other words electron deficiency of B stays &

hence it reduces its electron deficiency as BH3 dimerises to form B2H6.

iii) Carbon is able to form pπ – pπ bond with O atom and constitute a

stable non - polar molecule O = C = O. Due to weak inter particle

force its boiling point is low and it is gas at room temperature. Si

on the other hand is not able to from pπ – pπ bond with O atoms

because of its relatively large size. In order to complete its octet Si

is linked to four O atoms around it by sigma bond & these

constitutes network structure, which is responsible for its solid

Or

30.




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Or

i)

ii)




cbse class 11 chemistry sample paper 2013 (5) · 2020-05-24 · 1 xi - chemistry question paper –...

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