for Volume - I€¦ · (b)= 3, the electron has more negative energy For n than it does for n = 5...

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[1]

Kind Attention to the Students � From this year onwards, blue print system has been abolished. � Please note that questions will be framed from IN-TEXT portions ALSO. � Approximately 20% of the questions will be asked from IN-TEXT portions. � These questions will be based on Reasoning and Understanding of the lessons. � Further, Creative and Higher Order Thinking Skills questions will also be asked. It requires

the students to clearly understand the lessons. So the students have to think and answer such questions.

� It is instructed that henceforth if any questions are asked from ‘out of syllabus’, grace marks will not be given.

� Term Test, Revision Test and Model Exam will be conducted based on the above pattern only. � Concentrating only on the book-back questions and/or previous year questions, henceforth,

may not ensure to score 100% marks. � Also note that the answers must be written either in blue ink or in black ink. Avoid using both

the colour inks to answer the questions. � For MCQs, the answers should be written in full. Simply writing (a) or (b) etc. will not get full

marks. You have to write (a) or (b) etc., along with the answer given in the options.

Govt. Model Question Paper - 2018-19 for Volume - Iwith Answer Key

andSura’s Model Question Paper for Volume - I

with Answer Key

CHEMISTRYTime : 2½ hours Written Exam Marks : 70 Marks

On 21.08.2018, Model Question Paper is released by the Govt. We have given it along with Answer Key.

11thSTD.

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[email protected] Ph: 8124201000 / 9600175757

Namma Kalviwww.nammakalvi.org

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PART - AANSWER ALL THE QUESTIONS: (15 × 1 = 15)1. Which one of the following is a standard for atomic

mass?(a) 6C

12 (b) 6C14 (c) 6C

13 (d) 6C14

2. The equivalent mass of a divalent metal element is 10g eq–1. The molar mass of its anhydrous oxide is(a) 46 g (b) 36 g (c) 52 g (d) none of these

3. Consider the following sets of quantum numbers n l m s(i) 2 1 –1 3/2(ii) 1 1 1 +1/2(iii) 1 0 +1 –1/2(iv) 1 0 0 –1/2Which of the following sets of quantum numbers is not possible ?(a) (i) and (ii) (b) (ii) and (iv) (c) (i), (ii) and (iii) (d) (i), (ii), (iii) and (iv)

4. Based on equation E = –2.178 × 10–18 Z2

2n

J certain

conclusions are written. Which of them is not correct?(a) Equation can be used to calculate the energy change

when the electron changes orbit.(b) For n = 3, the electron has more negative energy

than it does for n = 5 which means that the electron is more tightly bound in the smallest allowed orbit.

(c) The negative sign in the equation simply means that the energy of electron bound to the nucleus is lower it would be if the electrons were at the infinite distance from nucleus.

(d) Smaller the value of n, the larger is the orbit radius.5. Which of the following pairs of elements exhibit

diagonal relationship?(a) Be and Mg (b) Be and Al (c) Be and B (d) C and Si

GOVT. MODEL QUESTION PAPER - I11th

STD.

6. The first ionization energy (IE1) and second ionization energy (IE2) of elements A, B and C are given below

Element A B CIE1 kJ mol–1 2370 522 1680IE2 kJ mol–1 5250 7298 3381

which one of the above elements is the most reactive metal?(a) A (b) B (c) C (d) A and C

7. Ionic hydrides are formed by(a) halogens (b) chalcogens (c) alkalimetals (d) inert gases

8. Volume strength of 0.5N H2O2 is(a) 2.8 (b) 8.4 (c) 5.6 (d) 16.8

9. Ionic radius of alkali metals are in the following order(a) Li < Na < K < Rb < Cs (b) Na < Li < K < Rb < Cs(c) Li > Na > K > Rb > Cs (d) Na < Li < Rb < K < Cs

10. Which one of the following is true?(a) Lithium on direct combination with nitrogen from

Li3N.(b) Magnesium on direct combination with nitrogen

from Mg3N.(c) Both (a) and (b)(d) Lithium and magnesium form bicarbonates.

11. Which of the following correctly represents Boyle’s Law?

(a) ( )P atm

31 dm−

υ

(b) ( )P atm

( )3V dm

(c) ( )3V dm

( )T k

(d) All of these

Time Allowed : 2.30 hours ChemisTrY mArks : 70

On 21.08.2018, Model Question Paper is released by the Govt. We have given it along with Answer Key

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Sura’s XI Std - Govt. Model Question Paper 2018-19 �� 3

12. What is the density of oxygen gas at 227°C and 4 atm pressure (R = 0.082 L atom K–1 mol–1)(a) 3.12 g/L (b) 3.41 g/L (c) 2.81 g/L (d) none of these

13. Which one of the following is an extensive property?(a) Molar Volume (b) Molality(c) Gibbs free energy (d) Free energy change

14. Pressure-Volume work involved in an isothermal compression is

(a) −

2 303. n f

iRT log

VV

(b) 2 303. n f

iRT log

VV

(c) − ∫ V VV

V

di

f

(d) ∆∆

VT

15. An ideal gas expands from the volume of 1 × 10–3 m3 to 1 × 10–2 m3 at 300K against a constant pressure at 1 × 105 Nm–2. The work done is(a) –900 J (b) 900 kJ (c) 270 kJ (d) –900 kJ

PART - BAnswer six questions. Question No. 24 is compulsory. Answer any five from the remaining: (6 × 2 = 12)16. Define equivalent mass.17. Consider the following electronic arrangement for p3

configuration.(a) (b)

(c) (d)

Which of these represents the ground state? Substantiate your answer with a proper reason.

18. Calculate the De-Broglie wavelength of a particle whose momentum is 66.26 × 10–28 kgms–1.

19. Is the definition given below for ionization enthalpy is correct? Why? “Ionisation enthalpy is defined as the energy required to remove the most loosely bound electron from the valence shell of an atom”.

20. What is meant by intramolecular hydrogen bond? Give one example.

21. Complete the following chemical reactions and classify them into(a) Hydrolysis (b) redox (c) hydration reactions(i) KMnO4 + H2O2 (ii) CaO + H2O

22. Give the reaction of sodium with ethyne.23. Distinguish between diffusion and effusion.24. One mole of an ideal gas is put through a series of

changes as shown below in a cyclic processName the process A Æ B, B Æ C and C Æ A.

PART - CAnswer six questions. Question No. 32 is compulsory. Answer any Five form the remaining : (6 × 3 = 18)25. Statement 1 : Two mole of glucose contains

12.044 × 1023 molecules of glucoseStatement 2 : Total number of entities present in

one mole of any substance is equal to 6.02 × 1022.

Whether the above statements are true? Is there any relation between these two statements?

26. Calculate the total number of electrons present in 17g of ammonia.

27. Match table -1 using the options given in table -2.Table - 1 Table-2

(a) Angular momention of an electron in

(i) 1/3

(b) Uncertainty principle (ii) zero(c) Cr3+ (iii) H Eψ ψ=

(d) Velocity of election in 3 orbitVelocity of election in 1

rd

stt orbit(iv) ∆ ∆ ≥E. t h

4π

(v) l l h+( )I4π

(vi) Half filled configuration

(vii) ∆ ∆ ≤x h. P4π

28. First ionization potential of carbon atom is greater than that of boron atom, where as the reverse is true for second ionization potential – give appropriate reason.

29. Atomic number of elements X,Y,Z and A are 4,8,7 and 12 respectively. Arrange them in the decreasing order of their electronegativity.

30. Give the uses of heavy water.31. How is Plaster of Paris prepared?32. At identical temperature and pressure, the rate of

diffusion of hydrogen gas is 3 3 times that of a hydrocarbon having molecular formula CnH2n–2. What is the value of n?

33. State the first law of thermodynamics.

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4 ��Sura’s XI Std - Govt. Model Question Paper 2018-19

PART - DAnswer all five questions : (5 × 5 = 25)

34. (a) Define oxidation number. Balance the following equation using oxidation number method.

As2S3 + HNO3 + H2O → H3AsO4 + H2SO4 + NO(OR)

(b) (i) Define limiting reagent. (2) (ii) In the reaction, 2Al + Fe2O3 → Al2O3 + 2Fe

324 g of aluminum is allowed to react with 1.12 kg of ferric oxide. How much of the excess reagent is left at the end of the reaction. (3)

35. (a) Describe Aufbau principle. Write the electronic configuration for Ni2+ using Aufbau principle

(OR)(b) (i) What is the de-Broglie wave length of an

election, which is accelerated from the rest, through a potential difference of 100V.

(ii) Give the relation between Bohr radius (r) and the de Broglie wavelength (l)

36. (a)

Electronaffinity

Atomic Number

F

ClBr

At

(i) Explain the above variations of electron affinity. (ii) Define electronegativity.

(OR)(b) (i) What is water gas shift reaction? (ii) NH3 has exceptionally high melting point

and boiling point as compared to those of the hydrides of the remaining element of group15. Explain.

37. (a) (i) Hydrogen peroxide can function as an oxidizing agent as well as reducing agent. Substantiate this statement with one example for each

(ii) Name the three types of covalent hydrides.(OR)

(b) (i) Alkaline earth metal (A) belongs to 3rd period reacts with oxygen and nitrogen to form compounds (B) and (C) respectively. It undergo metal displacement reation with AgNO3 solution to form compound (D).

Identify (A), (B), (C) and (D). Give the reactions.

(ii) Why sodium hydroxide is much more soluble than its chloride?

38. Write the Vander Waal's equation for a real gas. Explain the correction term for pressure and volume.

(OR)Give the relation between enthalpy (H) and internal energy (U).Calculate ΔH°f for the reactions.CO2(g) + H2(g) → CO(g) + H2O(g) given that ΔH°f for CO2(g), CO(g) and H2O(g) are −393.5, −111.31 and −242 kJ mol–1 respectively.

ANSWERSPART - A

1. (a) 6C12

2. (c) 52 g3. (c) (i), (ii) and (iii)4. (d) Smaller the value of n, the larger is the orbit

radius.5. (b) Be and Al6. (b) B7. (c) Alkali metals8. (a) 2.89. (a) Li < Na < K < Rb < Cs10. (c) Both (a) and (b)

11. (a) ( )P atm

31 dm−

υ

12. (a) 3.12 g/L13. (c) Gibbs free energy

14. (b) 2 303. n f

iRT log

VV

15. (a) –900 JPART - B

16. (i) Equivalent mass of an element, compound or ion is the mass that combines or displaces 1.008g hydrogen or 8g oxygen or 35.5g chlorine.

(ii) Equivalent mass has no unit.

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17. ⥮ ⥮ ⥮↿↿ ↿

(a)↿

(b)↿

(c)↿

(d)

� Among these the electronic configurations (a) represents the ground state.

� It is considered to be the most stable state.

18. λ

λ

= = ××

= ×

−

−

−

hp

m

6 626 1066 26 10

1 10

34

28

7

..

19. The definition given is incorrect.Reason : The energy required to remove the electron from the "valence shell of the isolated gaseous atom in its ground state".

20. Intramolecular hydrogen bonds are those which occur within one single molecule. This occurs between two functional groups within a molecule. Eg: ortho-nitrophenol and salicylaldehyde.

21. (i) KMnO4 + H2O2 + H2SO4 → 2MnSO4 + K2SO4 + 5O2 + 8H2O (Redox)

(ii) CaO + H2O → Ca(OH)2 (Hydrolysis)22. H C C H H C C Na Na C C NaNa Na− ≡ − → − ≡ − → − ≡ −

23. Most aerosol cans contain several ounces of a propellant carbon dioxide, propane or butane. When exposed to flame, aerosol containers experience temperatures and pressures significantly higher than they were designed to resist, causing them to rupture violently into large fire balls. This occurs when the gas and liquid inside expands when heated, increasing the pressure inside the container.

24. A Æ B : Isobaric processB Æ C : Isothermal processC Æ A : Isochoric process

PART - C25. The statements 1 & 2 are true. But there is no relation

between statement 1 and statement 2.26. No. of electrons present in one ammonia (NH3)

molecule (7 + 3) = 10No. of moles of NH3 =

MassMolar mass mol

mol= =−17

1711

gg

No. of molecules present in 1 mol of NH3 = 6.023 × 1023

No. of electrons present in 1 mol of NH3 = 10 × 6.023 × 1023

= 6.023 × 1024

27. Table - 1 Table-2

(a) Angular momention of an electron in (v) l l h+( )I

4π

(b) Uncertainty principle(vii) ∆ ∆ ≤x h. P

4π

(c) Cr3+ (vi) Half filled configuration

(d) Velocity of election in 3 orbitVelocity of election in 1

rd

stt orbit (i) 1/3

28. Electron configuration of carbon(Z = 6) 1s2 2s2 2p2.Electron configuration of boron(Z = 5) 1s2 2s2 2p1.The size of a carbon atom is smaller than boron. So the valence electron of carbon has greater nuclear charge than that of boron. Hence the first I.E of carbon is greater than that of boron.However, the second ionization enthalpy of boron is higher than that of carbon. This is because after losing 1 electron, Boron has a fully filled orbital (2s2) than carbon(2p1). Fully filled orbitals have more stability than partially filled orbitals so greater amount of energy will be needed to remove an electron from boron. So in this case, the second I.E of boron is higher than that of carbon.

29. Y > Z > X > A30. (i) Heavy water is widely used as moderator in

nuclear reactors.(ii) It is commonly used as a tracer to study element reaction mechanisms.(iii) It is used as a coolant in nuclear reactors.

31. Plaster of paris is obtained when gypsum, CaSO4.2H2O, is heated to 393K2(CaSO4.2H2O) → 2CaSO4.H2O + 3H2O

32. γ

γH

C H

C H

H

C H

2

2 2

2 2

2

2 23 32

n n

n n

n n

mm

m−

−

−

=

=

Squaring on both sides and rearranging

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27 2

54 2 2

54 12 2 254 14 2

54 2

2 2× =

= + −( )

= + −= −= +( )

−m

n n

n nn

n

n nC H

(12) (1)

// 145614

4= =

\n = 4.

33. The first law of thermodynamics, also known as the law of conservation of energy, states that “The total energy of an isolated system remains constant though it may change from one form to another.”The mathematical statement of the First Law is :

ΔU = q + wWhere q - the amount of heat supplied to the system; w - work done on the system

PART - D34. (a) Oxidation number : It is defined as the imaginary

charge left on the atom when all other atoms of the compound have been removed in their usual oxidation states that are assigned according to set of rules.

Solution : Step 1 : As2S3 + HNO3 + H2O → H3AsO4 +

H2SO4 + NO Step 2 : Assign the oxidation numbers and identify

the redox couples.

As2S3 + H N O3 + H2O → H3AsO4 + H2SO4 + NO+3 –2 +1 +5 –2 +1 –2 +1 +5 –2 +1 +6–2 +2–2

Decreased by 3

Increased by 2 Step 3 : Multiply As2S3 by 3 and HNO3 by 2.3As2S3 + 2HNO3 + H2O → H3AsO4 + H2SO4 + NO Step 4 : Balance all the elements in the equation (As, S and N) except H and O. 3As2S3 + 2HNO3 + H2O → 6H3AsO4 + 9H2SO4 + 2NO Step 5 : Balance the complete equation including

O & H. 3As2S3 + 28HNO3 + 4H2O → 6H3AsO4 + 9H2SO4 + 28NO

(OR)

(b) (i) Limiting reagent : When a reaction is carried out using non-stoichiometric quantities of the reactants, the product yield will be determined by the reactant that is completely consumed. It limits the further reaction from taking place and is called as the limiting reagent.

(ii) Given : 2Al + Fe2 O3 → Al2O3 + 2Fe

Reactants ProductsAl Fe2O3 Al2O3 Fe

Amount of reactant allowed to react

324 g 1.12 kg - -

Number of moles allowed to react

32427

12= mol 1 12 10160

73. × = mol - -

Stoichiometric Co-efficient 2 1 1 2

Number of moles consumed during reaction

12 mol 6 mol - -

Number of moles of reactant unreacted and number of moles of product formed

- 1 mol 6 mol 12 mol

Molar mass of Al2O3 formed = 6 mol × 102 g mol–1

Al Og

2 3

2 27 3 1654 48 102

612×( ) + ×( )+ =

=

Excess reagent = Fe2O3Amount of excess reagent left at the end of the reaction = 1 mol × 160 g mol–1

= 160 g Fe O

g2 3

2 56 3 16112 48 160

160×( ) + ×( )+ =

=

35. (a) Aufbau principle : The word Aufbau in German means 'building up'. In the ground state of the atoms, the orbitals are filled in the order of their increasing energies. That is the electrons first occupy the lowest energy orbital available to them.

Once the lower energy orbitals are completely filled, then the electrons enter the next higher energy orbitals. The order of filling of various orbitals as per the Aufbau principle which is in accordance with (n+l) rule.

The lower the value of (n+1) for an orbital, the lower is its energy. If two orbitals have the same (n+1) value, the orbital with lower value of n has the lower energy.

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n =1 1s

2s

3s 3p

4s

3d

4d 4f

6s 6p

7s

4p

5s 5p 5d 5f

6d

7p

2pn =2

n =3

n =4

n =5

n =6

n =7

Ni s s p s p s d

Ni s s p s

−

∴ −+

1 2 2 3 3 4 3

1 2 2 3

2 2 6 2 6 2 8

2 2 2 6 2 3 4 36 0 8p s d

(OR)(b) (i) Potential difference = 100V

= 100 × 1.6 × 10–19J

λ = hmev2

= 6 626 10

2 9 1 10 100 1 6 10

34 2 1

31 19

.

. .

×

× × × × ×

− −

− −

Kgm

Kg J

s

λ = 1.22 × 10–10 m (ii) Quantisation of angular momentum and

de-Broglie concept: According to the de-Broglie concept, the

electron that revolves around the nucleus exhibits both particle and wave character. In order for the electron wave to exist in phase, the circumference of the orbit should be an integral multiple of the wavelength of the electron wave. Otherwise, the electron wave is out of phase.

Circumference of the orbit = nλ 2πr = nλ (1) 2πr = nh/mv Rearranging, mvr = nh/2π (2) Angular momentum = nh/2π The above equation was already predicted by

Bohr. Hence, de-Broglie and Bohr’s concepts are in agreement with each other.

36. (a) (i) Variation of Electron affinity in a group: As we move down a group, generally the electron affinity decreases. It is due to increase in atomic size and the shielding effect of inner shell electrons. However, oxygen and fluorine have lower affinity than sulphur and chlorine respectively. The sizes of oxygen and fluorine atoms are comparatively small and they have high electron density. Moreover, the extra electron added to oxygen and fluorine has to be accommodated in the 2p orbital which is relatively compact compared to the 3p orbital of sulphur and chlorine so, oxygen and fluorine have lower electron affinity than their respective group elements sulphur and chlorine.

(ii) Electronegativity : It is defined as the relative tendency of an element present in a covalently bonded molecule, to attract the shared pair of electrons towards itself.

(OR)(b) (i) The carbon monoxide of the water gas can be

converted to carbon dioxide by mixing the gas mixture with more steam at 400°C and passed over a shift converter containing iron/copper catalyst. This reaction is called as water-gas shift reaction.

CO + H2O → CO2 + H2 The CO2 formed in the above process is

absorbed in a solution of potassium carbonate.CO2 + K2CO3 + H2O → 2KHCO3

(ii) NH3 has higher boiling and melting point compared to all other hydrides of group 15 elements due to intermolecular hydrogen bonding.

37. (a) (i) Hydrogen peroxide can act both as an oxidizing agent and a reducing agent. Oxidation is usually performed in acidic medium while the reduction reactions are performed in basic medium.

In acidic conditions : H2O2 + 2H+ +2e– → 2 H2O (E° = +1.77 V) For example 2FeSO4 + H2SO4 + H2O2 → Fe2(SO4)3 + 2H2O In basic conditions : HO2– + OH– → O2 + H2O + 2e–

(E° = + 0.08 V)

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For Example, 2KMnO4(aq) + 3H2O2(aq) → 2MnO2 + 2KOH + 2H2O + 3O2(g)

(ii) Covalent (Molecular) hydrides : They are compounds in which hydrogen is attached to another element by sharing of electrons.

Covalent hydrides are further divided into three categories, viz., electron precise (CH4, C2H6, SiH4, GeH4), electron-deficient (B2H6) and electron-rich hydrides (NH3, H2O).

(OR)(b) (i) Alkaline earth metal (A) belonging to 3rd

period is magnesium. Magnesium reacts with oxygen and

nitrogen to form magnesium oxide (B) and magnesium nitride (C).

2 2

3

2

2 3 2

Mg O MgO

Mg N Mg N

B

C

+ →

+ →

( )

( )

Magnesium undergoes metal displacement reaction with AgNO3 to form magnesium nitrate (D).

Mg AgNO Mg NO Ag

D+ → ( ) +2 23 3 2

( )

Compound/ Element Formula Name

A Mg MagnesiumB MgO Magnesium oxideC Mg3N2 Magnesium nitrideD Mg(NO3)2 Magnesium nitrate

(ii) NaOH is much more soluble than NaCl. Enthalpy of a solution can be expressed as

the sum of lattice enthalpy and enthalpy of hydration of a compound.ΔH solu NaCl = +3.9kJ/molΔH solu NaOH = –44.5 kJ/molΔH0 Lattice enthalpy of NaCl = +787.6 kJHydration enthalpy = –784.1kJTherefore ΔHsolu = +787.6 –784.1 = + 3.5kJ

Dissolution of NaCl is accompanied by very small heat change so solubility of NaCl is less than NaOH.

38. (a) The vander equation for a real gas is

P +Van2

2

(V – nb) = nRT

Pressure Correction : The pressure of a gas is directly proportional to the force created by the bombardment of molecules on the walls of the container. The speed of a molecule moving towards the wall of the container is reduced by the attractive forces exerted by its neighbours. Hence, the measured gas pressure is lower than the ideal pressure of the gas. Hence, van der Waals introduced a correction term to this effect.

(a) (b) (C)

Molecular attractionsbalanced

Inward pull

Inter-molecular forces of attraction Van der Waal's found out the forces of attraction

experienced by a molecule near the wall are directly proportional to the square of the density of the gas.

P′ a r2

r = nV

where n is the number of moles of gas and V is the volume of the container

fi P′ α nV

2

2

fi P′ = anV

2

2

where a is proportionality constant and depends on the nature of gas.

Therefore,

Pideal = P + anV

2

2

Volume Correction : As every individual molecule of a gas occupies a certain volume, the actual volume is less than the volume of the container, V. Van der Waals introduced a correction factor V′ to this effect. Let us calculate the correction term by considering gas molecules as spheres.

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Excluded volume

2rExcludedvolume

V = excluded volume Excluded volume for two molecules

= 43 p(2r)3

= 8 43

3πr

= 8Vm

where v is a volume of a single molecule Excluded volume for single molecule

= 8

2Vm = 4Vm

Excluded volume for n molecule = n(4Vm) = nb Where b is van der waaals constant which is equal to 4Vm Þ V = nb Videal = V – nb

Replacing the corrected pressure and volume in the ideal gas equation PV = nRT we get the van der Waals equation of state for real gases as below,

P anV

+

2

2 (V – nb) = nRT

The constants a and b are van der Waals constants and their values vary with the nature of the gas. It is an approximate formula for the non-ideal gas.

(OR)(b) (i) H = U + PV Considering the initial and final states of H, U

and V as H1, U1, and V1 and H2, U2 and V2 Change in enthalpy is (H2 – H1) = (U2 – U1) + P(V2 – V1) ΔH = ΔU + PΔV (ii) Solution : CO2(g) + H2(g) → CO(g) + H2O(g) ΔH°f (CO2) = –393.5 kJ mol–1

ΔH°f (CO) = –111.31 kJ mol–1

ΔH°f (H2O) = –242 kJ mol–1

ΔH°f = ΣΔH°f (product) – ΣΔH°f (reactant) = [–111.31 + (–242)] – [–393.5] = –111.31 – 242 + 393.5 ΔH°f = 40.19 kJ mol–1

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11thSTD.

Time : 2.30 Hours CHEMISTRY Marks : 70

Sura’S Sample QueStion paper

PART - AANSWER ALL THE QUESTIONS: (15 × 1 = 15)1. Carbon forms two oxides, namely carbon monoxide

and carbon dioxide. The equivalent mass of which element remains constant?

(a) Carbon (b) oxygen (c) both carbon and oxygen (d) neither carbon nor oxygen2. How many electrons in an atom with atomic number

105 can have (n + l) = 8 ? (a) 30 (b) 17 (c) 15 (d) unpredictable3. The correct order of decreasing electronegativity values

among the elements X, Y, Z and A with atomic numbers 4, 8 , 7 and 12 respectively

(a) Y > Z > X > A (b) Z > A > Y > X (c) X > Y > Z > A (d) X > Y > A > Z4. The hardness of water can be determined by

volumetrically using the reagent (a) sodium thio sulphate (b) potassium permanganate (c) hydrogen peroxide (d) EDTA5. Which is the correct sequence of solubility of

carbonates of alkaline earth metals ? (a) BaCO3 > SrCO3 > CaCO3 > MgCO3 (b) MgCO3 > CaCO3 > SrCO3 > BaCO3 (c) CaCO3 > BaCO3 > SrCO3 > MgCO3 (d) BaCO3 > CaCO3 > SrCO3 > MgCO3 6. 25g of each of the following gases are taken at 27°C

and 600 mm Hg pressure. Which of these will have the least volume ?

(a) HBr (b) HCl (c) HF (d) HI7. The correct thermodynamic conditions for the

spontaneous reaction at all temperature is (a) ΔH<0andΔS>0 (b) ΔH<0andΔS<0 (c) ΔH>0andΔS=0 (d) ΔH>0andΔS>0

[ 10 ]

8. The mass of a gas that occupies a volume of 612.5 ml at room temperature and pressure (25° c and 1 atm pressure) is 1.1g. The molar mass of the gas is

(a) 66.25 g mol–1 (b) 44 g mol–1

(c) 24.5 g mol–1 (d) 662.5 g mol–1

9. According to the Bohr Theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon ?

(a) n = 6 to n = 1 (b) n = 5 to n = 4 (c) n = 5 to n = 3 (d) n = 6 to n = 510. Assertion : Helium has the highest value of ionisation

energy among all the elements known Reason : Helium has the highest value of electron

affinityamongalltheelementsknown (a) Both assertion and reason are true and reason is

correct explanation for the assertion (b) Both assertion and reason are true but the reason

is not the correct explanation for the assertion (c) Assertion is true and the reason is false (d) Both assertion and the reason are false11. Water gas is (a) H2O(g) (b) CO + H2O (c) CO + H2 (d) CO + N2

12. In case of alkali metal halides, the ionic character increases in the order

(a) MF<MCl<MBr<MI (b) MI<MBr<MCl<MF (c) MI<MBr<MF<MCl (d) none of these13. The heat of formation of CO and CO2 are – 26.4 kcal

and – 94 kcal, respectively. Heat of combustion of carbon monoxide will be

(a) + 26.4 kcal (b) – 67.6 kcal (c) – 120.6 kcal (d) + 52.8 kcal

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Sura’s XI Std - Sura’s Sample Question Paper �� 11

14. A bottle of ammonia and a bottle of HCl connected through a long tube are opened simultaneously at both ends.Thewhiteammoniumchlorideringfirstformedwill be

(a) At the center of the tube (b) Near the hydrogen chloride bottle (c) Near the ammonia bottle (d) Throughout the length of the tube15. sodium is stored in (a) alcohol (b) water (c) kerosene (d) none of these

PART - BAnswer six questions. Question No. 24 is compulsory. Answer any five from the remaining: (6 × 2 = 12)16. What do you understand by the term mole.17. Thestabilisationofahalffilledd-orbitalismore

pronouncedthanthatofthep-orbitalwhy?18. In what period and group will an element with Z =

118 will be present?19. Predict which of the following hydrides is a gas on

a solid (a) HCl (b) NaH. Give your reason.20. An alkali metal (x) forms a hydrated sulphate, X2SO4.

10H2O. Is the metal more likely to be sodium (or) potassium.

21. Suppose there is a tiny sticky area on the wall of a container of gas. Molecules hitting this area stick there permanently. Is the pressure greater or less than on the ordinary area of walls?

22. DefineHess’slawofconstantheatsummation.23. Beryllium halides are Covalent whereas magnesium

halides are ionic why?24. What is lattice energy?

PART - CAnswer six questions. Question No. 32 is compulsory. Answer any Five form the remaining : (6 × 3 = 18)25. Mass of one atom of an element is 6.645 × 10–23g.

How many moles of element are there in 0.320 kg.26. Show that the circumference of the Bohr orbit for

the hydrogen atom is an integral multiple of the de Broglie wave length associated with the electron revolving arround the nucleus.

27. Why halogens act as oxidising agents?28. Whatiswater-gasshiftreaction?29. Which would you expect to have a higher melting

point,magnesiumoxide ormagnesiumfluoride?Explain your reasoning.

30. What are ideal gases? In what way real gases differ from ideal gases.

31. List the characteristics of internal energy.32. How would you explain the fact that the second

ionisation potential is always higher than first ionisation potential?

33. Distinguish between diffusion and effusion.

PART - DAnswer all five questions : (5 × 5 = 25)34. a) Calculate the empirical and molecular formula

of a compound containing 76.6% carbon, 6.38 % hydrogen and rest oxygen its vapour density is 47.

(or) b) How do you expect the metallic hydrides to be

useful for hydrogen storage ?35. a) Explain the important common features of Group

2 elements.(or)

b) A tank contains a mixture of 52.5 g of oxygen and 65.1 g of CO2 at 300 K the total pressure in the tanks is 9.21 atm. Calculate the partial pressure (in atm.) of each gas in the mixture.

36. a) A gas mixture of 3.67 lit of ethylene and methane on complete combustion at 25°C and at 1 atm pressure produce 6.11 lit of carbondioxide. Find out the amount of heat evolved in kJ, during this combustion. (ΔHC(CH4)=− 890 kJmol

−1 and (ΔHC(C2H4)=−1423kJmol

−1.(or)

b) Why the first ionisation enthalpy of sodium is lower than that of magnesium while its second ionisation enthalpy is higher than that of magnesium?

37. a) How fast must a 54g tennis ball travel in order to have a de Broglie wavelength that is equal to that of a photon of green light 5400Å?

(or) b) Explain whether a gas approaches on ideal

behavior or deviates from ideal behaviour if(a) it is compressed to a smaller volume at constant temperature.(b) the temperature is raised at while keeping the volume constant(c) more gas is introduced into the same volume and at the same temperature

38. a) The Li2+ ion is a hydrogen like ion that can be described by the Bohr model. Calculate the Bohr radius of the third orbit and calculate the energy of an electron in 4th orbit.

(or)

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12 �� Sura’s XI Std - Sura’s Sample Question Paper

b) Explain the following, give appropriate reasons.(i) Ionisation potential of N is greater than

that of O(ii) First ionisation potential of C-atom is

greater than that of B atom, where as the reverse is true is for second ionisation potential.

(iii) The electron affinity values of Be, Mgand noble gases are zero and those of N (0.02 eV) and P (0.80 eV) are very low

(iv) The formation of F–(g) from F(g) is exothermic while that of O2–(g) from O (g) is endothermic.

ANSWERSPART - A

1. (b) oxygen2. (b) 173. (a) Y > Z > X > A4. (d) EDTA5. (b) MgCO3 > CaCO3 > SrCO3 > BaCO36. (d) HI7. (a) ΔH<0andΔS>08. (b) 44 g mol–1

9. (d) n = 6 to n = 510. (c) Assertion is true and the reason is false11. (c) CO + H212. (b) MI<MBr<MCl<MF13. (b) – 67.6 kcal14. (b) Near the hydrogen chloride bottle15. (c) kerosene

PART - B16. One mole is the amount of substance of a system,

which contains as many elementary particles as thereareatomsin12gofcarbon-12isotope.Theelementary particles can be molecules, atoms, ions, electronsoranyotherspecifiedparticles.

17. (i)Thehalffilledd-orbitalismuchstabilisedthanthep-orbital.

(ii) This is due to its high exchange energy and symmetrythanthatofp-orbital.

18. Z = 118; [86Rn] 5f14 6d10 7s2 7p6

In the periodic table the element with Z = 118 is locatedinp-block.Period no. = 7 (as n = 7 for valence shell)

Group no. = 18 (group no = 10+ ns electrons + np electrons)(n-outermostshell)

19. (a) Gas - due to the absence of intermolecularhydrogen bonding.

(b) Solid Reason : NaH is a ionic hydride formed by

transfer of electrons from metal to hydrogen. They have high lattice enthalpy and high density since hydride ions occupy holes in the lattice of metal without distorting the metal lattice.

20. The metal more likely to form a hydrated sulphate is sodium of formula Na2SO4.10H2O.

Smaller the size of the ion greater is the degree of hydration. Hydration energy is in the order of, Li+ > Na+ > K+ > Rb+ > Cs+ so sodium is hydrated more easily than potassium.

21. If the gas molecules stick to the walls of the container, then the number of molecules striking the walls of the container decreases. An decrease in the number of gas molecules in the same volume container decreases pressure.

22. The enthalpy change of a reaction either at constant volume or constant pressure is the same whether it takes place in a single or multiple steps provided the initialandfinalstatesaresame.

A BΔHr

ΔHr = ΔH1 + ΔH2 + ΔH3

ΔH1 ΔH3ΔH2X Y

23. Due to small size of Be2+, the charge density is very high.AccordingtoFajan’sRule,cationwithsmallsize has high polarising power and prefers to form covalent bonds. So beryllium halides are Covalent in nature.

24. Latticeenergyisdefinedas“Theamountofenergyrequired to completely remove the constituent ions fromitscrystallatticetoaninfinitedistance.”Itisalso referred as lattice enthalpy.

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PART - C25. Given :

mass of one atom = 6.645 × 10–23 g\ mass of 1 mole of atom = 6.645 × 10–23 g × 6.022 × 1023

= 40 g\ number of moles of element in 0.320 kg

= 1

400 320 mole

gkg× .

= 1 320

40 mol g

g×

= 8 mol.26. Quantisation of angular momentum and de

Broglie concept : According to the de Broglie concept, the electron that revolves around the nucleus exhibits both particle and wave character. In order for the electron wave to exist in phase, the circumference of the orbit should be an integral multiple of the wavelength of the electron wave. Otherwise, the electron wave is out of phase.

Circumferenceoftheorbit=nλ 2πr=nλ 2πr=nh/mv Rearranging,mvr=nh/2π Angularmomentum=nh/2π The above equation was already predicted by Bohr. Hence,DeBroglie andBohr’s concepts are in

agreement with each other.27. (i) Due to low bond dissociation enthalpy, high

electronegativity and large negative electron gain enthalpy, halogen have a strong tendency to accept electrons and thus get reduced.

X2 + 2e– → 2X–

(ii) The ready acceptance of an electron is due to the strong oxidising nature of halogens.

(iii) F2 is the strongest oxidising halogen and it oxidises other halide ions in solution or even in the solid phase. In general, a halogen oxidises halide ions of higher atomic number.

(iv) Hence, the oxidising ability of halogens decreasesfromfluorinetoiodineas:

F2 > Cl2 > Br2 > I228. The carbon monoxide of the water gas can be

converted to carbon dioxide by mixing the gas mixture with more steam at 400°C and passed over a shift converter containing iron/copper catalyst.Thisreactioniscalledaswater-gasshiftreaction.

CO + H2O → CO2 + H2

The CO2 formed in the above process is absorbed in a solution of potassium carbonate.

CO2 + K2CO3 + H2O → 2KHCO329. MgO has higher melting point than MgF since charge

of oxygen is O2–andfluorineisF–. As the charge in the ionic lattice in magnesium oxide is two times larger, the ionic bonding is stronger so more energy is required to overcome the electrostatic attraction and break down the ionic lattice. So magnesium oxide would have higher melting point than magnesium fluoride.

30. Anidealgasisdefinedasoneinwhichallcollisionsbetween atoms or molecules are perfectly eleastic and in which there are no intermolecular attractive forces. An ideal gas is a gas that obeys the ideal gas law, PV = nRT, where n is the number of moles of the gas , R is the ideal gas constant,pressure P, volume V, and temperature T

Difference between ideal and real gas :Ideal Gas Real gas

(i) Ideal gases obey all gas laws under all conditions of temperature and pressure

Real obey gas law only at low pressures and high temperature.

(ii) The volume occupied by the molecules is negligible as compared to the total volume occupied by the gas.

The volume occupied by the molecules is not negligible as compared to the total volume of the gas.

(iii) The force of attraction among the molecules are negligible

The force of attraction are not negligible at all temperatures and pressures.

31. Characteristics of internal energy (U) : �� The internal energy of a system is an extensive property.�� The internal energy of a system is a state function. �� The change in internal energy of a system is expressedasΔU=Uf–Ui�� In a cyclic process, there is no internal energy change.ΔU(cyclic) = 0�� If the internal energy of the system in the finalstate (Uf) is less than the internal energy of the systeminitsinitialstate(Ui),thenΔUwouldbenegative.

ΔU=Uf−Ui=−ve(Uf<Ui)�� If the internal energy of the system in the finalstate(Uf) is greater than the internal energy of the systeminitsinitialstate(Ui),thenΔUwouldbepositive.

ΔU=Uf−Ui=+ve(Uf>Ui)

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32. The total number of electrons are less in the cation than the neutral atom while the nuclear charge remains the same. Therefore the effective nuclear charge of the cation is higher than the corresponding neutral atom. Thus the successive ionisation energies, always increase in the following order

IE1<IE2<IE3<.....33. Distinguish between diffusion and effusion

: Diffusion is the process by which molecules

move and travel from one place to another without requiring bulk motion. Diffusion results in molecules moving or mixing by only using kinetic energy. Effusion is the process by which molecules travel through a pinhole from a place of high concentration to low concentration. The process describes the ability of gas to travel through a small hole without collisions between molecules.

PART - D34. a)

Element Percentage Atomic mass Relative number of atoms Simple ratio Whole no

C 76.6 12 76 612

6 38. .= 6 381 06

6..

= 6

H 6.38 1 6 381

6 38. .= 6 381 06

6..

= 6

O 17.02 16 17 0216

1 06. .= 1 061 06

1..

= 1

Empirical formula = C6 H6 O

n = Molar

Calculated mass

empirical formula mass

= 2

942 47

941× = × =vapour density

, since Molar mass = 2 × Vapour density

molecular formula n × n empirical formula

\ molecular formula (C6H6O) × 1 = C6H6O.(or)

b) Insomeofthetransitionmetalhydrides,hydrogenisabsorbedasH-atoms.DuetotheinclusionofH-atoms,

the metal lattice expands and thus becomes less stable. Therefore, when such metallic hydride is heated, it decomposes to release hydrogen gas and very finely dividedmetal.The hydrogenevolved in this manner can be used as a fuel. Thus, transition metals or their alloys can act as sponge and can be used to store and transport hydrogen to be used as a fuel.

35. a) Group 2 elements are known as alkaline earth metals. It includes beryllium, magnesium, calcium, strontium, barium, and radium. They exist in +2 oxidation states.The general outer electronic configuration of

alkaline earth metal is ns2.

Physical Characteristics : �� They are silvery, white, and hard metals.

They are softer but harder than alkali metals.

�� Their melting and boiling points are higher compared to alkali metals.

�� They are strongly electropositive in nature. Alkaline earth metals give differentcolorwithflametest.

Chemical Properties : �� All alkaline earth metals forms monoxide

except beryllium.�� They have high electrical and thermal

conductivities as they have metallic bonding.

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�� The oxides of alkaline earth metals are basic but less basic than alkali metals.

�� Hydroxides of alkaline earth metals are basic in nature except beryllium hydroxide.

�� Group 2 metals forms solid carbonates. �� Alkaline earth metals also form sulphates

such as BeSO4, and MgSO4. �� Group 2 elements form hydrated,

crystallized nitrates. �� Alkaline earth metals forms halides after

reacting with halogens. (or)

b) mO2 = 52.5 g PO2

= ?mCO2

= 65.1 g PCO2 = ?

T = 300 K P = 9.21 atm PO2

= XO2 × Total Pressure

XO2 =

nn n

O

O CO

2

2 2+

nO2 =

MassMolar

of O mass of O

2

2

= 52 5

32 1. g

g mol− = 1.64 mol

nCO2 =

MassMolar

of CO mass of CO

2

2

= 65 1

44 1. g

g mol− = 1.48 mol

XO2 =

P VP1 1

2 =

1 643 12..

= 0.53

XCO2 =

nn n

CO

O CO

2

2 2+

= 1 483 12..

= 0.47

PO2 = XO2

× Total pressure

= 0.53 × 9.21 atm = 4.88 atm PCO2

= XCO2 × Total pressure

= 0.47 × 9.21 atm = 4.33 atm.36. a) ΔHC (CH4)=−890kJmol

−1

ΔHC (C2H4) = −1423kJmol−1

Let the mixture contain x lit of CH4 and (3.67−x)litofethylene.CH4 + 2O2

→ CO2 + 2H2Ox lit x litC2H4 + 3O2

→ 2CO2 + 2H2O

(3.67−x)lit2(3.67−x)litVolume of Carbondioxide formed = x+2(3.67−x)=6.11lit x+7.34−2x = 6.11 7.34−x = 6.11 x = 1.23 litGiven mixture contains 1.23 lit of methane and 2.44 lit of ethylene, hence

DHC = D ( )

×

+D ( )

× −H CH

xH C H

xC C4 2 4

22 4 22 43 67

.( )

.( . )

lit lit

lit llit

D ( )

×

+D ( )

× −H CH

xH C H

xC C4 2 4

22 4 22 43 67

.( )

.( . )

lit lit

lit llit

DHC = − ×

+ − × −−890

22 41 23 1423

22 43 67 1 23

1kJ lit mol lit lit.

..

. .(( )

lit

− ×

+ − × −−890

22 41 23 1423

22 43 67 1 23

1kJ lit mol lit lit.

..

. .(( )

lit

DHC = [–48.87 kJ mol–1] + [–155kJ mol–1]DHC = –203.87 kJ mol–1.

(or) b) TheelectronicconfigurationofSodium(Z=11)

1s2 2s2 2p6 3s1.Magnesium (Z = 12) 1s2 2s2 2p6 3s2.Magnesium atom has a smaller radius and higher nuclear charge than a sodium atom, thus more energy will be required to remove the electron from the same orbital (3s), making thefirstionisationenergyofmagnesiumhigherthan that of sodium.However, the second ionization enthalpy of sodium is higher than that of magnesium. This is because after losing 1 electron, sodium attains thestablenoblegasconfigurationofneon(1s2 2s2 2p6). On the other hand, magnesium, after losing 1 electron still has one electron in the 3s-orbital(1s2 2s2 2p6 3s1). In order to attain the stablenoblegasconfiguration,Thus,theenergyrequired to remove the second electron in case of sodium is much higher than that required in case of magnesium. Hence, the second ionization enthalpy of sodium is higher than that of magnesium.

37. a) Given : de Broglie wavelength of the tennis ball equal to 5400 Å.

m = 54 g v = ?

λ =h

mV

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V = h

mλ

V = 6 626 10

54 10 5400 10

34

3 10. ×

× × ×

−

− −JS

Kg m V = 2.27 × 10–26 ms–1.

(or) b) (a) It is compressed to a smaller volume at

constant temperature : In the isothermal compression of a gas work is done on the system to decrease the volume and increase the pressure. Increase in pressure increases the intermolecular force of attraction between the gas molecules and so deviate from ideal behaviour.

(b) The temperature is raised at while keeping the volume constant : In a isochoric process the quantity of the gas remains constant but rise in temperature increases the pressure so deviates from ideal gas behaviour.

(c) More gas is introduced into the same volume and at the same temperature : If more gas is introduced in the same volume and temperature is constant, more will be the pressure. Increase in pressure, increases the intermolecular force of attraction between the molecule, which makes the gases to deviate from Ideal behaviour.

38. a) rn = 0 529 2.( )n

zÅ En =

− ( ) −13 6 2

21. z

nev atom

for Li2+ z = 3Bohr raduis for the third orbit (r3)

= 0 529 3

3

2.( )( )

= 0.529 × 3 = 1.587 ÅEnergy of an electron in the fourth orbit

(E4) = − ( )

( )13 6 3

4

2

2.

= – 7.65 eV atom–1.(or)

b) (i) Electronconfigurationofnitrogen (Z = 7) 1s2 2s2 2p3. Electronconfigurationofoxygen

(Z = 8) 1s2 2s2 2p4.

Nitrogen has a half filled electronicconfigurationwhichismuchmorestablethan an incomplete p- orbital of oxygenwhichwouldneed togiveuponeof it’selectrons to attain the stability of nitrogen. Hence nitrogen would require more ionization energy to remove an electron fromit’soutershellthanoxygen.

(ii) Electronconfigurationofcarbon (Z = 6) 1s2 2s2 2p2.

ElectronconfigurationofBoron (Z = 5) 1s2 2s2 2p1.

The size of a carbon atom is smaller than boron. So the valence electron of carbon has greater nuclear charge than that of boron .Hence the first I.E of carbon isgreater than that of boron.

However, the second ionization enthalpy of boron is higher than that of carbon. This is because after losing 1 electron, Boron has afullyfilledorbital(2s2) than carbon(2p1). Fullyfilledorbitalshavemorestabilitythanpartiallyfilledorbitalssogreateramountofenergy will be needed to remove an electron from boron. So in this case, the second I.E of boron is higher than that of carbon.

(iii) The electron affinities of Be, Mg andnoble gases are almost zero because both Be (Z = 4; 1s2 2s2) and Mg (Z = 12; 1s2 2s2 2p6 3s2)arehavingsorbitalfullyfilledintheirvalenceshell.Fullyfilledorbitalsaremost stable due to symmetry. Therefore, these elements would be having least tendency to accept electron. Hence, Be and Mg would be having zero electron affinity.Whereas

N (Z = 7; 1s2 2s2 2px1 2py1 2pz1) and P (Z = 15) 1s2 2s2 2p6 3s2 3p3 is having half filled2p-subshell.Halffilledsubshellsaremoststableduetosymmetry(Hund’srule).Thus, nitrogen and phosphorous are having least tendency to accept electron. Hence, havelowelectronaffinity.

(iv) Fluorine is highly electro negative in nature therefore as it gains the electron its octet become stable and releases the energy so exothermic. while in oxygen theadditionoffirstelectronisexothermicin nature but addition of second electron experiences high repulsive force. So needs extra external energy to enter outer shell, hence endothermic in nature.

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