Practice Paper
TERM II (2021 – 2022)
Class – XI
Chemistry (043)
Time: 2 hours Maximum Marks: 35
General Instructions:
1. There are 12 questions in this question paper with internal choice.
2. SECTION A – Q. No. 1 to 3 are very short answer questions carrying 2 marks each.
3. SECTION B – Q. No. 4 to 11 are short answer questions carrying 3 marks each.
4. SECTION C- Q. No. 12 is case based question carrying 5 marks.
5. All questions are compulsory.
6. Use of log tables and calculators is not allowed
SECTION – A
1. Answer the following questions. (Any two)
a) According to which law volume is directly proportional to temperature at constant pressure and fixed moles of a gas.
Ans. Charle’s law
b) What will be the Van der Waal’s equation for 1 mol of a gas?
Ans. (P + a/V2)(v – b) = RT
c) Observe the following diagram and write value of Z for real gases
Ans. Z > 1 or Z < 1
2. If water vapour is assumed to be perfect gas, molar enthalpy changes for vaporaisation of 1 mol of water at 1 bar and 1000C is 41 KJ/mol. Calculate the internal energy change when 1 mol of water is vaporised at 1 bar pressure and 1000C.
Ans. H2O (l) — > H2O (g)
ΔH = ΔU + ΔngRT
ΔU = 41000KJ/mol – 1 x 8.3J/molK x 373K
= 41000 J/mol – 3096 J/mol
= 37.904 kJ/mol
3. Distinguish between homogenous equilibria and heterogeneous equilibria with example. (Write any 2 differences)
Ans. In an equilibrium system, all the reactants and products are in the same phase is known as a homogeneous system.
An Equilibrium system having more than one phase is called heterogeneous equilibrium. A familiar example of this type of system is the equilibrium between water vapour and liquid water in a closed container.
SECTION – B
4. a) State the Dalton’s law of partial pressure and write its equation.
Ans. The Dalton’s Law of Partial Pressures states that the total Pressure exerted by the mixture of non-reactive gases is equal to the sum of Partial Pressure of such individual gas.
Ptotal = P1 + P2 + P3 + ….
(b) Arrange the following forces of interaction in increasing order of their magnitude-
London forces, dipole induced dipole and hydrogen bond.
London forces < dipole induced dipole < hydrogen bond
OR
At 250C and 760 mm of Hg pressure a gas occupies 600mL volume. What will be its pressure at a height where temperature is 100C and the volume of the gas is 640mL?
Ans. P1V1/T1 = P2V2/T2
P2 = (760 mm Hg x 600mL x 283K) / (640mL x 298K)
P2 = 676.6 mm of Hg
5. a) What is the value of ΔG0 when water in equilibrium with ice at constant pressure.
Ans. ΔG0 = 0
b) For the reaction to be spontaneous at all temperature, what will be the sign of ΔH0 and ΔS0?
Ans. ΔH0 = NEGATIVE and ΔS0 = POSITIVE
c) Given N2(g) + 3H2(g) –> 2NH3(g) ΔrH0= -92.4 KJ/mol. What is the standard enthalpy of formation of NH3 gas?
Ans. ΔrH0= -92.4 KJ/mol/2 = -46.2 KJ/mol
6. (a) Classify the following species into Lewis acids and Lewis bases.
(i) NH3 (ii) OH– (iii) Cl+ (iv) H+
LEWIS acid – H+, Cl+
LEWIS base – OH–, NH3
(b) The concentration of hydrogen ion in a sample of soft drink is 3.8 x 10-3 M. What is its pH?
-log(3.8×10-3) = 0.5798
pH = -log[H+]
pH = -log(3.8×10-3) = -log 3.8+3
pH = 3 – 0.5798
pH = 2.4202
pH = 2.42
OR
(a) Define – common ion effect
Ans. Common ion effect – Decrease in the degree of dissociation of weak electrolyte by addition of strong electrolyte to it with common ion to that of weak electrolyte.
(b) Describe the effect:
(i) addition of H2
Ans. Forward
(ii) increasing pressure
Ans. Forward
(iii) removal of CO
Ans. Backward
(iv) use of catalyst on equilibrium of the reaction
Ans. No effect on equilibrium
7. Observe the following diagram and answer the following questions:
a) Why is sodium less reactive than potassium?
Ans. Due to higher ionisation enthalpy of sodium than potassium
b) Why are solutions of alkali metals in liquid ammonia are deep blue in colour?
Ans. Due to ammoniated electron
c) LiCl is predominantly covalent while NaCl is ionic. Why?
Ans. Due to small size and higher polarising power of lithium ion than sodium ion.
OR
(a) Why is KO2 paramagnetic?
Ans. Due to presence of one unpaired electron in π * 2p molecular orbital.
b) Why does solubility of alkaline earth metal carbonates and sulphates in water decreases down the group?
Ans. As the lattice enthalpy remain almost constant within a particular group and the hydration enthalpy decreases down the group.
c) Lithium resembles in properties with that of Magnesium. Why?
Due to almost similar charge to radius of both the ions
8.(a) Beryllium and magnesium do not impart colour to flame. Why?
Ans. Electrons of beryllium and magnesium are tightly held by the nucleus and the temperature of the flame is not sufficient to excite to higher energy level.
(b) Arrange the following ions in increasing order of their hydration enthalpy
Li+, Na+, K+, Rb+, Cs+
Ans. Cs+ < Rb+ < K+ < Na+ < Li+
(c) Rb and Cs, on being heated in presence of excess supply of oxygen form superoxide in preference to oxides. Why?
Ans. Rb+ and Cs+ are bigger cations and a bigger cation is stabilised by bigger anion.
9. a) Draw the structure of Cis and Trans But-2-ene.
Ans.
b) Write the IUPAC name of the following compound
CH3 – CH = CH – CH = CH2
Ans. Penta-1,3-diene
OR
a) What effect does branching of an alkene chain has on its boiling point?
Ans. Boiling point of alkenes decreases as branching increases because surface area decreases.
b) Arrange –benzene, n-hexane and ethyne in decreasing order of acidic behaviour.
Ans. ethyne > benzene > n-hexane
c) Write one example of Wurtz reaction (equation only)
Ans. CH3-Cl + CH3-Cl + 2Na — > CH3 – CH3 +2NaCl
10. a) Observe the structure of benzene and answer the following questions.
i) Why is benzene extraordinary stable though it contains three double bonds?
Ans. Due to resonance
ii) Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitution reactions with difficulty?
Ans. Due to attraction between pi electron cloud of benzene ring and electrophile and due to repulsion between pi electron cloud of benzene ring and nucleophile.
b) Arrange the following set of organic compounds in increasing order of reactivity towards electrophile.
Chlorobenzene,2,4-dinitrochlorobenzene, p-nitrochlorobenzene
Ans. 2, 4-dinitrochlorobenzene < p-nitrocholorobenzene < chlorobenzene
11. a) i) Draw the staggered and eclipsed conformation of ethane.
Ans.
First one is staggered conformation and second one is eclipsed conformation.
ii) Which one among the staggered and eclipsed conformation is more stable and why?
Staggered conformation is more stable than eclipsed conformation as there is less repulsion between bonded electrons of C-H bonds in staggered conformation than eclipsed conformation.
iii) Write the products of ozonolysis of ethane.
Ans. Two molecules of formaldehyde are obtained.
H2C=CH2 + O3 —Zn/H2O—> 2HCHO
SECTION C
Read the following passage and answer the questions that follow:
Group 13 and 14 are p-block elements, consisting of metals, non-metals and metalloids. These elements show variable oxidation states, lower oxidation state becomes more stable, when we go down the group. The combined effect of size and availability of d-orbitals influence their ability to form π-bonds. Lighter elements form p π-p π bonds whereas heavier elements form d π-p π or d π-d π bonds. Boron forms electron deficient compounds. Al exhibits +3 oxidation state. TI+ is more stable than TI3+ due to inert pair effect. Carbon is typical non-metal of group 14, shows property of catenation and forms large number of compounds. It also shows allotropy, diamond, graphite and fullerene are crystalline allotropes of carbon. Group 14 elements show +4 and +2 oxidation states, Pb2+ is more stable than Pb4+. CO and CO2 are oxides of carbon. CO2 is acidic and CO has lone pair, therefore, forms metal carbonyls. CO is deadly poisonous. CO2 is greenhouse gas. Silica, silicates and silicones are important classes of compounds of silicon.
12. (i) Why is Pb4+ good oxidising agent?
Ans. It is due to inert pair effect in Pb2+ which is more stable than Pb4+. Pb4+ gains 2 electrons to form Pb2+
(ii) Why is fullerene purest allotrope of carbons?
Ans. It has soccer ball like structure and does not have edges, therefore, impurities can’t enter.
(iii) Explain why graphite is good conductor than diamond?
(iv) Answer the following for group 14 elements:
(a) Name the element whose allotrope is used as abrasive
Ans. Diamond
(b) Name Thermodynamically most stable form of carbon.
Ans. Graphite
OR
(iv) Answer the following for group 13 elements:
(a) Write name of non-metal.
Ans. Boron
(b) Name the element which is used for transport of conc. HNO3
Ans. Aluminium