Students must start practicing the questions from CBSE Sample Papers for Class 11 Chemistry with Solutions Set 10 are designed as per the revised syllabus.
CBSE Sample Papers for Class 11 Chemistry Set 10 with Solutions
Time Allowed: 3 hours
Maximum Marks: 70
General Instructions:
- All questions are compulsory.
- This question paper contains 37 questions.
- Questions 1-20 in Section A are objective type-very short answer type questions carrying 1 mark each.
- Questions 21 – 27 in Section B are short answer type questions carrying 2 marks each.
- Questions 28 – 34 in Section C are long-answer I type questions carrying 3 marks each.
- Questions 35-37 in Section D are long-answer II type questions carrying 5 marks each.
- There is no overall choice. However, an internal choice has been provided in six questions of one mark, two questions of two marks, two questions of three marks and two questions of five marks. You must attempt only one of the choices in such questions.
- Use log tables, if necessary. Use of calculator is not allowed.
Section – A
Question 1.
The probability density plots of Is and 2s orbitals are given in figure : [1]
The density of dots in a region represents the probability density of finding electrons in the region.
On the basis of above diagram which of the following statement is incorrect ?
(A) Is and 2s orbitals are spherical in shape.
(B) The probability of finding the electron is maximum near the nucleus.
(C) The probability of finding the electron at a given distance is equal in all directions.
(D) The probability density of electrons for 2s orbital decreases uniformly as distance from the nucleus increases.
Answer:
(D) The probability density of electrons for 2s orbital decreases uniformly as distance from the nucleus increases.
Explanation:
The probability density of electrons in 2s orbital first increases and then decreases. It increases again as distance from nucleus increases.
Question 2.
Which of the following is not an actinoid ? [1]
(A) Curium (Z = 96)
(B) Californium (Z= 98)
(C) Uranium (Z = 92)
(D) Terbium (Z = 65)
OR
The order of screening effect of electrons of s, p, d and f orbitals of a given shell of an atom on its outer shell electrons is:
(A) s > p > d > f
(B) f > d > p > s
(C) p < d < s < f (D) f > p > s > f
Answer:
(D) Terbium (Z = 65)
Explanation:
Terbium belongs to lanthanide series. It is the ninth member in the lanthanide series.
OR
(C) p < d < s < f
Question 3.
Which of the following pair is expected to have the same bond order ? [1]
(A) O2, N2
(B) O2+, N2–
(C) O2–N2+
(D) O2–,N2––
Answer:
(B) O2+, N2–
Explanation:
Question 4.
What will be the molality of the solution containing 18.25 g of HC1 gas in 500 g of water? [1]
(A) 0.1m
(B) 1M
(C) 0.5 m
(D) 1 mA
Answer:
(D) 1 mA
Explanation:
Mass of HCl = 18.25 g Mass of solvent (water) = 500 g = 0.5 kg Molecular mass of HCl = 1 + 35.5 = 36.5
Number of moles of HCl = 18.25 36.5 = 0.5
Molality m = 0.5 ÷ 0.5 = 1 m
Question 5.
The last element of the p-block in 6th period is represented by the outermost electronic configuration. [1]
(A) 7s2 7p6
(B) 5f14 6d107s27p0
(C) 4f14 5d106s26p6
(D) 4f14 5d106s26p4
OR
The elements with atomic numbers 35, 53 and 85 are all.
(A) noble gases
(B) halogens
(C) heavy metals
(D) light metals
Answer:
(C) 4f14 5d106s26p6
OR
(B) halogens
Explanation:
The elements with atomic numbers 35, 53, 85 are all halogens as they have 7 electrons in their outermost shells.
Z = 35 Bromine-electronic configuration is
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5
Z = 53 Iodine-electronic configuration is
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p5
Z = 85 Astantine (-) electronic configuration is
[Xe] 4f10 5d10 6s2 6p5
Question 6.
What is the mass percent of carbon in carbon dioxide? [1]
(A) 0.034%
(B) 27.27%
(C) 3.4%
(D) 28.7%
Answer:
(B) 27.27%
Explanation:
Molecular mass of CO2 = 12 + 2(16) = 12 + 32 = 44 g 44 g of CO2 contains 12 g atoms of carbon. Mass percent of carbon = Mass of carbon in CO Molar mass of CO2 2 × 100 = 12 44 100 × = 27.27%
Read the passage given below and answer the questions given below it:
The existing large number of organic compounds and their ever-increasing numbers has made it necessary to classify them on the basis of their structures. Organic compounds are broadly classified as open-chain compounds which are also called aliphatic compounds. Aliphatic compounds further classified as homocyclic and heterocyclic compounds. Aromatic compounds are special types of compounds. Alicyclic compounds, aromatic compounds may also have heteroatom in the ring. Such compounds are called heterocyclic aromatic compounds. Organic compounds can also be classified on the basis of functional groups, into families or homologous series. The members of a homologous series can be represented by general molecular formula and the successive members differ from each other in molecular formula by a -CH2 unit.
Answer the questions from (7) to (10):
Question 7.
Which of the following is example of homocyclic compound? [1]
(A) Furan
(B) Thiophene
(C) Cyclohexane
(D) Pyrrole
Answer:
(C) Cyclohexane
Question 8.
Where do heterocyclic compounds find application? [1]
(A) Pharmaceuticals
(B) Agrochemicals
(C) Dye stuff
(D) All of the above
Answer:
(D) All of the above
Question 9.
What is the next member of homologous series for Ethanol? [1]
(A) Propanol
(B) Butanol
(C) Pentanol
(D) Hexanol
Answer:
(A) Propanol
Explanation:
The members of a homologous series can be represented by general molecular formula and the successive members differ from each other in molecular formula by a -CH2 unit. So adding -CH2 group in ethanol, next member is propanol.
Question 10.
Give the IUPAC name of: [1]
(A) Puran
(B) hexylcyclohexane
(C) Thiocyclohexane
(D) none of these
Answer:
(C) Thiocyclohexane
Question 11.
How many molecules of SO2 are present in 11.2 L at STP ? [1]
OR
Calculate the number of atoms in 32.0 u of He.
Answer:
∵ Number of molecules of SO2 present in 22.4 L at STP = 6.022 × 1023
∴ Number of molecules of SO2 present in 11.2 L at
STP = \(\frac{6.022 \times 10^{23} \times 11.2}{22.4}\)
= 3.011 × 1023
OR
1 atom of He = 4 u of He
∵ 4 u of He = 1 atom of He 1
∴ 32.0 u of He = \(\frac{1}{4}\) × 32 = 8 atoms
Question 12.
Can a moving cricket ball have a wave character ? Justify your answer. [1]
Answer:
Yes, a moving cricket ball has a wave character. It possesses wave length which is called de Broglie wave length and is given by the equation:
λ = \(\frac{h}{m v}=\frac{h}{p}\)
Question 13.
Write the IUPAC name and symbol of the element with atomic number 119. [1]
OR
Predict the position of the element in the periodic table satisfying the electronic configuration (n – 1)d1ns2 for n = 4.
Answer:
The roots for 1, 1 and 9 are un, un and enn respectively. Hence, the symbol and the IUPAC name are Uue and Ununennium respectively.
OR
(n – 1)d1ns2 for n = 4 becomes 3d14s2. It lies in the 4th period and in the 3rd group.
Commonly Made Error:
Students usually go wrong as they get confused with period and the group
Answering Tips:
- Students must remember the criteria on how to decide the group and period of an element based on electronic configuration.
- Learn the atomic number, element name, symbol, electron configurations, their group and period from the periodic table clearly.
Question 14.
What is enthalpy change at constant volume? Explain. [1]
Answer:
When reaction is carried out in a closed vessel so that volume remains constant, i.e., ΔV = 0, then qp = qv = ΔV or ΔH = ΔU. This is known as enthalpy change at constant volume.
Question 15.
Give IUPAC name of: [1]
Answer:
4-Ethyldeca-1, 5, 8-triene
Question 16.
Predict if the solutions of the following salts are neutral, acidic or basic : NaCl, NaCN ? [1]
Answer:
NaCl: Its solution is neutral as it is the salt of a strong acid HCl and a strong base NaOH.
KBr: Its solution is neutral as it is the salt of strong acid HBr and a strong base KOH.
Question 17.
Calculate the oxidation number of the underlined elements in the following compounds : [1]
(a) K2CrO4
(b) MnO4–
Answer:
(a) Let x be the oxidation no. of chromium in
K2CrO4
Oxidation no. of K = + 1
Oxidation no. of O = – 2
∴ 2 × (+1) + x + 4 × (- 2) = 0
or x – 6 = 0 or x = + 6
Hence, oxidation no. of Cr in K2CrO4 = +6.
(b) Let x be the oxidation no. of Mn in MnO4–
Oxidation of O = -2
∴ x + 4 × (- 2) = – 1
x – 8 = – 1
x = – 1 + 8 = +7.
Question 18.
Assertion (A): It is impossible to determine the exact position and exact momentum of an electron simultaneously. [1]
Reason (R): The path of an electron in an atom is clearly defined.
(A) Both A and R are correct and R is the correct explanation of A.
(B) Both A and R are correct but R is not the correct explanation of A.
(C) A is true but R is false.
(D) Both A and R are false.
Answer:
(C) A is true but R is false.
Question 19.
Assertion: The radius of the first orbit of hydrogen atom is 0.529A. [1]
Reason: Radius of each circular orbit (rn) – 0.529 Å (n2/Z), where n = 1,2, 3 and Z = atomic number.
(A) Both A and R are correct and R is the correct explanation of A.
(B) Both A and R are correct but R is not the correct explanation of A.
(C) A is true but R is false.
(D) Both A and R are false.
Answer:
(A) Both A and R are correct and R is the correct explanation of A.
Question 20.
Why 3° carbocation are more stable than 1° carbocation ? [1]
Answer:
Tertiary carbocation has three electron repelling alkyl groups. This increases +I effect on carbon and reduces the positive charge making it more stable.
Section B
Question 21.
(a) Arrange the following in the increasing order of their e/m values : electron, proton and neutron. [2]
(b) Calculate the number of molecules present in 22.0 g of CO2.
Answer:
(a) The increasing order of their e/m values of electron, proton and neutron is:
Neutron < proton < electron.
(b) Number of molecules
= \(=\frac{\text { Mass of } \mathrm{CO}_2}{\begin{array}{c}
\text { Molecular mass of } \\
\mathrm{CO}_2
\end{array}}\) × 6.022 × 1023
= \(\frac{22}{44}\) × 6.022 × 1023
= 3.011 × 1023 molecules
Question 22.
Account for the fact that the fourth period has 18 and not 8 elements. [2]
Answer:
The first element of fourth period is potassium (Z = 19) having electronic configuration 1s2 2s2 2p6 3s2 3p6 4s1. Thus it starts with the filling of 4s orbital. After the filling of 4s orbital, instead of 4p orbitals, there starts the filling 3d orbitals. This is in keeping with the fact that 3d orbitals have less energies than 4p orbitals. Thus, 10 elements are built up by the filling of 3d orbitals. After the filling of 3d orbitals, 4p orbitals are filled and that process is completed at krypton [Kr, Z = 36]. Hence the fourth period consists of 18 and not 8 elements.
Question 23.
Explain why the following systems are not aromatic?
Answer:
(i) For the given compound, the number of
p-electrons is 6.
By Huckel’s rule,
4n + 2 = 6
4n = 4
n = 1
For a compound to be aromatic, the value of n must be an integer (n = 0, 1, 2…). Since, the value of n is an integer, it should be aromatic but for aromatic p electron cloud is equally distributed on all the ring, that’s why it is non aromatic in nature.
(ii) For the given compound, the number of π-electrons is 4.
By Huckel’s rule,
4n + 2 = 4
4n = 2
n =1/2
For a compound to be aromatic, the value of n must be an integer (n = 0, 1, 2…), which is not true for the given compound. Hence, it is not aromatic in nature.
Question 24.
(i) Give one example of symmetric alkane.
(ii) How does dilution of water affect the pH of buffer solution?
OR
(i) How will you separate propene from propyne?
(ii) Define reaction quotient.
Answer:
(i) Butene is a symmetric alkane.
(ii) Rate of dilution shows no effect on pH of buffer solution as pH depends on ratio of salt, acid or salt, base and dilution don’t affect this ratio.
OR
(i) By passing mixture through ammoniacal silver nitrate solution when propyne reacts while propene pass over.
(ii) It is used to measure relative amount of product and reactant present in reaction at any time.
Question 25.
In an electrophilic substitution reaction of nitrobenzene, how the presence of nitro group affect the reaction ? [2]
Answer:
Nitro group by virtue of strong I-effect withdraws electrons from the ring and increases the charge and destabilises carbocation. In ortho, para-attack of electrophile on nitrobenzene, we get two structures in which positive charge appears on the carbon atom directly attached to the nitro group. As nitro group is electron withdrawing by nature, it decreases the stability of such product and hence meta attack is more feasible when electron withdrawing substituents are attached.
Commonly Made Error:
Students do mistake in placing +ve charge and drawing double bonds, when they write the resonating structure.
Answering Tip:
Understand drawing resonating structures and practice well.
Question 26.
Draw the cis- and fra ns- structures of but-2-ene. Which isomer will have a dipole moment ? [2]
Answer:
The cis-and-frans structures of but-2-ene are as follows:
Although both cis-but-2-ene and trans-but-2-ene has non- polar compounds but cis-but-2-ene has a small dipole moment. In trans-but-2-ene, the vectors of any small bond dipoles must cancel each other because of its shape. So, the dipole moment of trans-but-2-ene is zero.
Question 27.
Draw the resonating structure of : [2]
(a) Ozone molecule
(b) Nitrate ion
OR
Predict the shapes of the following molecules on the basis of hybridisation.
BCl3, CH4
Answer:
(a) The resonating structures of ozone molecule are
(b) The resonating structures of nitrate ion are
OR
BCl3 has sp2 hybridisation and CH4 has sp3 hybridisation.
BCl3 : 5B = 2, 3 = 1s2, 2s22p1
Since, BCl3 possesses sp2 hybridization. So, its shape is trigonal planar with bond angle of 120°.
The structure of BCl3 is shown as below :
Since, CH4 possesses sp3 hybridization with four bond pairs and zero lone pairs. So, its shape is tetrahedral with bond angle of 109.5°.
The structure of CH4 is shown as below :
Commonly Made Error:
Some students wrongly allocate electrons in ground and excited sate and thus write wrong hybridization.
Answering Tip:
- Understand the concept of predicting the shape of molecule by the state of hybridization of the central atom.
- Also practice drawing the structures of molecules.
Section – C
Question 28.
Complete the following blanks based on their respective equations :
(i) C4H10(g) + \(\frac{13}{2}\)O2(g) → 4CO2(g) + 5H2O ΔH = – 2878 kj
ΔH is the heat of butane gas.
(ii) HCl (aq) + NaOH(aq) → NaCl (aq) + H2O(l); ΔH = – 57.1 kJ
In real terms
H+ (aq) + Cl– (aq) + Na+ (aq) + OH– (aq) → Na+ (aq) + Cl– (aq) + H2O(I)
ΔH is the heat of of hydrochloric acid and sodium hydroxide solutions.
(iii) C(s) → C(g); ΔH = 716.7 kj
(graphite)
ΔH is the heat of …………….. of graphite. [3]
Answer:
(i) combustion
(ii) neutralisation
(iii) sublimation
Question 29.
Calculate the mass percent of different elements present in sodium sulphate (Na2SO4). [3]
Answer:
Molar mass of Na2SO4 = 2 (23.0) + 32.066 + 4(16.0) = 142.006 g
Mass % of an element
Question 30.
How much energy is required to ionise a H-atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H-atom (energy required to remove the electron from n = 1 orbit). [3]
Answer:
Energy for a hydrogen electron present in a particular energy shell
Ionization energy for hydrogen electron present in 5th orbit n = 5
IE5 = E∞ – E5
= 0 – \(\left(\frac{-2.18 \times 10^{-18}}{25}\right)\) J atom-1
= 8.72 × 10-20 J atom-1
Ionization energy for hydrogen electron present in 1st orbit n = 1
IE5 = E∞ – E5
= 0 – \(\left(\frac{-2.18 \times 10^{-18}}{1}\right)\)
= 2.18 × 10-18 J atom-1
∴ \(\frac{I E_5}{I E_1}\) = \(\frac{8.72 \times 10^{-20} \text { Jatom }^{-1}}{2.18 \times 10^{-18} \text { Jatom }^{-1}}\)
= 0.04
∴ IE1 is 25 times as compared to IE5.
Question 31.
(a) B2 has ten electrons but paramagnetic. Why ?
(b) Why KHF2 exist but KHCl2 does not ? [3]
OR
(a) Why do noble gas Ne2 does not exist but Ne2+ exists ?
(b) Give reason:
Water molecule has bent structure whereas BeCl2 has a linear structure.
Answer:
(a) B2 has ten electrons but it is paramagnetic in nature. It can be explained by its molecular orbital diagram.
5B = 2, 3 = 1s2, 2s22p1
5B = 2, 3 = 1s2, 2s22p1
Molecular Orbital Configuration
Since, it contains two unpaired electrons in π-bonding, molecular orbitals have equal energy. So, it is paramagnetic in nature.
Commonly Made Errors:
(a) Some students make mistake while writing MO electronic configuration and sometimes with MO diagram .
(b) They sometimes are not able to find Bonding and anti-bonding electrons correctly.
Answering Tips:
- Understand the basic concept of MO diagrams and learn how to write MO diagram, electronic configuration and finding bond order.
- Practice many examples.
(b) KF forms H-bonds with HF whereas KCl cannot form H-bond with HCl.
Hence, KHF2 can dissociate to give HF2– ion and therefore, KHF2 exists. However, KHCl2 cannot dissociate to HCl2– ion and therefore, KHCl2 does not exist.
OR
(a) Ne2 does not exist but Ne2+ exists. It can be explained by means of molecular orbital diagrams.
10Ne = 2, 8 = 1s2, 2s2 2p6
10Ne = 2, 8 = 1s2, 2s2 2p6
Molecular orbital configuration:
Since the bond order of Ne2 is zero, it does not exist as a molecule.
However, in case of Ne2+
Bond order = \(\frac{n_b-n_a}{2}=\frac{6-5}{2}=\frac{1}{2}\) = 0.5
(Due to positive charge, one electron is lost from σ*2pz antibonding orbital.)
Thus, it is clear that noble gas Ne2 does not exist but Ne2+ exists.
(b) Be (Atomic number 4) has a total of 4 electrons. Its electronic configuration is 1s2 2s2. It has two valence electrons and no lone pair of electrons. Each of the two electrons can form bonds with chlorine atoms. Chemical bonds involve electrons and because all electrons have a negative charge, the two bonds between the Be and Cl will arrange themselves around the central atom (Be) so that the electrons in the bonds are as far apart as possible. Thus, the molecule is linear with the chlorines on opposite sides of the Be.
In case of water, oxygen has a total of 8 electrons, with electronic configuration 1s2 2s2 2p4. Oxygen has two electron pairs in the outermost shell, as well as two valence electrons available for bonding. The two electrons can form bonds with two single electrons of hydrogen atoms and would like to again form a linear shape. However, the two electrons pairs exert a strong repulsive force on the O—H bonds, distorting them into the bent shape of the water molecule. This shape maximises the distance between all the electrons surrounding the central oxygen atom. That s why water molecule is bent in shape.
Question 32.
(i) Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per endothermic reaction :
CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g)
(a) Write an expression for Kp for the above reaction.
(b) How will the values of Kp and composition of the equilibrium mixture be affected by :
(a) increasing the pressure
(b) increasing the temperature
(c) using a catalyst ?
(ii) Write the expression for the equilibrium constant, Kc for each of the following reactions :
2NOCl(g) ⇌ 2NO(g) + Cl2(g)
OR
(i) Describe the effect of following conditions on the equilibrium
2H2(g) + CO(g) ⇌ CH3OH(g)
(a) addition of H2
(b) addition of CH3OH
(c) removal of CO
(d) removal of CH3OH
(ii) Write the expression for the equilibrium constant, BQ for each of the following reactions :
2CU(NO3)2(S) ⇌ 2CuO(s) + 4NO2(g) + O2(g)
Answer:
(i)
(a) kp = \(\frac{\left(p_{\mathrm{CO}}\right)\left(p_{\mathrm{H}_2}\right)^3}{\left(p_{\mathrm{CH}_4}\right)\left(p_{\mathrm{H}_2 \mathrm{O}}\right)}\)
(b) (A) By Le Chatelier’s principle, equilibrium will shift in the backward direction (as nr < np) on increasing pressure.
(B) As the given reaction is endothermic, by Le Chatelier’s principle, equilibrium will shift in the forward direction, on increasing the temperature.
(C) Equilibrium composition will not be disturbed but equilibrium will be attained quickly, by using a catalyst.
(ii) 2NOCl(g) ⇌ 2NO(g) + Cl2(g)
kc = \(\frac{[\mathrm{NO}(g)]^2\left[\mathrm{Cl}_2(g)\right]}{[N O C l(g)]^2}\)
OR
(i) (a) On addition of H2, equilibrium will shift in the forward direction. (H)
(b) On addition of CH3OH, equilibrium will shift in backward direction.
(c) On removal of CO, equilibrium will shift in the backward direction.
(d) On removal of CH3OH, equilibrium will shift in the forward direction.
(ii) 2CU(NO3)2(S) ⇌ 2CuO(s) + 4NO2(g) + O2(g)
Kc = [NO2(g)]4 [O2(g)] ∵ [Cu(NO3)2(s)] = 1 & [CuO(s)] = 1
Question 33.
Complete the following reactions :
(a) CaCO3(s) + 2HCl (aq) →
(b)
(c)
Answer:
(a) CaCO3(s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O(I)
(b)
(c)
Question 34.
Define:
(i) Salt bridge
(ii) electrode potential
(iii) Electrochemical series [3]
Answer:
(i) Salt Bridge: It is an inverted U-tube that contains an electrolyte and connects the two half cells in a galvanic cell.
(ii) Electrode Potential: The potential associated with each electrode is called electrode potential.
(iii) Electrochemical Series: The series in which different electrodes are arranged in decreasing order of their standard reduction potentials (EQ) is called electrochemical series.
Section – D
Question 35.
Define:
(i) Bond enthalpy
(ii) Calorific value
(iii) Explain the Born Haber Cycle to determine the lattice enthalpy of NaCl.
Answer:
(i) The average amount of energy required to break one mole of the bond of a particular type in gaseous molecules
(ii) It is the amount of heat produced on complete combustion of one gram of a fuel. Its unit is cal/g.
(iii) The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state.
Na+Cl–(s) → Na+(g) + Cl–(g), ∆latticeH°
=+788 kJ mol-1
The lattice enthalpy of NaCl can be determined by Born-Haber Cycle in following steps:
Step-I: Na(s) → Na(g), sublimation of sodium
metal (∆subH° = 108.4 kJ mol-1)
Step-II: Na(g) → Na+(g) + e–(g), Ionization of
sodium atom (ionization enthalpy ∆iH° = 496 kJ mol-1)
Step-III: Cl2(g) → Cl(g), dissociation of
chlorine, the reaction enthalpy is half the bond dissociation enthalpy.
(\(\frac{1}{2}\) ∆bondH° = 121 kJ mol-1)
Step-IV: Cl(g) + e–(g) → Cl–(g), electron gained by chlorine atoms
(Electron gain enthalpy ∆egH° = – 348.6 kJ mol-1)
Fig. Enthalpy diagram for lattice enthalpy of NaCl
Step-V: Na+(g) + Cl–(g) → Na+Cl–(s)
Applying Hess’s law,
∆lattice H° = 411.2 + 108.4 + 121 + 496 – 348.6 = + 788 kJ mol-1
Question 36.
An alkyl halide C5H11Br (A) reacts with ethanolic KOH to give an alkene ‘B’, which reacts with Br2 to give a
compound ‘C, which on dehydrobromination gives an alkyne ‘D’. On treatment with sodium metal in liquid ammonia one mole of ‘D’ gives one mole of the sodium salt of ‘D’ and half a mole of hydrogen gas. Complete hydrogenation of ‘D’ yields a straight chain alkane. Identify A,B, C and D. Give the reactions involved. [5]
OR
An unsaturated hydrocarbon A’ adds two molecules of H2 and on reductive ozonolysis gives butane-1,4-dial, ethanal and propanone. Give the structure of A’, write its IUPAC name and explain the reactions involved.
Answer:
The reactions suggest that (D) is a terminal alkyne. This means triple bond is at the end of the chain. It could be either (I) or (II).
Since alkyne ‘D’ on hydrogenation yields straight chain alkane, therefore structure (I) is the structure of alkyne (D).
Hence, the structures of A, B and C are as follows:
(A) CH3—CH2—CH2—CH2—CH2Br
(B) CH3—CH2—CH2—CH = CH2
(C) CH3—CH2—CH2—CH(Br)—CH2Br
Commonly Made Error:
Students get confuse, and many times miss reactions.
Answering Tips:
- Draw the structures of compounds for better understanding.
- Learn all the important conversion reactions in the chapter. Understand the problem given and then write the reactions in steps.
OR
Two molecules of hydrogen add on ‘A’ this shows that ‘A’ is either an alkadiene or an alkyne.
On reductive ozonolysis ‘A’ gives three fragments, one of which is dialdehyde. Hence, the molecule has broken down at two sites. Therefore, ‘A’ has two double bonds. It gives the following three fragments:
OHC—CH2—CH2—CHO, CH3CHO and CH3— CO—CH3
Hence, its structure as deduced from the three fragments must be
Question 37.
For the reaction :
N2(g) + 3H2(g) ⇌ 2NH3(g) + Heat; the equilibrium constant K = 0.50 at 673K
(a) Write the expression for Kc and Kp.
(b) What will be the equilibrium constant value for the reverse reaction at 673 K ?
(c) What is the effect of increasing temperature on the yield of NH3 ?
(d) What is the effect of adding N2(g) and H2(g) on the yield of NH3 ?
(e) What is the effect of increasing pressure on the yield of NH3? [5]
OR
(i) (a) The solubility of silver chloride (AgCl) in water at 25°C is 1.08 × 10-5 mol L-1. Calculate the solubility
product of AgCl at this temperature.
(b) The molar solubility of lead iodide is 4.0 × 10-5 mol/L at 25°C. Calculate the solubility product of lead iodate at this temperature.
(ii) In one litre saturated solution of AgCl [Ksp = 1.6 × 10-10], 0.1 mol of CuCl [Ksp = 1.0 × 10-6] is added. Find out the resultant concentration of Ag+ in the solution.
Answer:
(a) kc = \(\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}\)
and kp = \(\frac{\left[p_{\mathrm{NH}_3}\right]^2}{p_{\mathrm{N}_2} \times\left(p_{\mathrm{H}_2}\right)^3}\)
(b) Keq (for reverse reaction) = \(\frac{1}{K}\)
= \(\frac{1}{0.50}\)
(c) As this is an exothermic reaction, on increasing the temperature, the reaction will proceed in reverse direction. This results in decrease in concentration of NH3.
(d) On increasing the concentration of N2(g) and H2(g), the yield of NH3 will increase.
(e) On increasing the pressure, the reaction will proceed in forward reaction or pressure decreases.
OR
(i) (a) S = 1.08 × 10-5 mol L-1
Solubility product
Ksp = [Ag+] [Cl–]
= S × S
= (1.08 × 10-5) (1.08 × 10-5)
= 1.1664 ≈ 1.17 × 10-10
(b) S = 4.0 × 10-5 mol L-1
Solubility product
Ksp = [Pb2+] [IO3–]2
= (S) (2S)2
= S × 4S2 = 4S3
= 4(4 × 10-5)3
= 2.56 × 10-13
(ii) Let the concentration of AgCl is x mol/L
and concentration of CuCl is y mol/L
CuCl ⇌ Cu+ + Cl–
y = 0.1 y =0.1
Ksp of AgCl = [Ag+] [Cl–]
= x (x + y)
Ksp of CuCl = [Cu+] [Cl–]
= y(x + y)
On solving equations (i) and (ii), we get
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