XI MCQs Chapter 1 ………… Stoichiometry
1. If the volume occupied by oxygen gas (O2) at STP is 44.8dm3, the number molecules of O2 in the vessels are:
(a) 3.01 x 1023
(b) 6.02 x 1023
(c) 12.04 x 1023
(d) 24.08 x 1023
Explanation; (Answer, c)
22.4 dm3 (1 mole) of any gas at STP contains 6.02 x 1023 molecules. Hence 44.8 dm3 (2 mole) of any gas at STP would contain twice of 6.02 x 1023 molecules i.e. 12.04 x 1023 molecules.
No. of molecules = Vg/Vm x NA = 44.8/22.4 x 6.02 x 1023 = 12.04 x 1023 or 1.204 x 1024 molecules.
2. The number of carbon atoms in 1 mole of sugar (C12H22O11) are approximately:
(a) 6 x 1023
(b) 24 x 1023
(c) 60 x 1023
(d) 72 x 1023
Explanation; (Answer, d)
1 mole of sugar (C12H22O11) = 12 moles of carbon atoms
12 moles of carbon atoms = n x NA = 12 x 6.02 x 1023 = 72 x 1023 or 7.2 x 1024 carbon atoms.
3. In the reaction 2Na + 2H20 → 2NaOH + H2, if 23g of Na reacts with excess of water, the volume of hydrogen gas (H2) liberated at STP should be
(a) 11.2dm3
(b) 22.4dm3
(c) 33.6dm3
(d) 44.8dm3
Explanation; (Answer, a)
From stoichiometric ratio in the given equation:
46g (2 mole) Na liberates 22.4 dm3 of hydrogen gas at STP
23g (1 mole) Na liberates 22.4/46 x 23 dm3 of hydrogen gas at STP = 11.2 dm3 of hydrogen gas
4. Which of the following sample of substances contains the same number of atoms as that of 20g calcium:
(a) 16g S
(b) 20g C
(c) 19 g K
(d) 24g Mg
Explanation; (Answer, a)
Species containing same number of moles will have same number of particles (atoms, molecules or formula units). 20 g Ca and 16 g S constitute same number of moles (0.5 moles), hence they have same number of atoms.
No. of moles in 20 g Ca = mass/molar mass = 20/40 = 0.5 mol
No. of moles in 16 g S = mass/molar mass = 16/32 = 0.5 mol
5. The minimum number of moles are present in:
(a) 1 dm3 methane gas at STP
(b) 5 dm3 of helium gas at STP
(c) 10 dm3 of hydrogen gas at STP
(d) 22.4 dm3 of chlorine gas at STP
Explanation; (Answer, a)
For gases at STP, minimum number of moles are present in that gas which has least volume. Here 1 dm3 methane gas at STP has least number of moles as it has least volume.
6. Number of atoms in 60g carbon are:
(a) 3.01x 1023
(b) 3.01 x 1024
(c) 6.02 x 1023
(d) 6.02 x 1024
Explanation; (Answer, b)
No. of atoms = (Mass/molar mass) x NA = 60/12 x 6.02 x 1023 = 3.01 x 1024
7. Maximum number of molecules present in the following sample of gas:
(a) 100g O2
(b) 100g CH4
(c) 100g CO2
(d) 100g Cl2
Explanation; (Answer, b)
The species containing greatest moles will have greatest number of molecules.
For species having same mass, the specie having least molar mass would have greatest number of moles. Here CH4 has least molar mass of 16, so its 100 g would contain maximum number of moles and also maximum number of molecules.
8. Which of the following statement is incorrect?
(a) The mass of 1 mole Cl2 gas is 35.
(b) One mole of H2 gas contains 6.02 x 1023 molecules of H2
(c) Number of atoms in 23g Na and 24 g Mg are equal
(d) One moles of O2 at S.T.P occupy 22.4 dm3 volume
Explanation; (Answer, a)
The statement a is incorrect as the mass of 1 mole of Cl2 is 71 g (not 35).
9. For Avogadro's number, this statement is incorrect?
(a) It is the no. of particles in one mole of any substance
(b) Its numerical value is 6.02x 1023
(c) Its value change if temperature increases
(d) Its value change if number of moles increases.
Explanation; (Answer, c)
Avogadro’s number is independent of temperature and pressure. Its value only depends upon the number of moles and volume of gas at STP.
10. Generally actual yield is:
(a) Greater than theoretical yield
(b) Less than theoretical yield
(c) Equal to the theoretical yield
(d) Some times greater & some times less than theoretical yield
Explanation; (Answer, c)
Generally actual yield is Less than theoretical yield mainly due to side reactions and mechanical loss of product.
XI MCQs Chapter 2 …… Atomic Structure
1. Bohr's theory is not applicable to which of the following
(a) H
(b) H+
(c) He1+
(d) Li2+
Explanation; (Answer, b)
One of the limitations of Bohr's atomic model is that it does not explain the spectra of multi-electron atoms. Bohr's theory is applicable to hydrogen like atoms or hydrogenic ions (single electron system).
All the given species like H, He+ and Li2+ are isoelectronic and have only one electron. Their electronic configurations are same and so their spectra is explained by Bohr's atomic model. But H+ has no electron at all and hence cannot form spectrum.
2. Nitrogen has the electronic configuration 1s2 2s2 2px1 2py1 2pz1 and 1s2 2s2 2px2 2py1.This is determined by
(a) Aufbau principle
(b) Pauli’s rule
(c) Hund's rule
(d) n+l rule
Explanation; (Answer, c)
The filling of electrons in degenerate orbitals like p, d and f is governed by Hund’s rule of maximum electron multiplicity accordingly electrons in degenerate orbitals tend to remain singly with same spin until all of the degenerate orbitals become half filled then pairing of electrons are allowed in them.
3. Quantum number values for 3s orbital are
(a) n=0, l=1
(b) n=1, l=0
(c) n=3, l=1
(d) n=3, l=0
Explanation; (Answer, d)
In the notation 3s (nl), 3 stands for n while s represent l.
For 3s orbital, n =3 and l=0.
4. The radius of first orbit of hydrogen atom is .....
(a)529Å
(b) 52.9 Å
(c) 5.29 Å
(d) 0.529 Å
Explanation; (Answer, d)
Where ao is known as Bohr’s constant or Bohr radius and it the radius of first orbital of hydrogen atom. and its value is 0.529 x 10−10 m or 0.529 x 10−8 cm or 0.529 Å or 0.0529nm or 52.9 pm. This is the radius of the first orbit of H. This equation is used for the determination of nth orbit of hydrogen atom and hydrogen like ions like He+, Li2+ etc.
5. Line spectrum is used as a tool for the identification of ....
(a) Colors
(b) Electrons
(c) Elements
(d) Molecules
Explanation; (Answer, c)
samples of same element always produces same characteristic line spectrum. Each element emit of light of specific wavelength therefore the number of lines and the distance between them depends upon the nature of element, so line spectra is used as “Finger Print” for the identification of elements. For example, line spectrum of sodium contains two yellow coloured lines separated by a definite distance.
6. The shape of orbital for which l = 0 is
(a) Spherical
(b) Dumbbell
(c) Double dumbbell
(d) Complicated
Explanation; (Answer, a)
Its values show shape of orbitals. It has values l = 0 to (n – 1). e.g.
l = 0, used for s-orbital, spherical in shape
l = 1, used for p-orbital, dumb-bell in shape
l = 2, used for d-orbital, double dumb-bell in shape
l = 3, used for f-orbital, complicated shape
7. When 4d orbital is filled, the next electron enter into
(a) 5s
(b) 5p
(c) 5d
(d) 6s
Explanation; (Answer, b)
According to n+ l rule, electrons are filled in various orbitals in the increasing order of (n+l) value. Those orbitals which have lower value of (n+l) value filled first. In case, if two orbitals which have same (n+l) value, then orbital having lower ‘n’ value will be filled first.
The n+l value of 4d orbital is 4+2 =6. The next orbital to be filled must have n+l value either equal to 6 or greater than 6. The n+l value of 5d orbital is 5+2 =7, for 5s it is 5+0=5, for 5p it is 5+1=6 and for 6s it is 6+0=6. Since 5p and 6s orbitals have same n + l value, hence the electron first go to that orbital which has least n value i.e. 5p.
8. Which of the following is not an iso electronic pair?
(a) Na+ and Ne
(b) Na+ and F−
(c) Na and Ca
(d) Na+ and Mg2+
Explanation; (Answer, b)
The species having same number of total electrons and hence have same electronic configuration are called isoelectronic pair. Na+ and F− are isoelectronic pair as both of them contains 10 electrons.
9. Balmer series appears in the hydrogen spectrum if electron jumps from any appropriate higher energy orbit to
(a) Second orbit
(b) Third orbit
(c) Fourth orbit
(d) Fifth orbit
Explanation; (Answer, a)
Balmer noted first that the Hydrogen spectrum consisted of some well defined discrete lines in the visible region (i.e. having l between 4000 to 7000Å) of spectrum. This series is obtained by the transition of electrons from any higher orbit (n2 = 3, 4, 5, 6, 7 …. µ) to 2nd orbit (n1 = 2).
10. In 1935 A.D. James Chadwick was awarded Nobel Prize because ...
(a) He discovered proton
(b) He discovered neutron
(c) He determined the radius of hydrogen atom
(d) He gave the rules for electronic configuration
Explanation; (Answer, b)
James Chadwick won the 1935 Nobel Prize in Physics for the discovery of the neutron.
XI MCQs Chapter 3 … Chemical Bonding
1. If the bond angle is AB2 type molecule is 104.5o, it geometry should be;
(d) Planar Trigonal
Explanation; (Answer, c)
The bond angle of 104.5o is the characteristic of AB2 type of molecules with bent or V-shaped or angular geometry.
2. The highest bond energy in the following is:
(d) H–N
Explanation; (Answer, b)
Bond energy is directly proportional to the bond polarity or partial ionic character which in turn is proportional to the difference in electronegativity of the bonded atoms. Greater the ΔEN, greater is the bond energy. H – F bond has the greatest ΔEN (1.9) and hence it has the highest bond energy (in fact highest among all diatomic molecules containing single bond; 570 kJmol−1).
The second highest bond energy is for H – O bond with second highest ΔEN (1.4). The next highest bond energy is for H – N bond with third highest ΔEN (0.9). Cl–Cl being non-polar has the least bond energy.
3.The molecule which has zero dipole moment is:
(d) CCl4
Explanation; (Answer, d)
Symmetrical molecules with linear, trigonal or tetrahedral geometry show zero dipole moment. CCl4 has symmetrical tetrahedral structure in which all four bond moments for C – Cl bonds being directed in opposite directions are cancel out giving net zero dipole moment.
4. The molecule which has maximum bond angle
(d) BF3
Explanation; (Answer, a)
Linear molecules like CS2, CO2, BeCl2, C2H2, HgCl2, ZnCl2 etc. with AB2 formula has the largest bond angle of 180o.
5. The shape and hybridization of BCl3 molecule is:
(b) Linear and sp
(d) Angular and sp3
Explanation; (Answer, c)
Being AB3 molecule with no non-bonding orbital, BCl3 has sp2 hybridization with planar trigonal shape.
6. Amongst the following molecules which one has trigonal pyramidal shape?
(c) NH3
(d) C2H4
Explanation; (Answer, c)
The ÄB3 molecules like NH3, PH3, PCl3, NCl3 etc. with one lone pair and 3 bond pairs has trigonal pyramidal shape
7. A simple covalent molecule possesses two bond pairs and two lone pairs around the central atom, its shape should be:
(d) Tetrahedral
Explanation; (Answer, c)
The presence of two bond pairs and two lone pairs around the central atom gives rise to angular shape to the molecules. Such molecules have :ÄB2 type general formula.
8. The correct relation between Debye and coulomb meter is:
(b) 1D = 1.6x10−19Cm
(d) 1D = 1.23x10−8Cm
Explanation; (Answer, a)
The SI unit of dipole moment C-M and cgs unit debye is inter related as
1D = 3.33x10−30Cm
9. The bond order of N2 molecule is:
(d) 3
Explanation; (Answer, d)
The bond order of N2 molecule is 3 showing that it has triple bond between two nitrogen atoms.
10. The number of sigma and pi bonds in C2H4 molecules are respectively:
(d) 4 and 2
Explanation; (Answer, c)
There are total 5 sigma and one pi bond in in C2H4 molecule.
XI MCQs Chapter 4 State of Matter, Gas
1. According to Graham’s Law of diffusion, the ratio of diffusion of H2 and O2 are respectively:
(d) 4:1
Explanation; (Answer, d)
r1/r2= √M2/√M2 ⇒ r1/r2= √32/√2 ⇒ r1/r2= √16 ⇒ r1/r2= 4:1
2. Collection of gas over water is an example of:
(b) Dalton’s law
(d) Gay-Lussac law
Explanation; (Answer, b)
Collection of gas over water is an example of Dalton’s law. The pressure of dry gas is calculated by using Dalton’s law.
Pdry gas = Pmoist gas – Pwater vapours
3. The molar volume of oxygen gas is maximum at:
(b) 0°C and 2 atm
(d) 25°C and 2 atm
Explanation; (Answer, c)
The molar volume of a gas varies directly with temperature and varies inversely with pressure. The molar volume of a gas is maximum at highest temperature and least pressure. In option c and d, temperature is highest hence deciding factor will be pressure which is least in option c.
4. The volume of gas would be theoretically zero at:
(d) 273°C
Explanation; (Answer, b)
The volume of gas would be theoretically zero at 0 K (-273°C).
5. If the Kelvin temperature of ideal gas is increase to double and pressure is reduced to one half, the volume of gas will:
(b) Double
(d) Four times
Explanation; (Answer, d)
PV=nRT ⇒ PV = T (R and n = constant) ⇒ V = T/P ⇒ V = 2/ ½ = 2 x 2 = 4
Thus, doubling the Kelvin temperature and halving the pressure quadruples the volume.
6. The molar volume of oxygen gas is 22.4 dm3 at:
(a) 0°C and 1 atm
(b) 25°C and 0.5 atm
(d) 25 K and 0.5 atm
Explanation; (Answer, a)
The molar volume of a gas is 22.4 dm3 at STP i.e. at 0C and 1 atm.
7. Under similar condition CH4 gas diffuses........ times faster than SO2 gas:
(d) 16 times
Explanation; (Answer, b)
r1/r2= √M2/√M2 ⇒ r1/r2= √64/√16 ⇒ r1/r2= √4 ⇒ r1/r2= 2
8. Which one of the following statement is incorrect about the gas molecules?
(a) They have large spaces
(b) They possess kinetic energy
(c) Their collision is elastic
(d) Their molar mass depends upon temperature
Explanation; (Answer, d)
The molar mass of gas is independent of temperature.
9. The diffusion rate of C3H8 and CO2 are same because:
(a) Both are poly atomic gases
(b) Both are denser than air
(d) Both contains carbon atoms
Explanation; (Answer, c)
The rate of diffusion depends upon molar masses of gases. The gases with same molar masses would have same rate of diffusion. The given gases ethane (C3H8) and CO2 have same molar mass of 44 gmol−1.
10. Real gas reaches the ideal behavior at:
(a) Low temperature and low pressure
(b) High temperature and high pressure
(c) Low temperature and high pressure
(d) High temperature and low pressure
Explanation; (Answer, d)
High temperature and low pressure make the ideal.
XI MCQs Chapter 7 -Chemical Equilibrium
1. In the equilibrium system of PCl5(g) ⇌ PCl3(g)+ Cl2(g), the correct relationship between Kc and Kp is:
(a) Kp> Kc
(b) Kp< Kc
(c) Kp = Kc
(d) Kp/Kc=1
Explanation; (Answer, a)
2. In the reaction A(g)+ B(g) ⇌2C(g), the equilibrium constants Kp=Kc because:
(a) ∆n > 1
(b) ∆n < 1
(c) ∆n = 1
(d) ∆n = 0
Explanation; (Answer, d)
3. What happens to the value of Kc when a catalyst is added to a chemical system at equilibrium?
(a) It decreases
(b) it increases
(c) It becomes zero
(d) it remains unchanged
Explanation; (Answer, d)
4. The correct Ksp expression of a sparingly soluble salt Li2C2O4 among the following is
(a) Ksp=[Li+][C2O42−]
(b) Ksp=[Li+]2[C2O42−]
(c) Ksp=[2Li+][ C2O42−]
(d) Ksp=[2Li+]2[C2O42−]2
Explanation; (Answer, b)
5. If the equilibrium expression of a reversible reaction is
Kc = [C]2/[A][B]
The balanced equilibrium equation should be:
(a) C ⇌ A+B
(b) A+B ⇌ C
(c) 2C ⇌ A+ B
(d) A+B ⇌ 2C
Explanation; (Answer, d)
6. The term active mass use in law of mass action means:
(a) No. of mole
(b) No. of molecules
(c) mole per dm³
(d) gram per dm³
Explanation; (Answer, c)
7. The equilibrium of N2(g) +O2(g) ⇌ 2NO(g) (∆H = +ve) is affected by change in:
(a) Temperature only
(b) Pressure only
(c) Both temperature & pressure
(d) Neither temperature nor pressure
Explanation; (Answer, a)
8. The equilibrium of which of the following reaction would not be affected by an increase in pressure:
(a) PCl5(g) ⇌ PCl3(g)+ Cl2(g)
(b) 2NO(g) + Cl2(g) ⇌ 2NOCl(g)
(c) N2(g) + O2(g) ⇌ 2NO(g)
(d) 2SO2(g) + O2(g) ⇌ 2SO3(g)
Explanation; (Answer, b)
9. NaCl when added to an aqueous silver chloride solution:
(a) Decreases the solubility of AgCI
(b) Increases the solubility of AgCl
(c) Forms a clear solution
(d) Does not effect
Explanation; (Answer, a)
10. Some reactions are nearly to completion in the forward direction and identified by their:
(a) Very high value of Kc
(b) Very low value of Kc
(c) Very high value of ∆H
(d) Very low value of ∆H
Explanation; (Answer, a)
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