MCQs

Some basic concepts of Chemistry*************************
Q1 The dimensions of pressure are the same as that of a. force per unit volumeb. energy per unit volumec. forced. energyAns. b.energy per unit volumeQ2 Given the numbers: 161 cm, 0.161 cm, 0.0161 cm. The number of significant figures for the three numbers is a. 3,3, and 4 b. 3,4 and 4c. 3,4 and 5d. 3,3 and 3Ans. d. 3, 3, and 3Q3 Equal masses of H2, O2 and methane have been taken in a container of volume V at temperature 27'C in identical conditions. The ratio of the volumes of gases H2: O2: methane would bea. 8:16:1b. 16:8:1c. 16:1:2d. 8:1:2Ans. c.16:1:2Q4 What volume of oxygen gas (O2) measured at O'C and 1atm, is needed to burn completely 1L of propane gas (C3H8) measured under the same conditions?a. 5Lb. 10Lc. 7L d. 6LAns. a.5LExplanation: C3H8 + 5O2 = 3CO2 + 4H2OQ5 0.24g of a volatile gas, upon vaporisation, gives 45mL vapour at NTP. What will be the vapour density of the substance? (Density of H2 = 0.089g/L)a. 95.93b. 59.93c. 95.39d. 5.993Ans. b. 59.93Explanation: weight of the gas = 0.24 gvolume of the gas = 45 mL = 0.045 L and density of H2 = 0.089g/Lweight of 45mL of H2= density x volume = 0.089x0.045=4.005x10-3 Vapour density = wt of certain volume of substance/ wt of same vol of hydrogen= 0.24/4.005x10-3 = 59.93Q6 The molecular weight of O2 and SO2 are 32 and 64 respectively. At 15'C and 150mmHg pressure, one litre of O2 contains 'N' molecules. The number of molecules in two litres of SO2 under the same conditions of temperature and pressure will bea. N/2b. Nc. 2Nd. 4NAns. c.2NQ7 What is the weight of oxygen required for the complete combustion of 2.8kg of ethylene?a. 2.8kgb. 6.4kgc. 9.6 kgd. 96kgAns. c. 9.6kgExplanation: 96 x 2.8 x 103 /28 = 9.6kgQ8 An element X has the following isotopic composition: 200X: 90% and 199X : 8.0% and 202X : 2.0%The weighted average atomic mass of the naturally occurring element X is closest to a. 201 amub. 202 amuc. 199 amud. 200 amuAns. d. 200 amuExplanation: 200x90+199x8+202x2/90+8+2 = 199.96amu = 200 amuQ9. Boron has two stable isotopes, 10B (19%) and 11B (18%). Calculate average at. wt. of boron in the periodic table.a. 10.8b. 10.2c. 11.2d. 10.0Ans a. 10.81Q10 Which one of the followings has maximum number of atoms?a. 1g of Agb. 1g of Mgc. 1g of O2d. 1g of LiAns. d. 1g of LiExplanation: Smaller the atomic mass more the no. of atoms.

Structure of Atom*************************
Q1 The number of protons, neutrons and electrons in 175 71Lu respectively area. 71, 104 and 71b. 104, 71 and 71c. 71, 71 and 104d. 175, 104 and 71Ans.a. 71, 104 and 71Q2. Be2+ is isoelectronic with which of the following ions?a. H+b. Li+c. O2-d. N2-Ans. b. Li+Q3. Isoelectronic species area. CO, CN-, NO+, C2-b. CO-, CN, NO, C2-c. CO+ , CN-, NO-, C2d. CO, CN, NO, C2Ans.a. CO, CN-, NO+, C2-Q4. The ion that is isoelectronic with CO is a. CN- b. N2+c. O2-d. N2-Ans.a. CN- Q5 Which one of the following is not isoelectronic with O2-?a. Ti+b. Na+c. N3-d. F-Ans. a. Ti+Q6. Which of the following series of transition in the spectrum of hydrogen atom falls in the visible regions?a. Brackett seriesb. Lyman seriesc. Balmer series,d. Paschen SeriesAns. c. Balmer seriesExplanation: Lyman - UV region; Balmer - Visible Region; Paschen - IR region; Brackett - IR RegionQ7. Calculate the energy in joule corresponding to light of wavelength 45nm (Planck's constant h=6.63x10-34 Js, speed of light c=3x 108 m/s)a. 6.67x1015 b. 6.67x1011c. 4.42x10-15d. 4.42x10-18Ans. d. 4.42x10-18Explanation: E=hc/λ E=6.63x10-34 x 3x 108/45x10-9 = 4.42x10-18 J Q8. The value of planck's constant is 6.63x10-34 Js. The speed of light c=3x 108 m s-1. Which value is closest to the wavelength in nanometer of a quantum of light with frequency of 6x1015 s-1?a. 50b. 75c. 10d. 25Ans. a. 50Explanation: c=ν λ λ = c/ν = 3x 108 /6x 1015 = 50nm Q9. According to law of photochemical equivalence the energy absorbed (in ergs/ moles) is given as (h=6.62x10-27erg, c = 3x 108 m/s, NA = 6.62x1023 mol-1)a. 1.196x108/λ b. 2.859x105/λ c. 2.859x1016/λ d. 1.196x1016/λ Ans. a. 1.196x108/λ Explanation: E = hcNA /λ=6.62x10-27x3x 1010 cm s-1 x 6.62x1023 mol-1/ λ =1.196x1016/λ Q10. The energies E1 and E2 of two radiations are 25eV and 50eV respectively. The relation between their wavelengths i.e. λ1 and λ2 will bea. λ1 = λ2b. λ1 = 2 λ2c. λ1 = 4λ2d. λ1 = 1/2 λ2 Ans. b.λ1 = 2 λ2Explanation: - E1=hc/λ1 and E2 = hc/λ2 E1/E2 = 1/λ1/1/λ2 = λ2/λ1= 25eV/50eV = 1/2Q11. The value of Planck’s constant is 6.636×10–34 Js. The velocity of light is 3.0 ×108 m s–1. Which value is closest to the wavelength in nanometers of a quantum of light with frequency of 8×1015 s–1 ?(a) 2 × 10–25 (b) 5 × 10–18(c) 4×101(d) 3 × 107Ans.(c) 4×101Q12. For given energy, E=3.03×10-19 joules corresponding wavelength is (h = 6.626 × 10-34J sec, c = 3×108 m/sec)(a) 65.6 nm (b) 6.56 nm(c) 3.4 nm (d) 656 nmAns. (d) 656 nmQ13. What will be the longest wavelength line in Balmer series of spectrum?(a) 546 nm (b) 656 nm(c) 566 nm (d) 556 nmAns.(b) 656 nmQ14. Based on equation E=-2.178x10-18 J (Z2/n2) certain conclusions are written. Which of them is not correct? (a) Equation can be used to calculate the change in energy when the electron changes orbit.(b) For n = 1, the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit.(c) The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance from the nucleus.(d) Larger the value of n, the larger is the orbit radius.Ans.(b) For n = 1, the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit.Q15. 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 = 6 to n = 5 (d) n = 5 to n = 3Ans.(c) n = 6 to n = 5Q16. The energy of second Bohr orbit of the hydrogen atom is –328 kJ mol–1; hence the energy of fourth Bohr orbit would be(a) – 41 kJ mol–1 (b) –82 kJ mol–1(c) –164 kJ mol–1 (d) –1312 kJ mol–1 Ans.(b) –82 kJ mol–1Q17. The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be (Given ionization energy of H = 2.18 × 10–18 J atom–1 and h = 6.626 × 10–34 J s)a. 1.54 × 1015 s–1b. 1.03 × 1015 s–1c. 3.08 × 1015 s–1 d. 2.00 × 1015 s–1 Ans.c. 3.08 × 1015 s–1Q18. In hydrogen atom, energy of first excited state is–3.4 eV. Then find out K.E. of same orbit of hydrogen atom.(a) +3.4 eV (b) +6.8 eV(c) –13.6 eV (d) +13.6 eVAns.a) +3.4 eV Q19. Who modified Bohr's theory of introducing elliptical orbits for electron path?a. Rutherford b. Thomsonc. Hundd. SommerfeldAns.d. SommerfeldQ20. The Bohr orbit radius for the hydrogen atom (n = 1) is approximately 0.530 Å. The radius for the first excited state (n = 2) orbit is (in Å)(a) 4.77 (b) 1.06(c) 0.13 (d) 2.12Ans.(d) 2.12Q21. In a Bohr’s model of an atom, when an electron jumps from n = 1 to n = 3, how much energy will be emitted or absorbed?(a) 2.389 × 10–12 ergs (b) 0.239 × 10–10 ergs(c) 2.15 × 10–11 ergs (d) 0.1936 × 10–10 ergsAns.(d) 0.1936 × 10–10 ergsQ22. The radius of hydrogen atom in the ground state is 0.53 Å. The radius of Li2+ ion (atomic number = 3) in a similar state is(a) 0.53 Å (b) 1.06 Å(c) 0.17 Å (d) 0.265 ÅAns.(c) 0.17 Å Q23. The energy of an electron in the nth Bohr orbit of hydrogen atom isa.13.6/ n4 eVb. 13.6/ n3 eVc. 13.6/ n2 eVd. 13.6/ n eVAns.c. 13.6/ n2 eVQ24. The spectrum of He is expected to be similar to thata. H b. Li+c. Na d. He+ Ans.b. Li+Q25. If r is the radius of the first orbit, the radius of nth orbit of H-atom is given by(a)rn2(b) rn(c) r/n(d) r2 n2 Ans.(a)rn2Q26. In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is [Given that Bohr radius, a0=52.9 pm](a) 211.6 pm (b) 211.6 pm(c) 52.9 pm (d) 105.8 pmAns.(b) 211.6 pmQ27. A 0.66 kg ball is moving with a speed of 100 m/s. The associated wavelength will be (h = 6.6 × 10–34 J s)a.6.6 × 10–32 m b. 6.6 × 10–34 mc.1.0 × 10–35 m d. 1.0 × 10–32 mAns. c.1.0 × 10–35 mQ28 If uncertainty in position and momentum are equal, then uncertainty in velocity isa. √h/m √ 𝜋 b. √h/ √ 𝜋 c. √h/2m√ 𝜋 d.√h/√2𝜋 Ans.c. √h/2m√ 𝜋 Q29. The measurement of the electron position is associated with an uncertainty in momentum, which is equal to 1×10-18g cm s-1. The uncertainty in electron velocity is (mass of an electron is 9×10-28g)a.1 × 105cm s–1 b.1 × 1011cm s–1c.1 × 109 cm s–1 d. 1 ×106 cm s–1 Ans. c.1 × 109 cm s–1Q30.Given : The mass of electron is 9.11×10-31 kg, Planck constant is 6.626×10-34 J s, the uncertainty involved in the measurement of velocity within a distance of 0.1 Å is(a) 5.79 × 105 m s–1(b) 5.79 × 106 ms–1(c) 5.79 × 107 m s–1(d) 5.79 × 108 ms–1Ans.(b) 5.79 × 106 ms–1Q31 The uncertainty in momentum of an electron is 1x10-5 kg m/s. The uncertainty in its position will be (h=6.62 x 10-34 kg m2/s)a.5.27x10-30 mb. 1.05x10-26mc.1.05x10-28md.5.25x10-28mAns.a.5.27x10-30 mQ32. The de Broglie wavelength of a particle with mass 1 g and velocity 100 m/s is(a) 6.63 × 10–35 m (b) 6.63 × 10–34 m(c) 6.63 × 10–33 m (d) 6.65 × 10–35 mAns.c) 6.63 × 10–33 m Q33. The position of both, an electron and a helium atom is known within 1.0 nm. Further the momentum of the electron is known within 5.0 × 10-26 kg m s–1. The minimum uncertainty in the measurement of the momentum of the helium atom is(a) 8.0 × 10–26 kg m s–1 (b) 80 kg m s–1(c) 50 kg m s–1(d) 5.0 × 10–26 kg m s–1Ans.(d) 5.0 × 10–26 kg m s–1Q34. Uncertainty in position of an electron (Mass = 9.1 × 10-28 g) moving with a velocity of 3 × 104 cm/s accurate up to 0.001% will be (Use h/4𝜋 in uncertainty expression where h = 6.626 × 10–27 erg second)(a) 5.76 cm (b) 7.68 cm(c) 1.93 cm (d) 3.84 cm Ans.(c) 1.93 cmQ35. Which of the following statements do not form a part of Bohr’s model of hydrogen atom?(a) Energy of the electrons in the orbits are quantized.(b) The electron in the orbit nearest the nucleus has the lowest energy.(c) Electrons revolve in different orbits around the nucleus.(d) The position and velocity of the electrons in the orbit cannot be determined simultaneously.Ans.(d) The position and velocity of the electrons in the orbit cannot be determined simultaneously.Q36. 4d, 5p, 5f and 6p orbitals are arranged in the order of decreasing energy. The correct option is(a) 5f > 6p > 4d > 5p (b) 5f > 6p > 5p > 4d (c) 6p > 5f > 5p > 4d (d) 6p > 5f > 4d > 5pAns.(b) 5f > 6p > 5p > 4dQ37. Orbital having 3 angular nodes and 3 total nodes is(a) 5p (b) 3d (c) 4f (d) 6dAns.(c) 4f Q38. Which one is a wrong statement?a. Total orbital angular momentum of electron in s-orbital is equal to zero.b. An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers.c. The electronic configuration of N atom is 1s22s22p3 d. The value of m for dz2 is zeroAns.c. The electronic configuration of N atom is 1s22s22p3 Q39. Which one is the wrong statement?a. The uncertainty principle is ΔE x Δt ≥ h/4𝜋 b.Half filled and fully filled orbitals have greater stability due to greater exchange energy, greater symmetry and more balanced arrangement.c. The energy of 2s-orbital is less than the energy of 2p-orbital in case of hydrogen like atoms.d. de-Broglie’s wavelength is given by mvwhere m = mass of the particle, λ =h/mv; where m= mass of particle v=group velocity of the particle.Ans.c. The energy of 2s-orbital is less than the energy of 2p-orbital in case of hydrogen like atoms.Q40. How many electrons can fit in the orbital for whichn = 3 and l = 1?(a) 2 (b) 6(c) 10(d) 14Ans. (a) 2Q41. Which of the following pairs of d-orbitals will have electron density along the axes?a. dz2 , dzx b. dyz dxz c. dx2-y2 dz2 d. dxy dx2-y2 Ans.c. dx2-y2 dz2 Q42. Two electrons occupying the same orbital are distinguished by(a) azimuthal quantum number(b) spin quantum number(c) principal quantum number(d) magnetic quantum number. Ans.(b) spin quantum numberQ43. Which is the correct order of increasing energy of the listed orbitals in the atom of titanium?(At. no. Z = 22)(a)4s 3s 3p 3d (b) 3s 3p 3d 4s(c) 3s 3p 4s 3d(d) 3s 4s 3p 3dAns.(c) 3s 3p 4s 3dQ44. The number of d-electrons in Fe2+ (Z=26) is not equal to the number of electrons in which one of the following?a. d-electrons in Fe (z=26)b. p-electrons in Ne (z=10)c. s-electrons in Mg (z-12)d. p-electrons in Cl (z=17)Ans.d. p-electrons in Cl (z=17)Q45. The angular momentum of electron in 'd' orbital is equal to a. 2/3 √ ℏ b. 0ℏ c. 6ℏ d. √ 2ℏ Ans.c. 6ℏ Q46. What is the maximum no. of orbitals that can be identified with the following quantum numbers?n=3, l=1 ml=1a. 1b.2c. 3 d. 4 Ans.a. 1Q47. What is the maximum numbers of electrons that can be associated with the following set of quantum numbers? n = 3, l = 1 and m = –1(a) 4 (b) 2(c) 10(d) 6Ans.(b) 2Q48. The outer electronic configuration of Gd (At. No. 64) is(a) 4f 5 5d4 6s1 (b) 4f 7 5d1 6s2(c) 4f 3 5d5 6s2 (d) 4f 4 5d5 6s1Ans.(b) 4f7 5d1 6s2Q49. Maximum number of electrons in a subshell with l = 3 and n = 4 isa.14b.16c.10d.12Ans.a.14Q50. The correct set of four quantum no. for the valence electron configuration (a) 5, 1, 1, +1/2 (b) 6, 0, 0, +1/2(c) 5, 0, 0, +1/2 (d) 5, 1, 0, +1/2Ans.(c) 5, 0, 0, +1/2Q51 The orbital angular momentum of a p-electron is givena. h/√2 𝜋 b. √3 h / 2 𝜋 c. √3 h / √2 𝜋 d. √6 h/2 𝜋Ans.a. h/√2 𝜋 Q52 The total number of atomic orbitals in fourth energy level of an atom is(a) 8 (b) 16(c) 32 (d) 4Ans.(b) 16Q53. If n = 6, the correct sequence for filling of electrons will be(a) ns→ (n – 2)f → (n – 1)d → np(b) ns→ (n – 1)d → (n – 2)f → np(c) ns→ (n – 2)f → np → (n – 1)d(d) ns → np→ (n-1)d → (n-2)fAns.(a) ns→ (n – 2)f → (n – 1)d → npQ54. Maximum number of electrons in a subshell of an atom is determined by the following(a) 2l + 1 (b) 4l – 2(c) 2n2(d) 4l + 2Ans.(d) 4l + 2Q55. Which of the following is not permissible arrangement of electrons in an atom?(a) n = 5, l = 3, m = 0, s = +1/2 (b) n = 3, l = 2, m = –3, s =–1/2(c) n = 3, l = 2, m = –2, s = –1/2(d) n = 4, l = 0, m = 0, s = –1/2Ans.(b) n = 3, l = 2, m = –3, s =–1/2Q56. Which of the following sets of quantum number is not possible?..n...l....m......si)3...0.....0......+1/2ii)2..2.....1.......+1/2iii)4..3....-2.......-1/2iv)1...0....-1......-1/2v)3....2.....3.......+1/2Which of the following sets of quantum number is not possible?(a) (i), (ii), (iii) and (iv)(b) (ii), (iv) and (v)(c) (i) and (iii)(d) (ii), (iii) and (iv) Ans.(b) (ii), (iv) and (v)Q57. The orientation of an atomic orbital is governed by(a) principal quantum number(b) azimuthal quantum number(c) spin quantum number(d) magnetic quantum number. Ans.(d) magnetic quantum number. Q58. The following quantum numbers are possible for how many orbitals?n = 3, l = 2, m = +2(a) 1 (b) 2(c) 3 (d) 4Ans.(a) 1 Q59.For which of the following sets of four quantum numbers, an electron will have the highest energy?.....n......l.......m........sa.3........2.......1.......+1/2b.4........2......-1.......+1/2c.4........1.......0.......-1/2d.5........0.......0.......-1/2Ans.b.4........2......-1.......+1/2Q60. Electronic configuration of calcium atom can be written as(a) [Ne]4p2 (b) [Ar]4s2(c) [Ne]4s2 (d) [Kr]4p2 Ans.(b) [Ar]4s2Q61. In a given atom no two electrons can have the same values for all the four quantum numbers. This is called(a) Hund’s Rule(b) Aufbau principle(c) Uncertainty principle(d) Pauli’s Exclusion principle.Ans.(d) Pauli’s Exclusion principle.Q62. For azimuthal quantum number l = 3, the maximum number of electrons will be(a) 2 (b) 6(c) 0 (d) 14 Ans.(d) 14Q63. The order of filling of electrons in the orbitals of an atom will be(a) 3d, 4s, 4p, 4d, 5s (b) 4s, 3d, 4p, 5s, 4d(c) 5s, 4p, 3d, 4d, 5s (d) 3d, 4p, 4s, 4d, 5sAns.(b) 4s, 3d, 4p, 5s, 4dQ64. The electronic configuration of Cu (atomic number 29) is(a) 1s2 2s22p6 3s2 3p6 4s2 3d9 (b) 1s2 2s2 2p6 3s2 3p6 3d10 4s1(c) 1s2 2s2 2p6 3s2 3p6 4s2 4p6 5s2 5p1(d) 1s2 2s2 2p6 3s2 3p6 4s2 4p 63d3 Ans.(b) 1s2 2s2 2p6 3s2 3p6 3d10 4s1Q65. The total number of electrons that can be accommodated in all the orbitals having principal quantum number 2 and azimuthal quantum number 1 are(a) 2 (b) 4(c) 6 (d) 8 Ans.(c) 6Q66. An ion has 18 electrons in the outermost shell, it is(a)Cu+ (b) Th4+(c) Cs+ (d) K+ Ans.(a)Cu+Q67. Number of unpaired electrons in N2+ is/are(a) 2 (b) 0(c) 1 (d) 3 Ans.(c) 1 Q68. The maximum number of electrons in a subshell is given by the expression(a) 4l – 2 (b) 4l + 2(c) 2l + 2 (d) 2n2 Ans.(b) 4l + 2Q69. The number of spherical nodes in 3p orbitalsare/is(a) one (b) three(c) none (d) twoAns.a. one

Classification of Elements and Periodicity in Properties*************************
Q1 Identify the incorrect match.Name - IUPAC official Name(A) Unnilunium - (i) Mendelevium(B) Unniltrium - (ii) Lawrencium(C) Unnilhexium - (iii) Seaborgium (D) Unununium - (iv) Darmstadtiuma. (A) i b. (B) iic. (C) iiid. (D) ivAns. d. (D) iv Explanation: 0 - nil, 1-un, 2 - bi, 3-tri, 4-quad, 5-pent, 6-hex, 7-sept, 8-oct, 9-enn
101 Unnilunium Unu Mendelevium Md 102 Unnilbium Unb Nobelium No 103 Unniltrium Unt Lawrencium Lr 104 Unnilquadium Unq Rutherfordium Rf 105 Unnilpentium Unp Dubnium Db 106 Unnilhexium Unh Seaborgium Sg 107 Unnilseptium Uns Bohrium Bh 108 Unniloctium Uno Hassium Hs 109 Unnilennium Une Meitnerium Mt 110 Ununnillium Uun Darmstadtium Ds 111 Unununnium Uuu Rontgenium Rg 112 Ununbium Uub Copernicium Cn 113 Ununtrium Uut Nihonium Nh 114 Ununquadium Uuq Flerovium Fl 115 Ununpentium Uup Moscovium Mc 116 Ununhexium Uuh Livermorium Lv 117 Ununseptium Uus Tennessine Ts 118 Ununoctium Uuo Oganesson Og

Q2. The element Z = 114 has been discovered recently. It will belong to which of the following family/group and electronic configuration?(a) Carbon family, [Rn] 5f 14 6d10 7s2 7p2(b) Oxygen family, [Rn] 5f 14 6d10 7s2 7p4(c) Nitrogen family, [Rn] 5f14 6d10 7s2 7p6(d) Halogen family, [Rn] 5f 14 6d10 7s2 7p5Ans. (a) Carbon family, [Rn] 5f 14 6d10 7s2 7p2Q3. An atom has electronic configuration1s2 2s2 2p6 3s2 3p6 3d3 4s2, you will place itin(a) fifth group (b) fifteenth group(c) second group (d) third group. Ans.(a) fifth groupQ4. The electronic configuration of an element is 1s2 2s2 2p6 3s2 3p3. What is the atomic number of the element, which is just below the above element in the periodic table?(a) 36 (b) 49(c) 33 (d) 34 Ans.(c) 33Q5. If the atomic number of an element is 33, it will be placed in the periodic table in the(a) first group (b) third group(c) fifth group (d) seventh group. Ans. (c) fifth groupQ6. The electronic configuration of four elements are given below. Which elements does not belong to the same family as others?(a) [Xe]4f 145d104s2 (b) [Kr]4d105s2(c) [Ne]3s23p5 (d) [Ar]3d104s2 Ans. (c) [Ne]3s23p5 Q7. For the second period elements the correct increasing order of first ionization enthalpy is(a) Li < Be < B < C < O < N < F < Ne(b) Li < Be < B < C < N < O < F < Ne(c) Li < B < Be < C < O < N < F < Ne(d) Li < B < Be < C < N < O < F < Ne Ans.(c) Li < B < Be < C < O < N < F < NeQ8. Match the oxide given in column I with its property given in column II.Column I.................Column II(i)Na2O ....................A. Neutral(ii)Al2O3 ..................B. Basic(iii)N2O ....................C. Acidic(iv)Cl2O7.................D. AmphotericWhich of the following options has all correct pairs?(a) (i)-B, (ii)-A, (iii)-D, (iv)-C(b) (i)-C, (ii)-B, (iii)-A, (iv)-D(c) (i)-A, (ii)-D, (iii)-B, (iv)-C(d) (i)-B, (ii)-D, (iii)-A, (iv)-C Ans.(d) (i)-B, (ii)-D, (iii)-A, (iv)-C Q9. Which of the following oxides is most acidic in nature?(a) MgO (b) BeO(c) BaO (d) CaO Ans.(b) BeOQ10. In which of the following options the order of arrangement does not agree with the variation of property indicated against it?(a)I < Br < Cl < F (increasing electron gain enthalpy)(b)Li < Na < K < Rb (increasing metallic radius)(c)Al3+ < Mg2+ < Na+ < F– (increasing ionic size)(d)B < C < N < O (increasing first ionisation enthalpyAns.(a)I < Br < Cl < F (increasing electron gain enthalpy)(d)B < C < N < O (increasing first ionisation enthalpyQ11.The formation of the oxide ion, O2– from oxygen atom requires first an exothermic and then an endothermic step as shown below :O2- + e - → O- Δ H = - 141 kJ mol-1O- + e - → O2- Δ H = +780 kJ mol-1Thus, process of formation of O2– in gas phase is unfavourable even though O2– is isoelectronic with neon. It is due to the fact that,(a)O- ion has comparatively smaller size than oxygen atom(b)oxygen is more electronegative(c)addition of electron in oxygen results in larger size of the ion(d)electron repulsion outweighs the stability gained by achieving noble gas configuration. Ans.(d)electron repulsion outweighs the stability gained by achieving noble gas configuration. Q12. Which of the following orders of ionic radii is correctly represented?(a) H- > H+ > H (b) Na+ > F- > O2- (c) F- > O2- >Na+(d) Al3+ > Mg2+ > N3- Ans. NoneQ13. Which one of the following arrangements represents the correct order of least negative to most negative electron gain enthalpy for C, Ca, Al, F and O?(a) Al < Ca < O < C < F(b) Al < O < C < Ca < F(c) C < F < O < Al < Ca(d) Ca < Al < C < O < F Ans. (d) Ca < Al < C < O < F Q14. In which of the following arrangements the given sequence is not strictly according to the property indicated against it?(a)HF < HCl < HBr < HI : increasing acidic strength(b)H2O < H2S < H2Se < H2Te : increasing pK values(c)NH3 < PH3 < AsH3 < SbH3 : increasing acidic character(d)CO2 < SiO2 < SnO2 < PbO2 : increasing oxidising powerAns. (b)H2O < H2S < H2Se < H2Te : increasing pK valuesQ15. Identify the wrong statement in the following.(a)Amongst isoelectronic species, smaller the positive charge on the cation, smaller is the ionic radius.(b)Amongst isoelectronic species, greater the negative charge on the anion, larger is the ionic radius.(c)Atomic radius of the elements increases as one moves down the first group of the periodic table.(d)Atomic radius of the elements decreases as one moves across from left to right in the 2nd period of the periodic tableAns. (a)Amongst isoelectronic species, smaller the positive charge on the cation, smaller is the ionic radius.Q16. What is the value of electron gain enthalpy of Na+ ifIE1 of Na = 5.1 eV?(a) –5.1 eV (b) –10.2 eVc) +2.55 eV (d) +10.2 eVAns.(a) –5.1 eVQ17 Which of the following oxides is amphoteric?a. SnO2b. CaOc. SiO2d. CO2Ans.a. SnO218. The correct order of the decreasing ionic radii among the following isoelectronic species is(a) Ca2+ > K+ > S2- > Cl- (b) Cl- > S2- > Ca2+ > K+(c) S2- > Cl- > K+ > Ca2+(d) K+ > Ca2+ > Cl– > S2- Ans. (c) S2- > Cl- > K+ > Ca2+Q19. Which of the following represents the correct order of increasing electron gain enthalpy with negative sign for the elements O, S, F and Cl?(a)Cl < F < O < S (b) O < S < F < Cl(c) F < S < O < Cl (d) S < O < Cl < FAns.(b) O < S < F < ClQ20. Among the elements Ca, Mg, P and Cl, the order of increasing atomic radii is(a)Mg < Ca < Cl< P (b) Cl < P < Mg < Ca(c) P < Cl < Ca < Mg (d) Ca < Mg < P < ClAns.(b) Cl < P < Mg < CaQ21. Among the following which one has the highest cation to anion size ratio?(a) CsI (b) CsF(c) LiF (d) NaF Ans.(b) CsFQ22. Amongst the elements with following electronic configurations, which one of them may have the highest ionisation energy?(a) Ne [3s2 3p2] (b) Ar [3d10 4s2 4p3](c) Ne [3s2 3p1] (d) Ne [3s2 3p3] Ans.(d) Ne [3s2 3p3] Q23. Identify the correct order of the size of the following.(a)Ca2+ < K+ < Ar < Cl- < S2- (b)Ar < Ca2+ < K+ < Cl- < S2- (c)Ca2+ < Ar < K+ < Cl- < S2- (d) Ca2+ < K+ < Ar < S2- < Cl- Ans.(a)Ca2+ < K+ < Ar < Cl- < S2- Q24. With which of the following electronic configuration an atom has the lowest ionisation enthalpy?(a) 1s2 2s2 2p3 (b) 1s2 2s2 2p5 3s1(c) 1s2 2s2 2p6 (d) 1s2 2s2 2p5Ans.(b) 1s2 2s2 2p5 3s1Q25. Which one of the following ionic species has the greatest proton affinity to form stable compound?a. NH2- b. F- c. I- d. HS- Ans. a. NH2Q26. Which of the following is the most basic oxide?(a) SeO2 (b) Al2O3 (c) Sb2O3 (d) Bi2O3 Ans.(d) Bi2O3Q27. What is the correct relationship between the pH of isomolar solutions of sodium oxide, Na2O (pH1), sodium sulphide, Na2S (pH2 ), sodium selenide, Na2Se (pH3) and sodium telluride Na2Te (pH4)?(a) pH1 > pH2 > pH3 > pH4(b) pH1 > pH2 pH3 > pH4(c) pH1 < pH2 < pH3 < pH4(d) pH1 < pH2 < pH3 pH4 Ans. (a) pH1 > pH2 > pH3 > pH4Q28. Ionic radii are(a) inversely proportional to effective nuclear charge(b) inversely proportional to square of effective nuclear charge(c) directly proportional to effective nuclear charge(d) directly proportional to square of effective nuclear charge.Ans. (a) inversely proportional to effective nuclear chargeQ29. The ions O2- F- Na+ Mg2+ and Al3+ are isoelectronic.Their ionic radii showa. a significant increase from O2- to Al3+b. a significant decrease from O2- to Al3+c. an increase from O2- to F– and then decrease from Na+ to Al3+d. a decrease from O2- to F– and then increase from Na+ to Al3+.Ans. b. a significant decrease from O2- to Al3+Q30. Which of the following order is wrong?(a)NH3 <PH3 <AsH3 – acidic(b)Li < Be < B < C – 1st IP(c)Al2O3 < MgO < Na2O < K2O – basic(d)Li+ < Na+ < K+ < Cs+ – ionic radius. Ans. (b)Li < Be < B < C – 1st IPQ31. Correct order of 1st ionisation potential among following elements Be, B, C, N, O is(a)B < Be < C < O < N(b)B < Be < C < N < O(c) Be < B < C < N < O(d) Be < B < C < O < N Ans.(a)B < Be < C < O < NQ32. Which of the following elements has the maximum electron affinity?(a) I (b) Br(c) Cl (d) FAns.(c) ClQ33. The first ionization potentials (eV) of Be and B respectively are(a) 8.29, 8.29 (b) 9.32, 9.32(c) 8.29, 9.32 (d) 9.32, 8.29Ans.(d) 9.32, 8.29Q34. Which one of the following is correct order of the size of iodine species?(a) I+ > I– > I (b) I– > I > I+(c) I > I– > I+ (d) I > I+ > I– Ans.(b) I– > I > I+Q35. Which of the following ions is the largest in size?(a)K+ (b) Ca2+ (c) Cl- (d) S2- Ans. (d) S2- Q36. Which of the following has the smallest size?(a) Al3+ (b) F–(c) Na+ (d) Mg2+ Ans.(a) Al3+ Q37. Among the following oxides, the one which is most basic isa. ZnO b. MgOc. Al2O3 d. N2O5 Ans.b. MgOQ38. Which of the following has largest size?(a) Na (b) Na+(c) Na– (d) Can’t be predicted.Ans.(c) Na–Q39. Na+, Mg2+, Al3+ and Si4+ are isoelectronic. The order of their ionic size is(a) Na+ > Mg2+ < Al3+ < Si4+ (b) Na+ < Mg2+ > Al3+ > Si4+(c) Na+ > Mg2+ > Al3+ > Si4+(d) Na+ < Mg2+ > Al3+ < Si4+ Ans.(c) Na+ > Mg2+ > Al3+ > Si4+Q40. In the periodic table from left to right in a period, the atomic volume(a) decreases(b) increases(c) remains same(d) first decreases then increases. Ans. (d) first decreases then increases.Q41. Which electronic configuration of an element has abnormally high difference between second and third ionization energy?(a) 1s2, 2s2, 2p6, 3s1(b) 1s2, 2s2, 2p6, 3s1, 3p1 (c) 1s2, 2s2, 2p6, 3s2, 3p2(d) 1s2, 2s2, 2p6, 3s2 Ans.(d) 1s2, 2s2, 2p6, 3s2Q42. One of the characteristic properties of non-metals is that they(a) are reducing agents(b) form basic oxides(c) form cations by electron gain(d) are electronegative. Ans. (d) are electronegative.Q43. Which one of the following has minimum value of cation/anion ratio?(a) NaCl (b) KCl(c) MgCl2(d) CaF2Ans. (c) MgCl2Q44. Which of the following sets has strongest tendency to form anions?(a) Ga, Ni, Tl(b) Na, Mg, Al(c) N, O, F(d) V, Cr, Mn Ans.(c) N, O, FQ45. Elements of which of the following groups will form anions most readily?(a) Oxygen family (b) Nitrogen family(c) Halogens (d) Alkali metalsAns. (c) Halogens Q46. In the periodic table, with the increase in atomic number, the metallic character of an element(a) decreases in a period and increases in a group(b) increases in a period and decreases in a group(c) increases both in a period and the group(d) decreases in a period and the group.Ans.(a) decreases in a period and increases in a groupQ47. Which of the following atoms will have the smallest size?(a) Mg (b) Na(c) Be (d) LiAns. (c) Be

Thermodynamics

≠ ΔrG0 = -nFE0cell →

Q1. Which of the following are not state functions?

(I) q + w

(II) q

(III) w

(IV) H – TS

(a) (I), (II) and (III)

(b) (II) and (III)

(d) (II), (III) and(IV)

Ans. b.

Q2. In a closed insulated container a liquid is stirred with a paddle to increase the temperature, which of the following is true?

(a) E = W≠0, q = 0 (b) E=W=q≠ 0 (c) E = 0, W = q≠0 (d) W = 0, E = q≠0

Ans. a.

Q3. Which of the following is the correct equation?

(a) U = W + Q (b) U = Q – W

Ans. b

Q4. The correct option for free expansion of an ideal gas under adiabatic condition is

(a) q = 0, ΔT = 0 and w = 0

(b) q = 0, ΔT < 0 and w > 0

(d) q > 0, ΔT > 0 and w > 0

Ans. a.

Q5. Under isothermal conditions, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is [Given that 1 L bar = 100 J]

(a) 30 J (b) –30 J

Ans. b.

Explanation: W = - Pext x V = -30J

Q6. Reversible expansion of an ideal gas under isothermal and adiabatic conditions are as shown in the figure.

AB → Isothermal expansion

AC → Adiabatic expansion

Which of the following options is not correct ?

(a) ΔSisothermal > ΔSadiabatic

(b) TA = TB

(d) Tc > TA

Ans. d.

Q7. An ideal gas expands isothermally from 10–3 m3 to 10–2 m3 at 300 K against a constant pressure of 105 N m–2. The work done on the gas is

(a) +270 kJ (b) –900 J

Ans. b.

Q8. A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy U of the gas in joules will be

(a) –500 J (b) –505 J

Ans.b.

Q9. Equal volumes of two monatomic gases, A and B at same temperature and pressure are mixed. The ratio of specific heats (CP/CV) of the mixture will be

(a) 0.83 (b) 1.50

Ans. d.

Q10. Which of the following is correct option for free expansion of an ideal gas under adiabatic condition?

(a) q = 0, ΔT ≠ 0, w = 0

(b) q ≠ 0, ΔT = 0, w = 0

(d) q = 0, ΔT < 0, w≠ 0

Ans. c

Q11. Three moles of an ideal gas expanded spontaneously into vacuum. The work done will be

(a) infinite (b) 3 Joules

Ans. d.

Q12. Assume each reaction is carried out in an open container. For which reaction will ΔH = ΔE?

(a) 2CO(g) + O2(g) → 2CO2(g)

(b) H2(g) + Br2(g) → 2HBr(g)

(d) PCl5(g) PCl3(g) + Cl2(g)

Ans. b.

Q13. The work done during the expansion of a gas from a volume of 4 dm3 to 6 dm3 against a constant external pressure of 3 atm is (1 L atm = 101.32 J)

(a) – 6 J (b) – 608 J

Ans. b.

Q14. For the reaction,

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

at constant temperature, H – E is

(a) + RT (b) –3RT

Ans. b.

Q15. The molar heat capacity of water at constant pressure, C , is 75 J K–1 mol–1. When 1.0 kJ of heat is supplied to 100 g of water which is free to expand, the increase in temperature of water is

(a) 1.2 K (b) 2.4 K

Ans. b.

Q16 When 1 mol of gas is heated at constant volume temperature is raised from 298 to 308K. Heat supplied to the gas is 500 J. Then which statement is correct?

(a) q = w = 500 J, Δ E = 0 (b) q = Δ E = 500 J, w = 0 (c) q = w = 500 J, ΔE = 0

(d) ΔE = 0, q = w = –500 J

Ans. b.

Q17. For the reaction,

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

which one is true?

(a) Δ H = Δ E – RT (b) Δ H = Δ E + RT

Ans. a.

Q18. In an endothermic reaction, the value of H is

(a) negative (b) positive

Ans. b.

Q19. One mole of an ideal gas at 300 K is expanded isothermally from an initial volume of 1 litre to 10 litres. The E for this process is

(R = 2 cal mol–1 K–1)

(a) 1381.1 cal (b) zero

Ans. b.

Q20. During isothermal expansion of an ideal gas, its

(a) internal energy increases

(b) enthalpy decreases

(d) enthalpy reduces to zero.

Ans.c.

Q21. For the reactionm, N2 + 3H2 ⇌ 2NH3 , ΔH = ?

(a) ΔE + 2RT

(b) ΔE – 2RT

(d) ΔE – RT

Ans. b.

Q22. If ΔH is the change in enthalpy and ΔE, the change in internal energy accompanying a gaseous reaction, then

(a) ΔH is always greater than ΔE

(b) Δ H < Δ E only if the number of moles of the products is greater than the number of moles of the reactants

(d) Δ H < Δ E only if the number of moles of products is less than the number of moles of the reactants.

Ans. d.

Q24. Standard enthalpy of vaporisation ΔvapH° for water at 100°C is 40.66 kJ mol–1. The internal energy of vaporisation of water at 100°C (in kJ mol–1) is

(a) +37.56 (b) – 43.76

(Assume water vapour to behave like an ideal gas)

Ans.a.

Q25. Consider the following processes :

..............................H (kJ/mol)

1/2A → B............+150

3B → 2C + D ........125

E + A → 2D.........+350

For B + D → E + 2C, ΔH will be

(a) 525 kJ/mol (b) –175 kJ/mol

Ans. b.

Q26. The following two reactions are known

Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g); H = –26.8 kJ

FeO(s) + CO(g) → Fe(s) + CO2(g); H = – 16.5 kJ

The value of H for the following reaction

Fe2O3(s) + CO(g) 2FeO(s) + CO2(g) is

Equilibrium

Equilibrium acts/ works / comes into force in reversible reactions.

A + B ⇌ C + D

If concentration of reactants

-increases then the reaction proceeds in the forward direction

-decrease then the reaction proceeds in the backward direction

If concentration of products

- increases, then the reaction proceeds in the backward direction

- decreases, then the reaction proceeds in the forward direction

Dynamic Equilibrium - the rates of forward and reverse reactions are equal and there is no net change in composition

A + B ⇌ C + D

Kc = [C] [D] / [A] [B]

Kc is equilibrium constant.

4NH3 (g) + 5O2 (g) ⇌ 4NO(g) + 6H2O (g)

Kc = [NO]4[H2O]6/[NH3 ]4[O2 ]5

Equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward direction.

4NH3 (g) + 5O2 (g) ⇌ 4NO(g) + 6H2O (g)

Kc = [NO]4[H2O]6/[NH3 ]4[O2 ]5

but for the reverse reaction/ backward reaction the K'c=1/Kc

pv = nRT

p = n/v RT and we can write n/V = C and C is concentration

p = CRT and R = 0.0831 bar litre/mol K

Kp=Kc(RT)Δn ∆n = (number of moles of gaseous products) – (number of moles of gaseous reactants)

1pascal, Pa=1Nm-2, and 1bar = 105 Pa

If Kc > 103 , products predominate over reactants, i.e., if Kc is very large, the reaction proceeds nearly to completion.

If Kc < 10-3, reactants predominate over products, i.e., if Kc is very small, the reaction proceeds rarely.

If Kc is in the range of 10-3 to 103 , appreciable concentrations of both reactants and products are present.

If Qc < Kc , net reaction goes from left to right

If Qc > Kc , net reaction goes from right to left.

If Qc = Kc , no net reaction occurs

-∆G is negative, then the reaction is spontaneous and proceeds in the forward direction.

-∆G is positive, then reaction is considered non-spontaneous. Instead, as reverse reaction would have a negative ∆G, the products of the forward reaction shall be converted to the reactants.

-∆G is 0, reaction has achieved equilibrium; at this point, there is no longer any free energy left to drive the reaction.

∆G0 = – RT ln Kc

Le Chatilier's Principle: - change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change.

Conjugate is difference in H+ ion

HF is an acid, then its conjugate base is F-

NH3 is a base, then its conjugate acid is NH4+

HCl (acid) + H2O(base) ⇌ H3O+ (conjugate acid) + Cl- (conjugate base)

*strong acids have very weak conjugate bases and vice versa.

H2O(base) + H2O(acid) ⇌ H3O+ (conjugate acid) + OH- (conjugate base)

Kw = [H3O+] [OH-] /[H2O]

Kw is ionic product of water and its value is equal to 10-14

Acidic: [H3O+] > [OH-]

Neutral: [H3O+] = [OH-]

Basic : [H3O+] < [OH-]

pH = - log [H+]

ionisation constant for weak acid is Ka and for weak base is Kb

Q1. In liquid-gas equilibrium, the pressure of vapours above the liquid is constant at

(a) constant temperature

(b) low temperature

(d) none of these.

Ans. a. constant temperature

Q20 20. If the value of equilibrium constant for a particular reaction is 1.6 × 1012, then at equilibrium the system will contain

(a) mostly products

(b) similar amounts of reactants and products

(d) mostly reactants.

Ans. (a) mostly products.

Q21. In Haber process, 30 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the composition of gaseous mixture under the aforesaid condition in the end?

(a) 20 litres ammonia, 20 litres nitrogen, 20 litres hydrogen

(b) 10 litres ammonia, 25 litres nitrogen, 15 litres hydrogen

(d) 20 litres ammonia, 25 litres nitrogen, 15 litres hydrogen

Ans. N2 + 3H2 ⇌ 2NH3

b. 10 L ammonia, 25 L nitrogen, 15 L hydrogen

Q26. Which one of the following conditions will favour maximum formation of the product in the reaction

A2(g) + B2(g) ⇌ X2(g), rH = –X kJ ?

(a) Low temperature and high pressure

(b) Low temperature and low pressure

(d) High temperature and low pressure

Ans. a. low temp and high pressure.

PDPs

Q1 The equilibrium constant for the reaction N2 + 3H2 = 2NH3 is K. The equilibrium constant for

1/2 N2 + 3/2 H2 ⇌ NH3 will be

a. K/2

b. 2K

c. √ K

d. K2

Ans. c. √ K

Q2 If an inert gas is added to the equilibrium mix of the dissociation of PCl5 in a closed vessel,

a. Conc of Cl2 increases

b. Conc of PCl3 increases

c. Conc of PCl5 increases

d. the equilibrium conc will remain unaffected.

Ans. d. the equilibrium conc will remain unaffected.

PCl5 ⇌ PCl3 + Cl2

Q3 If pressure is increased on the equilibrium N2 + O2 ⇌ 2NO, the equilibrium will

a. shift in the forward direction

b. shift in the backward direction

c. remain undisturbed

d. may shift in the forward or backward direction

Ans. c. remain undisturbed.

Q4 For the reaction: PCl3 (g) + Cl2 (g) ⇌ PCl5(g) and the value of Kc at 250 oC is 26. The value of Kp at this temp will be?

a. 0.61

b. 0.57

c. 0.83

d. 0.46

Ans. a. 0.61

Solution: Kp=Kc(RT)Δn

Δn= no of moles of gaseous product - number of moles of gaseous reactant =1−2=−1

R = 0.0821, T = 250 + 273.15 = 523.15 and Kc= 26

Kp=0.61

Q5 According to Le Chatelier's Principle - adding heat to a solid and liquid in equilibrium will cause the:

a. amount of solid to decrease

b. amount of liquid to decrease

c. temperature to rise

d. temperature to fall.

Ans. a. amount of solid to decrease

Q6 In a reaction A2(g) + 4B2(g) ⇌ 2AB4(g), ΔH<0. The formation of AB4(g) will be favoured by

a. low temp and high pressure

b. high temp and low pressure

c. low temp and low pressure

d. high temp and high pressure

Ans. a. low temp and high pressure

Solution: as the reaction is exothermic, when temp is lowered; according to Le Chatelier's principle the reactants will react to release heat so decreasing the temp will favour the forward reaction. Also when we put pressure in reactant side we have 5 molecules in total however 2 molecules in product side. High pressure will make reactant molecules which are larger in no. to collide and form product.

Q7 The reaction which proceeds in the forward direction is:

a. Fe3O4 + 6HCl ⇌ 2FeCl3 + 3H2O

b. NH3 + H2O + NaCl ⇌ NH4Cl + NaOH

c. SnCl4 + Hg2Cl ⇌ SnCl2 + 2HgCl2

d. 2CuI + I2 + 4K+ ⇌ 2Cu2+ + 4KI

Ans. a. Fe3O4 + 6HCl ⇌ 2FeCl3 + 3H2O

Due to absence of hydrolysis of FeCl3, backward reaction will not take place.

Q8 For which of the following reaction Kp = Kc

a. 2NOCl(g) ⇌ 2NO(g) + Cl2(g)

b. N2(g) + 3H2(g) ⇌ 2NH3(g)

c. H2(g) + Cl2(g) ⇌ 2HCl(g)

d. 2N2O4(g) ⇌ 2NO2(g)

Ans.c. H2(g) + Cl2(g) ⇌ 2HCl(g)

Solution: if we get change in no. of moles as zero then Kp = Kc

Kp=Kc(RT)Δn

Kp=Kc(RT)0

Q9 In a vessel containing SO3, SO2 and O2 at equilibrium some helium gas is introduced so that the total pressure increases while temp and vol remain constant. According to Le Chatelier's Principle, the dissociation of SO3;

a. increases

b. decreases

c. remains unaltered

d. changes unpredictably.

Ans. c. remains unaltered

Solution: SO3 ⇌ SO2 + O2

Q10 An equilibrium mixture for the reaction

2H2S (g) ⇌ 2H2(g) + S2(g) has 1 mole of H2S, 0.2 mole of H2 and 0.8 mole of S2 in a 2L flask. The value of Kc in mol / L is

a. 0.004

b. 0.08

c. 0.016

d. 0.160

Ans. d.0.16

Redox Reactions****
Q1. Which of the following is redox reaction?(a) Evaporation of H2O(b) Both oxidation and reduction(c) H2SO4 with NaOH(d) In atmosphere O3 from O2 by lightningAns. b. Both oxidation and reductionQ2. Without losing its concentration, ZnCl2 solution cannot be kept in contact with(a) Pb (b) Al(c) Au (d) AgAns. b. AlQ3. What is the change in oxidation number of carbon in the following reaction?CH4 + 4Cl2 ----> CCl4 + 4HCla. +4 to +4b. 0 to +4c. -4 to +4d. 0 to -4Ans. c. -4 to +4Q4 The correct structure of tribomooctaoxide is Ans. daily hwQ5. Which of the following reactions are disproportionation reactions?a. 2Cu+ → Cu2+ + Cub. 3MnO2- + 4H+ → 2MnO- + MnO + 2H2O c. 2KMnO4 → K2MnO4 + MnO2 + O2 d. 2MnO4- + 3Mn2+ + 2H2O → 5MnO2 + 4H+ Select the correct option from the followinga. i, iv onlyb. i, ii onlyc. i, ii and iiid. i, iii, ivAns. b. i, ii onlyQ6 The oxidation state of Cr in CrO5 is (a) –6 (b) +12(c) +6 (d) +4Ans. c. +67. The correct order of N-compounds in its decreasing order of oxidation states is(a) HNO3, NO, N2, NH4Cl(b) HNO3, NO, NH4Cl, N2(c) HNO3, NH4Cl, NO, N2(d) NH4Cl, N2, NO, HNO3Ans. (a) HNO3, NO, N2, NH4ClQ8. For the redox reaction, MnO4- + C2O4-2 + H+ → Mn2+ + CO2 + H2O The correct coefficients of the reactants for the balanced equation areMnO4- , C2O4-2, H+a. 16, 5, 2b. 2, 5, 16c. 2, 16, 5d.5, 16, 2Ans. b. 2, 5, 16Q9. Hot concentrated sulphuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidizing behaviour?(a) Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O(b) S + 2H2SO4 → 3SO2 + 2H2 O(c) C + 2H2SO4 → CO2 + 2SO2 + 2H2 O(d) CaF2 + H2SO4 → CaSO4 + 2HF Ans. (d) CaF2 + H2SO4 → CaSO4 + 2HF Q10. (I) H2O2 + O3 → H2O + 2O2(II) H2O2 + Ag2O → 2Ag + H2O + O2Role of hydrogen peroxide in the above reactions is respectively(a) oxidizing in (I) and reducing in (II)(b) reducing in (I) and oxidizing in (II)(c) reducing in (I) and (II)(d) oxidizing in (I) and (II)Ans. c. reducing in I and II11. The pair of compounds that can exist together is(a) FeCl3, SnCl2(b) HgCl2, SnCl2(c) FeCl2, SnCl2 (d) FeCl3, KIAns. (c) FeCl2, SnCl212. A mixture of potassium chlorate, oxalic acid and sulphuric acid is heated. During the reaction which element undergoes maximum change in the oxidation number?(a) S (b) H(c) Cl (d) CAns. Cl13. Oxidation numbers of P in PO43-, of S in SO42- and that of Cr in Cr2 O72- are respectively (a) +3, +6 and +5(b) +5, +3 and +6(c) –3, +6 and +6(d) +5, +6 and +6Ans. (d) +5, +6 and +6Q14. Number of moles of MnO4- required to oxidize one mole of ferrous oxalate completely in acidic medium will be(a) 7.5 moles (b) 0.2 moles(c) 0.6 moles (d) 0.4 moles. Ans. (d) 0.4 moles.Q15. Which is the best description of the behaviour of bromine in the reaction given below?H2O + Br2 → HOBr + HBr(a) Proton acceptor only(b) Both oxidised and reduced(c) Oxidised only(d) Reduced onlyAns. (b) Both oxidised and reducedQ16. The oxidation states of sulphur in the anions SO32-, S2O42- ,S2O62- a. S2O42-<SO32- < S2O62- b. SO32- < S2O42- < S2O62- c. S2O42- < S2O62- < SO32-d. S2O62-< S2O42- < SO32- Ans. a. S2O42-<SO32- < S2O62- Q17 Oxidation state of Fe in Fe3O4 isa. 5/4b. 4/5c. 3/2d. 8/3Ans. d. 8/3Q18. Reaction of sodium thiosulphate with iodine gives(a) tetrathionate ion (b) sulphide ion(c) sulphate ion (d) sulphite ion. Ans. a. tetrathionate ionQ19. The oxide, which cannot act as a reducing agent is(a) CO2(b) ClO2(c) NO2 (d) SO2 Ans. (a) CO220. Which substance is serving as a reducing agent in the following reaction?14H+ + Cr2O72- + 3Ni → 7H2O + 2Cr3+ + 3Ni2+ a. H+b. Cr2O72-c. H2O d. NiAns. d. NiQ21. The oxidation state of I in H2IO- isa. +1b. -1 c. +7 d. +5Ans. c. +7Q22. Consider the change in oxidation state of bromine corresponding to different emf values as shown in the given diagram: BrO4- + 1.82V ---> BrO3- + 1.5V -----> HBrO + 1.595V ----> Br2 + 1.0652V ----> Br- Then the species undergoing disproportionation is a. BrO3-b. BrO4-c. Br2 d. HBrOAns. d. HBrO

HydrogenQ1. One would expect proton to have very large(a) charge(b) ionization potential(c) hydration energy(d) radius.Ans. bQ2. The ionization of hydrogen atom would give rise to(a) hydride ion(b) hydronium ion(c) proton(d) hydroxyl ion.Ans. c.Q3.Which one of the following pairs of substances on reaction will not evolve H2 gas?(a)Copper and HCl (aqueous)(b)Iron and steam(c)Iron and H2SO4 (aqueous)(d)Sodium and ethyl alcoholAns. a Q4.Water gas is produced by(a)passing steam through a red hot coke(b)saturating hydrogen with moisture(c)mixing oxygen and hydrogen in the ratio of 1 : 2(d)heating a mixture of CO2 and CH4 in petroleum refineries.Ans. aQ5.Which of the following metal evolves hydrogen on reacting with cold dilute HNO3?(a)Mg (b) Al(c) Fe (d) CuAns. a6. Which of the following statements about hydrogen is incorrect?(a) Hydronium ion, H3O+ exists freely in solution.(b) Dihydrogen does not act as a reducing agent.(c) Hydrogen has three isotopes of which tritium is the most common.(d) Hydrogen never acts as cation in ionic salts.Ans. b and cQ7. Which of the following is electron-deficient?(a) (BH3)2 (b) PH3(c) (CH3)2 (d) (SiH3)2Ans. aQ8. The method used to remove temporary hardness of water is(a) synthetic resins method(b) Calgon’s method(c) Clark’s method(d) ion-exchange method. Ans. cQ9. The number of hydrogen bonded water molecule(s) associated with CuSO4. 5H2O is(a) 3 (b) 1(c) 2 (d) 5Ans. dQ10. Which of the following groups of ions makes the water hard?(a) Sodium and bicarbonate(b) Magnesium and chloride(c) Potassium and sulphate(d) Ammonium and chlorideAns.bQ11. At its melting point, ice is lighter than water because(a) H2O molecules are more closely packed in solid state(b) ice crystals have hollow hexagonal arrangement of H2O molecules(c) on melting of ice the H2O molecules shrinks in size(d) ice forms mostly heavy water on first meltingAns. b13. The structure of H2O2 is(a) spherical (b) non-planar(c) planar (d) linearAns. b 14. The volume strength of 1.5 N H2O2 solutionis (a) 8.8(b) 8.4(c) 4.8(d) 5.2Ans. b15. The O – O – H bond angle in H2O2 is (a) 106° (b) 109°28(c) 120° (d) 97°Ans. d16. Hydrogen peroxide molecules are(a) monoatomic and form X 2– ions(b) diatomic and form X– ionsc. diatomic and form X- ionsd. monoatomic and form X- ionsAns. bQ17. In diborane how many 3c-2e- are therea. oneb. twoc. threed. fourAns. b. two18. The reaction of H2O2 with H2S is an example of ......... reaction.(a) addition (b) oxidation(c) reduction (d) acidic Ans. b19. Some statements about heavy water are given below.(i) Heavy water is used as a moderator in nuclear reactors.(ii) Heavy water is more associated than ordinary water.(iii) Heavy water is more effective solvent than ordinary water.Which of the above statements are correct?(a) (i) and (ii) (b) (i), (ii) and (iii)(c) (ii) and (iii) (d) (i) and (iii)Ans. a

Q1 Which of the following statements is most applicable to hydrogen? It can act

a. as a reducing agent

b. as an oxidising agent

c. both as oxidising agent and reducing agent

d. neither as an oxidising nor as a reducing agent

Ans. c. both as oxidising agent and reducing agent.

Explanation: hydrogen when gets bonded with metals, it oxidises metals example Na+H-

Q2 Hydrogen combines with other elements by

a. losing an electron

b. gaining an electron

c. sharing an electron

d. losing, gaining and sharing of an electron

Ans. d. losing, gaining and sharing of an electron.

Q3. The first ionisation enthalpy (kJ/mol) for H, Li, F, Na has one of the following values: 1681, 520, 1312, 495. Which of these values corresponds to that of hydrogen.

a. 1681

b. 1312

c. 520

d. 495

Ans. b. 1312 kJ/mol

Q4 Which of the following metals cannot be used for liberating hydrogen from dil HCl?

a. Zn

b. Copper

c. Iron

d. Magnesium

Ans. b. Copper

Explanation: Electrode potential of Copper is more positive than other metals. Hence, it does not substitute H in HCl.

Q5. Which of the following is used as a moderator in nuclear reactors?

a. Hard water

b. Heavy Water

c. Deionised water

d. Mineral water.

Ans. b. Heavy water

Q6 The temporary hardness of water due to Calcium bicarbonate can be removed by adding.

a. CaCO3

b. Ca(OH)2

c. CaCl2

d. HCl

Ans. b. Ca(OH)2

Explanation: Permanent hardness is due to the presence of Chlorides & Sulfates of Ca and Mg.

Temporary hardness is due to the presence of bicarbonates of Ca and Mg.

Q7 In the calgon process of softening of water, which of the following is used?

a. Sodium polymetaphosphate

b. Hydrated sodium aluminium silicate

c. Cation exchange resins

d. Anion exchange resins

Ans. a. Sodium polymetaphosphate

Q8 Hydrogen will not reduce

a. heated cupric oxide

b. heated ferric oxide

c. heated stannic oxide

d. heated aluminium oxide

Ans. d. heated aluminium oxide will never be reduced by hydrogen as Reduction potential of aluminium is more negative than hydrogen.

Q9 The oxidation states exhibited by hydrogen in its various compounds are

a. -1 only

b. 0 only

c. +1, -1, 0

d. +1 only

Ans. c. +1, -1, 0

Q10 The oxidation states of the most electronegative element in the products of the reaction of BaO2 with dil. H2SO4 are

a. 0 and -1

b. -1 and -2

c. -2 and 0

d. -2 and +1

Ans. b. -1 and -2.

Explanation: BaO2 + H2SO4 = BaSO4 + H2O2

Here in this reaction oxidation state of the most electronegative element i.e. oxygen is -2 everywhere besides in H2O2 where the oxidation state of oxygen is -1.

Q11. 30 volumes H2O2 means.

a. 30% H2O2

b. 30cm3 of the solution contains 1g of H2O2

c. 1cm3 of the solution liberates 30 cm3 of O2 at STP

d. 30 cm3 of the solution contains one mole of H2O2

Ans. c. 1cm3 of the solution liberates 30cm3 of O2 at STP.

Explanation.

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