Practice Questions

Q31.For the following reaction, the mass of water produced from 445 g of C57H110O6 is: 2 C57H110O6(s) + 163O2(g) →114 CO2(g) + 110H2O(l) (1) 490 g (2) 890 g (3) 445 g (4) 495 g

Q31.The minimum amount of O2g consumed per gram of reactant is for the reaction: (Given atomic mass: Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1 ) (1) P4s + 5O2g →P4O10s (2) 2 Mgs + O2g →2MgOs (3) 4Fes + 3O2g →2Fe2O3s (4) C3H8g + 5O2g →3CO2g + 4H2O ( l )

Q31.The quantum number of four electrons are given below: I . n = 4, l = 2, ml = – 2, ms = – 1 / 2 II . n = 3, l = 2, ml = 1, ms = + 1 / 2 III . n = 4, l = 1, ml = 0, ms = + 1 / 2 IV . n = 3, l = 1, ml = 1, ms = – 1 / 2 The correct order of their increasing energies will be: (1) I < III < II < IV (2) IV < II < III < I (3) I < II < III < II (4) IV < III < II < I

Q31.For a reaction, N2(g) + 3H2(g) ⟶2NH3(g), identify di-hydrogen (H2) as a limiting reagent in the following reaction mixtures. (1) 28 g of N2 + 6 g of H2 (2) 35 g of N2 + 8 g of H2 (3) 56 g of N2 + 10 g of H2 (4) 14 g of N2 + 4 g of H2

Q31.What would be the molality of 20% (mass/mass) aqueous solution of KI ? (molar mass of KI = 166 g mol−1 ) (1) 1.35 (2) 1.51 (3) 1.08 (4) 1.48

Q31.A 10mg effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 mL of CO2 at T = 298.15 K and P = 1 bar. If molar volume of CO2 is 25.0 L under such condition, what is the percentage of sodium bicarbonate in each tablet? [Molar mass of NaHCO3 = 84 g mol−1] (1) 0.84 (2) 33.6 (3) 16.8 (4) 8.4

Q31.Which of the graphs shown below does not represent the relationship between incident light and the electron ejected from metal surface? JEE Main 2019 (10 Jan Shift 1) JEE Main Previous Year Paper (1) (2) (3) (4)

Q31.For emission line of atomic hydrogen from ni = 8 to nf = n, the plot of wave number -v against n2 will be: (The Rydberg constant, RH is in wave number unit) (1) Linear with slope -RH (2) Non linear (3) Linear with slope RH (4) Linear with intercept -RH

Q31.What is the work function of the metal if the light of wavelength 4000 A generates photoelectrons of velocity 6 × 105 ms−1 from it? (Mass of electron = 9 × 10−31 kg, velocity of light = 3 × 108 ms−1 , Planck's constant = 6 .626 ×10−34 Js, Charge of electron = 6 .626 ×10−34 Js) (1) 0 .9 eV (2) 4 .0 eV (3) 3 .1 eV (4) 2 .1 eV

Q31.The 71st electron of an element X with an atomic number of 71 enters the orbital: (1) 5d (2) 4f (3) 6p (4) 6s

Q32.The electrons are more likely to be found: (1) Only in the region a (2) In the region a and b (3) In the region a and c (4) Only in the region c

Q32.Which of the following combination of statements is true regarding the interpretation of the atomic orbitals? (A) An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the orbital of lower angular momentum. JEE Main 2019 (09 Jan Shift 2) JEE Main Previous Year Paper (B) For a given value of the principal quantum number, the size of the orbit is inversely proportional to the azimuthal quantum number. (C) According to wave mechanics, the ground state angular momentum is equal to 2πh . (D) The plot of ψ Vs r for various azimuthal quantum numbers, shows peak shifting towards higher r value. (1) (B), (C) (2) (A), (B) (3) (A), (C) (4) (A), (D)

Q32.The type of hybridization and no. of lone pair(s) of electron of Xe in XeOF4 , respectively, are: (1) sp3d2 and 1 (2) sp3d2 and 2 (3) sp3d and 2 (4) sp3d and 1

Q32.The ground state energy of a hydrogen atom is −13 .6 eV. The energy of second excited state of He+ ion in eV is: (1) −27.2 (2) −6.04 (3) −3.4 (4) −54.4

Q32.The graph between |ψ|2 and r (radical distance) is shown below. This represents: (1) 3s orbital (2) 2p orbital (3) 2s orbital (4) 1s orbital

Q32.The de Broglie wavelength (λ) associated with a photoelectron varies with the frequency (v) of the incident radiation as, [v0 is threshold frequency]: 1 (1) λ ∝ 1 (2) λ ∝ 1 (v−v0) 4 (v−v0) (3) λ ∝ 1 3 (4) λ ∝ 1 1 (v−v0) 2 (v−v0) 2

Q32.In general, the properties that decrease and increase down a group in the periodic table, respectively, are : (1) Atomic radius and electronegativity (2) Electronegativity and atomic radius (3) Electronegativity and electron gain enthalpy (4) Electron gain enthalpy and electronegativity

Q32.Among the following, the energy of 2s orbital is lowest in: JEE Main 2019 (12 Apr Shift 2) JEE Main Previous Year Paper (1) K (2) H (3) Na (4) Li

Q32.Heat treatment of muscular pain involves radiation of wavelength of about 900nm. Which spectral line of H atom is suitable for this purpose? [RH = 1 × 105 cm−1 ⋅h = 6.6 × 10−34Js, c = 3 × 108 ms−1] (1) Paschen, ∞→3 (2) Paschen, 5 →3 (3) Balmer, ∞→2 (4) Lyman, ∞→1

Q32.The size of the iso-electronic species Cl- , Ar and Ca2 + is affected by: (1) nuclear charge (2) azimuthal quantum (3) electron –electron (4) Principal quantum number of valence interaction in the outer number of valence shell orbitals shell

Q32.Which one of the following about an electron occupying the 1s orbital in a hydrogen atom is incorrect? (The Bohr radius is represented by a0 ). (1) The total energy of the electron is maximum when it is at a distance a0 from the nucleus. (2) The electron can be found at a distance 2a0 from the nucleus. (3) The probability density of finding the electron is maximum at the nucleus. (4) The magnitude of the potential energy is double that of its kinetic energy on an average.

Q32.The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9. The spectral series are: (1) Paschen and Pfund (2) Balmer and Brackett (3) Lyman and Paschen (4) Brackett and Pfund

Q32.If the de Broglie wavelength of the electron in nth Bohr orbit in a hydrogenic atom is equal to 1.5πa0 (a0 is Bohr radius ), then the value of nz is: (1) 1.50 (2) 1.0 (3) 0.4 (4) 0.75

Q32.The element with Z = 120 (not yet discovered) will be a/an (1) transition metal. (2) alkaline earth metal. (3) alkali metal. (4) inner-transition metal.

Q32.For any given series of spectral lines of atomic hydrogen, let Δ−v = −v max −−v min be the difference in maximum and minimum wave number in cm−1 . The ratio Δ−v Lyman/Δ−v Balmar is (1) 5 : 4 (2) 27 : 5 (3) 4 : 1 (4) 9 : 4