Question 4: The Tyndall effect is observed only when the following conditions are satisfied :

(a) The diameter of the dispersed particles is much smaller than the wavelength of the light used.

(b) The diameter of the dispersed particle is not much smaller than the wavelength of the light used.

(c) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude.

(d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

1. 1) (a) and (c)
2. 2) (b) and (c)
3. 3) (a) and (d)
4. 4) (b) and (d)

Question 5: A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is ‘a’, the closest approach between two atoms in metallic crystal will be :

1. 1) √2 a
2. 2) a/√2
3. 3) 2a
4. 4) 2√2 a

Solution:

1. Answer: (2)

   In FCC, one of the faces is like

   

   By △ABC,

   2a² = 16r²

   r² = (1 / 8) a²

   r = 1 / 2√2 a

   Distance of closest approach = 2r = a / √2

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Question 6: Given

Among the following, the strongest reducing agent is:

1. 1) Cr³⁺
2. 2) Cl⁻
3. 3) Cr
4. 4) Mn²⁺

Solution:

1. Answer: (3)

   

   Positive E° is for Cr, hence it is the strongest reducing agent.

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Question 7: The freezing point of benzene decreases by 0.45^oC when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be : (K_f for benzene = 5.12 K kg mol⁻¹)

1. 1) 74.6%
2. 2) 94.6%
3. 3) 64.6%
4. 4) 80.4%

Solution:

1. Answer: (2)

   0.45 = i (5.12) [(0.2 / 60) / 20] * 1000

   i = 0.527

   2CH₃ COOH ⇋ [CH₃ COOH]₂

   (1 - α) α / 2

   i = (1 – α) / 2

   0.527 = 1 – (α / 2)

   α / 2 = 0.473

   α = 0.946

   % association = 94.6%

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Question 8: The radius of the second Bohr orbit for the hydrogen atom is : (Planck’s Constant. h = 6.6262 × 10⁻³⁴ Js; the mass of electron = 9.1091 × 10⁻³¹ kg; charge of electron e = 1.60210 × 10⁻¹⁹ C; permittivity of vacuum ∈₀ = 8.854185 × 10⁻¹² kg⁻¹m⁻³A²)

1. 1) 0.529 Å
2. 2) 2.12 Å
3. 3) 1.65 Å
4. 4) 4.76 Å

Solution:

1. Answer: (2)

   r = a₀ (n² / Z)

   = 0.529 × 4

   = 2.12 Å

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Question 9: Two reactions R₁ and R₂ have identical pre-exponential factors. The activation energy of R₁ exceeds that of R₂ by 10 kJ mol⁻¹. If k₁ and k₂ are rate constants for reactions R₁ and R₂ respectively at 300 K, then ln (k₂/k₁) is equal to (R = 8.314 J mol⁻¹K⁻¹)

1. 1) 6
2. 2) 4
3. 3) 8
4. 4) 12

Solution:

1. Answer: (2)

   

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Question 10: pK_a of a weak acid (HA) and pK_b of a weak base (BOH) are 3.2 and 3.4, respectively. The pH of their salt (AB) solution is :

1. 1) 7.0
2. 2) 1.0
3. 3) 7.2
4. 4) 6.9

Solution:

1. Answer: (4)

   pH = 7 + [(1 / 2) (pK_a - pK_b)]

   = 7 + (1 / 2) [3.2 - 3.4]

   = 6.9

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Question 11: Both lithium and magnesium display several similar properties due to the diagonal relationship; however, the one which is incorrect, is :

1. 1) both form nitrides
2. 2) nitrates of both Li and Mg yield NO₂ and O₂ on heating
3. 3) both form basic carbonates
4. 4) both form soluble bicarbonates

Solution:

1. Answer: (3)

   Mg forms basic carbonate 3MgCO₃ · Mg (OH)₂ · 3H₂O but no such basic carbonate is formed by Li.

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Question 12: Which of the following species is not paramagnetic?

1. 1) O₂
2. 2) B₂
3. 3) NO
4. 4) CO

Solution:

1. Answer: (4)

   CO has 14 electrons (even) ∴ it is diamagnetic.

   NO has 15e^– (odd) ∴ it is paramagnetic and has 1 unpaired electron in π * 2p molecular orbital.

   B₂ has 10e^– (even) but still paramagnetic and has two unpaired electrons in π * 2p_x and π * 2p_y (s-p mixing).

   O₂ has 16 e^– (even) but still paramagnetic and has two unpaired electrons in π * 2p_x and π * 2p_y molecular orbitals.

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Question 13: Which of the following reactions is an example of a redox reaction?

1. 1) XeF₆ + H₂O → XeOF₄ + 2HF
2. 2) XeF₆ + 2H₂O → XeO₂ F₂ + 4HF
3. 3) XeF₄ + O₂ F₂ → XeF₆ + O₂
4. 4) XeF₂ + PF₅ → [XeF]⁺ PF6 ⁻

Solution:

1. Answer: (3)

   Xe is oxidised from +4 (in XeF₄) to +6 (in XeF₆)

   Oxygen is reduced from +1 (in O₂F₂) to zero (in O₂)

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Question 14: A water sample has ppm level concentration of following anions

F^- = 10; SO^2-₄ = 100; NO₃^- = 50

The anion/anions that make/makes the water sample unsuitable for drinking is/ are :

1. 1) only F⁻
2. 2) only SO^2-₄
3. 3) only NO₃^-
4. 4) both SO^2-₄ and NO₃^-

Solution:

1. Answer: (1)

   The permissible limit of F ^– in drinking water is up to 1 ppm.

   Excess concentration of F ^– > 10 ppm causes decay of bones.

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Question 15: The group having isoelectronic species is :

1. 1) O2⁻, F⁻, Na, Mg²⁺
2. 2) O⁻, F⁻, Na⁺, Mg²⁺
3. 3) O²⁻, F⁻, Na⁺, Mg²⁺
4. 4) O⁻, F⁻, Na, Mg⁺

Solution:

1. Answer: (3)

   Mg²⁺, Na⁺, O^2– and F^– all have 10 electrons each.

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Question 16: The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are :

1. 1) Cl⁻ and ClO⁻
2. 2) Cl⁻ and ClO₂ ⁻
3. 3) ClO⁻ and ClO₃⁻
4. 4) ClO₂⁻ and ClO₃

Solution:

1. Answer: (1)

   

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Question 17: In the following reactions, ZnO is respectively acting as a/an :

(a) ZnO + Na₂O → Na₂ZnO₂

(b) ZnO + CO₂ → ZnCO₃

1. 1) acid and acid
2. 2) acid and base
3. 3) base and acid
4. 4) base and base

Solution:

1. Answer: (2)

   In (a), ZnO acts as acidic oxide as Na₂O is a basic oxide.

   In (b), ZnO acts as basic oxide as CO₂ is an acidic oxide.

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Question 18: Sodium salt of an organic acid ‘X’ produces effervescence with conc. H₂SO₄. ‘X’ reacts with the acidified aqueous CaCl₂ solution to give a white precipitate which decolourises the acidic solution of KMnO₄. ‘X’ is :

1. 1) CH₃COONa
2. 2) Na₂C₂O₄
3. 3) C₆H₅COONa
4. 4) HCOONa

Solution:

1. Answer: (2)

   

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Question 19: The most abundant elements by mass in the body of a healthy human adult are: Oxygen (61.4%); Carbon (22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight which a 75 kg person would gain if all ¹H atoms are replaced by ²H atoms is :

1. 1) 7.5 kg
2. 2) 10 kg
3. 3) 15 kg
4. 4) 37.5 kg

Solution:

1. Answer: (1)

   Mass of hydrogen = (10/100 ) * 75 = 7.5 kg

   Replacing ¹H by ²H would replace 7.5 kg with 15 kg.

   Net gain = 7.5 kg

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Question 20: On treatment of 100 mL of 0.1 M solution of CoCl₃.6H₂O with excess AgNO₃; 1.2 × 10²² ions are precipitated. The complex is :

1. 1) [Co(H₂O)₆]Cl₃
2. 2) [Co(H₂O)₅Cl]Cl₂.H₂O
3. 3) [Co(H₂O)₄Cl₂]Cl.2H₂O
4. 4) [Co(H₂O)₃Cl₃].3H₂O

Solution:

1. Answer: (2)

   Millimoles of AgNO₃ = [(1.2 x 10²²) / (6 * 10²³)] * 1000 = 20

   Millimoles of CoCl₃.6H₂O = 0.1 × 100 = 10

   Each mole of CoCl₃.6H₂O gives two chloride ions.

   [Co(H₂O)₅Cl]Cl₂.H₂O

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Question 21: Which of the following compounds will form a significant amount of meta product during the mono-nitration reaction?

Solution:

1. Answer: (1)

   

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Question 22: Which of the following, upon treatment with tert-BuONa followed by addition of bromine water, fails to decolourize the colour of bromine?

Solution:

1. Answer: (3)

   

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Question 23: The formation of which of the following polymers involves hydrolysis reaction?

1. 1) Nylon 6, 6
2. 2) Terylene
3. 3) Nylon 6
4. 4) Bakelite

Solution:

1. Answer: (3)

   Caprolactam is hydrolysed to produce caproic acid which undergoes condensation to produce Nylon-6.

   

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Question 24: Which of the following molecules is the least resonance stabilized?

Solution:

1. Answer: (2)

   However, all molecules given in options are stabilised by resonance but compound given in option (2) is least resonance stabilised (other three are aromatic)

   

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Question 25: The increasing order of the reactivity of the following halides for the S_N1 reaction is :

1. 1) (I) < (III) < (II)
2. 2) (II) < (III) < (I)
3. 3) (III) < (II) < (I)
4. 4) (II) < (I) < (III)

Solution:

1. Answer: (4)

   Rate of S_N1 reaction ∝ stability of carbocation

   

   So, II < I < III

   Increase in the stability of carbocation and hence increase the reactivity of halides.

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Question 26: The major product obtained in the following reaction is:

1. 1) (+) C₆H₅CH(O^tBu)CH₂C₆H₅
2. 2) (−) C₆H₅CH(O^tBu)CH₂C₆H₅
3. 3) (±) C₆H₅CH(O^tBu)CH₂C₆H₅
4. 4) C₆H₅CH = CHC₆H₅

Solution:

1. Answer: (4)

   

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Question 27: Which of the following compounds will behave as reducing sugar in an aqueous KOH solution?

Solution:

1. Answer: (3)

   Sugars in which there is free anomeric –OH group are reducing sugars.

   

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Question 28: 3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is :

1. 1) Two
2. 2) Four
3. 3) Six
4. 4) Zero

Solution:

1. Answer: (2)

   

   Since the product (X) contains two chiral centres and it is unsymmetrical.

   So, its total stereoisomers = 2² = 4.

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Question 29: The correct sequence of reagents for the following conversion will be :

Solution:

1. Answer: (3)

   

   

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Question 30: The major product obtained in the following reaction is :

Solution:

1. Answer: (4)

   DIBAL — H reduces esters and carboxylic acids into aldehydes

   

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Video Lessons - April 2^nd - Chemistry

JEE Main 2017 Chemistry Paper With Solutions 2nd April

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