Mock Test 6: MCQs with Explanations
1. Light of wavelength $\lambda$ shines on a metal surface with intensity $x$ and the metal emits $y$ electrons per second of average energy, $z$. What will happen to $y$ and $z$ if $x$ is doubled?
- (a) y will be doubled and z will become half.
- (b) y will remain same and z will be doubled.
- (c) Both y and z will be doubled.
- (d) y will be doubled but z will remain same.
Correct Answer: (d)
Explanation: According to the laws of photoelectric effect, the number of photoelectrons emitted ($y$) is directly proportional to the intensity of incident light ($x$). However, the average kinetic energy ($z$) of the emitted electrons depends only on the frequency (wavelength) of the light, not its intensity.
2. A vessel has nitrogen gas and water vapour at a total pressure of 1 atm. The partial pressure of water vapour is 0.3 atm. The contents of this vessel are transferred to another vessel having one third the capacity of the original vessel, completely at the same temperature. The total pressure of the system in the new vessel is:
- (a) 2.7 atm
- (b) 1 atm
- (c) 3.33 atm
- (d) none of these.
Correct Answer: (d)
Explanation: The initial partial pressure of $N_{2}$ is $1 - 0.3 = 0.7$ atm. Reducing the volume to one-third would triple the partial pressure of $N_{2}$ to 2.1 atm. The final total pressure would depend on whether the water remains in the vapour phase or reaches saturation; regardless, the calculated result does not match options a, b, or c.
3. The molar ionic conductivities of $NH_{4}^{+}$ and $OH^{-}$ at infinite dilution are 72 and 198 $ohm^{-1} cm^{2}$ respectively. The molar conductivity of a centinormal $NH_{4}OH$ solution at the same temperature is found to be 9 $ohm^{-1} cm^{2}$. The percentage dissociation of $NH_{4}OH$ at this concentration will be:
- (a) 3.33%
- (b) 7.14%
- (c) 12.5%
- (d) 4.54%
Correct Answer: (a)
Explanation: Limiting molar conductivity $\Lambda_{m}^{\circ}(NH_{4}OH) = 72 + 198 = 270$ $\Omega^{-1} cm^{2}$. The degree of dissociation $\alpha = \Lambda_{m}^{c} / \Lambda_{m}^{\circ} = 9 / 270 = 1/30$. In percentage, this is $(1/30) \times 100 = 3.33\%$.
4. The equation for Langmuir adsorption isotherm under high pressure is:
- (a) $x/m = a/b$
- (b) $x/m = aP$
- (c) $x/m = 1/aP$
- (d) $x/m = b/a$
Correct Answer: (a)
Explanation: In the Langmuir equation $x/m = ap / (1 + bp)$, at very high pressure, $bp \gg 1$. The equation thus reduces to the limiting form $x/m = a/b$, which is a constant.
5. Which of the following defect is also known as dislocation defect?
- (a) Frenkel defect
- (b) Schottky defect
- (c) Non-stoichiometric defect
- (d) Simple interstitial defect.
Correct Answer: (a)
Explanation: In a Frenkel defect, a smaller ion (usually the cation) is dislocated from its normal lattice site to an interstitial site.
6. Which of the following alkenes on ozonolysis gives a mixture of ketones only?
- (a) $CH_{3}-CH=CH-CH_{3}$
- (b) $CH_{3}-CH(CH_{3})-CH=CH_{2}$
- (c) 2,3-Dimethylbut-2-ene [Structure (c) in source]
- (d) $H_{2}C=CH_{2}$
Correct Answer: (c)
Explanation: Alkene (c) is tetrasubstituted. Ozonolysis of $(CH_{3})_{2}C=C(CH_{3})_{2}$ cleaves the double bond and adds oxygen to each side, yielding two molecules of acetone (a ketone).
7. For the reduction of FeO at the temperature near top of the furnace, which of the following statements is correct?
- (a) $\Delta G$ value for the overall reduction reaction with carbon monoxide is zero.
- (b) $\Delta G$ value for the overall reduction reaction with a mixture of 1 mole carbon and 1 mole oxygen is positive.
- (c) $\Delta G$ value for the overall reduction reaction with a mixture of 2 moles carbon and 1 mole oxygen will be positive.
- (d) $\Delta G$ value for the overall reduction reaction with carbon monoxide is negative.
Correct Answer: (a)
Explanation: At the specific temperature conditions near the top of the blast furnace, the free energy change ($\Delta G$) for the reduction of FeO by CO is effectively zero.
8. In $\alpha$-helix structure, polypeptide chains are folded in a:
- (a) right hand side
- (b) left hand side
- (c) both way
- (d) none of the above.
Correct Answer: (a)
Explanation: In the $\alpha$-helix secondary structure, the polypeptide chain is coiled into a right-handed screw (helix) stabilized by intramolecular hydrogen bonds.
9. Which of the following compounds does not show geometrical isomerism?
- (a) $C_{6}H_{5}N=NC_{6}H_{5}$
- (b) $C_{6}H_{5}CH=CHC_{6}H_{5}$
- (c) $C_{6}H_{5}-C(CH_{3})=NOH$
- (d) $(C_{6}H_{5})_{2}C=NCH_{3}$
Correct Answer: (d)
Explanation: Geometrical isomerism requires that the atoms or groups attached to the double-bonded atoms are different. In (d), there are two identical phenyl groups attached to the same carbon atom.
10. The water pollutants mainly responsible for eutrophication are:
- (a) Cd, Pb and Hg present in industrial waste
- (b) heavy metals present in mining waste
- (c) detergents and fertilizers containing phosphate anion
- (d) polychlorinated biphenyls.
Correct Answer: (c)
Explanation: Eutrophication is the nutrient enrichment of water bodies, primarily caused by phosphates from detergents and fertilizers, which leads to excessive algal growth.
11. Amongst the following, the most stable complex is:
- (a) $[Fe(H_{2}O)_{6}]^{3+}$
- (b) $[Fe(NH_{3})_{6}]^{3+}$
- (c) $[Fe(C_{2}O_{4})_{3}]^{3-}$
- (d) $[FeCl_{6}]^{3-}$
Correct Answer: (c)
Explanation: Stability is enhanced by the "chelate effect." The oxalate ion ($C_{2}O_{4}^{2-}$) is a bidentate chelating ligand, making its complex significantly more stable than those with unidentate ligands.
12. The polymer of natural rubber is:
- (a) all trans-isoprene
- (b) buna-N
- (c) all cis-isoprene
- (d) none of these.
Correct Answer: (c)
Explanation: Natural rubber is a linear polymer of isoprene (2-methyl-1,3-butadiene) where every double bond has the cis-configuration.
13. At some temperature and under a pressure of 4 atm, $PCl_{5}$ is 10% dissociated. Calculate the pressure at which $PCl_{5}$ will be 20% dissociated, temperature remaining same.
- (a) 0.96 atm
- (b) 1.02 atm
- (c) 2.34 atm
- (d) 0.86 atm
Correct Answer: (a)
Explanation: Using the formula $K_{p} = P\alpha^{2} / (1 - \alpha^{2})$, we first find $K_{p}$ at 10% dissociation ($\alpha=0.1$): $K_{p} = (4 \times 0.01) / (1 - 0.01) \approx 0.0404$ atm. Keeping $K_{p}$ constant for 20% dissociation ($\alpha=0.2$): $0.0404 = (P \times 0.04) / (1 - 0.04) \Rightarrow P \approx 0.96$ atm.
14. Which order of basicity is correct?
- (a) Aniline > m-toluidine > o-toluidine
- (b) Aniline > o-toluidine > m-toluidine
- (c) o-toluidine > aniline > m-toluidine
- (d) o-toluidine < aniline < m-toluidine.
Correct Answer: (d)
Explanation: Methyl groups are electron-donating (+I and hyperconjugation), increasing basicity. m-Toluidine is more basic than aniline. However, due to the "ortho effect" (steric and electronic factors), o-toluidine is less basic than aniline despite the methyl group.
15. In the compound, $CH_{2}=CH-CH_{2}-CH_{2}-C \equiv CH$, the $C_{2}-C_{3}$ bond is of the type:
- (a) $sp-sp^{2}$
- (b) $sp^{3}-sp^{3}$
- (c) $sp-sp^{3}$
- (d) $sp^{2}-sp^{3}$
Correct Answer: (d)
Explanation: C2 is part of a double bond and is $sp^{2}$ hybridised. C3 is an alkyl carbon (single bonds only) and is $sp^{3}$ hybridised.
16. At low pressure, the fraction of the surface covered follows:
- (a) zero order kinetics
- (b) first order kinetics
- (c) second order kinetics
- (d) fractional order kinetics.
Correct Answer: (b)
Explanation: According to the Freundlich and Langmuir isotherms, at low pressure, the amount of gas adsorbed is directly proportional to the pressure ($x/m \propto P^{1}$), indicating first-order kinetics.
17. The term 'anode mud' refers to:
- (a) impure metal acting as anode
- (b) anode coated with mud
- (c) insoluble matter collecting under the anode during electrolytic refining
- (d) calcium silicate.
Correct Answer: (c)
Explanation: During electrorefining, insoluble impurities from the impure anode fall to the bottom of the cell and are collected as anode mud (or sludge).
18. The aqueous solution of an unknown sodium salt gives: (i) white turbidity with dilute HCl, (ii) decolourises iodine solution, and (iii) white ppt. with $AgNO_{3}$ that becomes black on standing. The salt is:
- (a) sodium sulphite
- (b) sodium sulphide
- (c) sodium bisulphite
- (d) sodium thiosulphate.
Correct Answer: (d)
Explanation: Sodium thiosulphate ($Na_{2}S_{2}O_{3}$) reacts with HCl to precipitate sulphur (white turbidity). It reduces $I_{2}$ to $I^{-}$, and forms $Ag_{2}S_{2}O_{3}$ with silver nitrate, which hydrolyses to black $Ag_{2}S$.
19. When $NaBH_{4}$ is dissolved in water:
- (a) $Na^{+}$ and $BH_{4}^{-}$ ions are formed which are stable
- (b) it decomposes with the evolution of $H_{2}$
- (c) $BH_{4}^{-}$ is formed initially decomposed to give $OH^{-}$ ions
- (d) $NaH$ and $B_{2}H_{6}$ are produced.
Correct Answer: (b)
Explanation: Sodium borohydride is unstable in water and reacts to produce hydrogen gas and sodium borate.
20. Match the geometry: $IF_{7}$ (A), $ClF_{3}$ (B), $I_{3}^{-}$ (C), $BrF_{5}$ (D).
- (a) A-2, B-3, C-4, D-1
- (b) A-3, B-2, C-4, D-1
- (c) A-2, B-4, C-3, D-1
- (d) A-2, B-3, C-1, D-4
Correct Answer: (a)
Explanation: $IF_{7}$ is pentagonal bipyramidal (2); $ClF_{3}$ is T-shaped (3); $I_{3}^{-}$ is linear (4); $BrF_{5}$ is square pyramidal (1).
21. $CO + NaOH \xrightarrow{200^{\circ}C, 5-10 atm} A \xrightarrow{\Delta} B$. Compounds A and B are:
- (a) $NaHCO_{3}$, $Na_{2}CO_{3}$
- (b) $HCOONa$, $(COONa)_{2}$
- (c) $HCOONa$, $NaOH$
- (d) $NaHCO_{3}$, $NaOH$
Correct Answer: (b)
Explanation: Carbon monoxide reacts with sodium hydroxide under pressure to form sodium formate (A). Heating sodium formate causes it to decompose into sodium oxalate (B) and hydrogen.
22. Identify products in Lanthanoid (Ln) reaction sequence:
- (a) $A-LnO_{3}, B-LnS_{3}, C-LnC_{2}$
- (b) $A-LnO_{3}, B-LnX_{3}, C-LnC_{2}$
- (c) $A-H_{2}, B-LnX_{3}, C-LnC_{2}, D-LnN, E-Ln_{2}S_{3}, F-Ln_{2}O_{3}$
- (d) $A-H_{2}, B-LnX_{3}, C-Ln_{2}C_{3}$
Correct Answer: (c)
Explanation: Lanthanoids react with acids to liberate $H_{2}$ (A), with halogens to form $LnX_{3}$ (B), with carbon to form $LnC_{2}$ (C), and with other elements as described in (c).
23. $CCl_{4}$ is also used as a:
- (a) fertilizer
- (b) antiseptic
- (c) dry cleaning agent
- (d) paint remover.
Correct Answer: (c)
Explanation: Carbon tetrachloride is a common industrial solvent and has historically been used as a dry cleaning agent and fire extinguisher.
24. Which of the following pairs represents anomers?
- (a) [Open chain sugars]
- (b) [Open chain sugars]
- (c) [$\alpha$-D-glucose and $\beta$-D-glucose]
- (d) [Structure d]
Correct Answer: (c)
Explanation: Anomers are cyclic monosaccharides that differ in configuration specifically at the hemiacetal or hemiketal (anomeric) carbon.
25. The IUPAC name for $[Co(NH_{3})_{6}][Cr(CN)_{6}]$ is:
- (a) hexaamminecobalt(III) hexacyanochromium(VI)
- (b) hexaamminecobalt(III) hexacyanochromate(III)
- (c) hexacyanochromium(III) hexaamminecobalt(III)
- (d) hexacyanochromiumcobalthexaammine(VI)
Correct Answer: (b)
Explanation: Both metal centers are in the +3 oxidation state. The cationic complex comes first (hexaamminecobalt(III)), followed by the anionic complex ending in "-ate" (hexacyanochromate(III)).
26. Match: Vitamin $B_{12}$ (A), Codon (B), Proteins (C), Biotin (D).
- (a) A-P, B-Q, C-R, D-S
- (b) A-R, B-Q, C-S, D-P
- (c) A-Q, B-R, C-S, D-P
- (d) A-S, B-R, C-Q, D-P
Correct Answer: (b)
Explanation: Vitamin $B_{12}$ is Riboflavin (R); [Note: $B_{2}$ is Riboflavin, $B_{12}$ is cyanocobalamin, but the key matches $B_{12}$ to R in this context]. Codon is a three-base sequence (Q). Proteins involve pepsin/amino acids (S). Biotin is for fatty acid synthesis (P).
27. Which of the following processes are used for extraction of Ag from $Ag_{2}S$?
- (a) Cyanide process
- (b) Lead process
- (c) Mexican amalgamation process
- (d) All of these
Correct Answer: (d)
Explanation: All three methods—cyanidation (leaching with CN-), smelting with lead, and amalgamation—are established techniques for recovering silver from sulphide ores.
28. The slag obtained during the extraction of copper from copper pyrites is:
- (a) $Cu_{2}S$
- (b) $FeSiO_{3}$
- (c) $CuSiO_{3}$
- (d) $SiO_{2}$
Correct Answer: (b)
Explanation: During the smelting of copper pyrites, iron oxide (impurity) reacts with silica flux ($SiO_{2}$) to form a fusible slag of ferrous silicate ($FeSiO_{3}$).
29. Find the equilibrium constant for the oxidation of $NH_{3}$ to NO using given $K_{1}, K_{2}, K_{3}$:
- (a) $K_{1}K_{2} / K_{3}$
- (b) $K_{2}K_{3}^{3} / K_{1}$
- (c) $K_{2}K_{3}^{2} / K_{1}$
- (d) $K_{2}^{2}K_{3} / K_{1}$
Correct Answer: (b)
Explanation: The target reaction is the combination of the given steps. Multiplying step 3 by three and step 2 by one, then subtracting step 1, yields the relation $K = K_{2} \times K_{3}^{3} / K_{1}$.
30. $HC \equiv N + HCl \xrightarrow{AlCl_{3}} A \xrightarrow{C_{6}H_{6}} B \xrightarrow{H_{2}O} C$. A, B, and C are:
- (a) $CHCl=NH, C_{6}H_{5}Cl, C_{6}H_{5}OH$
- (b) $CHCl=NH, C_{6}H_{5}CH=NH, C_{6}H_{5}CHO$
- (c) $CHCl=NH, C_{6}H_{5}Cl, C_{6}H_{5}OH$
- (d) $CH_{2}O, C_{6}H_{5}CH_{2}Cl, C_{6}H_{5}CH_{2}OH$
Correct Answer: (b)
Explanation: Reaction of HCN and HCl forms an imido-chloride (A). This reacts with benzene (Gattermann aldehyde synthesis logic) to form an imine (B), which hydrolyses to benzaldehyde (C).
31. $HCHO + CH_{3}MgI \rightarrow A \xrightarrow{H_{2}O} B + Mg(OH)I$. A and B are:
- (a) $CH_{3}OMgI$ and $CH_{3}OH$
- (b) $CH_{3}CH_{2}OMgI$ and $C_{2}H_{5}OC_{2}H_{5}$
- (c) $CH_{3}CH_{2}OMgI$ and $CH_{3}CH_{2}OH$
- (d) $CH_{3}CH_{2}I$ and $CH_{3}CH_{2}OH$
Correct Answer: (c)
Explanation: Formaldehyde reacts with a Grignard reagent to form an addition adduct (A, ethylmagnesium iodide salt), which upon hydrolysis yields a primary alcohol (B, ethanol).
32. Energy of 2nd Bohr orbit = -328 kJ/mol. The energy of 4th orbit is:
- (a) 41 kJ/mol
- (b) -82 kJ/mol
- (c) -164 kJ/mol
- (d) -1312 kJ/mol
Correct Answer: (b)
Explanation: $E_{n} \propto 1/n^{2}$. $E_{4} / E_{2} = 2^{2} / 4^{2} = 4 / 16 = 1/4$. Thus, $E_{4} = -328 / 4 = -82$ kJ/mol.
33. Four quantum numbers for the valence electron in Copper (Cu) are:
- (a) n=4, l=0, m=0, s=+1/2
- (b) n=4, l=2, m=0, s=+1/2
- (c) n=4, l=-2, m=-2, s=+1/2
- (d) n=4, l=2, m=+2, s=-1/2
Correct Answer: (a)
Explanation: The valence electron for Cu is in the 4s orbital ($4s^{1}$). This corresponds to $n=4, l=0, m=0$, and spin $s=\pm 1/2$.
34. Sucrose on hydrolysis gives:
- (a) glucose and fructose
- (b) galactose and glucose
- (c) galactose and maltose
- (d) none of the above.
Correct Answer: (a)
Explanation: Sucrose is a disaccharide that breaks down into equal parts D-glucose and D-fructose.
35. Rock salt structure $A^{+}$ (radius x) and $B^{-}$ (radius y). Edge length is:
- (a) $(2x+y)$ pm
- (b) $(x+y)$ pm
- (c) $2(x+y)$ pm
- (d) $(x+2y)$ pm
Correct Answer: (c)
Explanation: In the rock salt (NaCl) structure, the edge length $a = 2 \times (r_{+} + r_{-}) = 2(x + y)$.
36. Halogen compound X on hydrolysis followed by acidification gives acetic acid. X is:
- (a) $ClCH_{2}CH_{2}Cl$
- (b) $CH_{3}CHCl_{2}$
- (c) $ClCH_{2}CHCl_{2}$
- (d) $CH_{3}CCl_{3}$
Correct Answer: (d)
Explanation: 1,1,1-Trichloroethane ($CH_{3}CCl_{3}$) hydrolyses to form $CH_{3}C(OH)_{3}$, which then loses water to become acetic acid.
37. Best method to prepare 3-methylbutan-2-ol from 3-methylbut-1-ene:
- (a) addition of water with dil. $H_{2}SO_{4}$
- (b) HBr/peroxide then hydrolysis
- (c) hydroboration-oxidation
- (d) oxymercuration-demercuration.
Correct Answer: (d)
Explanation: This specific alcohol is the Markovnikov addition product. Oxymercuration-demercuration ensures Markovnikov addition without the rearrangements that occur with acid catalysts.
38. Match: $V_{2}O_{5}$ (A), Ziegler-Natta (B), Peroxide (C), Fe (D).
- (a) A-iv, B-i, C-ii, D-iii
- (b) A-iv, B-iii, C-ii, D-i
- (c) A-iii, B-i, C-ii, D-iv
- (d) A-iv, B-ii, C-i, D-iii
Correct Answer: (a)
Explanation: $V_{2}O_{5}$ is for $H_{2}SO_{4}$ (iv); Ziegler-Natta is for HDPE (i); Peroxide is for polyacrylonitrile (ii); Fe is for $NH_{3}$ (iii).
39. Under Wolff-Kishner reduction, possible conversions: (1) benzophenone to diphenylmethane, (2) benzaldehyde to benzyl alcohol, (3) cyclohexanone to cyclohexane, (4) cyclohexanone to cyclohexanol.
- (a) 2, 3
- (b) 1, 2
- (c) 3, 4
- (d) 1, 3
Correct Answer: (d)
Explanation: Wolff-Kishner reduction specifically converts carbonyl groups ($>C=O$) into methylene groups ($>CH_{2}$), thus 1 and 3 are correct.
40. Dumas method: 0.35 g organic compound, 55 mL nitrogen at 300 K and 715 mm (Aq. tension 15 mm). Percentage of Nitrogen is:
- (a) 15.45%
- (b) 16.45%
- (c) 17.45%
- (d) 14.45%
Correct Answer: (b)
Explanation: First calculate the volume of $N_{2}$ at STP ($V_{2} = 46.1$ mL). %N = $V_{2} / (8 \times W) = 46.1 / (8 \times 0.35) = 16.46\%$.
41. Geometrical shapes of complexes of $Ni^{2+}$ with $Cl^{-}, CN^{-}, H_{2}O$:
- (a) octahedral, tetrahedral, square planar
- (b) tetrahedral, square planar, octahedral
- (c) square planar, tetrahedral, octahedral
- (d) octahedral, square planar, octahedral
Correct Answer: (b)
Explanation: $Cl^{-}$ (weak field) gives tetrahedral ($sp^{3}$); $CN^{-}$ (strong field) gives square planar ($dsp^{2}$); $H_{2}O$ (octahedral coord) gives octahedral ($sp^{3}d^{2}$).
42. Order of basicity of ethylamines and $NH_{3}$:
- (a) $NH_{3} > 1^\circ > 2^\circ > 3^\circ$
- (b) $3^\circ > 2^\circ > 1^\circ > NH_{3}$
- (c) $2^\circ > 1^\circ > 3^\circ > NH_{3}$
- (d) $2^\circ > 3^\circ > 1^\circ > NH_{3}$
Correct Answer: (d)
Explanation: In aqueous solution, the basicity order for ethylamines is $secondary > tertiary > primary > ammonia$.
43. Reagent used to distinguish acetophenone from benzophenone:
- (a) 2,4-DNP
- (b) Benedict solution
- (c) Tollen's reagent
- (d) $I_{2}$ and $Na_{2}CO_{3}$
Correct Answer: (d)
Explanation: Acetophenone is a methyl ketone and will give a positive iodoform test (yellow ppt. of $CHI_{3}$), whereas benzophenone will not.
44. $K_{sp}$ of $CaF_{2} = 3.2 \times 10^{-11}$. Solubility is:
- (a) $2.0 \times 10^{-4}$ M
- (b) $12.0 \times 10^{-3}$ M
- (c) $0.2 \times 10^{-4}$ M
- (d) $2.0 \times 10^{-3}$ M
Correct Answer: (a)
Explanation: $K_{sp} = 4s^{3}$. $3.2 \times 10^{-11} = 4s^{3} \Rightarrow s^{3} = 8 \times 10^{-12} \Rightarrow s = 2 \times 10^{-4}$ mol/L.
45. pH of mixture of 50 mL 0.01 N acetic acid and 2.5 mL 0.1 N NaOH:
- (a) 4.8
- (b) 3.4
- (c) 6.5
- (d) 7.6
Correct Answer: (a)
Explanation: This creates a buffer where $[salt] = [acid]$. Thus, $pH = pK_{a} = - \log(1.7 \times 10^{-5}) \approx 4.8$.