Which of the following statements is not true for halogens?
a.
All form monobasic oxyacids.
Correct Answer
b.
Chlorine has the highest electron-gain enthalpy.
c.
All but fluorine show positive oxidation states.
d.
All are oxidizing agents.
Step-by-Step AI Explanation
Verified Method
Not all halogens form monobasic oxyacids. For example, chlorine forms HClO, HClO₂, HClO₃, HClO₄ — all monobasic, but fluorine does not form oxyacids (except HOF). However, the key issue is that iodine forms polybasic oxyacids (e.g., H₅IO₆ is polybasic), so the statement 'All form monobasic oxyacids' is not true. Fluorine has higher electron affinity per se, but chlorine has the highest electron-gain enthalpy due to smaller atomic size of F causing more electron-electron repulsion — statement (2) is correct. Statement (3) is correct. Statement (4) is correct.
Q2mcq
The correct order of atomic radii in group 13 elements is
a.
B < Al < In < Ga < Tl
b.
B < Ga < Al < In < Tl
Correct Answer
c.
B < Ga < Al < Tl < In
d.
B < Al < Ga < In < Tl
Step-by-Step AI Explanation
Verified Method
In Group 13, the atomic radius generally increases down the group, but Ga has a smaller atomic radius than Al due to the poor shielding effect of the 3d electrons (d-block contraction). Thus the order is B < Ga < Al < In < Tl.
Q3mcq
In the structure of ClF₃, the number of lone pairs of electrons on the central atom 'Cl' is
a.
one
b.
three
c.
four
d.
two
Correct Answer
Step-by-Step AI Explanation
Verified Method
Cl in ClF₃ has 7 valence electrons. It forms 3 bonds with F, using 3 electrons, leaving 4 electrons = 2 lone pairs on Cl. The geometry is T-shaped with 2 lone pairs on the central atom.
Q4mcq
The correct order of N-compounds in its decreasing order of oxidation states is
a.
HNO₃, NO, N₂, NH₄Cl
Correct Answer
b.
NH₄Cl, N₂, NO, HNO₃
c.
HNO₃, NH₄Cl, NO, N₂
d.
HNO₃, NO, NH₄Cl, N₂
Step-by-Step AI Explanation
Verified Method
Oxidation states of N: HNO₃ = +5, NO = +2, N₂ = 0, NH₄Cl = −3. Decreasing order: +5 > +2 > 0 > −3, i.e., HNO₃ > NO > N₂ > NH₄Cl.
Q5mcq
Which one of the following elements is unable to form MF₆³⁻ ion?
a.
Ga
b.
In
c.
B
Correct Answer
d.
Al
Step-by-Step AI Explanation
Verified Method
Boron (B) cannot form MF₆³⁻ because it lacks d-orbitals in its valence shell (it is in the second period) and cannot expand its octet to accommodate 6 F atoms. Al, Ga, and In have available d-orbitals and can form hexafluoro complexes.
Q6mcq
Considering the Ellingham diagram, which of the following metals can be used to reduce alumina?
a.
Fe
b.
Cu
c.
Mg
Correct Answer
d.
Zn
Step-by-Step AI Explanation
Verified Method
In the Ellingham diagram, a metal whose oxide line lies below that of Al₂O₃ can reduce alumina. Mg forms MgO with a more negative ΔG° of formation than Al₂O₃ at relevant temperatures, so Mg can reduce Al₂O₃. Fe, Cu, and Zn have oxide lines above that of Al₂O₃ and cannot reduce it.
Q7mcq
The compound A on treatment with Na gives B, and with PCl₅ gives C. B and C react together to give diethyl ether. A, B and C are in the order
a.
C₂H₅OH, C₂H₆, C₂H₅Cl
b.
C₂H₅OH, C₂H₅ONa, C₂H₅Cl
Correct Answer
c.
C₂H₅Cl, C₂H₆, C₂H₅OH
d.
C₂H₅OH, C₂H₅Cl, C₂H₅ONa
Step-by-Step AI Explanation
Verified Method
A = C₂H₅OH (ethanol). Ethanol + Na → C₂H₅ONa (sodium ethoxide) = B. Ethanol + PCl₅ → C₂H₅Cl (chloroethane) = C. Then C₂H₅ONa + C₂H₅Cl → C₂H₅OC₂H₅ (diethyl ether) by Williamson's synthesis.
Q8mcq
Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is
a.
CH≡CH
b.
CH₄
Correct Answer
c.
CH₃–CH₃
d.
CH₂=CH₂
Step-by-Step AI Explanation
Verified Method
CH₄ reacts with Br₂ by free-radical substitution to give CH₃Br. Wurtz reaction: 2CH₃Br + 2Na → C₂H₆ + 2NaBr. C₂H₆ is a gaseous hydrocarbon with less than 4 carbon atoms. CH₃–CH₃ would give CH₃CH₂Br, and Wurtz reaction gives C₄H₁₀ (butane), which is also gaseous but has more than 2 carbons; however C₄H₁₀ is still less than 4? No — butane has 4 carbon atoms, which is not less than 4. So CH₄ is the correct answer giving ethane (2 carbons < 4).
Q9mcq
The compound C₇H₈ undergoes the following reactions. The product 'C' is
a.
m-bromotoluene
b.
p-bromotoluene
c.
3-bromo-2,4,6-trichlorotoluene
Correct Answer
d.
o-bromotoluene
Step-by-Step AI Explanation
Verified Method
C₇H₈ is toluene. When toluene is first treated with Cl₂/AlCl₃ (EAS), chlorination occurs at ortho and para positions. With excess Cl₂, all three activated positions (2, 4, 6) get chlorinated to give 2,4,6-trichlorotoluene. Subsequent bromination then occurs at position 3 (only remaining activated ring position), giving 3-bromo-2,4,6-trichlorotoluene as product C.
Q10mcq
Which oxide of nitrogen is not a common pollutant introduced into the atmosphere both due to natural and human activity?
a.
N₂O₅
Correct Answer
b.
NO
c.
N₂O
d.
NO₂
Step-by-Step AI Explanation
Verified Method
N₂O₅ (dinitrogen pentoxide) is not a common atmospheric pollutant from both natural and human sources. NO is produced by lightning (natural) and combustion engines (human). N₂O is produced by denitrifying bacteria (natural) and fertilizers (human). NO₂ is produced by lightning and industrial/vehicular emissions. N₂O₅ is mainly formed as a secondary atmospheric species and is not directly emitted as a primary pollutant from both natural and human activities.
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Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations:
a. 60 mL 10⁻¹ M HCl + 40 mL 10⁻¹ M NaOH
b. 55 mL 10⁻¹ M HCl + 45 mL 10⁻¹ M NaOH
c. 75 mL 5×10⁻² M HCl + 25 mL 5×10⁻² M NaOH
d. 100 mL 10⁻¹ M HCl + 100 mL 10⁻¹ M NaOH
pH of which one of them will be equal to 1?
a.
b
b.
c
c.
d
d.
a
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Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations:
a. 60 mL 10⁻¹ M HCl + 40 mL 10⁻¹ M NaOH
b. 55 mL 10⁻¹ M HCl + 45 mL 10⁻¹ M NaOH
c. 75 mL 5×10⁻² M HCl + 25 mL 5×10⁻² M NaOH
d. 100 mL 10⁻¹ M HCl + 100 mL 10⁻¹ M NaOH
pH of which one of them will be equal to 1?
a.
b
b.
c
c.
d
d.
a
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The solubility of BaSO₄ in water is 2.42 × 10⁻³ g L⁻¹ at 298 K. The value of its solubility product (Ksp) will be (Given molar mass of BaSO₄ = 233 g mol⁻¹)
a.
1.08 × 10⁻¹⁰ mol² L⁻²
b.
1.08 × 10⁻⁸ mol² L⁻²
c.
1.08 × 10⁻¹⁴ mol² L⁻²
d.
1.08 × 10⁻¹² mol² L⁻²
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Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code:
Column I: a. Co³⁺, b. Cr³⁺, c. Fe³⁺, d. Ni²⁺
Column II: i. √8 B.M., ii. √35 B.M., iii. √3 B.M., iv. √24 B.M., v. √15 B.M.
a.
a-iv, b-v, c-ii, d-i
b.
a-iii, b-v, c-i, d-ii
c.
a-iv, b-i, c-ii, d-iii
d.
a-i, b-ii, c-iii, d-iv
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Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is 1s² 2s² 2p³, the simplest formula for this compound is
a.
Mg₂X₃
b.
Mg₃X₂
c.
Mg₂X
d.
MgX₂
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Iron exhibits bcc structure at room temperature. Above 900°C, it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900°C (assuming molar mass and atomic radii of iron remains constant with temperature) is
a.
√2/√3
b.
√2/1
c.
2√4/3√3
d.
2√3/3√4
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Which one is a wrong statement?
(1) Total orbital angular momentum of electron in 's' orbital is equal to zero.
(2) The value of m for d(z²) is zero.
(3) The electronic configuration of N atom is [He] 2s² 2p¹ₓ 2p¹ᵧ 2p¹_z (half-filled 2p subshell with parallel spins).
(4) An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers.
a.
Total orbital angular momentum of electron in 's' orbital is equal to zero.
b.
The value of m for d(z²) is zero.
c.
The electronic configuration of N atom is [He] 2s² 2p¹ₓ 2p¹ᵧ 2p¹_z.
d.
An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers.
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Which one is a wrong statement?
(1) Total orbital angular momentum of electron in 's' orbital is equal to zero.
(2) The value of m for d_z² is zero.
(3) The electronic configuration of N atom is [shown with all three 2p orbitals each having one electron with same spin, per Hund's rule].
(4) An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers.
a.
Total orbital angular momentum of electron in 's' orbital is equal to zero.
b.
The value of m for d_{z²} is zero.
c.
The electronic configuration of N atom violates Hund's rule.
d.
An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers.
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The correct difference between first- and second-order reactions is that
a.
the rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations
b.
the rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations
c.
a first-order reaction can be catalyzed; a second-order reaction cannot be catalyzed
d.
the half-life of a first-order reaction does not depend on [A]₀; the half-life of a second-order reaction does depend on [A]₀
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A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H₂SO₄. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be
a.
1.4
b.
4.4
c.
2.8
d.
3.0
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A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H₂SO₄. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be
a.
1.4
b.
4.4
c.
2.8
d.
3.0
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For the redox reaction MnO₄⁻ + C₂O₄²⁻ + H⁺ → Mn²⁺ + CO₂ + H₂O, the correct coefficients of the reactants for the balanced equation are:
(MnO₄⁻ : C₂O₄²⁻ : H⁺)
a.
16 : 5 : 2
b.
5 : 16 : 2
c.
2 : 16 : 5
d.
2 : 5 : 16
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For the redox reaction MnO₄⁻ + C₂O₄²⁻ + H⁺ → Mn²⁺ + CO₂ + H₂O, the correct coefficients of the reactants for the balanced equation are:
MnO₄⁻ C₂O₄²⁻ H⁺
a.
16 5 2
b.
5 16 2
c.
2 16 5
d.
2 5 16
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The bond dissociation energies of X₂, Y₂ and XY are in the ratio of 1 : 0.5 : 1. ΔH for the formation of XY is −200 kJ mol⁻¹. The bond dissociation energy of X₂ will be
a.
200 kJ mol⁻¹
b.
400 kJ mol⁻¹
c.
800 kJ mol⁻¹
d.
100 kJ mol⁻¹
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