The correct difference between first- and second-order reactions is that
a.
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
Correct Answer
b.
a first-order reaction can be catalyzed; a second-order reaction cannot be catalyzed
c.
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
d.
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
Step-by-Step AI Explanation
Verified Method
For a first-order reaction, t₁/₂ = ln2/k, which is independent of initial concentration [A]₀. For a second-order reaction, t₁/₂ = 1/(k[A]₀), which depends on [A]₀. Both first- and second-order reactions can be catalyzed, and both have rates that depend on reactant concentrations.
Q137mcq
Among CaH₂, BeH₂, BaH₂, the order of ionic character is
a.
CaH₂ < BeH₂ < BaH₂
b.
BeH₂ < BaH₂ < CaH₂
c.
BeH₂ < CaH₂ < BaH₂
Correct Answer
d.
BaH₂ < BeH₂ < CaH₂
Step-by-Step AI Explanation
Verified Method
Ionic character of metal hydrides increases as the electropositive character of the metal increases. Down Group 2: Be < Ca < Ba in electropositive character (as atomic size increases, ionization energy decreases). Therefore ionic character order is BeH₂ < CaH₂ < BaH₂.
Q138mcq
In which case is the number of molecules of water maximum?
a.
0.18 g of water
b.
0.00224 L of water vapours at 1 atm and 273 K
c.
18 mL of water
Correct Answer
d.
10⁻³ mol of water
Step-by-Step AI Explanation
Verified Method
Option A: 0.18 g / 18 g/mol = 0.01 mol. Option B: 0.00224 L / 22.4 L/mol = 10⁻⁴ mol. Option C: 18 mL × 1 g/mL = 18 g / 18 g/mol = 1 mol. Option D: 10⁻³ mol. Comparing moles: C (1 mol) >> A (0.01 mol) >> D (10⁻³ mol) >> B (10⁻⁴ mol). Maximum molecules are in 18 mL of water.
Q140mcq
In the structure of ClF₃, the number of lone pairs of electrons on central atom 'Cl' is
a.
two
Correct Answer
b.
four
c.
one
d.
three
Step-by-Step AI Explanation
Verified Method
Cl has 7 valence electrons. In ClF₃, it forms 3 bonds with F atoms, using 3 electrons. Remaining 4 electrons form 2 lone pairs. Total electron pairs = 5 (3 bonding + 2 lone pairs), giving a T-shaped geometry (sp³d hybridization). Hence, the central Cl has 2 lone pairs.
Q140mcq
In the structure of ClF₃, the number of lone pairs of electrons on central atom 'Cl' is
a.
two
Correct Answer
b.
four
c.
one
d.
three
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 Cl atom.
Q141mcq
The correct order of N-compounds in its decreasing order of oxidation states is
a.
HNO₃, NO, NH₄Cl, N₂
b.
HNO₃, NH₄Cl, NO, N₂
c.
HNO₃, NO, N₂, NH₄Cl
Correct Answer
d.
NH₄Cl, N₂, NO, HNO₃
Step-by-Step AI Explanation
Verified Method
Oxidation states of N: HNO₃ = +5, NO = +2, N₂ = 0, NH₄Cl = −3. So decreasing order is HNO₃ > NO > N₂ > NH₄Cl, which is option (3).
Q141mcq
The correct order of N-compounds in its decreasing order of oxidation states is
a.
HNO₃, NO, NH₄Cl, N₂
b.
HNO₃, NH₄Cl, NO, N₂
c.
HNO₃, NO, N₂, NH₄Cl
Correct Answer
d.
NH₄Cl, N₂, NO, HNO₃
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.
Q142mcq
Which one of the following elements is unable to form MF₆³⁻ ion?
a.
Al
b.
B
Correct Answer
c.
Ga
d.
In
Step-by-Step AI Explanation
Verified Method
Boron (B) cannot form BF₆³⁻ because it lacks d-orbitals and cannot expand its octet beyond 4 bonds. Al, Ga, and In have available d-orbitals and can form MF₆³⁻ by expanding their coordination number to 6.
Q142mcq
Which one of the following elements is unable to form MF₆³⁻ ion?
a.
Al
b.
B
Correct Answer
c.
Ga
d.
In
Step-by-Step AI Explanation
Verified Method
To form MF₆³⁻, the central atom must be able to expand its octet (accommodate 6 F atoms). Boron (B) has no d-orbitals available (it is in period 2) and cannot expand its octet beyond 4. Al, Ga, and In are in periods 3, 4, and 5 respectively and have available d-orbitals for expansion. Hence B cannot form BF₆³⁻.
Q143mcq
The correct order of atomic radii in group 13 elements is
a.
B < Al < Ga < In < Tl
b.
B < Ga < Al < Tl < In
c.
B < Al < In < Ga < Tl
d.
B < Ga < Al < In < Tl
Correct Answer
Step-by-Step AI Explanation
Verified Method
Due to the poor shielding by d-electrons, Ga has a smaller atomic radius than Al (d-block contraction). So the correct order is B < Ga < Al < In < Tl.
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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.
3.0
b.
2.8
c.
1.4
d.
4.4
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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.
CH₂=CH₂
b.
CH₃–CH₃
c.
CH≡CH
d.
CH₄
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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.
2 : 5 : 16
b.
2 : 16 : 5
c.
16 : 5 : 2
d.
5 : 16 : 2
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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.
100 kJ mol⁻¹
b.
800 kJ mol⁻¹
c.
200 kJ mol⁻¹
d.
400 kJ mol⁻¹
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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 / 3√4
b.
2√4 / 3√3
c.
√2 / √3
d.
2 / 1
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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.
MgX₂
b.
Mg₂X
c.
Mg₂X₃
d.
Mg₃X₂
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Which one is a wrong statement?
(1) An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers.
(2) The electronic configuration of N atom is [He] 2s² 2p¹ 2p¹ 2p¹ (with all three 2p electrons in separate orbitals with same spin).
(3) Total orbital angular momentum of electron in 's' orbital is equal to zero.
(4) The value of mₗ for d_{z²} is zero.
a.
An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers.
b.
The electronic configuration of N atom is [He] 2s² 2p¹↑ 2p¹↑ 2p¹↑ (half-filled, all parallel spins — this is correct per Hund's rule).
c.
Total orbital angular momentum of electron in 's' orbital is equal to zero.
d.
The value of mₗ for d_{z²} is zero.
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Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations:
a. 60 mL (M/10) HCl + 40 mL (M/10) NaOH
b. 55 mL (M/10) HCl + 45 mL (M/10) NaOH
c. 75 mL (M/5) HCl + 25 mL (M/5) NaOH
d. 100 mL (M/10) HCl + 100 mL (M/10) NaOH
The pH of which one of them will be equal to 1?
a.
a
b.
d
c.
b
d.
c
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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 × 10⁻¹² mol² L⁻²
b.
1.08 × 10⁻¹⁴ mol² L⁻²
c.
1.08 × 10⁻¹⁰ mol² L⁻²
d.
1.08 × 10⁻⁸ mol² L⁻²
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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 × 10⁻¹² mol² L⁻²
b.
1.08 × 10⁻¹⁴ mol² L⁻²
c.
1.08 × 10⁻¹⁰ mol² L⁻²
d.
1.08 × 10⁻⁸ 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-i, b-ii, c-iii, d-iv
b.
a-iv, b-i, c-ii, d-iii
c.
a-iv, b-v, c-ii, d-i
d.
a-iii, b-v, c-i, d-ii
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