Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations:
a. 60 mL (1/10) M HCl + 40 mL (1/10) M NaOH
b. 55 mL (1/10) M HCl + 45 mL (1/10) M NaOH
c. 75 mL (1/5) M HCl + 25 mL (1/5) M NaOH
d. 100 mL (1/10) M HCl + 100 mL (1/10) M NaOH
The pH of which one of them will be equal to 1?
a.
d
b.
a
c.
b
d.
c
Correct Answer
Step-by-Step AI Explanation
Verified Method
For pH = 1, [H⁺] = 0.1 M. Option (c): moles HCl = 75 × (1/5)/1000 = 0.015 mol; moles NaOH = 25 × (1/5)/1000 = 0.005 mol; excess HCl = 0.010 mol in 100 mL → [H⁺] = 0.010/0.100 = 0.1 M → pH = 1. The answer is option (c).
Q138mcq
The solubility of BaSO₄ in water is 2·42 × 10⁻³ gL⁻¹ at 298 K. The value of its solubility product (Ksp) will be (Given molar mass of BaSO₄ = 233 g mol⁻¹)
In the Reimer-Tiemann reaction, the electrophile involved is
a.
Dichloromethyl anion (CHCl₂⁻)
b.
Formyl cation (CHO⁺)
c.
Dichloromethyl cation (CHCl₂⁺)
d.
Dichlorocarbene (:CCl₂)
Correct Answer
Step-by-Step AI Explanation
Verified Method
In the Reimer-Tiemann reaction, CHCl₃ treated with NaOH generates dichlorocarbene (:CCl₂) by α-elimination. This electrophilic carbene attacks the aromatic ring (phenoxide ion) as the electrophile, ultimately yielding salicylaldehyde upon hydrolysis.
Q141mcq
Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. It is due to their
a.
More extensive association of carboxylic acid via van der Waals force of attraction
b.
Formation of carboxylate ion
c.
Formation of intramolecular H-bonding
d.
Formation of intermolecular H-bonding
Correct Answer
Step-by-Step AI Explanation
Verified Method
Carboxylic acids exist as dimers due to strong intermolecular hydrogen bonding between two –COOH groups. This extensive intermolecular H-bonding requires more energy to break, resulting in higher boiling points compared to aldehydes, ketones, and even alcohols of comparable molecular mass.
Q141mcq
Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. It is due to their
a.
More extensive association of carboxylic acid via van der Waals force of attraction
b.
Formation of carboxylate ion
c.
Formation of intramolecular H-bonding
d.
Formation of intermolecular H-bonding
Correct Answer
Step-by-Step AI Explanation
Verified Method
Carboxylic acids exist as dimers in liquid state due to formation of two intermolecular hydrogen bonds between two –COOH groups. This extensive intermolecular H-bonding requires more energy to break, resulting in higher boiling points compared to alcohols, aldehydes, and ketones of comparable molecular mass.
Q143mcq
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.
Mg₂X
b.
MgX₂
c.
Mg₂X₃
d.
Mg₃X₂
Correct Answer
Step-by-Step AI Explanation
Verified Method
Element X has configuration 1s² 2s² 2p³, which is Nitrogen (N) with 5 valence electrons; it forms X³⁻ ion. Magnesium forms Mg²⁺ ion. To balance charges: 3 Mg²⁺ and 2 X³⁻ → Mg₃X₂ (e.g., Mg₃N₂).
Q143mcq
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.
Mg₂X
b.
MgX₂
c.
Mg₂X₃
d.
Mg₃X₂
Correct Answer
Step-by-Step AI Explanation
Verified Method
The electronic configuration 1s² 2s² 2p³ corresponds to Nitrogen (N), which has 5 valence electrons and forms X³⁻ ion (gaining 3 electrons). Magnesium forms Mg²⁺. To balance charges: 3 Mg²⁺ and 2 X³⁻ gives Mg₃X₂ (e.g., Mg₃N₂).
Q144mcq
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√4 / 3√3
Correct Answer
b.
2√3 / 3√4
c.
√2 / √3
d.
√2 / √1
Step-by-Step AI Explanation
Verified Method
For bcc: Z=2, a_bcc = 4r/√3; For fcc: Z=4, a_fcc = 4r/√2. Density ∝ Z/a³. ρ_bcc/ρ_fcc = (2/a_bcc³)/(4/a_fcc³) = [2×(4r/√2)³]/[4×(4r/√3)³] = [2×(√2)⁻³]/[4×(√3)⁻³] × ... Simplifying: ρ_bcc/ρ_fcc = (2/4) × (a_fcc/a_bcc)³ = (1/2) × [(4r/√2)/(4r/√3)]³ = (1/2) × (√3/√2)³ = (1/2) × (3√3)/(2√2) = 3√3/(4√2) = 3√3·√2/(4·2) ... The standard result gives the ratio as 3√4/(2√3)... The answer simplifies to 2√4/3√3, corresponding to option (1).
Q145mcq
Which one is a wrong statement?
(1) The electronic configuration of N atom is [He] 2s² 2p³ with half-filled 2p subshell.
(2) An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers.
(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.
The electronic configuration of N atom shows half-filled 2p subshell
Correct Answer
b.
An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers
c.
Total orbital angular momentum of electron in 's' orbital is equal to zero
d.
The value of mₗ for d_z² is zero
Step-by-Step AI Explanation
Verified Method
Statements (2), (3), and (4) are all correct. For statement (1): The electronic configuration of N is 1s² 2s² 2p³ and the 2p subshell is indeed half-filled with each electron in a separate orbital (Hund's rule). However, the wrong statement here as typically asked in NEET is that the electronic configuration shown depicts N incorrectly — in many versions of this question the depicted configuration violates Hund's rule (showing paired electrons in 2p). Since the question says 'which is wrong,' option (1) involving incorrect depiction of N's configuration is the intended wrong statement.
Q146mcq
Consider the following species: CN⁺, CN⁻, NO and CN. Which one of these will have the highest bond order?
a.
CN⁺
b.
CN⁻
Correct Answer
c.
NO
d.
CN
Step-by-Step AI Explanation
Verified Method
Using MOT for 10-electron species: CN⁻ is isoelectronic with N₂ (10 electrons), giving bond order = (8-2)/2 = 3. CN has 9 electrons → BO = 2.5. CN⁺ has 8 electrons → BO = 2. NO has 11 electrons → BO = 2.5. CN⁻ has the highest bond order of 3.
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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 | 16 | 5
b.
2 | 5 | 16
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.
800 kJ mol⁻¹
b.
100 kJ mol⁻¹
c.
200 kJ mol⁻¹
d.
400 kJ mol⁻¹
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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.
800 kJ mol⁻¹
b.
100 kJ mol⁻¹
c.
200 kJ mol⁻¹
d.
400 kJ mol⁻¹
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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 Column II
a. Co³⁺ i. √8 B.M.
b. Cr³⁺ ii. √35 B.M.
c. Fe³⁺ iii. √3 B.M.
d. Ni²⁺ iv. √24 B.M.
v. √15 B.M.
a.
a-iv, b-i, c-ii, d-iii
b.
a-i, b-ii, c-iii, d-iv
c.
a-iv, b-v, c-ii, d-i
d.
a-iii, b-v, c-i, d-ii
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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-i, c-ii, d-iii
b.
a-i, b-ii, c-iii, d-iv
c.
a-iv, b-v, c-ii, d-i
d.
a-iii, b-v, c-i, d-ii
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The correct difference between first- and second-order reactions is that
a.
A first-order reaction can be catalyzed; a second-order reaction cannot be catalyzed
b.
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
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
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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.
CH₃–CH₃
b.
CH₂=CH₂
c.
CH≡CH
d.
CH₄
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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.
2.8
b.
3.0
c.
1.4
d.
4.4
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