Nitration of aniline in strong acidic medium also gives m-nitroaniline because
a.
In spite of substituents nitro group always goes to only m-position.
b.
In electrophilic substitution reactions amino group is meta directive.
c.
In absence of substituents nitro group always goes to m-position.
d.
In acidic (strong) medium aniline is present as anilinium ion.
Correct Answer
Step-by-Step AI Explanation
Verified Method
In strong acidic medium, the amino group (-NH2) is protonated to form the anilinium ion (-NH3+), which is meta-directing due to its electron-withdrawing nature.
Q48mcq
Which of the following oxides is most acidic in nature?
a.
MgO
b.
BeO
Correct Answer
c.
BaO
d.
CaO
Step-by-Step AI Explanation
Verified Method
Among Group 2 alkaline earth metal oxides, basicity increases down the group. BeO is amphoteric, while MgO, CaO, and BaO are basic, making BeO relatively the most acidic.
Q49mcq
The difference between amylose and amylopectin is
a.
Amylopectin have 1 → 4 α-linkage and 1 → 6 α-linkage
Correct Answer
b.
Amylose have 1 → 4 α-linkage and 1 → 6 β-linkage
c.
Amylopectin have 1 → 4 α-linkage and 1 → 6 β-linkage
d.
Amylose is made up of glucose and galactose
Step-by-Step AI Explanation
Verified Method
Amylose is a linear polymer with α-1,4-glycosidic linkages, while amylopectin is a branched polymer with α-1,4-linkages and α-1,6-glycosidic linkages at branch points.
Q50mcq
Regarding cross-linked or network polymers, which of the following statements is incorrect?
a.
They contain covalent bonds between various linear polymer chains.
b.
They are formed from bi- and tri-functional monomers.
c.
Examples are bakelite and melamine.
d.
They contain strong covalent bonds in their polymer chains.
Correct Answer
Step-by-Step AI Explanation
Verified Method
Statements (a), (b), and (c) are standard characteristics of cross-linked polymers. Statement (d) is vague/incorrect because cross-linked polymers contain cross-links between chains, not just strong bonds within chains compared to others.
Q52mcq
Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. It is due to their
a.
formation of intramolecular H-bonding
b.
formation of carboxylate ion
c.
more extensive association of carboxylic acid via van der Waals force of attraction
d.
formation of intermolecular H-bonding
Correct Answer
Step-by-Step AI Explanation
Verified Method
Carboxylic acids form stronger and more extensive intermolecular hydrogen bonding (forming dimers) compared to alcohols and other carbonyl compounds, leading to significantly higher boiling points.
Q54mcq
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 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
c.
a first-order reaction can be catalyzed; a second-order reaction cannot be catalyzed
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, t(1/2) = 0.693/k (independent of initial concentration). For a second-order reaction, t(1/2) = 1/(k[A]0), which depends on initial concentration.
Q54mcq
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 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
c.
a first-order reaction can be catalyzed; a second-order reaction cannot be catalyzed
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 = 0.693/k (independent of initial concentration). For a second-order reaction, t1/2 = 1/(k[A]0) (dependent on initial concentration).
Q55mcq
Among CaH2, BeH2, BaH2, the order of ionic character is
a.
BeH2 < CaH2 < BaH2
Correct Answer
b.
CaH2 < BeH2 < BaH2
c.
BeH2 < BaH2 < CaH2
d.
BaH2 < BeH2 < CaH2
Step-by-Step AI Explanation
Verified Method
Ionic character increases as electronegativity of the metal decreases (or electropositivity increases) down the group. Be is the least electropositive, Ba is the most.
Q55mcq
Among CaH2, BeH2, BaH2, the order of ionic character is
a.
BeH2 < CaH2 < BaH2
Correct Answer
b.
CaH2 < BeH2 < BaH2
c.
BeH2 < BaH2 < CaH2
d.
BaH2 < BeH2 < CaH2
Step-by-Step AI Explanation
Verified Method
Ionic character increases as the electropositivity of the metal increases. Since electropositivity increases down group 2 (Be < Mg < Ca < Sr < Ba), the ionic character follows BeH2 < CaH2 < BaH2.
Q57mcq
In which case is the number of molecules of water maximum?
a.
18 mL of water
Correct Answer
b.
0.18 g of water
c.
0.00224 L of water vapours at 1 atm and 273 K
d.
10^-3 mol of water
Step-by-Step AI Explanation
Verified Method
18 mL water = 18 g = 1 mol = 6.022e23 molecules (Maximum). 0.18 g = 0.01 mol. 0.00224 L = 0.0001 mol. 10^-3 mol = 0.001 mol.
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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.
MgX2
c.
Mg2X
d.
Mg3X2
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Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations: a. 60 mL 10^-3 M HCl + 40 mL 10^-3 M NaOH, b. 55 mL 10^-3 M HCl + 45 mL 10^-3 M NaOH, c. 75 mL 5*10^-3 M HCl + 25 mL 5*10^-3 M NaOH, d. 100 mL 10^-3 M HCl + 100 mL 10^-3 M NaOH. pH of which one of them will be equal to 1?
a.
b
b.
a
c.
d
d.
c
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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 HCl 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.
a
c.
d
d.
c
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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)
a.
1.08 * 10^-10 mol^2 L^-2
b.
1.08 * 10^-12 mol^2 L^-2
c.
1.08 * 10^-14 mol^2 L^-2
d.
1.08 * 10^-8 mol^2 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^-10 mol2 L^-2
b.
1.08 * 10^-12 mol2 L^-2
c.
1.08 * 10^-14 mol2 L^-2
d.
1.08 * 10^-8 mol2 L^-2
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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.
CH ≡ CH
b.
CH2 = CH2
c.
CH3 – CH3
d.
CH4
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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.
100 kJ mol^-1
c.
800 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: 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.
(1) iv v ii i
b.
(2) i ii iii iv
c.
(3) iv i ii iii
d.
(4) iii v i ii
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