The correct order of N-compounds in decreasing order of oxidation states is
a.
HNO3, NH4Cl, NO, N2
b.
HNO3, NO, NH4Cl, N2
c.
HNO3, NO, N2, NH4Cl
Correct Answer
d.
NH4Cl, N2, NO, HNO3
Step-by-Step AI Explanation
Verified Method
Oxidation states of nitrogen are: HNO3 = +5, NO = +2, N2 = 0, NH4Cl = -3. Hence decreasing order is HNO3 > NO > N2 > NH4Cl.
Q46mcq
The correct order of N-compounds in decreasing order of oxidation states is
a.
HNO3, NH4Cl, NO, N2
b.
HNO3, NO, NH4Cl, N2
c.
HNO3, NO, N2, NH4Cl
Correct Answer
d.
NH4Cl, N2, NO, HNO3
Step-by-Step AI Explanation
Verified Method
Oxidation states of N are: HNO3 = +5, NO = +2, N2 = 0, NH4Cl = -3. Hence decreasing order is HNO3 > NO > N2 > NH4Cl.
Q47mcq
Which one of the following elements is unable to form MF6^3− ion?
a.
B
Correct Answer
b.
Al
c.
Ga
d.
In
Step-by-Step AI Explanation
Verified Method
Boron has no vacant d-orbitals in its valence shell and cannot expand its coordination number beyond 4, so it cannot form BF6^3−.
Q47mcq
Which one of the following elements is unable to form MF6^3- ion?
a.
B
Correct Answer
b.
Al
c.
Ga
d.
In
Step-by-Step AI Explanation
Verified Method
Boron has no vacant d-orbitals and cannot expand its covalency to form hexafluoro complex BF6^3-. Hence B cannot form MF6^3-.
Q48mcq
Considering Ellingham diagram, which of the following metals can be used to reduce alumina?
a.
Mg
Correct Answer
b.
Zn
c.
Fe
d.
Cu
Step-by-Step AI Explanation
Verified Method
A metal more reactive than aluminium can reduce alumina. Among the given options, Mg is the strongest reducing metal.
Q49mcq
The correct order of atomic radii in group 13 elements is
a.
B < Ga < Al < Tl < In
b.
B < Al < Ga < In < Tl
c.
B < Al < In < Ga < Tl
d.
B < Ga < Al < In < Tl
Correct Answer
Step-by-Step AI Explanation
Verified Method
Due to poor shielding by d- and f-electrons, the size trend is irregular, but the accepted order is B < Ga < Al < In < Tl.
Q50mcq
Which of the following statements is not true for halogens?
a.
All but fluorine show positive oxidation states
Correct Answer
b.
All are oxidizing agents
c.
All form monobasic oxyacids
d.
Chlorine has the highest electron-gain enthalpy
Step-by-Step AI Explanation
Verified Method
Fluorine does not show positive oxidation states; in HOF it has oxidation state +1 for oxygen’s sake? Actually the standard NCERT point is that fluorine shows only -1 oxidation state in its compounds and no positive oxidation state. Hence statement (a) is not true as written.
Q51mcq
In the structure of ClF3, the number of lone pairs of electrons on the central atom Cl is
a.
Four
b.
Two
Correct Answer
c.
One
d.
Three
Step-by-Step AI Explanation
Verified Method
ClF3 has 5 electron pairs around chlorine: 3 bond pairs and 2 lone pairs, giving a T-shaped molecule.
Q52mcq
Identify the major products P, Q and R in the following sequence of reactions: benzene reacts with propyl chloride in presence of anhydrous AlCl3, then (i) O2, (ii) H3O+/Δ.
a.
CH3CH(OH)CH3, ...
b.
CH3CH2CHO, CH3COOH, ...
c.
CH3CH2CHO, CH3CH2OH, ...
d.
(CH3)2CHOH, CH3COCH3, ...
Correct Answer
Step-by-Step AI Explanation
Verified Method
Propyl chloride with AlCl3 undergoes rearrangement to give isopropylbenzene type intermediate; oxidation followed by acidic hydrolysis of the hydroperoxide gives phenol/acetone-type products as indicated. The correct option given is (4), matching the rearrangement products shown in the source.
Q53mcq
Which of the following compounds can form a zwitterion?
a.
Benzoic acid
b.
Acetanilide
c.
Aniline
d.
Glycine
Correct Answer
Step-by-Step AI Explanation
Verified Method
Glycine contains both acidic -COOH and basic -NH2 groups and exists as a zwitterion in solution.
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A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with concentrated 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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Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to a 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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Compound A, C8H10O, is found to react with NaOI (produced by reacting Y with NaOH) and yields a yellow precipitate with characteristic smell. A and Y are respectively
a.
C6H5CH(OH)CH3 and I2
b.
CH3CH2CH(OH)CH3 and I2
c.
C2H5OH and I2
d.
C6H5OH and I2
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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 BM
ii. 35 BM
iii. 3 BM
iv. 24 BM
v. 15 BM
a.
iv i ii iii
b.
i ii iii iv
c.
iv v ii i
d.
iii v i ii
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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
pH of which one of them will be equal to 1?
a.
d
b.
a
c.
b
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 M HCl + 40 mL 10 M NaOH
b. 55 mL 10 M HCl + 45 mL 10 M NaOH
c. 75 mL 5 M HCl + 25 mL 5 M NaOH
d. 100 mL 10 M HCl + 100 mL 10 M NaOH
pH of which one of them will be equal to 1?
a.
d
b.
a
c.
b
d.
c
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Given van der Waals constant for NH3, H2, O2 and CO2 are respectively 4.17, 0.244, 1.36 and 3.59, which one of the following gases is most easily liquefied?
a.
O2
b.
H2
c.
NH3
d.
CO2
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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^-14 mol^2 L^-2
b.
1.08 × 10^-12 mol^2 L^-2
c.
1.08 × 10^-10 mol^2 L^-2
d.
1.08 × 10^-8 mol^2 L^-2
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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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Consider the change in oxidation state of bromine corresponding to different emf values as shown below: BrO4− / BrO3− / HBrO / Br2 / Br−. The species undergoing disproportionation is
a.
Br2
b.
BrO4−
c.
BrO3−
d.
HBrO
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Consider the change in oxidation state of bromine corresponding to different emf values as shown below:
BrO4−/BrO3− : 1.82 V
BrO3−/HBrO : 1.5 V
HBrO/Br2 : 1.595 V
Br2/Br− : 1.0652 V
Then the species undergoing disproportionation is
a.
Br2
b.
BrO4−
c.
BrO3−
d.
HBrO
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The bond dissociation energies of X2, Y2 and XY are in the ratio 1 : 0.5 : 1. If 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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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
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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.
Mg2X
b.
MgX2
c.
Mg2X3
d.
Mg3X2
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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 remain constant with temperature, is
a.
3/4
b.
4/3
c.
3/2
d.
1/2
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