The two functional groups characteristic of sugars are
a.
carbonyl and phosphate
b.
hydroxyl and methyl
c.
carbonyl and hydroxyl
Correct Answer
d.
carbonyl and methyl
Step-by-Step AI Explanation
Verified Method
Carbohydrates (sugars) are polyhydroxy aldehydes or polyhydroxy ketones, meaning they contain multiple hydroxyl groups and a carbonyl group.
Q125mcq
The two functional groups characteristic of sugars are
a.
carbonyl and phosphate
b.
hydroxyl and methyl
c.
carbonyl and hydroxyl
Correct Answer
d.
carbonyl and methyl
Step-by-Step AI Explanation
Verified Method
Sugars (carbohydrates) are polyhydroxy aldehydes or ketones, meaning they contain hydroxyl (-OH) groups and a carbonyl (C=O) group.
Q136mcq
In which case is the number of molecules of water maximum?
a.
0.00224 L of water vapours at 1 atm and 273 K
b.
18 mL of water
Correct Answer
c.
10^-3 mol of water
d.
0.18 g of water
Step-by-Step AI Explanation
Verified Method
(a) 0.00224 L / 22.4 L/mol = 0.0001 mol. (b) 18 g / 18 g/mol = 1 mol. (c) 0.001 mol. (d) 0.18 g / 18 g/mol = 0.01 mol. Thus, (b) has the maximum molecules.
Q137mcq
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 rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations
c.
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
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
Correct Answer
Step-by-Step AI Explanation
Verified Method
For first-order reactions, t1/2 = 0.693/k (independent of concentration). For second-order reactions, t1/2 = 1/(k[A]0), which depends on initial concentration.
Q138mcq
Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below: Then the species undergoing disproportionation is
a.
Br2
b.
Br-3O
c.
HBrO
Correct Answer
d.
Br-4O
Step-by-Step AI Explanation
Verified Method
In Latimer diagrams for bromine, HBrO (hypobromous acid) has a positive potential to both disproportionate to Br2 and to BrO3- in acidic medium.
Q139mcq
Among CaH2, BeH2, BaH2, the order of ionic character is
a.
BeH2 < BaH2 < CaH2
b.
BeH2 < CaH2 < BaH2
Correct Answer
c.
BaH2 < BeH2 < CaH2
d.
CaH2 < BeH2 < BaH2
Step-by-Step AI Explanation
Verified Method
Ionic character increases as the electropositivity of the metal increases. Going down group 2 (Be to Ba), atomic size increases and ionization energy decreases, so ionic character increases: BeH2 < CaH2 < BaH2.
Q140mcq
Which one of the following ions exhibits d-d transition and paramagnetism as well?
a.
MnO4-
b.
CrO4-2
c.
MnO4-2
Correct Answer
d.
Cr2O7-2
Step-by-Step AI Explanation
Verified Method
MnO4-2 (manganate ion) has Mn in +6 state with d1 configuration, allowing d-d transition and resulting in paramagnetism.
Q140mcq
Which one of the following ions exhibits d-d transition and paramagnetism as well?
a.
MnO4-
b.
CrO4-2
c.
MnO4-2
Correct Answer
d.
Cr2O7-2
Step-by-Step AI Explanation
Verified Method
MnO4-2 (manganate) has a d1 configuration (Mn is +6). It has one unpaired electron (paramagnetic) and can exhibit d-d transitions.
Q141mcq
Iron carbonyl, Fe(CO)5 is
a.
trinuclear
b.
tetranuclear
c.
dinuclear
d.
mononuclear
Correct Answer
Step-by-Step AI Explanation
Verified Method
Fe(CO)5 contains only one iron atom, thus it is mononuclear.
Q141mcq
Iron carbonyl, Fe(CO)5 is
a.
trinuclear
b.
tetranuclear
c.
dinuclear
d.
mononuclear
Correct Answer
Step-by-Step AI Explanation
Verified Method
Fe(CO)5 contains only one iron atom, making it mononuclear.
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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 i ii iii
b.
(2) iv v ii i
c.
(3) iii v i ii
d.
(4) i ii iii iv
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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.
1.4
c.
4.4
d.
3.0
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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.
CH ≡ CH
c.
CH4
d.
CH2 = CH2
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The solubility of BaSO4 in water is 2.42 x 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 x 10^-14 mol^2 L^-2
b.
1.08 x 10^-10 mol^2 L^-2
c.
1.08 x 10^-8 mol^2 L^-2
d.
1.08 x 10^-12 mol^2 L^-2
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Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is 1s^2 2s^2 2p^3, the simplest formula for this compound is
a.
Mg2X
b.
Mg2X3
c.
Mg3X2
d.
MgX2
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