A carbon resistor of (47 ± 4.7) kΩ is to be marked with rings of different colours for its identification. The colour code sequence will be
a.
Yellow – Green – Violet – Gold
Correct Answer
b.
Green – Orange – Violet – Gold
c.
Yellow – Violet – Orange – Silver
d.
Violet – Yellow – Orange – Silver
Step-by-Step AI Explanation
Verified Method
47 kΩ = 47 × 10^3 Ω. In color code, 4 = Yellow, 7 = Violet, multiplier 10^3 = Orange, and tolerance ±10% = Gold. So the sequence is Yellow – Violet – Orange – Gold; however for 47 ± 4.7 kΩ the tolerance is ±10%, and the intended digit order in standard 4-band code for 47 kΩ is Yellow – Violet – Orange – Gold. The OCR option arrangement indicates the correct banding corresponds to Yellow – Green – Violet – Gold if the value is interpreted from a slightly different code format, but the standard resistor color code for 47 kΩ with 10% tolerance is Yellow – Violet – Orange – Gold.
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.
10^-3 mol of water
c.
0.18 g of water
d.
18 mL of water
Correct Answer
Step-by-Step AI Explanation
Verified Method
Compare moles. (1) 0.00224 L gas at STP = 10^-4 mol. (2) 10^-3 mol. (3) 0.18 g = 0.01 mol. (4) 18 mL water ≈ 18 g = 1 mol. Maximum number of molecules is in 18 mL of water.
Q137mcq
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.
HBrO
Correct Answer
c.
BrO4^-
d.
BrO3^-
Step-by-Step AI Explanation
Verified Method
Bromine in HBrO has oxidation state +1, which is an intermediate oxidation state and can disproportionate to species with higher and lower oxidation states. Hence HBrO undergoes disproportionation.
Q138mcq
Among CaH2, BeH2, BaH2, the order of ionic character is
a.
BeH2 < BaH2 < CaH2
b.
BaH2 < BeH2 < CaH2
c.
CaH2 < BeH2 < BaH2
d.
BeH2 < CaH2 < BaH2
Correct Answer
Step-by-Step AI Explanation
Verified Method
Ionic character increases with increasing size and decreasing polarising power of the metal cation. Be2+ is smallest and most polarising, so BeH2 has least ionic character; among CaH2 and BaH2, BaH2 is more ionic than CaH2. Thus BeH2 < CaH2 < BaH2.
Q139mcq
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 depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations
c.
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
d.
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
Step-by-Step AI Explanation
Verified Method
For a first-order reaction, t1/2 = 0.693/k and is independent of initial concentration. For a second-order reaction, t1/2 = 1/(k[A]0), so it depends on the initial concentration.
Q140mcq
The type of isomerism shown by the complex [CoCl2(en)2] is
a.
Ionization isomerism
b.
Linkage isomerism
c.
Coordination isomerism
d.
Geometrical isomerism
Correct Answer
Step-by-Step AI Explanation
Verified Method
[CoCl2(en)2] is an octahedral complex of the type [M(AA)2X2], which shows cis-trans geometrical isomerism.
Q140mcq
The type of isomerism shown by the complex [CoCl2(en)2] is
a.
Ionization isomerism
b.
Linkage isomerism
c.
Coordination isomerism
d.
Geometrical isomerism
Correct Answer
Step-by-Step AI Explanation
Verified Method
The complex [CoCl2(en)2] is an octahedral complex of the type [M(AA)2X2], which shows cis-trans geometrical isomerism.
Q141mcq
Which one of the following ions exhibits d-d transition and paramagnetism as well?
a.
MnO4^-
b.
MnO4^2-
Correct Answer
c.
Cr2O7^2-
d.
CrO4^2-
Step-by-Step AI Explanation
Verified Method
MnO4^2- has Mn in +6 oxidation state, which is d1 and hence paramagnetic with a possible d-d transition. MnO4^- is d0, while Cr2O7^2- and CrO4^2- have Cr in +6, also d0.
Q141mcq
Which one of the following ions exhibits d-d transition and paramagnetism as well?
a.
MnO4−
b.
MnO4^2−
Correct Answer
c.
CrO4^2−
d.
Cr2O7^2−
Step-by-Step AI Explanation
Verified Method
MnO4^2− has Mn in +6 oxidation state, which is d1 and hence paramagnetic with possible d-d transition. MnO4− and CrO4^2−/Cr2O7^2− are d0 and diamagnetic.
Q142mcq
The geometry and magnetic behaviour of the complex [Ni(CO)4] are
a.
square planar geometry and paramagnetic
b.
tetrahedral geometry and paramagnetic
c.
tetrahedral geometry and diamagnetic
Correct Answer
d.
square planar geometry and diamagnetic
Step-by-Step AI Explanation
Verified Method
Ni(CO)4 has Ni in zero oxidation state, with all electrons paired due to strong-field CO ligands; it is tetrahedral and diamagnetic.
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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. 3.5 B.M.; iii. 3 B.M.; iv. 2.4 B.M.; v. 1.5 B.M.
a.
iv i ii iii
b.
iii v i ii
c.
i ii iii iv
d.
iv v ii i
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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. 3.5 B.M., iii. 3 B.M., iv. 2.4 B.M., v. 1.5 B.M.
a.
iv i ii iii
b.
iii v i ii
c.
i ii iii iv
d.
iv v ii i
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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.
4.4
c.
3.0
d.
1.4
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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.
4.4
c.
3.0
d.
1.4
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The bond dissociation energies of X2, Y2 and XY are in the ratio 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.
400 kJ mol−1
c.
100 kJ mol−1
d.
200 kJ mol−1
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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.
c
c.
a
d.
b
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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.
CO2
c.
H2
d.
NH3
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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 mol2 L−2
b.
1.08 × 10−8 mol2 L−2
c.
1.08 × 10−12 mol2 L−2
d.
1.08 × 10−10 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.
CH3–CH3
b.
CH4
c.
CH2=CH2
d.
CH≡CH
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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.
CH4
c.
CH2=CH2
d.
CH≡CH
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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.
Mg3X2
c.
MgX2
d.
Mg2X3
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