The type of isomerism shown by the complex [CoCl2(en)2] is
a.
Geometrical isomerism
Correct Answer
b.
Linkage isomerism
c.
Ionization isomerism
d.
Coordination isomerism
Step-by-Step AI Explanation
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In [CoCl2(en)2], the coordination number of Co is 6 and it has an octahedral geometry. It exhibits geometrical isomerism (cis and trans forms).
Q2mcq
Which one of the following ions exhibits d-d transition and paramagnetism as well?
a.
CrO4^2-
b.
MnO4^2-
Correct Answer
c.
MnO4^-
d.
Cr2O7^2-
Step-by-Step AI Explanation
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MnO4^2- is Mn(VI) with [Ar] 3d^1 configuration, having one unpaired electron, making it paramagnetic.
Q3mcq
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. Co^3+, b. Cr^3+, c. Fe^3+, d. Ni^2+
Column II: i. √8 BM, ii. √35 BM, iii. √3 BM, iv. √24 BM, v. √15 BM
Fe(CO)5 contains only one iron atom, therefore it is classified as mononuclear.
Q5mcq
The geometry and magnetic behaviour of the complex [Ni(CO)4] are
a.
Square planar geometry and diamagnetic
b.
Tetrahedral geometry and paramagnetic
c.
Square planar geometry and paramagnetic
d.
Tetrahedral geometry and diamagnetic
Correct Answer
Step-by-Step AI Explanation
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Ni(CO)4 has sp3 hybridization due to the strong field ligand CO, leading to a tetrahedral geometry and diamagnetic behavior as all electrons are paired.
Q6mcq
The correct order of N-compounds in its decreasing order of oxidation states is
a.
HNO3, NO, N2, NH4Cl
Correct Answer
b.
NH4Cl, N2, NO, HNO3
c.
HNO3, NH4Cl, NO, N2
d.
HNO3, NO, NH4Cl, N2
Step-by-Step AI Explanation
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Oxidation states: HNO3 (+5), NO (+2), N2 (0), NH4Cl (-3). Decreasing order is +5, +2, 0, -3.
Q6mcq
The correct order of N-compounds in its decreasing order of oxidation states is
a.
HNO3, NO, N2, NH4Cl
Correct Answer
b.
NH4Cl, N2, NO, HNO3
c.
HNO3, NH4Cl, NO, N2
d.
HNO3, NO, NH4Cl, N2
Step-by-Step AI Explanation
Verified Method
Oxidation states are: HNO3(+5), NO(+2), N2(0), NH4Cl(-3). The decreasing order is +5, +2, 0, -3.
Q7mcq
Which one of the following elements is unable to form MF6^3- ion?
a.
Ga
b.
In
c.
B
Correct Answer
d.
Al
Step-by-Step AI Explanation
Verified Method
Boron has no vacant d-orbitals in its valence shell, so it cannot extend its covalency beyond 4.
Q7mcq
Which one of the following elements is unable to form MF6^3- ion?
a.
Ga
b.
In
c.
B
Correct Answer
d.
Al
Step-by-Step AI Explanation
Verified Method
Boron (B) has no vacant d-orbitals in its valence shell and cannot expand its octet/covalency beyond 4.
Q8mcq
Considering Ellingham diagram, which of the following metals can be used to reduce alumina?
a.
Fe
b.
Cu
c.
Mg
Correct Answer
d.
Zn
Step-by-Step AI Explanation
Verified Method
The metal which is more reactive than aluminium can reduce alumina; Mg is the correct option here.
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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.
CH3CH2OH and I2
b.
CH3CH(CH3)OH and I2
c.
C6H5CH(OH)CH3 and I2
d.
CH2=CH-CH2OH and I2
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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.
CH4
c.
CH3 – CH3
d.
CH2 = CH2
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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.
1.4
b.
4.4
c.
2.8
d.
3.0
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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. pH of which one of them will be equal to 1?
a.
b
b.
c
c.
d
d.
a
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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^–8 mol^2 L^–2
c.
1.08 × 10^–14 mol^2 L^–2
d.
1.08 × 10^–12 mol^2 L^–2
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For the redox reaction MnO4^- + C2O4^2- + H+ -> Mn^2+ + CO2 + H2O. The correct coefficients of the reactants for the balanced equation are (MnO4^-, C2O4^2-, H+):
a.
16, 5, 2
b.
5, 16, 2
c.
2, 16, 5
d.
2, 5, 16
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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.
400 kJ mol–1
c.
800 kJ mol–1
d.
100 kJ mol–1
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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 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.
A first-order reaction can be catalyzed; a second-order reaction cannot be catalyzed
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
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Consider the change in oxidation state of Bromine corresponding to different emf values as shown: BrO4- (1.82V) -> BrO3- (1.5V) -> HBrO (1.595V) -> Br2 (1.0652V) -> Br-. Then the species undergoing disproportionation is:
a.
BrO3-
b.
HBrO
c.
Br2
d.
BrO4-
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Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below: BrO4- (1.82 V) -> BrO3- (1.5 V) -> HBrO (1.595 V) -> Br2 (1.0652 V) -> Br-. Then the species undergoing disproportionation is
a.
BrO3-
b.
HBrO
c.
Br2
d.
BrO4-
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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.
Mg3X2
c.
Mg2X
d.
MgX2
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