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NEET Previous Year Question Papers
Quiz 6
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Q.1
Among \(CaH_2, BeH_2, BaH_2,\) the order of ionic character is
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\(BeH_2
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\(CaH_2
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\(BaH_2
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\(BeH_2
Explanation
Ionic character of these hydrides increases down the group as the metal becomes more electropositive (larger, less tightly held valence electrons): BeH₂ < CaH₂ < BaH₂.
Q.2
The bond dissociation energies of \(X_2,Y_2\) and XY are in the ratio of \(1:05 :1\). \(\Delta H\) for the formation of XY is \(-200 kJ mol^{-1}\) The bond dissociation energy of \(X_2\) will be
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\(100 kJ mol^{-1}\)
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\(800 kJ mol^{-1}\)
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\(200 kJ mol^{-1}\)
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\(400 kJ mol^{-1}\)
Explanation
Let bond energy of X₂ = E, so Y₂ = 0.5E and XY = E (matching the given 1:0.5:1 ratio). For ½X₂ + ½Y₂ → XY: ΔH = [½E + ½(0.5E)] − E = 0.5E + 0.25E − E = −0.25E Setting −0.25E = −200: E = 800 kJ/mol.
Q.3
The correct order of N-compounds in its decreasing order of oxidation states is
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\(HNO_3,NH_4Cl,NO,N_2\)
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\(NH_4Cl,N_2,NO,HNO_3\)
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\(HNO_3,NO,NH_4Cl,N_2\)
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\(HNO_3,NO,N_2,NH_4Cl\)
Explanation
Oxidation states of nitrogen: HNO₃ (+5), NO (+2), N₂ (0), NH₄Cl (−3). In decreasing order: HNO₃ > NO > N₂ > NH₄Cl.
Q.4
Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. It is due to their
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formation of intermolecular H-bonding
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formation of intramolecular H-bonding
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formation of carboxylate ion
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more extensive association of carboxylic acid via van der Waals force of attraction
Explanation
Carboxylic acid molecules form strong intermolecular hydrogen bonds — pairing up into stable dimers held together by two H-bonds simultaneously — which takes considerably more energy to break apart than the single H-bonds alcohols form, or the weaker dipole interactions in aldehydes/ketones, giving carboxylic acids notably higher boiling points.
Q.5
Given van der wails’ constant for \(NH_3, H_2, O_2, CO_2\) are respectively 4.17, 0.244, 1.36 and 3.59, which one of the following gases is most easily liquefied?
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\(H_2\)
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\(NH_3\)
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\(CO_2\)
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\(O_2\)
Explanation
A larger van der Waals constant 'a' reflects stronger intermolecular attraction, which makes a gas easier to liquefy. NH₃ has the largest given 'a' value (4.17), so it's the most easily liquefied of the four.
Q.6
Which one of the following ions exhibits d-d transition and paramagnetism as well?
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\( Cr_2O_{7}^{2-}\)
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\( CrO_{4}^{2-}\)
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\( MnO_{4}^{-}\)
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\( MnO_{4}^{2-}\)
Explanation
MnO₄²⁻ has manganese in the +6 oxidation state, giving a d¹ configuration — one unpaired electron makes it both paramagnetic and capable of a d-d transition. MnO₄⁻ (Mn +7) and CrO₄²⁻/Cr₂O₇²⁻ (Cr +6) are all d⁰ — no d electrons means no d-d transition and no paramagnetism.
Q.7
The correct order of atomic radii in group 13 elements is
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B < A1 < Ga < In < Tl
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B < Ga < A1 < In < Tl
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B < A1 < In < Ga < Tl
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B < Ga < A1 < Tl < In
Explanation
Atomic radii in group 13 follow B < Ga < Al < In < Tl — gallium's radius is anomalously smaller than aluminium's due to poor shielding by the filled 3d subshell just before it (the d-block contraction effect), breaking the otherwise expected steady increase down the group.
Q.8
A mixture of 2.3g formic acid and 4.5g oxalic acid is treated with conc. \(H_2SO_4\). The evolved gaseous mixture is passed through \(KOH\) pellets. Weight (in g) of the remaining product at STP will be
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1.4
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3.0
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4.4
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2.8
Explanation
With conc. H₂SO₄: formic acid dehydrates to CO (HCOOH → CO + H₂O), while oxalic acid decomposes to both CO and CO₂ ((COOH)₂ → CO + CO₂ + H₂O). Moles formic acid = 2.3/46 = 0.05 mol → 0.05 mol CO. Moles oxalic acid = 4.5/90 = 0.05 mol → 0.05 mol CO + 0.05 mol CO₂. Total CO = 0.1 mol; total CO₂ = 0.05 mol. KOH absorbs the acidic CO₂ but not CO, leaving 0.1 mol CO. Mass of remaining CO = 0.1 × 28 = 2.8 g.
Q.9
The difference between amylase and amylopectin is
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Amylose is made up of glucose and galactose
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Amylopectin have 1 → 4 α–linkage and 1 → 6 β–linkage
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Amylopectin have 1 → 4 α–linkage and 1 → 6 β–linkage
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Amylose have 1 → 4 α–linkage and 1 → 6 β–linkage
Explanation
Amylopectin is a branched polysaccharide, built from straight chains linked by α-1,4 glycosidic bonds with additional α-1,6 glycosidic bonds forming the branch points — unlike amylose, which is unbranched, containing only α-1,4 linkages. (The source option's "β" for the 1→6 linkage is a typo — amylopectin's branch linkages are α-1,6, not β.)
Q.10
Which one of the following elements is unable to form \(MF_{6}^{3-}\) ion?
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B
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In
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Al
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Ga
Explanation
Boron is too small to accommodate six fluoride ions around it (it lacks accessible d-orbitals and typically maxes out at 4-coordination, as in BF₄⁻), so it cannot form BF₆³⁻. The larger group 13 elements (Al, Ga, In) readily achieve 6-coordination and form MF₆³⁻ ions.
Q.11
Which of the following compounds can form a zwitterion?
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Acetanilide
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Glycine
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Benzoic acid
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Aniline
Explanation
Glycine, an amino acid with both a basic −NH₂ group and an acidic −COOH group, forms an internal salt (zwitterion, ⁺H₃N-CH₂-COO⁻) in which the amino group is protonated and the carboxyl group is deprotonated simultaneously.
Q.12
Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is \(1s^2\,\,2s^2\,\,2p^2\) the simplest formula for this compound is
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\(MgX_2\)
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\(Mg_2X\)
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\(Mg_3X_2\)
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\(Mg_2X_3\)
Explanation
The configuration 1s²2s²2p² corresponds to carbon, which can form the carbide ion C⁴⁻ (gaining 4 electrons to complete its octet) in ionic carbide compounds. Combined with Mg²⁺, simple charge balance would give Mg₂C — though the marked answer Mg₃X₂ (implying X³⁻) may reflect a different valence convention used in this particular question's source.
Q.13
Iron exhibits bcc structure at room temperature. Above \(900^0\text{C,}\) it transforms to fcc structure. The ratio of density of iron at room temperature to that at \(900^0\text{C,}\) (assuming molar mass and atomic radii of iron remains constant with temperature) is
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\( \frac{3\sqrt{3}}{4\sqrt{2}}\)
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\( \frac{4\sqrt{3}}{3\sqrt{2}}\)
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\( \frac{1}{2}\)
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\( \frac{\sqrt{3}}{\sqrt{2}}\)
Explanation
Density ρ = ZM/(N_A·a³). For BCC: Z=2, edge length a = 4r/√3. For FCC: Z=4, edge length a = 2√2r. ρ(bcc)/ρ(fcc) = [Z_bcc/Z_fcc] × [a_fcc/a_bcc]³ = (2/4) × [(2√2r)/(4r/√3)]³ = (1/2) × (√6/2)³ = (1/2)(3√6/4) = 3√6/8, which is equivalent to 3√3/(4√2) (same value, written with a rationalised denominator).
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