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NEET Chemistry MCQ
Quiz 1
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Q.1
The free space in a BCC unit cell is
0%
48%
0%
32%
0%
26%
0%
23%
Explanation
The packing efficiency of a BCC unit cell is π√3/8 ≈ 68%, so the free (empty) space is 100% − 68% ≈ 32%.
Q.2
The number of carbon atoms per unit cell of diamond unit cell is
0%
1
0%
4
0%
6
0%
8
Explanation
Diamond's structure is an FCC lattice of carbon atoms with 4 more carbon atoms occupying alternate tetrahedral voids. Counting: FCC contributes 4 atoms per unit cell, plus 4 more in the tetrahedral voids, giving 4 + 4 = 8 carbon atoms per unit cell.
Q.3
The number of Octahedral void per atom present in cubic close packed structure is
0%
1
0%
2
0%
4
0%
3
Explanation
In a cubic close-packed (CCP/FCC) structure, the number of octahedral voids exactly equals the number of atoms in the lattice — a 1:1 ratio, so there is 1 octahedral void per atom.
Q.4
A metal M crystallizes in FCC and BCC structures depending upon the temperature. The ratio of its densities in FCC and BCC structures would be
0%
$\frac {1}{2}$
0%
$\frac {8}{3 \sqrt 6}$
0%
$\frac {8}{ \sqrt 6}$
0%
$\frac {16}{3 \sqrt 6}$
Explanation
Density = (Z × M) / (Nₐ × a³), where Z is atoms per unit cell and a is the edge length in terms of atomic radius r. FCC: Z = 4, and (face diagonal) 4r = a√2 → a = 2√2 r BCC: Z = 2, and (body diagonal) 4r = a√3 → a = 4r/√3 Density ratio (FCC/BCC) = (Z_FCC/Z_BCC) × (a_BCC/a_FCC)³ = (4/2) × [(4r/√3) / (2√2 r)]³ = 2 × [2/√6]³ = 2 × 8/(6√6) = 8/(3√6)
Q.5
The ionic radius of the A+ is $.98 \times 10^{-10} \ m$ and that of B- is $ 1.81\times 10^{-10} \ m$. The coordination number of each ion in AB is
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2
0%
6
0%
4
0%
8
Explanation
Radius ratio = r(A+)/r(B−) = 98 pm / 181 pm ≈ 0.54. A radius ratio between about 0.414 and 0.732 corresponds to octahedral coordination, i.e. a coordination number of 6 for each ion.
Q.6
A substance $A_xB_y$ crystallizes in a face centred cubic (FCC) lattice in which atoms A occupy each corner of the cube and atoms B occupy the centre of each face of the cube.The correct composition of the substance $A_xB_y$ is
0%
$A_3B$
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$A_4B_3$
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$AB_3$
0%
Composition cannot be specified
Explanation
Corner atoms (A) contribute 8 × 1/8 = 1 atom per unit cell. Face-centre atoms (B) contribute 6 × 1/2 = 3 atoms per unit cell. Ratio A : B = 1 : 3, giving the formula AB₃.
Q.7
Match the column
0%
p ->iii, q ->i, r -> iv , s -> ii
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p ->ii, q ->iii, r -> iv , s -> i
0%
p ->i, q ->ii, r -> I , s -> iv
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p ->iii, q ->ii, r -> I , s -> iv
Explanation
Matching each defect type to its description: (p) Simple vacancy defect — occurs in non-ionic solids, decreases density (mass is missing, volume roughly unchanged) → (iii) (q) Simple interstitial defect — occurs in non-ionic solids, increases density (extra particles added in the gaps) → (i) (r) Schottky defect — occurs in ionic solids as paired cation/anion vacancies, decreases density → (iv) (s) Frenkel defect — occurs in ionic solids as a displaced ion moving to an interstitial site, leaving density essentially unchanged (nothing is actually lost) → (ii) So: p → iii, q → i, r → iv, s → ii.
Q.8
Match the column
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p -> ii, q -> i, r -> iv, s-> iii
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p -> ii, q -> i, r -> iii, s-> iv
0%
p -> i, q -> iii, r -> ii, s-> iv
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p -> iii, q -> ii, r -> iv, s-> i
Explanation
Matching conduction/semiconductor type: (p) Mg in solid state — a metal, conducts via free electrons → Electronic conductor (ii) (q) MgCl₂ in molten state — an ionic compound, conducts via mobile ions → Electrolytic conductor (i) (r) Silicon doped with phosphorus — phosphorus (Group 15) has one more valence electron than silicon (Group 14), donating an extra electron → n-Type semiconductor (iv) (s) Germanium doped with boron — boron (Group 13) has one fewer valence electron than germanium, creating an electron deficiency/hole → p-Type semiconductor (iii) So: p → ii, q → i, r → iv, s → iii.
Q.9
Match the column
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(p) -> (i), (q) ->(iii), r ->(iv) , (s) -> (ii)
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(p) -> (iv), (q) ->(ii), r ->(i) , (s) -> (iii)
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(p) -> (iv), (q) ->(iii), r ->(ii) , (s) -> (i)
0%
(p) -> (iv), (q) ->(iii), r ->(i) , (s) -> (ii)
Explanation
Matching each crystal system to its standard textbook example: (p) Trigonal → HgS (cinnabar) (iv) (q) Cubic → NaCl (iii) (r) Hexagonal → ZnO (i) (s) Tetragonal → CaSO₄ (ii) So: p → iv, q → iii, r → i, s → ii.
Q.10
The total volume of atoms present in BCC unit cell of metal(atomic radius =r) is given by?
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$\frac {24}{3} \pi r^3$
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$\frac {16}{3} \pi r^3$
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$\frac {32}{3} \pi r^3$
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$\frac {8}{3} \pi r^3$
Explanation
A BCC unit cell contains Z = 2 atoms. Total atomic volume = Z × (4/3)πr³ = 2 × (4/3)πr³ = (8/3)πr³.
Q.11
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b)Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Assertion: Semiconductors are solids with conductivities in the intermediate range from $10^{–6}$ – $10^{4}$ ohm–1m–1 Reason: Intermediate conductivity in semiconductor is due to partially filled valence band
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(a)
0%
(b)
0%
(c)
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(d)
Explanation
The Assertion's stated conductivity range for semiconductors is the standard textbook figure and is correct. But the Reason is wrong: a semiconductor's intermediate conductivity arises from a SMALL energy gap between a filled valence band and an empty conduction band (allowing some electrons to be thermally excited across it) — not from a "partially filled valence band", which isn't the actual mechanism.
Q.12
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b)Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Assertion : The total number of atoms present in a simple cubic unit cell is one. Reason: In a Simple cubic arrangement ,there are atoms at the corners of each cube
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(a)
0%
(b)
0%
(c)
0%
(d)
Explanation
In a simple cubic unit cell, atoms sit only at the 8 corners, and each corner atom is shared among 8 adjacent unit cells, so each contributes 1/8 to this cell: 8 × 1/8 = 1 atom total. The Reason correctly identifies why — corner-only occupancy is exactly what produces the count of 1 — so it is the correct explanation of the Assertion.
Q.13
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b)Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Assertion : band gap in Germanium is small Reason: The energy spade of each germanium atomic energy level is infinitesimally small
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(a)
0%
(b)
0%
(c)
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(d)
Explanation
Germanium's band gap genuinely is small, so the Assertion is true. The Reason, that individual atomic energy levels split into an almost continuous range of closely spaced levels when atoms come together in a solid, is also a true general description of how bands form — but this happens in ALL solids alike, so it doesn't specifically explain why germanium's gap in particular is small. Both statements are true, but the Reason is not the correct explanation of the Assertion.
Q.14
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b)Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Assertion: Total number of tetrahedral voids present in unit cell of cubic close packing is 8 Reason : FCC has 2 atoms per unit cell
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(a)
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(b)
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(c)
0%
(d)
Explanation
The Assertion (8 tetrahedral voids in an FCC/CCP unit cell) is correct — tetrahedral voids number twice the atoms in the cell, i.e. 2 × 4 = 8. But the Reason is false: FCC actually has 4 atoms per unit cell (8 corners × 1/8 + 6 faces × 1/2 = 1 + 3 = 4), not 2.
Q.15
When Zinc converts from liquid state to solid state, it has hcp structure, then find the number of nearest atoms
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6
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4
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8
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12
Explanation
In a hexagonal close-packed (HCP) structure — the arrangement zinc adopts on solidifying — each atom has 12 nearest neighbours, the maximum possible for close packing.
Q.16
The appearance of colour in solid alkali atom halides is generally due to
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interstitial positions
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F centers
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Schottky defect
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Frenkel defect
Explanation
Colour in alkali metal halide crystals typically arises from F-centres — anion vacancies that have trapped an electron in their place, which absorbs visible light and produces colour (e.g. NaCl heated in sodium vapour turns yellow this way).
Q.17
Which of the following molecules has three-fold axis of symmetry
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$NH_3$
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$C_2H_4$
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$CO_2$
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$SO_2$
Explanation
NH₃ is pyramidal, with a three-fold (C₃) rotation axis passing through the nitrogen atom and the centre of the triangle formed by the three hydrogens. C₂H₄ (planar) and SO₂ (bent) have only two-fold symmetry, and linear CO₂ has no three-fold axis.
Q.18
The edge length of a face centered cubic cell of an ionic substance is 508 pm. If the radius of the cation is 110 pm, the radius of the anion is
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288 pm
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398 pm
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618 pm
0%
144 pm
Explanation
In this FCC (rock-salt type) ionic structure, cations and anions touch along the cell edge: edge length a = 2 × (r₊ + r₋). 508 = 2 × (110 + r₋) → 254 = 110 + r₋ → r₋ = 144 pm.
Q.19
A compound AB crystallizes in the BCC lattice with Unit cell edge length of 380 pm. Calculate the distance between the oppositely charged ions in lattice
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351 pm
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288 pm
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300 pm
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329.1 pm
Explanation
In a BCC (CsCl-type) ionic lattice, the oppositely charged ions touch along the body diagonal. The centre-to-centre distance between them is half the body diagonal: (√3/2) × a. (√3/2) × 380 = 0.866 × 380 ≈ 329.1 pm.
Q.20
What is the radius of B- if the radius of A+ is 154.1 pm
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175 pm
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162 pm
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112 pm
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none of these
Explanation
Continuing from the previous question: the centre-to-centre distance between the ions was found to be 329.1 pm, and that distance equals r(A+) + r(B−). 329.1 − 154.1 = 175 pm.
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