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Jee Chemistry Chapter Wise Mock Test
Quiz 1
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Q.1
(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 octahedral voids present in unit cell of cubic close packing is 4 Reason : We have one octahedral void at the body centre of the cube and 12 octahedral voids located at each edge , each of which is shared between four unit cells
0%
(a)
0%
(b)
0%
(c)
0%
(d)
Explanation
Solution:
Assertion: true -- a cubic close-packed (FCC) unit cell has 1 octahedral void at its body centre plus one at the centre of each of its 12 edges, and each edge-centre void is shared between 4 neighbouring unit cells.
Reason: true, and it is exactly the reasoning that gives the count: 1 (body centre) + 12 × 1/4 (edges) = 1 + 3 = 4 octahedral voids per cell -- the reason directly explains the assertion.
Q.2
Potassium crystallizes in the body centered cubic lattice with edge length $a= 5.2 A^o$ The distance between nearest neighbors is
0%
$1.5 A^o$
0%
$2.5 A^o$
0%
$5.2 A^o$
0%
$4.5 A^o$
Explanation
Solution:
In a BCC lattice, the nearest neighbour to a corner atom is the atom at the cube's body centre, at a distance of (√3/2)·a along the body diagonal.
(√3/2) × 5.2 ≈ 0.866 × 5.2 ≈ 4.5 Å.
Q.3
The distance between next nearest neighbors
0%
$1.3 A^o$
0%
$2.5 A^o$
0%
$5.2 A^o$
0%
$4.5 A^o$
Explanation
Solution:
The next-nearest neighbours in a BCC lattice are the corner atoms directly along the cube's edges, at a distance equal to the edge length itself, a = 5.2 Å.
Q.4
How many nearest neighbors does each K atom have
0%
2
0%
4
0%
6
0%
8
Explanation
Solution:
In a BCC lattice, each corner atom's nearest neighbours are the body-centre atoms of the 8 cubes that meet at that corner, giving a coordination number of 8.
Q.5
How many next nearest neighbors does each K atom have
0%
2
0%
4
0%
6
0%
8
Explanation
Solution:
A BCC lattice's next-nearest neighbours are the 6 corner atoms reached by moving one edge length along each of the +x, -x, +y, -y, +z, -z directions.
Q.6
The coordination number of Al in the crystalline state of AlCl3
0%
3
0%
4
0%
6
0%
8
Explanation
Solution:
In crystalline AlCl
3
, the structure is built from a close-packed arrangement of Cl
-
ions with Al
3+
ions occupying octahedral voids, so each Al
3+
is surrounded by 6 Cl
-
ions -- a coordination number of 6.
Q.7
A metallic element crystallizes into a lattice containing a sequence of layers of ABABAB. Any packing of sphere leaves out voids in the lattice. What percentage by volume of this lattice is occupied by voids?
0%
74%
0%
26%
0%
68%
0%
32%
Explanation
Solution:
An ABAB... hexagonal close-packed arrangement (and equally, an ABCABC... cubic close-packed one) fills 74% of the available space with spheres, leaving the remaining 26% as empty voids between them.
Q.8
The composition of a sample of wustite is $Fe_{0.93} O_{1.00}$. What % of the iron is present in the form of Fe(III)?
0%
15.05%
0%
21%
0%
74%
0%
None of these
Explanation
Solution:
In Fe
0.93
O
1.00
, let x be the fraction of iron present as Fe
3+
and (0.93 − x) the fraction as Fe
2+
. For the compound to be electrically neutral against 1.00 mol of O
2-
(total negative charge 2.00):
2(0.93 − x) + 3x = 2.00 ⇒ 1.86 + x = 2.00 ⇒ x = 0.14 mol.
Fraction of iron as Fe
3+
= 0.14 ÷ 0.93 ≈ 15.05%.
Q.9
$AgI$ crystallizes in the Cubic close packed zinc blende structure. Assuming that the iodide ions occupy lattice points
0%
75%
0%
100%
0%
33.3%
0%
50%
Explanation
Solution:
In the zinc blende (cubic ZnS-type) structure, the anions (here I
-
) form a cubic close-packed lattice, which creates 8 tetrahedral voids per unit cell for every 4 lattice points. The cations (Ag
+
) occupy only the alternate tetrahedral voids -- 4 out of the 8 -- which is 50% of them.
Q.10
The radius of the Na+ is 95 pm and that of Cl- is 181 pm. The probable coordination number of Na+ is
0%
2
0%
6
0%
4
0%
8
Explanation
Solution:
The radius ratio r
+
/r
-
= 95 ÷ 181 ≈ 0.525.
A radius ratio between 0.414 and 0.732 corresponds to octahedral coordination, i.e. a coordination number of 6.
Q.11
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%
$A_3B$
0%
$A_4B_3$
0%
$AB_3$
0%
Composition cannot be specified
Explanation
Solution:
Atoms A sit at the 8 corners of the cube, each shared among 8 unit cells, contributing 8 × 1/8 = 1 A atom per cell.
Atoms B sit at the centre of each of the 6 faces, each shared between 2 unit cells, contributing 6 × 1/2 = 3 B atoms per cell.
The ratio A:B = 1:3, giving the formula AB
3
.
Q.12
Match the column
0%
p ->ii, q ->iii, r -> i , s -> iv
0%
p ->ii, q ->iii, r -> iv , s -> i
0%
p ->i, q ->ii, r -> i , s -> iv
0%
p ->iii, q ->ii, r -> i , s -> iv
Explanation
Solution:
Square close packing in 2D gives each sphere 4 touching neighbours, so its coordination number is 4 (p→ii).
Cubic close packing in 3D (ABCABC... stacking) repeats its layer pattern only after returning to the same layer type further down the stack (q→iii).
Hexagonal close packing in 2D leaves triangular gaps between the touching spheres (r→i).
Hexagonal close packing in 3D (ABAB... stacking) repeats its pattern every alternate layer (s→iv).
Q.13
Match the column
0%
p -> ii, q -> i, r -> iii, s-> iv
0%
p -> i, q -> iii, r -> ii, s-> iv
0%
p -> iii, q -> ii, r -> iv, s-> i
0%
p -> ii, q -> i, r -> iv, s-> iii
Explanation
Solution:
Mg metal conducts by the free movement of delocalised electrons through its lattice -- an electronic conductor (p→ii).
Molten MgCl
2
conducts because its Mg
2+
and Cl
-
ions are free to move -- an electrolytic conductor (q→i).
Silicon doped with phosphorus (group 15, one extra valence electron) gives an n-type semiconductor (r→iv).
Germanium doped with boron (group 13, one electron short) gives a p-type semiconductor (s→iii).
Q.14
Match the column
0%
(p) -> (i), (iii) , (q) ->(i) .(iv), r ->(ii), (v) , (s) -> (ii) ,(iv)
0%
(p) -> (i), (iv) , (q) ->(i) .(iv), r ->(ii), (iv) , (s) -> (iii) ,(v)
0%
(p) -> (i), (v) , (q) ->(i) .(iv), r ->(ii), (v) , (s) -> (iii) ,(iv)
0%
(p) -> (i), (iv) , (q) ->(i) .(iv), r ->(ii), (v) , (s) -> (iii) ,(iv)
Explanation
Solution:
Matching each crystal system to its defining edge-length and angle relations:
Trigonal: a=b=c with α=β=γ≠90° (p→i, iii).
Cubic: a=b=c with α=β=γ=90° (q→i, iv).
Hexagonal: a=b≠c with α=β=90°, γ=120° (r→ii, v).
Tetragonal: a=b≠c with α=β=γ=90° (s→ii, iv).
Q.15
The compound in which cation occupy alternate tetrahedral voids in Cubic close packing?
0%
ZnS
0%
NaCl
0%
$Na_2 O$
0%
$CaF_2$
Explanation
Solution:
In ZnS (zinc blende), the S
2-
ions form a cubic close-packed lattice and the smaller Zn
2+
cations occupy only the alternate tetrahedral voids of that lattice -- the structure this question describes.
Q.16
A compound is formed by elements X and P. This crystallizes in the cubic structure where the X atoms are at the corners of the cube and Y atoms are at the body centers. The simplest formula of the compound is:
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$X_6Y$
0%
$XY_6$
0%
$XY$
0%
$X_2Y$
Explanation
Solution:
X atoms at the 8 corners contribute 8 × 1/8 = 1 atom per cell. Y atoms at the body centre contribute 1 × 1 = 1 atom per cell (a body-centre position is not shared with any neighbouring cell).
The ratio X:Y = 1:1, giving the simplest formula XY.
Q.17
In a FCC unit cell? (More than one correct)
0%
effective number of atoms in FCC is 4
0%
8 tetrahedral per unit cellcorrect
0%
coordination number is 12correct
0%
Rank of unit cell is 3
Explanation
Solution:
Several facts about an FCC (cubic close-packed) unit cell are true at once: it has 4 effective atoms per cell (8 corners × 1/8 + 6 faces × 1/2 = 4), 8 tetrahedral voids per cell, and a coordination number of 12 (each atom touches 12 neighbours). "Rank of unit cell" is not a standard crystallography term, so that option doesn't apply. The number-of-atoms fact is the one marked here.
Q.18
(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: Graphite is a good conductor of electricity however diamond belongs to the category of insulators Reason: Graphite is soft in nature on the other hand diamond is very hard and brittle
0%
(a)
0%
(b)
0%
(c)
0%
(d)
Explanation
Solution:
Assertion: true -- graphite conducts because each carbon is only bonded to 3 others within a layer, leaving one delocalised electron per atom free to move, while diamond's carbons are all bonded to 4 neighbours with no free electrons, making it an insulator.
Reason: also true -- graphite's layers are held together only by weak van der Waals forces (making it soft), while diamond's rigid 3D covalent network makes it very hard.
But hardness/softness is not what makes one conduct and the other not -- that comes down to electron delocalisation, not mechanical strength. So both statements are correct, but the reason does not explain the assertion.
Q.19
(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 body centered cubic unit cell is four. Reason: In a FCC arrangement ,there are atoms at the corners and at the faces of each cube
0%
(a)
0%
(b)
0%
(c)
0%
(d)
Explanation
Solution:
Reason (about FCC): true -- an FCC cell does have atoms at every corner and at the centre of every face.
Assertion (about BCC): a body-centred cubic cell has atoms only at the 8 corners and 1 at the body centre, giving 8 × 1/8 + 1 = 2 atoms per cell, not four -- four atoms per cell is the count for an FCC cell, not a BCC one. So, by the standard definition, this assertion is inconsistent with the usual BCC atom count.
Q.20
(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
0%
(a)
0%
(b)
0%
(c)
0%
(d)
Explanation
Solution:
Assertion: true -- germanium's band gap (≈0.66 eV) is small compared to an insulator's, which is exactly what makes it a semiconductor.
Reason: also a true general fact -- when atoms come together in a crystal, each atomic energy level splits into a huge number of extremely closely spaced levels, which is why solids have continuous "bands" of energy at all.
But that band-formation mechanism applies equally to every crystalline solid (silicon, diamond, metals...), not just germanium, so it doesn't specifically explain why germanium's gap in particular is small. Both statements are correct, but the reason isn't the correct explanation for the assertion.
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