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NEET • CHEMISTRY • SOLID STATE

Solid state practice · Mixed difficulty

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Showing 121-150 of 238 questions

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121 Medium
121 of 238
Understanding +4 | -1
An example of a non-stoichiometric compound is
View answer and explanation
Correct option: D
Fe3O4\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}
Option D (Fe3O4\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}) is correct because Fe3O4\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4} (magnetite) is a non-stoichiometric compound, meaning the ratio of elements is not a simple whole number ratio.
122 Easy
122 of 238
Understanding +4 | -1
A pure crystalline substance, on being heated gradually, first forms a turbid looking liquid and then the turbidity completely disappears. This behavior is the characteristic of substances forming
View answer and explanation
Correct option: B
liquid crystals
Option B (liquid crystals) is correct because liquid crystals exhibit a turbid liquid phase before clearing upon heating.
123 Easy
123 of 238
Understanding +4 | -1
The hardness of metals increases with increase in number of ……involved in metallic bonding.
View answer and explanation
Correct option: C
Electrons
Option C (Electrons) is correct because the hardness of metals increases with the number of delocalized electrons involved in metallic bonding.
124 Medium
124 of 238
Application +4 | -1
A solid compound contains X,YX,Y and ZZ atoms in a cubic lattice with XX atom occupying the corners. YY atoms in the body centred positions and ZZ atoms at the centres of faces of the unit cell. What is the empirical formula of the compound?
View answer and explanation
Correct option: B
XYZ3XY{Z_3}
Option B (XYZ3XY{Z_3}) is correct because it accurately represents the ratio of X, Y, and Z atoms based on their positions in the cubic lattice.
125 Medium
125 of 238
Remembering +4 | -1
In the calcium fluoride structure, the coordination number of the cation and anion are respectively
View answer and explanation
Correct option: D
8,48,\mathrm{ }4
Option D (8, 4) is correct because in the CaF2CaF_2 structure, each Ca2+Ca^{2+} ion is surrounded by 8 F−F^- ions, and each F−F^- ion is surrounded by 4 Ca2+Ca^{2+} ions.
126 Easy
126 of 238
Remembering +4 | -1
A solid with high electrical and thermal conductivity from the following is :
View answer and explanation
Correct option: B
Li\mathrm{Li}
Option B (Li) is correct because Lithium is a metal and metals are generally good conductors of heat and electricity.
127 Easy
127 of 238
Understanding +4 | -1
The vacant space in BCC lattice unit cell is
View answer and explanation
Correct option: C
32%32\%
Option C (32%32\%) is correct because the packing efficiency in a BCC lattice is 68%68\%, leaving 32%32\% as vacant space.
128 Medium
128 of 238
Remembering +4 | -1
The unit cell with crystallographic dimensions a≠b≠c,   α=γ=90oa \ne b \ne c,\,\,\,\alpha = \gamma = {90^\mathrm{o}} and β≠90o\beta \ne {90^\mathrm{o}} is
View answer and explanation
Correct option: A
Monoclinic
Option A (Monoclinic) is correct because the given crystallographic dimensions match the definition of a monoclinic unit cell.
129 Medium
129 of 238
Remembering +4 | -1
p-type semiconductor is formed when germanium is doped with
View answer and explanation
Correct option: C
Aluminium
Option C (Aluminium) is correct because Aluminium is a trivalent impurity, and doping Germanium with a trivalent impurity creates a p-type semiconductor.
130 Medium
130 of 238
Analyzing +4 | -1
What is common in HCP and FCC unit cell?
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because there is no commonality among the given options for HCP and FCC unit cells.
131 Medium
131 of 238
Application +4 | -1
The number of atoms in 100 g of an fcc crystal with density d=10g/cm3d = 10\mathrm{g/c}{\mathrm{m}^\mathrm{3}} and cell edge equal to 100 pm, is equal to
View answer and explanation
Correct option: D
4×10254 \times {10^{25}}
Option D (4×10254 \times {10^{25}}) is correct because it's the calculated number of atoms based on the given parameters.
132 Medium
132 of 238
Application +4 | -1
A,B\mathrm{A}, \mathrm{B} and C\mathrm{C} have the oxidation numbers of +6,−2+6,-2 and −1-1 respectively, the possible molecular formula when these atoms combine will be:
View answer and explanation
Correct option: B
AB2C2\mathrm{AB}_2 \mathrm{C}_2
Option B is correct because the charges must balance in a neutral molecule. With A having +6, B having -2, and C having -1, the formula AB2C2AB_2C_2 gives a net charge of +6 + 2(-2) + 2(-1) = +6 - 4 - 2 = 0.
133 Easy
133 of 238
Understanding +4 | -1
In face centred cubic unit cell edge length is
View answer and explanation
Correct option: D
42r\frac{4}{{\sqrt 2 }}r
Option D (42r\frac{4}{{\sqrt 2 }}r) is correct because in a face-centered cubic (FCC) unit cell, the atoms touch along the face diagonal, and the relationship between the edge length (a) and the atomic radius (r) is a=4r2a = \frac{4r}{\sqrt{2}}.
134 Easy
134 of 238
Remembering +4 | -1
The ABAB.....ABAB..... packing and ABCABC.....ABCABC..... packing are respectively called
View answer and explanation
Correct option: A
HCP  and  CCPHCP\,\,and\,\,CCP
Option A (HCP  and  CCPHCP\,\,and\,\,CCP) is correct because ABAB.....ABAB..... packing is characteristic of Hexagonal Close Packing (HCP) and ABCABC.....ABCABC..... packing is characteristic of Cubic Close Packing (CCP).
135 Easy
135 of 238
Application +4 | -1
A cation is in the centre touches three anions.
Assume that the anions also touch each
other. The limiting radius ratio, r+/r−r^+ / r^- is
View answer and explanation
Correct option: A
0.1547
Option A (0.1547) is correct because it represents the limiting radius ratio for a coordination number of 3, where the cation touches three anions that also touch each other.
136 Medium
136 of 238
Understanding +4 | -1
Given below are two statements.
Statement I Frenkel defects are vacancy as well as interstitial defects.
Statement II Frenkel defects leads to colour in ionic solids due to presence of F - centres.
Choose the most appropriate answer from the options given below.
View answer and explanation
Correct option: C
Statement I is true but statement II is false.
Option C (Statement I is true but statement II is false) is correct because Frenkel defects are both vacancy and interstitial defects, but they don't directly lead to color due to F-centers.
137 Easy
137 of 238
Remembering +4 | -1
Which of the following solid is non-crystalline solid
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Correct option: D
Glass
Option D (Glass) is correct because glass lacks a long-range ordered structure, making it a non-crystalline solid.
138 Easy
138 of 238
Application +4 | -1
A metal XX crystallises in a face-centred cubic arrangement with the edge length 862pm862 \mathrm{pm}. What is the shortest separation of any two nuclei of the atom? (2=1.4)(\sqrt{2}=1.4)
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Correct option: D
616 pm616 \mathrm{~pm}
Option D (616 pm) is correct because the shortest separation in an FCC lattice is half the face diagonal, which is calculated as 22a\frac{{\sqrt{2}}}{{2}}a.
139 Easy
139 of 238
Application +4 | -1
A solid compound XY has NaClNaCl structure. If the radius of the cation is 100pm100pm, the radius of the anion (Y−)\left( {{Y^ - }} \right) is
View answer and explanation
Correct option: C
241.5 pm241.5\,pm
Option C (241.5 pm241.5\,pm) is correct because the radius ratio for NaCl structure falls within a specific range, allowing calculation of the anion radius.
140 Hard
140 of 238
Application +4 | -1
In a mixed oxide, O2−{O}^{2-} has fcc arrangement, univalent B⊕{B}^{\oplus} is present in all octahedral voids and divalent A2⊕A^{2 \oplus} is present in 14\frac{1}{4} th of tetrahedral voids, then formula of the oxide should be
View answer and explanation
Correct option: D
AB2O2{AB}_2 {O}_2
Option D (AB2O2{AB}_2 {O}_2) is correct because it represents the correct ratio of A, B, and O based on their positions in the lattice.
141 Easy
141 of 238
Application +4 | -1
The fraction of total volume occupied by the atoms present in the simple cube is
View answer and explanation
Correct option: D
π6\frac{\pi }{6}
Option D (π6\frac{\pi }{6}) is correct because the packing fraction for a simple cubic cell is calculated as the volume occupied by the atoms divided by the total volume of the unit cell, which simplifies to π6\frac{\pi }{6}.
142 Medium
142 of 238
Remembering +4 | -1
Close packing is maximum in the crystal lattice of :
View answer and explanation
Correct option: B
Face centred
Option B (Face centred) is correct because face-centered cubic (FCC) structures have the highest packing efficiency (74%).
143 Hard
143 of 238
Understanding +4 | -1
Statement-I: The close packing of atoms in cubic structure is in the order fcc>bcc>sc.\text{fcc} > \text{bcc} > \text{sc}.

Statement-II : Packing efficiency =Volume occupied by atom in unit cellTotal volume of the unit cell.=\frac{\text {Volume occupied by atom in unit cell}}{\text {Total volume of the unit cell}}.
View answer and explanation
Correct option: A
Both statement I and statement II are true
Option A is correct because the packing efficiency order is indeed fcc > bcc > sc, and the packing efficiency formula is correctly stated as the ratio of the volume occupied by atoms in the unit cell to the total volume of the unit cell.
144 Hard
144 of 238
Analyzing +4 | -1
A solid has a structure in which X{X} atoms are located at cubic corners of unit cell, O{O} atoms are at the edge centers and Y{Y} atoms at cube centre. Then the formula of compound is XaYbOc{X}_{{a}} {Y}_{{b}} {O}_{{c}}. If two atoms of O{O} are missing from any of two edge centers per unit cell, then the molecular formula is XxYyOz{X}_{{x}} {Y}_{{y}} {O}_z. Then, find the value of (x+y+z)−({x}+{y}+{z})- (a+b+c)(a+b+c).
View answer and explanation
Correct option: A
4
Option A (4) is correct because the change in the number of oxygen atoms due to the missing atoms at the edge centers affects the overall formula, leading to a difference of 4 in the sum of the subscripts.
145 Medium
145 of 238
Understanding +4 | -1
The example of orthosilicate is :
View answer and explanation
Correct option: B
Mg2SiO4\mathrm{M}{\mathrm{g}_2}\mathrm{Si}{\mathrm{O}_4}
Option B (Mg2SiO4\mathrm{M}{\mathrm{g}_2}\mathrm{Si}{\mathrm{O}_4}) is correct because orthosilicates are characterized by isolated SiO44−SiO_4^{4-} tetrahedra, which is represented by the formula Mg2SiO4\mathrm{M}{\mathrm{g}_2}\mathrm{Si}{\mathrm{O}_4}.
146 Hard
146 of 238
Analyzing +4 | -1
In a fcc lattice A,B,CA, B, C and DD atoms are arranged at corners, face centres octahedral voids and half of tetrahedral voids respectively and if the crystal deposited is defected such that all the particles at one body diagonal of each unit cell are missing, correspondingly then what is the resulted formula of the compound?
View answer and explanation
Correct option: B
AB4C4D4{AB}_4 {C}_4 {D}_4
Option B (AB4C4D4{AB}_4 {C}_4 {D}_4) is correct because it accurately reflects the atom ratios after accounting for the missing atoms along the body diagonal.
147 Hard
147 of 238
Analyzing +4 | -1
A compound of formula A2B3{A_2}{B_3} has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms?
View answer and explanation
Correct option: C
hcp lattice-BB, 13\frac{1}{3} tetrahedral voids-AA
Option C (hcp lattice-BB, 13\frac{1}{3} tetrahedral voids-AA) is correct because in an hcp lattice, the number of atoms is 6, and the number of tetrahedral voids is 12. The formula A2B3A_2B_3 implies that B forms the hcp lattice, and A occupies 1/3 of the tetrahedral voids.
148 Easy
148 of 238
Understanding +4 | -1
With which one of the following elements silicon should be doped so as to give P−typeP - type of semiconductor
View answer and explanation
Correct option: B
BB
Option B (BB) is correct because doping silicon with Boron (B), a group 13 element, creates a p-type semiconductor due to the introduction of holes.
149 Easy
149 of 238
Analyzing +4 | -1
Which is the incorrect statement?
View answer and explanation
Correct option: B
FeO0.98\mathrm{Fe}{\mathrm{O}_{0.98}} has non stoichiometric metal deficiency defect
Option B (FeO0.98\mathrm{Fe}{\mathrm{O}_{0.98}} has non stoichiometric metal deficiency defect) is incorrect because while FeO can have a metal deficiency defect, the statement implies it *always* does, which isn't true; it depends on the specific composition.
150 Medium
150 of 238
Remembering +4 | -1
The correct option for the number of body centred unit cells in all 14 types of Bravais lattice unit cells is:
View answer and explanation
Correct option: B
3
Option B (3) is correct because there are three Bravais lattices that have body-centered unit cells: body-centered cubic, body-centered tetragonal, and body-centered orthorhombic.
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