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31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - NEET MCQ


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30 Questions MCQ Test Chemistry Class 12 - 31 Years NEET Previous Year Questions: The Solid State (Old NCERT)

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) for NEET 2024 is part of Chemistry Class 12 preparation. The 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) questions and answers have been prepared according to the NEET exam syllabus.The 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) below.
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31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 1

Right option for the number of tetrahedral and octahedral voids in hexagonal primitive unit cell are:   [2021]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 1

No. of atoms in Hexagonal primitive unit cell (N) = 6
No. of Tetrahedral voids = 2 × No. of atoms per unit cell
= 2 × 6 = 12
No. of Octahedral voids = No. of atoms per unit cell = 6

Therefore, the correct answer is B

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 2

The number of carbon atoms per unit cell of diamond unit cell is : [NEET 2013]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 2

Diamond is like ZnS. In a diamond cubic unit cell, there are eight corner atoms, six face-centred atoms and four more atoms inside the structure.

No.  of atoms contributed by the corner atoms to a unit cell is 1/8×8 =1

 No. of atoms contributed by the face centred atoms to the unit cell is 1/2 × 6 = 3

  And Atoms inside the structure =4

 So total number of atoms present in a diamond cubic unit cell is 1 + 3 + 4 = 8

 Since each carbon atom is surrounded by four more carbon atoms, the coordination number is 4

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31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 3

A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm-3.The molar mass of the metal is?

(NA Avogadro's constant= 6.02 x 1023 mol-1) [NEET 2013]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 3

Given, cell is fcc, So no. of effective constituent particles in one unit cell (Z) =4
Edge length, a = 404 pm = 4.04 x 10-8 cm
Density of metal, d = 2.72 g cm-3
NA = 6.02 x 1023 mol-1
Molar mass of the metal, M =?
We know that

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 4

Structure of a mixed oxide is cubic close-packed (c.c.p). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide is : [2012 M]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 4

It is given that the structure of mixed oxide is cubic closed packed.

Therefore, the number of O2- ions according to ccp =4
So, the number of tetrahedral voids =8 and octahedral voids =4
Ions occupy 1/4​th of tetrahedral voids = 2
B ions occupy all octahedral voids = 4
A:B:O = 2:4:4 = 1:2:2   

Hence, the formula is AB2O2.

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 5

The number of octahedral void(s) per atom present in a cubic close-packed structure is : [2012]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 5

The number of octahedral voids in ccp, is equal to an effective number of atoms.
In ccp, the effective number of atoms are 4 so, 4 octahedral voids.
So, 1 octahedral void per atom. 

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 6

A metal crystallizes with a face-centered cubic lattice. The edge length of the unit cell is 408 pm.The diameter of the metal atom is : [2012]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 6

For fcc lattice,

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 7

A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion (Y) will be : [2011 M]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 7

Radius ratio of NaCl like crystal

Radius of cation = 100 pm

Radius of anion = ?

=100/0.414= 241.5= r-

Therefore, Radius of anion = 241.5 pm

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 8

Among the following which one has the highest cation to anion size ratio? [2010]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 8
  • The cation to anion size ratio will be maximum when the cation is of the largest size and the anion is of the smallest size.
  • Among the given species, Cs+ has the maximum size in given cations and F- has the smallest size among given anions, thus CsF has the highest Rc/ Ra ratio.

 is highest in CsF

∴    Correct choice : (c)

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 9

AB; crystallizes in a body centred cubic lattice with edge length ‘ ’ equal to 387 pm . The distance between two oppositely charged ions in the lattice is : [2010]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 9

The distance between the two oppositely charged ions for BCC: 

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 10

Copper crystallises in a face-centred cubic lattice with a unit cell length of 361 pm. What is the radius of copper atom in pm? [2009]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 10

Since Cu  metal crystallises in a face centred cubic lattice

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 11

Lithium metal crystallises in a body centred cubic crystal. If the length of the side of the unit cell of lithium is 351 pm, the atomic radius of the lithium will be:[2009]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 11

Since lithium metal crystallises in a body centred cubic crystal

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 12

With which one of the following elements silicon should be doped so as to give p-type of semiconductor ? [2008]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 12
  • The semiconductors formed by the introduction of impurity atoms containing one electron less than the parent atoms of insulators are termed p-type semiconductors.
  • Therefore silicon-containing 14 electrons are to be doped with boron-containing 13 electrons to give a p-type semi conductor.
31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 13

Percentage of free space in a body centred cubic unit cell is : [2008]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 13

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 14

The fraction of total volume occupied by the atoms present in a simple cube is

[2007]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 14

Let the edge length of the cube be a and radius of the atom be r.

Now, in the simple cube, the number of atoms present =1/8​×8=1

So, the volume occupied = 1* 4/3πr3

Now, in simple cube atoms at corners will be touching each other.

a=2r

Therefore, r= a/2

The volume of cube =a3

So, a fraction of the total volume occupied = (4/3πr3) / (a3)

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 15

Which of the folloiwng anions is present in the chain structure of silicates? [2007]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 15

Chain silicates are formed by sharing two oxygen atoms with each tetrahedron.

Anions of chain silicate have two general formulas: (i) (SiO32−)n​        (ii) (Si4​O116−)n​

For example, spodumene LiAl(SiO3​)2​; enstatite MgSiO3​ are pyroxene type chain silicates.

Hence, option B is correct.

Additional Information: Spodumene LiAl (SiO3)2, Diposide CaMg(SiO3)2.

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 16

If NaCl is doped with 10– 4 mol % of SrCl2, the concentration of cation vacancies will be (NA = 6.02 × 1023 mol–1) [2007]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 16

NaCl is doped with 10-4 moles of SrCl2, i.e., 100 moles of NaCl are dropped in 10-4 moles of SrCl2.
That means, 1 mole of NaCl is dropped in 10-6 mole of SrCl2.
Thus, one Sr2+ creates one cation vacancy.
So, the number of cation vacancies = the Number of the mole of SrCl2 × Avogadro’s number
= 10-6 × 6.023 × 1023
= 6.023 × 1017

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 17

CsBr crystallises in a body centered cubic lattice.The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 and that of Br = 80 amu and Avogadro number being 6.02 × 1023 mol–1, the density of CsBr is

[2006]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 17

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 18

The appearance of colour in solid alkali metal halides is generally due to [2006]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 18
  • Alkali halides like NaCl and KCl show metal excess defects due to anionic vacancies.
  • When crystals of NaCl are heated in an atmosphere of sodium vapour the sodium atoms are deposited on the surface of the crystal.
  • The Cl- ions diffuse to the surface of the crystal and combine with Na atoms to give NaCl.
  • This happens by the loss of Valance electrons by Na atoms to form Na+ ions.
  • The released electrons diffuse into the crystal and occupy anionic sites.
  • As a result the crystal now has an excess of sodium.
  • The anionic sites occupied by unpaired electrons are called Fcentres (from the German word, Farbenzenter for colour centre).
  • They impact yellow colour to the crystals of NaCl.
  • The colour results from the excitation of these electrons when they absorb energy from the visible light falling on the crystals.
  • Similarly excess of Li makes LiCi crystals pink, and excess of K makes KCl crystal violet (or lilac).
31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 19

In face-centred cubic lattice, a unit cell is shared equally by how many unit cells? [2005]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 19

An isolated fcc cell is shown here.

  • Each face of the cell is common to two adjacent cells.
  • Therefore, each face centre atom contributes only half of its volume and mass to one cell.
  • Arranging six cells each sharing the remaining half of the face-centred atoms, constitutes fcc cubic lattice. e.g., Cu and Al.

Therefore a cubic unit cell has six faces. In a cubic lattice irrespective of its nature, a cubic unit cell is shared equally by 6 unit cells. 

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 20

A compound for med by elemen ts X an d Y crystallizes in a cubic structure in which the X atoms are at the corners of a cube and the Y atoms are at the face centres. The formula of the compound is [2004]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 20

As only 1 atom can accommodate on 1 corner of a cube,

No. of X - atoms per unit cell = 8/8 = 1

As 2 atoms can accommodate on 1 face of a cube,

No. of Y - atoms per unit cell = 6/2 = 3

Hence Compound is XY3

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 21

When molten zinc is converted into solid state, it acquires HCP structure. The number of nearest neighbours will be [2001]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 21

hcp is a closed packed arrangement in which the unit cell is hexagonal and coordination number is 12.

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 22

A compound is formed by elements A and B. The crystalline cubic structure has the A atoms at the corners of the cube and B atoms at the body centre. The simplest formula of the compound is [2000]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 22

Given:

Atoms are present in the corners of cube = A

Atom present at body centre = B.

We know that a cubic unit cell has 8 corners.

Therefore the contribution of each atom at the corner =

Since the number of atoms per unit cell is 8, therefore the total contribution

We also know that atoms are in the body centre, therefore the number of atoms per unit cell = 1.

Thus the formula of the compound is AB.

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 23

The second order Bragg diffraction of X-rays with = 1.00 Å from a set of parallel planes in a metal occurs at an angle 60º. The distance between the scattering planes in the crystal is [1998]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 23

Given :

  • Order of Bragg diffraction (n) = 2
  • Wavelength (λ) = 1 Å and angle (θ) = 60º.

We know from the Bragg’s equation nλ = 2d sin θ or 2 × 1 = 2d sin 60º

(where d = Difference between the scattering planes)

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 24

Schottky defect in crystals is observed when [1998]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 24

If in an ionic crystal of the type A+, B, an equal number of cations and anions are missing from their lattice sites so that the electrical neutrality is maintained. The defect is called Schottky defect.

Some common examples of salts where the Schottky defect is prominent include Sodium Chloride (NaCl), Potassium Chloride (KCl), Potassium Bromide (KBr), Caesium Chloride (CsCl) and Silver Bromide (AgBr).

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 25

For two ionic solids CaO and KI, identify the wrong statement amongst the following : [1997]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 25
  • KI is an ionic compound (polar compound), so, it won't be soluble in a non-polar solvent. Hence, it will not get dissolved in benzene because benzene is a non-polar solvent. 
  • Lattice energy is directly proportional to the product of charge and CaO has more charge compared to KI.
  • Lattice energy leads to a high melting point.

Therefore, the correct answer is C. 

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 26

A solid with high electrical and thermal conductivity from the following is [1994]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 26

Out of the given substances, only Li has high electrical and thermal conductivity.

Li being an alkali metal has high electrical and thermal conductivity due to the presence of one electron in the outermost orbit which is readily available for delocalisation and contribution.

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 27

When electrons are trapped into the crystal in anion vacancy, the defect is known as : [1994]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 27

When electrons are trapped in ion vacancies, these are called F-centres.

The lattice sites containing the electrons trapped in the anion vacancies are called F-centres because they are responsible for imparting colour to the crystals (F=Farbe which is a German word for colour).

The presence of F - centres in a crystal make it

31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 28

The pure crystalline substance on being heated gradually first forms a turbid liquid at constant temperature and still at higher temperature turbidity completely disappears. The behaviour is a characteristic of substance forming [1993]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 28
  • Liquid crystals are matter in a state that has properties between those of conventional liquid and those of solid crystal.
  • For instance, a liquid crystal may flow like a liquid but its molecules may be oriented in a crystal-like way.
31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 29

In graphite electrons are : [1993]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 29

In graphite, the electrons are spread out between the sheets.

Explanation: 

  • Each carbon atom uses three of its electrons to form simple bonds with its three close neighbours.
  • That leaves the fourth electron in the bonding level.
  • These "spare" electrons in each carbon atom become delocalized over the whole of the sheet of atoms in one layer.
  • They are no longer associated directly with any particular atom or pair of atoms but are free to wander throughout the whole sheet. 
31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 30

On doping Ge metal with a little of In or Ga, one gets[1993]

Detailed Solution for 31 Years NEET Previous Year Questions: The Solid State (Old NCERT) - Question 30

p-type semiconductors are produced

(a) due to metal deficiency defects

(b) by adding impurity containing fewer electrons (i.e. atoms of group 13)

Ge belongs to Group 14 and In to Group 13.

Hence on doping, a p-type semiconductor is obtained.

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