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

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

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

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
N_A = 6.02 x 10²³ mol^-1
Molar mass of the metal, M =?
We know that

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

Therefore, the number of O^2- 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.

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. 

For fcc lattice,

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

• 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 R_c/ R_a ratio.

 is highest in CsF

∴    Correct choice : (c)

The distance between the two oppositely charged ions for BCC: 

Since Cu  metal crystallises in a face centred cubic lattice

Since lithium metal crystallises in a body centred cubic crystal

• 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.

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)

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

Anions of chain silicate have two general formulas: (i) (SiO₃​²⁻)n​        (ii) (Si_4​O₁₁​⁶⁻)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.

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

• 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 F^- centres (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).

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. 

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

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

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.

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)

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).

• 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. 

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.

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).

• 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.

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. 

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.