NCERT Exemplar Class 12 Chemistry Solutions Chapter 1 Solid State

Question 1 Which of the following conditions favours the existence of a substance in the solid state?
(i) High temperature
(ii) Low temperature
(iii) High thermal energy
(iv) Weak cohesive forces

Answer:

The answer is option (ii).

At low temperature, the substance exists in solid state due to low thermal energy and hence decreased molecular motion, which in turn leads to strong intermolecular cohesive i.e., forces, which hold the constituent particles together. Hence, option b is correct.

Question 2 Which of the following is not a characteristic of a crystalline solid?
(i) Definite and characteristic heat of fusion.
(ii) Isotropic nature.
(iii) A regular periodically repeated pattern of arrangement of constituent particles in the entire crystal.
(iv) A true solid
Answer:

The answer is the option (ii).

Anisotropy: Crystalline solids are anisotropic in nature. In Anisotropic substances, physical properties like electrical resistance or refractive index show different values when measured along different directions in the same crystal. This is primarily due to different arrangement of particles in different directions arrangement of particles along different directions



Isotropy: In isotropic substances, physical properties like electrical conductivity, refractive index etc. show the same value when measured in different directions just like in glass or liquid. Amorphous substances exhibit Isotropy.

Question 3 Which of the following is an amorphous solid?
(i) Graphite (C)
(ii) Quartz glass $(Si{O}_2)$
(iii) Chrome alum
(iv) Silicon carbide (SiC)

Answer:

The answer is option (ii).

${\mathrm{SiO}}_2$ is an amorphous solid. ${\mathrm{SiO}}_2$ has a regular and repeating arrangement of atoms, ${\mathrm{SiO}}_2$ lacks long-range order in its atomic structure. Due to this disordered arrangement, it is classified as an amorphous solid. It does not exhibit a sharp melting point and behaves more like a supercooled liquid, which is a typical characteristic of amorphous substances.

Question 4 Which of the following arrangements shows the schematic alignment of magnetic moments of antiferromagnetic substances?

Answer:

The answer is the option (iv).
Antiferro-magnetic moment arises when net dipole is zero. In this, the dipoles are aligned in equal and opposite direction to cancel out each other. For example: MnO, FeO, CoO, NiO

Question 5 Which of the following is true about the value of the refractive index of quartz glass?
(i) Same in all directions
(ii) Different in different directions
(iii) Cannot be measured
(iv) Always zero
Answer:

The answer is the option (i).

Quartz glass is an amorphous solid. Thus it will exhibit an identical value of refractive index in all the directions.

Question 6 Which of the following statement is not true about amorphous solids?
(i) On heating, they may become crystalline at a certain temperature.
(ii) They may become crystalline on keeping for a long time.
(iii) Amorphous solids can be moulded by heating.
(iv) They are anisotropic in nature.
Answer:

The answer is the option (iv).

Amorphous solids are isotropic because they show physical, thermal and optical properties that are identical in all directions.

Question 7 The sharp melting point of crystalline solids is due to ___________.
(i) a regular arrangement of constituent particles observed over a short distance in the crystal lattice.
(ii) a regular arrangement of constituent particles observed over a long distance in the crystal lattice.
(iii) same arrangement of constituent particles in different directions.
(iv) different arrangement of constituent particles in different directions.
Answer:

The answer is the option (ii).

A solid is said to be crystalline whose constituents such as atoms, molecules or ions are arranged in highly ordered microscopic structure.
Crystalline solids have sharp melting point is due to this regular arrangement of constituent particles over a long distance in crystal lattice.

Question 8 Iodine molecules are held in the crystals lattice by ____________.
(i) London forces
(ii) Dipole-dipole interactions
(iii) Covalent bonds
(iv) Coulombic forces
Answer:

The answer is the option (i).

$I_2$ is a molecular solid. Molecular solids are held together by the Dispersion or London forces and have very low melting point. Some other examples are $Ar,C{O}_2,H_2.$

Question 9 Which of the following is a network solid?
(i) $S{O}_2$ (Solid)
(ii) $I_2$
(iii) Diamond
(iv)$H_{2}O$(Ice)
Answer:

The answer is the option (iii).

Diamond has a 3 dimensional tetrahedral arrangement of carbon atoms. Every carbon atom is bonded to 4 other carbon atoms.

Question 10 Which of the following solids is not an electrical conductor?
(A) Mg (s)

(B) TiO (s)

(C) $I_2$ (s)

(D) $H_{2}O$ (s)
(i) (A) only
(ii) (B) Only
(iii) (C) and (D)
(iv) (B), (C) and (D)
Answer:

The answer is the option (iii).
In $I_2$ molecules are soft and non-conductor of electricity.
Hydrogen is covalently bonded to more electronegative oxygen atom in water. Due to non-ionic nature, they do not dissociate in ions and cannot conduct electricity.

Question 11 Which of the following is not the characteristic of ionic solids?
(i) A very low value of electrical conductivity in the molten state.
(ii) Brittle nature.
(iii) Very strong forces of interactions.
(iv) Anisotropic nature.
Answer:

The answer is the option (i).

A very low value of electrical conductivity in the molten state. Ionic solids are hard but brittle in appearance. They have high melting point and act as insulators in solid state but conductors in molten state or aqueous solutions. They have very strong forces of attraction. Hence, option A is incorrect.

Question 12 Graphite is a good conductor of electricity due to the presence of __________.
(i) lone pair of electrons
(ii) free valence electrons
(iii) cations
(iv) anions
Answer:

The answer is the option (ii).

Every carbon atom is bonded to 3 other carbon atoms in graphite by covalent bonds. One valence electron of each carbon atom is free. Graphite is a good conductor of electricity due to the presence of this free electron.

Question 13 Which of the following oxides behaves as conductor or insulator depending upon temperature?
(i) TiO
(ii) $Si{O}_2$
(iii)$Ti{O}_3$
(iv) MgO
Answer:

The answer is the option (iii).
$Ti{O}_3$ behaves as conductor or insulator based on temperature. The variation of energy gap between valence band and conduction band with the variation of temperature plays an important role in it. When heated, electrons gain energy which helps them to cross this band gap.

Question 14 Which of the following oxides shows electrical properties like metals?
(i) $Si{O}_2$
(ii) MgO
(iii) $S{O}_2$(s)
(iv) $Cr{O}_2$
Answer:

The answer is the option (iv).

$Cr{O}_2$, TiO and$Re{O}_3$ are some metal oxides that exhibit electrical properties. $S{O}_2$ and MgO do not show electrical properties.

Question 15 The lattice site in a pure crystal cannot be occupied by _________.
(i) molecule
(ii) ion
(iii) electron
(iv) atom
Answer:

The answer is the option (iii).
Pure crystals have constituents i.e., atoms, molecules, or ions arranged in a highly ordered microscopic structure in fixed stoichiometric ratio. Lattice sites can be occupied by electron only when there is imperfection in solid. They do not occur in pure crystals. Hence, the lattice site in a pure crystal cannot be occupied by electron. It is feasible only in case of imperfection giving rise to Crystal defects.

Question 16 Graphite cannot be classified as __________.
(i) conducting solid
(ii) network solid
(iii) covalent solid
(iv) ionic solid
Answer:

The answer is the option (iv).

Graphite has a covalent structure wherein each C atom is bonded to the 3 remaining Carbon atoms by covalent bonds. Despite being a covalent solid, its electrical conductivity is very high.

Question 17 Cations are present in the interstitial sites in __________.
(i) Frenkel defect
(ii) Schottky defect
(iii) Vacancy defect
(iv) Metal deficiency defect
Answer:

The answer is the option (i).
Frenkel effect is when a cation is missing from the normal lattice position and is occupying interstitial site. It generally occurs in Ionic solids where the anion is larger than the cation.

Question 18 Schottky defect is observed in crystals when __________.
(i) some cations move from their lattice site to interstitial sites.
(ii) the equal number of cations and anions are missing from the lattice.
(iii) some lattice sites are occupied by electrons.
(iv) some impurity is present in the lattice.
Answer:

The answer is the option (ii).
Schottky defect is when a crystal has equal number of cations and anions missing from their normal lattice site by creating vacancies or holes. In this case, the electrical neutrality is maintained.

Question 19 Which of the following is true about the charge acquired by p-type semiconductors?
(i) positive
(ii) neutral
(iii) negative
(iv) depends on the concentration of p impurity
Answer:

The answer is the option (ii).
p-Type semiconductors have no charge and are neutral. They conduct electricity through positive holes.

Question 20 To get a n-type semiconductor from silicon, it should be doped with a substance with valence__________.
(i) 2
(ii) 1
(iii) 3
(iv) 5
Answer:

The answer is the option (iv).
To get n-type semiconductor impurity of higher group is doped. Thus, to get a n-type semiconductor, silicon (valency = 4) should be doped with the element with valency equal to 5.

Question 21 The total number of tetrahedral voids in the face-centred unit cell is __________.
(i) 6
(ii) 8
(iii) 10
(iv) 12
Answer:

The answer is the option (ii).

Fee unit cell contains 8 tetrahedral voids which are located midway between each corner and centre of unit cell. The coordination number in face centred cubic is 12.

Question 22 Which of the following point defects are shown by AgBr(s) crystals?
(A) Schottky defect
(B) Frenkel defect
(C) Metal excess defect
(D) Metal deficiency defect
(i) (A) and (B)
(ii) (C) and (D)
(iii) (A) and (C)
(iv) (B) and (D)
Answer:

The answer is the option (i).
Silver bromide shows Schottky as well as Frenkel defects. In AgBr, both the silver cation and bromide ion are absent from the lattice causing Schottky defect. But the silver ions are smaller in size and mobile, so they have a tendency to occupy interstitial sites when removed from lattice point. Hence cause Frenkel defect. ‘

Question 23 In which pair most efficient packing is present?
(i) hcp and bcc
(ii) hcp and ccp
(iii) bcc and ccp
(iv) bcc and simple cubic cell
Answer:

The answer is the option (ii).
Packing efficiency is defined as the percentage of space occupied by constituent particles in a unit cell.

$\text{Packing efficiency}=\frac{\text{Volume occupied by four spheres in the unit cell}}{\text{Total volume of unit cell}}\times 100$

Packing efficiency of various types of unit cell are:

Question 24 The percentage of empty space in a body centred cubic arrangement is ________.
(i) 74
(ii) 68
(iii) 32
(iv) 26
Answer:

The answer is the option (iii).
Packing efficiency for bcc arrangement is 68%, which means that out of 100cc (let us assume) 68 cc of volume is occupied. So (100- 68) = 32 cc is empty space which gives us 32/100 i.e. 32%

Question 25 Which of the following statement is not true about the hexagonal close packing?
(i) The coordination number is 12.
(ii) It has 74% packing efficiency.
(iii) Tetrahedral voids of the second layer are covered by the spheres of the third layer.
(iv) In this arrangement spheres of the fourth layer are exactly aligned with those of the first layer.
Answer:

The answer is the option (iv).
Hexagonal packing has layers arranged in ABAB pattern. The spheres of first layer are aligned with the third layer and not the fourth layer.
The 1st layer and 4th layer are not exactly aligned. Thus, statement (d) is not correct while other statements (a), (b) and (c) are true.

Question 26 In which of the following structures coordination number for cations and anions in the packed structure will be same?
(i) $C{l}^{-}$ ion form fcc lattice and $N{a}^{+}$ ions occupy all octahedral voids of the unit cell.
(ii) $C{a}^{2+}$ ions form fcc lattice and $F^{-}$ ions occupy all the eight tetrahedral voids of the unit cell.
(iii) $O^{2-}$ions form fcc lattice and $N{a}^{+}$ ions occupy all the eight tetrahedral voids of the unit cell.
(iv) $S^{2-}$ ions form fcc lattice and Zn²⁺ ions go into alternate tetrahedral voids of the unit cell.
Answer:

The answer is the option (i).
In NaCl (Rock salt) structure, the anions have face centered cubic arrangement and all the octahedral voids of given unit cell are occupied by the cation $N{a}^{+}$
Where, C.N of $N{a}^{+}$ = 6
C.N of $C{l}^{-}$= 6

Question 27 What is the coordination number in a square close-packed structure in two dimensions?
(i) 2
(ii) 3
(iii) 4
(iv) 6

Answer:

The answer is the option (iii).

Coordination number in a square closed packed structure in two dimensions is equal to 4.

Question 28 Which kind of defects are introduced by doping?
(i) Dislocation defect
(ii) Schottky defect
(iii) Frenkel defects
(iv) Electronic defects
Answer:

The answer is the option (iv).
Doping introduces electronic defect. Electronic defect occurs due to the addition of electron rich or electron deficient impurity to a perfect crystal. There are 2 types of dopants- n-type and p-type.

Question 29 Silicon doped with electron-rich impurity forms ________.
(i) p-type semiconductor
(ii) n-type semiconductor
(iii) intrinsic semiconductor
(iv) insulator
Answer:

The answer is the option (ii).

Silicon has four valence electrons. The conductivity increases on doping with an electron-rich impurity as the extra electron becomes delocalised. The increase in conductivity is due to negatively charged electrons. That is why silicon doped with electron-rich impurity is called an n-type semiconductor.

Question 30 Which of the following statements is not true?
(i) Paramagnetic substances are weakly attracted by the magnetic field.
(ii) Ferromagnetic substances cannot be magnetised permanently.
(iii) The domains in antiferromagnetic substances are oppositely oriented with respect to each other.
(iv) The pairing of electrons cancels their magnetic moment in the diamagnetic substances.
Answer:

The answer is the option (ii).
Ferromagnetic materials such as iron are strongly attracted to the magnetic field and can be permanently magnetised. Hence, choice (ii) is not true, while the other three choices are correct.

Question 31 Which of the following is not true about the ionic solids?
(i) Bigger ions form the close-packed structure.
(ii) Smaller ions occupy either the tetrahedral or the octahedral voids depending upon their size.
(iii) Occupation of all the voids is not necessary.
(iv) The fraction of octahedral or tetrahedral voids occupied depends upon the radii of the ions occupying the voids.
Answer:

The answer is the option (iv).

The fraction of Octahedral or tetrahedral voids is dependent on the radii of the ions occupying the voids. Hence, option (iv) is not true about ionic solids.

Question 32 A ferromagnetic substance becomes a permanent magnet when it is placed in a magnetic field because ________.
(i) all the domains get oriented in the direction of the magnetic field.
(ii) all the domains get oriented in the direction opposite to the direction of the magnetic field.
(iii) domains get oriented randomly.
(iv) domains are not affected by the magnetic field.
Answer:

The answer is option (i).
A ferromagnetic substance becomes a permanent magnet when placed in a magnetic field as the domains get aligned in the direction of the applied external magnetic field.

Question 33 The correct order of the packing efficiency in different types of unit cells is ________.
(i) fcc < bcc < simple cubic
(ii) fcc > bcc > simple cubic
(iii) fcc < bcc > simple cubic
(iv) bcc < fcc > simple cubic

Answer:

The answer is option (ii).

The correct order of the packing efficiency in different types of unit cells is as: fcc (74%) > bcc (68%) > simple cubic (52%).

Question 34 Which of the following defects is also known as dislocation defect?
(i) Frenkel defect
(ii) Schottky defect
(iii) Non-stoichiometric defect
(iv) Simple interstitial defect
Answer:

The answer is option (i).
In the Frenkel defect, some cations create a vacancy and occupy an interstitial site. That is why it is also called a dislocation defect.

Question 35 In the cubic close packing, the unit cell has ________.
(i) 4 tetrahedral voids, each of which is shared by four adjacent unit cells.
(ii) 4 tetrahedral voids within the unit cell.
(iii) 8 tetrahedral voids, each of which is shared by four adjacent unit cells.
(iv) 8 tetrahedral voids within the unit cells.
Answer:

The answer is option (iv).

In the cubic close packing, the unit cell has 8 tetrahedral voids within the unit cell.

Question 36 The edge lengths of the unit cells in terms of the radius of spheres constituting fcc, bcc and simple cubic unit cell are respectively________

(i) $2\sqrt{2}r$, $\frac{4r}{\sqrt{3}}$, 2r

(ii) $\frac{4r}{\sqrt{3}},2\sqrt{2}r,2r$

(iii) $2r,2\sqrt{2}r,\frac{4r}{\sqrt{3}}$

(iv) $2r,\frac{4r}{\sqrt{3}},2\sqrt{2}r$
Answer:

The answer is option (i).

Distance between two atoms is always measured from their centres. (i) If the crystal lattice consists of SCC, the atoms which is present at the corners touch each other

(ii) In the case of FCC, the atom present at the corner and the centre of the face touch each other.

(iii)In case of BCC atom present at the corner and centre of the body, touch each other
Edge lengths for fcc, bcc and scc types of unit cell are $2\sqrt{2}r,\frac{4}{\sqrt{3}}r$ and 2r respectively.

Question 37 Which of the following represents correct order of conductivity in solids?
(i) K _metals >> K _insulators< K_{semiconductors}
(ii) K_metals << K_insulators < K_{semiconductors}
(iii) K_metals$\simeq$ K_{semiconductors}> K_insulators = zero
(iv) K_metals < K_{semiconductors} >K _insulators ≠ zero
Answer:

The answer is the option (i).
The relative values of conductivity of solid is represented in terms of k. The value for k for metals >> semiconductor>> Insulator

NCERT Exemplar Class 12 Chemistry Solutions Chapter 1: MCQ (Type 2)

MCQ-type questions are covered to improve your conceptual thinking and problem-solving ability:

Question 38 Which of the following is not true about the voids formed in 3-dimensional hexagonal close-packed structure?
(i) A tetrahedral void is formed when a sphere of the second layer is present above triangular void in the first layer.
(ii) All the triangular voids are not covered by the spheres of the second layer.
(iii) Tetrahedral voids are formed when the triangular voids in the second layer lie above the triangular voids in the first layer and the triangular shapes of these voids do not overlap.
(iv) Octahedral voids are formed when the triangular voids in the second layer exactly overlap with similar voids in the first layer
Answer:

The answer is the option (iii, iv).

Tetrahedral voids are formed when the triangular void in the second layer lie exactly above the triangular voids in the first layer and the triangular shape of these voids oppositely overlap.
Octahedral voids are formed when triangular void of second layer is not exactly overlapping with similar void in first layer.

Question 39 The value of the magnetic moment is zero in the case of antiferromagnetic substances because the domains ________.
(i) get oriented in the direction of the applied magnetic field.
(ii) get oriented opposite to the direction of the applied magnetic field.
(iii) are oppositely oriented with respect to each other without the application of the magnetic field.
(iv) cancel out each other’s magnetic moment.
Answer:

The answer is the option (iii, iv).

In the case of antiferromagnetic substances like manganese oxide, the magnetic moment becomes zero because the magnetic moments of atoms align in a pattern with neighbouring spins pointing in opposite directions, which cancel out each other, resulting in a 0 net dipole moment.

Question 40 Which of the following statements are not true?
(i) A vacancy defect results in a decrease in the density of the substance.
(ii) Interstitial defects increase the density of the substance.
(iii) Impurity defect does not affect the density of the substance.
(iv) Frankel defect results in an increase in the density of the substance.
Answer:

The answer is the option (iii, iv).
Impurity defect involves the replacement of ions by an impurity which may have a different density than the ion present on perfect crystal. Frenkel defect results in neither a decrease nor an increase in the density of the substance as it involves only the migration of ions within the crystal.

Question 41 Which of the following statements are true about metals?
(i) Valence band overlaps with conduction band.
(ii) The gap between the valence band and the conduction band is negligible.
(iii) The gap between the valence band and the conduction band cannot be determined.
(iv) Valence band may remain partially filled.
Answer:

The answer is the option (i, ii, iv).

The electrical conduction through metals is dependent on the type of valence band and its gap with the conduction band. This gap determines the conducting property of metal.

Question 42 Under the influence of the electric field, which of the following statements is true about the movement of electrons and holes in a p-type semiconductor?
(i) Electron will move towards the positively charged plate through electron holes.
(ii) Holes will appear to be moving towards the negatively charged plate.
(iii) Both electrons and holes appear to move towards the positively charged plate.
(iv) Movement of electrons is not related to the movement of holes.
Answer:

The answer is the option (i, ii).

In p-type semiconductor, holes act as the charge carriers. On exposure of P-type semiconductor to electric field, positively charged hole starts moving towards negatively charged plate while the electron towards positively charged plate.

Question 43 Which of the following statements are true about semiconductors?
(i) Silicon doped with an electron-rich impurity is a p-type semiconductor.
(ii) Silicon doped with an electron-rich impurity is an n-type semiconductor.
(iii) Delocalised electrons increase the conductivity of doped silicon.
(iv) An electron vacancy increases the conductivity of the n-type semiconductor.
Answer:

The answer is the option (ii, iii).

Silicon (valence electron – 4) doped with electron rich impurity is an n-type semiconductor due to extra electron. The conductivity of doped siliconis increased by the delocalised electrons.

Question 44 An excess of potassium ions makes KCl crystals appear violet or lilac in colour since ________.
(i) some of the anionic sites are occupied by an unpaired electron.
(ii) some of the anionic sites are occupied by a pair of electrons.
(iii) there are vacancies at some anionic sites.
(iv) F-centres are created which impart colour to the crystals.
Answer:

The answer is the option (i, iv).
When KCl is heated in vapour of K, anionic vacancies are created as some of the $C{l}^{-}$ ions leave their lattice site. This chloride ion wants to combine with K vapour to form KCl. For this to happen, loss of electron by potassium atom to form potassium ion occurs. The unpaired free electron gets entrapped in the anion vacancy called F-centre that are responsible for imparting color to crystals. This entrapped electron gains energy due to the visible light falling on the crystal goes to the higher level and when it comes back to the ground state, energy is released in the form of light.

Question 45 The number of tetrahedral voids per unit cell in NaCl crystal is ________.
(i) 4
(ii) 8
(iii) twice the number of octahedral voids.
(iv) four times the number of octahedral voids.
Answer:

The answer is the option (ii, iii).

NaCl has FCC arrangement
Number of atoms in packing per unit cell = 4
Number of tetrahedral voids per unit cell = 2 $\times$ No. of particles present in close packing
$=2\times 4=8$
Number of tetrahedral voids per unit cell = 2 $\times$ No. of octahedral voids
Hence, option ii and iii are correct.

Question 46 Amorphous solid can also be called ________.
(i) pseudo solids
(ii) true solids
(iii) supercooled liquids
(iv) supercooled solids
Answer:

The answer is the option (i, iii).

Amorphous solid has short-range order which has a tendency to flow very slowly. Thus, although they are solids, they resemble liquid in many respects. Therefore amorphous solid can also be called as pseudo solids or super cooled liquids. Some examples of amorphous solid includes rubber, glass, plastic and gels.

Question 47 A perfect crystal of silicon (Figure) is doped with some elements as given in the options. Which of these options show n-type semiconductors?

Answer:

The answer is the option (i, iii).

N-type semiconductor are formed when a crystal is doped with pentavalent impurity elements i.e. elements with 5 electrons in valence shell.
In option a and c, the crystal has been doped with elements of group 15 like P and As.

Question 48 Which of the following statements are correct?
(i) Ferrimagnetic substances lose ferrimagnetism on heating and become paramagnetic.
(ii) Ferrimagnetic substances do not lose ferrimagnetism on heating and remain ferrimagnetic.
(iii) Antiferromagnetic substances have domain structures similar to ferromagnetic substances and their magnetic moments are not cancelled by each other.
(iv) In ferromagnetic substances, all the domains get oriented in the direction of the magnetic field and remain as such even after removing magnetic field.
Answer:

The answer is the option (i, iv).

There is a loss of ferrimagnetism of Ferrimagnetic substances on heating changing them into paramagnetic.
All the domains get oriented in the direction of magnetic field in ferromagnetic substances and remain as such even on removal of the magnetic field.

Question 49 Which of the following features are not shown by quartz glass?
(i) This is a crystalline solid.
(ii) Refractive index is the same in all the directions.
(iii) This has a definite heat of fusion.
(iv) This is also called supercooled liquid
Answer:

The answer is the option (i, iii).

Since quartz glass is an amorphous solid, it has no definite heat of fusion. This is due to short-range order of molecule. Also known as super cooled liquid, quartz glass is isotropic in nature.

Question 50 Which of the following cannot be regarded as molecular solid?
(i) SiC (Silicon carbide)
(ii) AlN
(iii) Diamond
(iv) $I_2$
Answer:

The answer is the option (i, ii, iii).

Silicon carbide, AIN and diamond are examples of network solid instead of molecular solids as they have three-dimensional structure while; $I_2$ is a molecular solid, because such solid particles are held together by dipole-dipole interactions. SiC and AIN are interstitial solids.

Question 51 In which of the following arrangements octahedral voids are formed?
(i) hcp
(ii) bcc
(iii) simple cubic
(iv) fcc
Answer:

The answer is the option (i, iv).

In hcp and FCC arrangement, octahedral voids are formed. In an fcc unit cell, the octahedral voids are observed at edge centre and body centre whereas in bcc and simple cubic, no octahedral voids are observed. Cubic voids are formed in bcc arrangement.

Question 52 Frenkel defect is also known as ________.
(i) stoichiometric defect
(ii) dislocation defect
(iii) impurity defect
(iv) non-stoichiometric defect
Answer:

The answer is the option (i, ii).

Dislocation of cations takes place from one place to another in frenkel defect and the stoichiometry of the crystal remains unchanged.

Question 53 Which of the following defects decrease the density?
(i) Interstitial defect
(ii) Vacancy defect
(iii) Frankel defect
(iv) Schottky defect
Answer:

The answer is the option (ii, iv).

Vacancy and Schottky defect lead to decrease the density. In Vacancy defect, an atom goes missing from one of the lattice site thus decreasing the density. In Schottky defect, an atom moves from inside of crystal to its surface. The density remains the same in case of Frenkel defect and interstitial defect.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 1: Short Answer Type

Short-answer type questions are covered to improve your problem solving ability.

Question 54 Why are liquids and gases categorised as fluids?
Answer:

The liquids and gases have an ability to flow. Liquid and gas molecules can easily move fast and tumble over one another freely. These have been categorized as fluids because of their tendency to flow.

Question 55 Why are solids incompressible?
Answer:

The particles in solid are already packed closely together. There will be a large repulsive force between electron clouds of these particles if brought closer. Therefore, solids are incompressible.

Question 56 Inspite of long range order in the arrangement of particles why are the crystals usually not perfect?
Answer:

Despite of having long range in the arrangement of particles, crystals are not perfect. The reason behind this is that the constituent particles may not get enough time to arrange themselves in a perfect order when crystallisation occurs at a fast rate or moderate rate.

Question 57 Why does table salt, NaCl, sometimes appear yellow in colour ?
Answer:

The yellow colour of NaCl crystals can be attributed to metal excess defect as some chlorine ions leave their lattice sites and are occupied by unpaired electrons. These sites are called F-centres. On absorption of energy, electrons gets excited and falls on the crystals imparting yellow color.

Question 58 Why is FeO (s) not formed in stoichiometric composition?
Answer:

Iron oxide (FeO) has rock salt structure.
In the crystals of FeO, $O_2$ ions adopt on sites.
FeO is always non-stoichiometric i.e., the composition of $F{e}^{2+}$ and $O_2$ ions is 0.95 : 1 instead of 1:1 .
This non stoichiometric composition can be obtained if a small number of $F{e}^{2+}$ ions are replaced by $\frac{2}{3}$ of $F{e}^{3+}$ ions in OH sites to maintain the positive charge.

Question 59 Why does white ZnO (s) become yellow upon heating?
Answer:

On heating, ZnO gives $Z{n}^{2+}$, electrons and colour.
The excess Zn ions thus formed move to the interstitial sites and electron in the neighbourhood vacant interstitial sites for neutralisation. This electron is responsible for the yellow colour and electrical conductivity in crystals. The color reverts back to white on cooling.

Question 60 Why does the electrical conductivity of semiconductors increase with a rise in temperature?
Answer:

The energy gap between valence and conduction band is small in semiconductors. Thus, they do not conduct electricity at normal temperature but with a rise in temperature large number of electrons get sufficient energy to jump from valence band to conduction band. Therefore, the electrical conductivity of a semi-conductor will increase exponentially with an increase in temperature.

Question 61 Explain why does conductivity of germanium crystals increase on doping with gallium?
Answer:

p-type semiconductor:


When germanium is doped with gallium or any other group 13 elements ,some of the positions of lattice of germanium atom are occupied by gallium. Gallium contains only 3 valence electrons. Thus the 4th valency of nearby germanium atom is not satisfied leaving this place vacant. Electron from neighbouring atom comes and fills the gap and creates a hole in its original position.
The electrons move towards positively charged plates through these under the influence of electric field for conduction of electricity. The holes appear to move towards negatively charged plates.

Question 62 In a compound, nitrogen atoms (N) make cubic close packed lattice and metal atoms (M) occupy one-third of the tetrahedral voids present. Determine the formula of the compound formed by M and N?
Answer:

Let the no. of atoms per unit cell be n. For ccp lattice the value of n is 4
Number of tetrahedral voids generated $=2\times n=2\times 4=8$
We know that only one-third of tetrahedral voids are occupied by metal M.
Number of metal atoms $M=13\times 8$
$M:N=83:4=2:3$
Thus formula is $M_2{N}_3$

Question 63 Under which situations can an amorphous substance change to crystalline form?
Answer:

To convert from amorphous to crystalline, one must heat it till its melting point and then quickly cool it down rapidly. Heating beyond its crystallization temperature will lead to recrystallization.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 1: Matching Type

Matching-type questions are covered to improve conceptual clarity and topic awareness:

Question 64 Match the defects given in Column I with the statements given in Column II.

Answer:

(i) $⟶$ (c); (ii) $⟶$ (a); (iii) $⟶$(d); (iv) $⟶$(b)

Vacancy defect occurs when some of the lattice sites are vacant resulting in decrease in density. In case of Schottky defect, an equal number of cations and anions are missing from their normal lattice site by creating vacancies or holes.
In case of Frenkel defect, the density remains same as the cation missing from the normal lattice position occupies interstitial site. Thus, maintaining the same density. In Interstitial defect, some of the constituent atoms occupy the interstitial site. Hence, no change in density.

Question 65 Match the type of unit cell given in Column I with the features given in Column II.

Answer:

(i) $⟶$(b, c); (ii) $⟶$(c, d); (iii) $⟶$(c, e); (iv) $⟶$ (a, d)
(i) In primitive unit cell each of the 8 corners contribute 1/8th to the unit cell
Total number of atoms per unit cell = $\frac{1}{8}\times 8=1$

(ii) Body centered cubic unit cell contains atoms at corner as well as body centre.
each of the 8 corners contribute 1/8th to the unit cell and the contribution due to the atom at the centre of the body=1

(iii) In face centred unit cell, total constituent ions per unit cell present at corners $=\frac{1}{8}\times 8=1$
Total contribution of atoms per unit cell present at face centre $=\frac{1}{2}\times 6=3$
(iv) For end centred orthorhombic unit cell $a\ne b\ne c$
each of the 8 corners contribute 1/8th to the unit cell $=\frac{1}{8}\times 8=1$
whereas at end centre the total contribution is $=\frac{1}{2}\times 2=1$
Thus, other than corner it contains total one atom per unit cell.

Question 66 Match the types of defect given in Column I with the statement given in Column II.

Answer:

i) $⟶$ (c); (ii) $⟶$ (a); (iii) $⟶$ (b)
(A) (i) Impurity defects: The defects introduced in the crystal lattice due to presence of the certain impurity are called impurity defects.
Example: Substitution of $N{a}^{+}$ions in NaCl by $S{r}^{2+}$ ions.
Structure with defect:

Impurity defect due to substitution of $N{a}^{+}$ ions in NaCl by $S{r}^{2+}$ ions (Cation vacancy) ‘Schottky Defect’

(B) On heating NaCl in vapour of sodium, some of the Cl ions leave their lattice site and create anion vacancies. For chloride ions to combine with sodium vapour to form sodium chloride, sodium atom loses electrons to form $N{a}^{+}$ ions. This released electron on diffusion into the crystal gets entrapped in the anion vacancy called F-centre.

(C) Metal deficiency is caused when a cation is missing from its lattice site and a nearby metal ion acquires 2 positive charge to maintain electrical neutrality.
This type of defect can be seen in compounds where metal exhibit variable valency.
Example: FeO, FeS, NiO

Question 67 Match the items given in Column I with the items given in Column II.

Answer:

$(i)\to (d);(ii)\to (c);(iii)\to (b);(iv)\to (a)$
(i) electronic conductivity is shown by Mg in solid state due to presence of free electrons.
(ii) electrolytic conductivity is shown by $MgC{l}_2$ in molten state due to the presence of electrolytes in molten state.
(iii) Silicon doped with phosphorus is p-type semiconductor. It contain one extra electron due to which it shows conductivity under the influence of electric field.
(iv) Germanium doped with boron is p-type semiconductor. It shows conductivity under the influence of electric field due to the presence of a hole.

Question 68 Match the type of packing given in Column I with the items given in Column II.

Answer:

(i) $⟶$ (c); (ii) $⟶$ (a); (iii) $⟶$ (d); (iv) $⟶$ (b)
(i) In Square close packing in two dimensions, each sphere has coordination number 4, as shown in the diagram below.

(ii)In hexagonal close packing, in two dimensions, each sphere has coordination number 6 and a triangular void is created.

(iii) hcp in 3 dimensions is a repetitive pattern of ABAB…..

(iv) In ccp spheres are repeated at every 4th layer as in ABCABCA….

NCERT Exemplar Class 12 Chemistry Solutions Chapter 1: Assertion and Reason Type

Assertion and Reason type questions are covered to improve your conceptual thinking and problem-solving solving ability

Question 69 In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices.
(a) Assertion and Reason both are correct statements and Reason is the correct explanation for Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation for Assertion.
(c) Assertion is correct but Reason is wrong.
(d) Assertion is wrong but Reason is correct.
Assertion (A): The total number of atoms present in a simple cubic unit cell is one.
Reason (R): Simple cubic unit cell has atoms at its comers, each of which is shared between eight adjacent unit cells.
Answer:

The answer is the option (a) Assertion and Reason both are correct statements, and Reason is the correct explanation for Assertion.
In simple cubic unit cell since only corners are occupied by atoms, the total number of atoms is $Z=\frac{1}{8}\times 8=1$ and Simple cubic unit cell has atoms at its comers, each of which is shared between eight adjacent unit cells.

Question 70 In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices.
(a) Assertion and Reason both are correct statements and Reason is the correct explanation for Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation for Assertion.
(c) Assertion is correct but Reason is wrong.
(d) Assertion is wrong but Reason is correct.
Assertion (A): Graphite is a good conductor of electricity, however, diamond belongs to the category of insulators.
Reason (R): Graphite is soft in nature on the other hand diamond is very hard and brittle.
Answer:

The answer is the option (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Diamond is a non-conductor of electricity because all valence electrons of carbon are involved in bonding and thus there is no delocalised electron in its structure. In graphite, one valence electron is free to move between adjacent layers making it a good conductor.
In Graphite, parallel layers are held together by week van der Waals force. That is why it is soft. However, diamond is hard since C atom is covalently bonded with the adjacent 3 carbon atoms forming a 3-dimensional network.

Question 71 In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices.
(a) Assertion and Reason both are correct statements and Reason is the correct explanation for Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation for Assertion.
(c) Assertion is correct but Reason is wrong.
(d) Assertion is wrong but the Reason is correct.
Assertion (A): Total number of octahedral voids present in unit cell of cubic close packing including the one that is present at the body centre, is four.
Reason (R): Besides the body centre, there is one octahedral void present at the centre of each of the six faces of the unit cell and each of which is shared between two adjacent unit cells.
Answer:

The answer is the option (c) Assertion is correct, but Reason is wrong.
There are 12 octahedral voids at the edge centre and 1 in the body centre.
So total number of octahedral voids present in unit cell of cubic close packing = $12\times \frac{1}{2}+1=4$

Question 72 In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices.
(a) Assertion and Reason both are correct statements and Reason is the correct explanation for Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation for Assertion.
(c) Assertion is correct but Reason is wrong.
(d) Assertion is wrong but Reason is correct.
Assertion (A): The packing efficiency is maximum for the fcc structure.
Reason (R): The coordination number is 12 in fcc structures.
Answer:

The answer is the option (b) Assertion and Reason both are correct statements but Reason is not the correct explanation for Assertion.
In FCC unit cell, there is CCP arrangement with packing efficiency of 74.01% which is maximum. In CCP arrangement, coordination number is 12.

Question 73 In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices.
(a) Assertion and Reason both are correct statements and Reason is the correct explanation for Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation for Assertion.
(c) Assertion is correct but Reason is wrong.
(d) Assertion is wrong but Reason is correct.
Assertion (A): Semiconductors are solids with conductivities in the intermediate range from ${10}^{-6}-{10}^{4}oh{m}^{-1}{m}^{-1}$

Reason (R): Intermediate conductivity in semiconductor is due to partially filled valence band.
Answer:

The answer is the option (c) Assertion is correct but Reason is wrong.
Conductance of semiconductors lies between metals and insulators, i.e., in the range of ${10}^{-6}-{10}^{4}oh{m}^{-1}{m}^{-1}$
Valence band is either filled or empty in semiconductors, it is the gas which is in between conductors and insulators.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 1: Long Answer Type

Long-answer type questions are covered to improve your subject knowledge and conceptual thinking

Question 74 With the help of a labelled diagram show that there are four octahedral voids per unit cell in a cubic close packed structure.
Answer:

In CCP, one octahedral void is present at the centre of body surrounded by six atoms at the centre of respective face.
Where N_c=Number of atoms at corner
$N_f$ =Number of atoms at face centre
= $N_c$ 'contribution+$N_f^{\prime }$contribution
$=8\times 18+6\times 12=4$
Position of octahedral voids=Edge centre and body centre
Number of octahedral voids per unit cell in cubic close packing$=Ne\times \frac{1}{4}+Nb\times 1$

$=12\times \frac{1}{4}+1\times 1$

$=4$
Number of octahedral voids=4

Question 75 Show that in a cubic close-packed structure, eight tetrahedral voids are present per unit cell.
Answer:

In ccp, each unit cell consists of eight cubic components and the number of atoms per unit cell is given by
$N_c^{\prime }$contribution+$N_f^{\prime }$ contribution
$=8\times \frac{1}{}8+6\times \frac{1}{2}=4$

The tetrahedral voids are positioned at the centre of cubic cell.
In cubic close packing number of tetrahedral voids per unit cell $=8\times 1=8$

Question 76 How does doping increase the conductivity of semiconductors?
Answer:

The conductivity of semiconductors are exceptionally low for practical use. Semiconductors are doped to generate either a surplus or deficiency in valence electrons depending on whether n-type semiconductors are required or p-type. An electron rich or electron-deficient impurity as compared to the intrinsic semiconductor, silicon or germanium is used for doping. Electronic defects are introduced in them by such impurities.

1. On doping silicon with electron rich impurities, the extra electron becomes delocalized increasing the conductivity of doped silicon. Hence silicon doped with electron-rich impurity is called n-type semiconductor due to the negatively charged electron. For ex. When group 15 elements are added to silicon

2. When silicon is doped with electron-deficit impurities to increase the conductivity through positive holes, this type of semiconductors are called p-type semiconductors. Ex. Addition of group 13 elements to group 14 element

Question 77 A sample of ferrous oxide has actual formula $F{e}_{0.93}{O}_{1.00}$. In this sample, what fraction of metal ions are $F{e}^{2+}$ ions? What is the type of non-stoichiometric defect present in this sample?
Answer:

Let the number of $O^{2-}$ be 100
Then number of $F{e}^{2+}=x$
Number of $F{e}^{3+}=93-x$
To maintain electrical neutrality net positive charge is equal to net negative charge
$2x+393-x=2\times 100$
$2x+279-3x=200$
$x=79$
$F{e}^{2+}=79$
$\frac{F{e}^{2+}}{F{e}^{2+}+F{e}^{3+}}=799$
$=0.849$
Metal deficiency defect is present in the sample because iron is in less amount than that required for stoichiometric composition.