CBSE Class 12 Chemistry Chapter 1 The Solid State Competency Based Questions Set 01

Competency Based Questions (MCQs)

Question 1. Which of the following conditions favours the existence of a substance in the solid state?
(a) High temperature
(b) Low temperature
(c) High thermal energy
(d) Weak cohesive forces
Answer: (b) Low temperature
In simple words: At lower temperatures, thermal motion is minimized, which allows powerful attractive forces to bind particles together in stable, fixed positions to form a solid.

Exam Tip: Remember that solid formation is favored when cohesive intermolecular forces are much stronger than thermal disruptive energy, which happens at cold temperatures.

Question 2. Which of the following is not a characteristic of a crystalline solid?
(a) Definite and characteristic heat of fusion.
(b) Isotropic nature.
(c) A regular periodically repeated pattern of arrangement of constituent particles in the entire crystal.
(d) A true solid
Answer: (b) Isotropic nature.
In simple words: Crystalline solids are actually anisotropic, meaning their electrical, optical, and physical properties change depending on the direction in which they are measured.

Exam Tip: Amorphous substances are isotropic because of their random layouts, whereas crystalline solids are anisotropic due to their highly ordered molecular lattices.

Question 3. Which of the following is an amorphous solid?
(a) Graphite (C)
(b) Quartz glass (SiO2)
(c) Chrome alum
(d) Silicon carbide (SiC)
Answer: (b) Quartz glass (SiO2)
In simple words: Quartz glass is an amorphous solid because its constituent particles lack a long-range orderly repeating arrangement.

Exam Tip: Be sure to distinguish between quartz (which is crystalline) and quartz glass (which is amorphous and disordered).

Question 4. Which of the following is true about the value of refractive index of quartz glass?
(a) Same in all directions
(b) Different in different directions
(c) Cannot be measured
(d) Always zero
Answer: (a) Same in all directions
In simple words: Since quartz glass is amorphous, it is isotropic, meaning its physical properties like refractive index are identical in all directions.

Exam Tip: Properties such as refractive index and electrical conductivity remain identical along any axis in any isotropic (amorphous) material.

Question 5. Which of the following statement is not true about amorphous solids?
(a) On heating they may become crystalline at certain temperature.
(b) They may become crystalline on keeping for long time.
(c) Amorphous solids can be moulded by heating.
(d) They are anisotropic in nature.
Answer: (d) They are anisotropic in nature.
In simple words: Amorphous solids are isotropic because their random, disordered arrangement of molecules looks the same in every direction.

Exam Tip: Amorphous solids are also called pseudo-solids or supercooled liquids because they can flow extremely slowly over long periods.

Question 6. The sharp melting point of crystalline solids is due to ................ .
(a) a regular arrangement of constituent particles observed over a short distance in the crystal lattice:
(b) a regular arrangement of constituent particles observed over a long distance in the crystal lattice.
(c) same arrangement of constituent particles in different directions.
(d) different arrangement of constituent particles in different directions.
Answer: (b) a regular arrangement of constituent particles observed over a long distance in the crystal lattice.
In simple words: The long-range order of a crystalline solid ensures that all intermolecular bonds break at the exact same temperature, causing it to melt sharply.

Exam Tip: Long-range order leads to sharp melting points in crystals, whereas short-range order causes amorphous materials to soften slowly over a range of temperatures.

Question 7. Iodine molecules are held in the crystals lattice by ............... .
(a) london forces
(b) dipole-dipole interactions
(c) covalent bonds
(d) coulombic forces
Answer: (a) london forces
In simple words: Iodine consists of non-polar molecules, so the forces holding them in the solid lattice are weak, temporary dispersion forces.

Exam Tip: Always categorize molecular solids based on their polarity; non-polar molecules like iodine, carbon dioxide, and helium are held by weak London dispersion forces.

Question 8. Which of the following is a network solid?
(a) SO2 (Solid)
(b) I2
(c) Diamond
(d) H2O (Ice)
Answer: (c) Diamond
In simple words: Diamond is a covalent network solid where carbon atoms are bound to each other in a continuous three-dimensional network of strong covalent bonds.

Exam Tip: Covalent network solids like diamond, quartz, and silicon carbide have exceptionally high melting points because you must break strong covalent bonds to melt them.

Question 9. Which of the following solids is not an electrical conductor?
I. Mg (s)
II. TiO (s)
III. I2 (s)
IV. H2O (s)
(a) (A) only
(b) (B) Only
(c) (C) and (D)
(d) (B), (C) and (D)
Answer: (c) (C) and (D)
In simple words: Iodine and ice are molecular solids. They do not conduct electricity because they have no free electrons or moving ions to carry electrical charge.

Exam Tip: Metallic solids (like Mg) and certain transition metal oxides (like TiO) conduct electricity, but molecular solids are strictly electrical insulators.

Question 10. Which of the following is not the characteristic of ionic solids?
(a) Very low value of electrical conductivity in the molten state.
(b) Brittle nature.
(c) Very strong forces of interactions.
(d) Anisotropic nature.
Answer: (a) Very low value of electrical conductivity in the molten state.
In simple words: When melted, the rigid crystal lattice of an ionic solid breaks down, allowing the ions to move freely and conduct electricity very well.

Exam Tip: Ionic solids are insulators in their solid form because the ions are locked in place, but they become excellent conductors when dissolved in water or melted.

Assertion-Reason Questions

Direction: Mark the option which is most suitable:
(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.

Question 1. Assertion: The total number of atoms present in a simple cubic unit cell is one.
Reason: Simple cubic unit cell has atoms at its corners, each of which is shared between eight adjacent unit cells.
Answer: (a) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion.
In simple words: In a simple cubic cell, atoms only occupy the corners. Because each corner atom is shared among eight neighboring cells, the net contribution is exactly one atom per cell.

Exam Tip: Calculate the atom count of a simple cubic unit cell using the formula: \( 8 \times \frac{1}{8} = 1 \) atom.

Question 2. Assertion: The number of NaCl units per unit cell is 2.
Reason: There are four chloride ions per unit Cell of NaCl.
Answer: (d) Assertion is wrong statement but Reason is correct statement.
In simple words: Since NaCl crystallizes in a face-centered cubic lattice, there are actually four formula units of NaCl in each unit cell, which makes the assertion false.

Exam Tip: In an FCC unit cell of NaCl, there are 4 sodium ions and 4 chloride ions, giving a total of 4 NaCl formula units per unit cell.

Question 3. Assertion: Semiconductors are solids with conductivities in the intermediate range from 10–6 – 104 ohm–1m–1.
Reason: Intermediate conductivity in semiconductor is due to partially filled valence band.
Answer: (c) Assertion is correct statement but Reason is wrong statement.
In simple words: Semiconductors have intermediate electrical flow because of a very narrow energy gap between their full valence band and empty conduction band, rather than a partially filled valence band.

Exam Tip: In semiconductors, the valence band is completely filled and the conduction band is completely empty at absolute zero; conductivity arises as temperature increases.

Question 4. Assertion: Diamond and graphite do not have the same crystal structure.
Reason: Diamond is crystalline while graphite is amorphous.
Answer: (c) Assertion is correct statement but Reason is wrong statement.
In simple words: Both diamond and graphite are crystalline forms of carbon. However, they have different structures because carbon is \( sp^3 \) hybridized in diamond and \( sp^2 \) hybridized in graphite.

Exam Tip: Remember that both diamond and graphite are crystalline allotropes of carbon; graphite is definitely not amorphous.

Question 5. Assertion: Na+ and Al3+ are isoelectronic but the magnitude of ionic radius of Al3+ is less than that of Na+.
Reason: The magnitude of an effective nuclear charge on the outer shell electrons in Al3+ is greater than that of Na+.
Answer: (a) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion.
In simple words: In isoelectronic species, the ion with more protons has a stronger pull on the same number of electrons. This makes the Al\(^{3+}\) ion smaller and increases its effective nuclear charge.

Exam Tip: For any set of isoelectronic ions, the radius always decreases as the positive charge of the nucleus increases.

Question 6. Assertion: In caesium chloride crystal, Cs+ ion is present on the centre of cube of the unit cell.
Reason: For N-atoms adopting fcc arrangement, there are 2N tetrahedral voids.
Answer: (b) Assertion and Reason both are correct statements but Reason is not correct explanation for Assertion.
In simple words: Both statements are true. However, the rule about tetrahedral voids in an FCC lattice does not explain why CsCl forms a body-centered cubic-like arrangement.

Exam Tip: The CsCl lattice is structurally treated as a simple cubic arrangement of anions with a cation in the center, giving it an 8:8 coordination ratio.

Question 7. Assertion: The packing efficiency is maximum for the fcc structure.
Reason: The coordination number is 12 in fcc structures.
Answer: (b) Assertion and Reason both are correct statements but Reason is not correct explanation for Assertion.
In simple words: Both statements are correct on their own. However, having a coordination number of 12 is not the direct mathematical reason why the packing efficiency is 74%.

Exam Tip: FCC and HCP are the most closely packed crystal structures possible for identical spheres, both reaching 74% packing efficiency.

Question 8. Assertion: White ZnO becomes yellow upon heating.
Reason: On heating ZnO loses oxygen and free electrons go into exited stated and upon returning imparts yellow radiation.
Answer: (a) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion.
In simple words: Heating drives oxygen out of the ZnO lattice, creating metal excess defects. The trapped free electrons get excited by absorbing light, which makes the solid appear yellow.

Exam Tip: This yellowing of zinc oxide is a classic textbook example of a metal excess defect caused by anionic vacancies (also known as F-centers).

Question 9. Assertion: In p-type semiconductor holes will appear to be moving towards the negatively charged plate.
Reason: Delocalised electrons increase the conductivity of doped silicon.
Answer: (b) Assertion and Reason both are correct statements but Reason is not correct explanation for Assertion.
In simple words: Both statements are correct. However, the general statement about delocalized electrons boosting conductivity does not explain why positive holes travel towards a negative plate in p-type silicon.

Exam Tip: Under an electric field in a p-type semiconductor, valence electrons jump into neighboring vacancies (holes), which makes it look like the holes are moving in the opposite direction.

Question 10. Assertion: Valence band may remain partially filled.
Reason: The gap between valence band and conduction band cannot be determined.
Answer: (c) Assertion is correct statement but Reason is wrong statement.
In simple words: A valence band can be partially filled, but the energy difference between the valence and conduction bands can easily be determined using spectroscopic methods.

Exam Tip: In metals, electrical conductivity is high because the valence band overlaps with the conduction band, leaving no energy gap between them.

Case Based Questions

1. The adjective, ‘crystalline’ when applied to solids, implies an ideal crystal in which the structural units, termed as unit cells, are repeated regularly and indefinitely in three dimensions in space. The unit cell, containing at least one molecule has definite orientation and shape defined by the translational vectors, a, b and c. The unit cell therefore has a definite volume, V that contains the atoms and molecules necessary for generating the crystal. Every crystal can be classified as a member of one of the seven possible crystal systems or crystal classes that are defined by the relationships between the individual dimensions, a, b and c of the unit cell and between the individual angles, α, β, and γ of the unit cell. The structure of the given crystal may be assigned to one of the 7 crystal systems, to one of the 14 Brevais lattices, and to one of the 230 space groups. These uniquely define the possible ways of rearranging atoms in a three-dimensional solid. Based on these observations, seven crystal systems were identified: triclinic, monoclinic, trigonal or rhombohedral, tetragonal, hexagonal, rhombic or orthorhombic and cubic.

Question (i) The unit cell with the structure given below represents ............ crystal system.
(a) cubic
(b) orthorhombic
(c) tetragonal
(d) trigonal
Answer: (a) cubic
In simple words: This unit cell represents a cubic system because all three axes are equal in length and meet at ninety-degree angles.

Exam Tip: The relations \( a = b = c \) and \( \alpha = \beta = \gamma = 90^\circ \) uniquely define the cubic crystal system, which has the highest spatial symmetry.

Question (ii) The crystal system of a compound with unit cell dimensions, a = 0.387 nm, b = 0.387 nm and c = 0.504 nm and α = β = 90º and γ = 120° is.
(a) cubic
(b) hexagonal
(c) orthorhombic
(d) rhombohedral
Answer: (b) hexagonal
In simple words: This is a hexagonal system because two of the sides are identical in length, two of the angles are ninety degrees, and the third angle is exactly one hundred and twenty degrees.

Exam Tip: Look out for \( \gamma = 120^\circ \) along with \( a = b \neq c \); these parameters are the classic signature of a hexagonal crystal system.

Question (iii) In a triclinic crystal
(a) a = b = c, α = β = γ ≠ 90°
(b) a ≠ b = c, α = β = γ = 90°
(c) a ≠ b ≠ c, α ≠ β ≠ γ ≠ 90°
(d) a ≠ b ≠ c, α = γ = 90°, β ≠ 90°
Answer: (c) a ≠ b ≠ c, α ≠ β ≠ γ ≠ 90°
In simple words: A triclinic crystal is completely asymmetrical. None of its edge lengths are equal, and none of its angles are equal or ninety degrees.

Exam Tip: Triclinic is the most unsymmetrical crystal system because it has no equal sides or equal angles, and none are \( 90^\circ \).

Question (iv) The unit cell with dimensions α = β = γ = 90°, a = b ≠ c is
(a) cubic
(b) triclinic
(c) hexagonal
(d) tetragonal
Answer: (d) tetragonal
In simple words: A tetragonal crystal system is like a stretched cube. It has two equal side lengths and all three angles are ninety degrees.

Exam Tip: Tetragonal systems have \( a = b \neq c \) with all right angles, which is similar to the shape of a standard square prism.

Question (v) An example of orthorhombic crystal system is
(a) SnO2
(b) KNO3
(c) ZnO
(d) K2Cr2O7
Answer: (b) KNO3
In simple words: Potassium nitrate (\( \text{KNO}_3 \)) is a standard substance that crystallizes in an orthorhombic shape, where all side lengths are different but all angles are ninety degrees.

Exam Tip: Memorize popular examples of orthorhombic crystals like rhombic sulfur, \( \text{KNO}_3 \), and \( \text{BaSO}_4 \).

2. In contrast to the disorders of gases and liquids, there is translational order in crystals. However, disordered or amorphous solids also exist which lack such order, they are really highly viscous liquids. In translational order entire structure or lattice, can be generated by repeated replication of a small regular figure, termed as unit cell. The planes of any crystalline structure can be specified using Miller Indices, which is also serve to identify single crystal faces. The ordered structure or lattice, of a solid can be determined by X-ray or neutron diffraction studies, in which a beam of X-rays of neutrons is scattered from the sample to produce a diffraction pattern which can be analyzed to reveal the crystal structure of the sample. All crystal lattices can be classified into 14 Bravais lattices belonging to 7 systems. For example, the simple cubic, face centred cubic and body-centred cubic lattices are the 3 lattices of the cubic system. Cubic and hexagonal close-packed structures have the structure of tightly packed spheres where each sphere touches 12 neighbours, 6 in the same plane and 3 above and 3 below. These two close-packed structures differ in the placement of successive planes or layers. For the hexagonal close packing, a third layer is laid down to reproduce the first layer, so the structure could be represented by ABABAB... . For cubic close packing, third layer is again displaced, corresponding to ABCABC... .

Question (i) In hexagonal close packing, a sphere has coordination number of
(a) 4
(b) 6
(c) 8
(d) 12
Answer: (d) 12
In simple words: In HCP, each sphere is surrounded by twelve other spheres: six in its own plane, three sitting directly above it, and three sitting directly below it.

Exam Tip: The coordination number is 12 for both hexagonal close-packed (HCP) and cubic close-packed (CCP/FCC) structures.

Question (ii) Which of the following arrangements correctly represents hexagonal and cubic close packed structure?
(a) ABCABC... and ABAB...
(b) ABAB... and ABCABC...
(c) Both have ABAB... arrangement.
(d) Both have ABCABC... arrangement.
Answer: (b) ABAB... and ABCABC...
In simple words: Hexagonal close packing repeats every two layers (ABAB...), while cubic close packing repeats every three layers (ABCABC...).

Exam Tip: Note that HCP uses an ABAB... pattern because the third layer aligns with the first, whereas CCP uses an ABCABC... pattern because the third layer sits over octahedral voids.

Question (iii) The arrangement of the first two layers, one above the other in hcp and ccp arrangements is
(a) exactly same in both cases.
(b) partly same and partly different.
(c) different from each other.
(d) nothing definite.
Answer: (a) exactly same in both cases.
In simple words: Both HCP and CCP start with the exact same two close-packed layers of spheres; the structures only begin to differ when you place the third layer.

Exam Tip: Since both arrangements begin with a close-packed base, the first two layers (A and B) are completely identical in both structures.

Question (iv) Which of the following statements is not correct?
(a) The amorphous solids have a random, disordered arrangement of constituents.
(b) The sime cubic, face-centred and body centred are the three lattices of the cubic system.
(c) The number of Bravais lattice in which a crystal can be categorized is 7.
(d) A metal that crystallizes in hcp structure has coordination number 12.
Answer: (c) The number of Bravais lattice in which a crystal can be categorized is 7.
In simple words: This is incorrect because there are 14 unique Bravais lattices, which are sorted into 7 different crystal systems.

Exam Tip: Make sure you do not confuse the 7 basic crystal systems (shapes) with the 14 Bravais lattices (variations of those shapes).

Question (v) Which of the following statements about amorphous solids is incorrect?
(a) They melt over a range of temperature.
(b) There is no orderly arrangement of particles.
(c) They are anisotropic.
(d) They are rigid and incompressible.
Answer: (c) They are anisotropic.
In simple words: This is wrong because amorphous solids are actually isotropic, meaning their physical properties are the same no matter which direction you measure them.

Exam Tip: Remember that amorphous solids have no organized long-range layout, which naturally makes them isotropic rather than anisotropic.

3. In an ideal crystal, there must be regular repeating arrangement of the constituting particles and its entropy must be zero at absolute zero temperature. However, it is impossible to obtain an ideal crystal and it suffers from certain defects called imperfections. In pure crystal, these defects arises either due to disorder or dislocation of the constituting particles from the normal positions or due to the movement of the particles even at absolute zero temperature. Such defects increase with rise in temperature. In addition to this certain defects arise due to the presence of some impurities. Such defects not only modify the existing properties of the crystalline solids but also impart certain new characteristics to them.

Question (i) AgCl is crystallized from molten AgCl containing a little CdCl2. The solid obtained will have
(a) cationic vacancies equal to number of Cd2+ ions incorporated.
(b) cationic vacancies equal to double the number of Cd2+ ions.
(c) anionic vacancies.
(d) neither cationic nor anionic vacancies.
Answer: (a) cationic vacancies equal to number of Cd2+ ions incorporated.
In simple words: Each divalent Cd\(^{2+}\) ion replaces two monovalent Ag\(^+\) ions to keep the lattice neutral. One of those empty spots is filled by the Cd\(^{2+}\) ion, which leaves exactly one vacant silver site.

Exam Tip: In impurity defects involving divalent cations in monovalent salts, the number of cation vacancies created is always equal to the number of divalent ions added.

Question (ii) Lattice defect per 1015 NaCl is 1. What is the number of lattice defects in a mole of NaCl?
(a) 6.02 × 1023
(b) 6.02 × 108
(c) 1014
(d) None of the options
Answer: (b) 6.02 × 108
In simple words: Since there is 1 defect for every \( 10^{15} \) formula units, a mole of NaCl (\( 6.02 \times 10^{23} \) units) has \( \frac{6.02 \times 10^{23}}{10^{15}} = 6.02 \times 10^8 \) defects.

Exam Tip: Set up a direct proportion of defects per unit to quickly solve defect concentration problems for a mole of material.

Question (iii) The ionic substances in which the cation and anion are of almost similar size shows
(a) non-stoichiometric defect
(b) Schottky defect
(c) Frenkel defect
(d) all of these
Answer: (b) Schottky defect
In simple words: Schottky defects occur in highly ionic compounds where the cations and anions are similar in size, creating equal numbers of missing positive and negative ions.

Exam Tip: Compounds like NaCl, KCl, and CsCl undergo Schottky defects because their cations and anions are of comparable sizes.

Question (iv) If Al3+ ions replace Na+ ions at the edge centres of NaCl lattice, then the number of vacancies in 1 mole of NaCl will be
(a) 3.01 × 1023
(b) 6.02 × 1023
(c) 9.03 × 1023
(d) 12.04 × 1023
Answer: (a) 3.01 × 1023
In simple words: The 12 edge centers contribute 3 Na\(^+\) ions. Since one Al\(^{3+}\) replaces three Na\(^+\) ions and creates two empty spots, replacing all edge center sodium ions results in a total of \( 3.01 \times 10^{23} \) vacancies.

Exam Tip: Carefully compute the fractional contribution of atoms at edge centers (\( 12 \times \frac{1}{4} = 3 \)) to find the exact number of ions being replaced.

Question (v) Which of the following gives both Frenkel and Schottky defect?
(a) AgCl
(b) CsCl
(c) KCl
(d) AgBr
Answer: (d) AgBr
In simple words: Silver bromide is an exception because the Ag\(^+\) ion is intermediate in size, allowing AgBr to show both Schottky and Frenkel defects.

Exam Tip: AgBr is a very common exam question because it is unique in showing both Frenkel and Schottky defects.

4. In hexagonal system of crystals, a frequency encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them. A space-filling model of this structure, called hexagonal close packed (hcp), is constituted of a spheres on a flat surface surrounded in the same plane by six identical spheres as closely as possible. Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of these three sphere touches three spheres of the bottom layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position.

Question (i) The volume of this hcp unit cell is
(a) 24 2 3r
(b) 16 2 3r
(c) 12 2 3r
(d) 64 33 3r
Answer: (a) 24 2 3r
In simple words: The volume of an HCP cell is derived by multiplying its base area by its height, which yields the final geometric formula of \( 24\sqrt{2}r^3 \).

Exam Tip: Note the height of the HCP unit cell, \( 4r\sqrt{\frac{2}{3}} \), and use it along with the hexagonal base area to find the total volume.

Question (ii) The number of atoms in this hcp unit cell is
(a) 4
(b) 6
(c) 12
(d) 17
Answer: (b) 6
In simple words: An HCP unit cell has an effective total of 6 atoms. This includes corner contributions, face-center contributions, and three completely enclosed middle atoms.

Exam Tip: Remember that HCP has an effective atom count of Z = 6, while FCC has Z = 4 and BCC has Z = 2.

Question (iii) The empty space in this hcp unit cell is
(a) 74%
(b) 47.6%
(c) 32%
(d) 26%
Answer: (d) 26%
In simple words: Since HCP has a maximum packing efficiency of 74%, the remaining empty space is \( 100\% - 74\% = 26\% \).

Exam Tip: Subtracting the packing fraction (0.74) from 1 gives you the empty space fraction (0.26, or 26%) in a close-packed arrangement.

Question (iv) Which of the following statements is correct about hexagonal close packing?
(a) In this arrangement, third layer is identical to the first layer.
(b) The coordination number in this arrangement is 6.
(c) It is as closely packed as body centered cubic packing
(d) It has 32% empty space.
Answer: (a) In this arrangement, third layer is identical to the first layer.
In simple words: HCP uses an ABAB... sequence, meaning the spheres of the third layer align directly over the spheres of the first layer.

Exam Tip: In HCP, the spheres of the third layer cover the tetrahedral voids of the second layer, making the third layer completely identical to the first.

Question (v) In hexagonal close packing of spheres in three dimensions.
(a) In one unit cell there are 12 octahedral voids and all are completely inside the unit cell.
(b) In one unit cell there are six octahedral voids and all are completely inside the unit cell.
(c) In one unit cell there are six octahedral voids out of which three are completely inside the unit cell and other three are from contributions of octahedral voids which are partially inside the unit cell.
(d) In one cell unit there are 12 tetrahedral voids, all are completely inside the unit cell.
Answer: (b) In one unit cell there are six octahedral voids and all are completely inside the unit cell.
In simple words: Since there are 6 atoms per HCP unit cell, there are exactly 6 octahedral voids, and all of them are located entirely within the unit cell boundaries.

Exam Tip: For any close-packed lattice containing N atoms, there are always N octahedral voids and 2N tetrahedral voids.

5. In ideally ionic structures, the coordination numbers of the ions are determined by electrostatic considerations. Cations surround themselves with as many anions as possible and vice versa. This maximizes the attractions between neighbouring ions of opposite charge and hence maximizes the lattice energy of the crystal. This requirement led to the formulation of the radius ratio rule for ionic structures in which the ions and the structure adopted for a particular compound depend on the relative sizes of the ions. Thus, for the stable ionic crystalline structures, there is definite radius ratio limit for a cation to fit perfectly in the lattice of anions called radius ratio rule. This depends upon the ratio of radii of two types of ions, r+/r–. This ratio for coordination numbers 3, 4, 6 and 8 respectively 0.155–0.225, 0.225–0.414, 0.414–0.732 and 0.732–1.000. The coordination number of ionic solids also depends upon temperature and pressure. On applying high pressure, coordination number increases. On the other hand, on applying high temperature, it decreases.

Question (i) The ionic radii K+, Rb+ and Br– are 137, 148 and 195 pm. The coordination number of cation in RbBr and KBr structures are respectively.
(a) 8, 6
(b) 6, 4
(c) 6, 8
(d) 4, 6
Answer: (a) 8, 6
In simple words: The radius ratio of RbBr (\( \frac{148}{195} = 0.76 \)) points to a coordination number of 8, while KBr (\( \frac{137}{195} = 0.702 \)) points to a coordination number of 6.

Exam Tip: Use the radius ratio rule: ratios from 0.414 to 0.732 have a coordination number of 6, and ratios from 0.732 to 1.000 have a coordination number of 8.

Question (ii) For a coordination number 4, the maximum limiting radius ratio is
(a) 0.414
(b) 0.732
(c) 0.225
(d) 0.155
Answer: (a) 0.414
In simple words: For a tetrahedral coordination number of 4, the radius ratio ranges from 0.225 to 0.414. Therefore, the maximum limit is 0.414.

Exam Tip: Memorize the standard radius ratio limits: 0.155 (triangular), 0.225 (tetrahedral), 0.414 (octahedral), and 0.732 (cubic).

Question (iii) If the radius of Na+ ion is 95 pm and that of Cl– ion is 181 pm, the coordination number of Na+ ion is
(a) 6
(b) 4
(c) 8
(d) 12
Answer: (a) 6
In simple words: The radius ratio is \( \frac{95}{181} = 0.524 \). Since this falls between 0.414 and 0.732, sodium has an octahedral coordination number of 6.

Exam Tip: The classic rock salt (NaCl) structure has a 6:6 coordination ratio, which is predicted by its radius ratio of 0.524.

Question (iv) Which is not the correct statement for ionic solids in which positive and negative ions are held by strong electrostatic attractive forces?
(a) The radius ratio r+/r– increases as coordination number increases.
(b) As the difference in size of ions increases, coordination number increases.
(c) When coordination number is eight, r+/r– ratio lies between 0.225 to 0.414.
(d) In ionic solid of the type AX(ZnS, wurtzite), the coordination number of Zn2+ and S2– respectively are 4 and 4.
Answer: (c) When coordination number is eight, r+/r– ratio lies between 0.225 to 0.414.
In simple words: This statement is false because a coordination number of 8 requires a larger radius ratio between 0.732 and 1.000.

Exam Tip: Remember that as the cation size approaches the anion size (radius ratio increases towards 1), the coordination number increases.

Question (v) If the pressure of CsCl is increased, then its coordination number will
(a) increase
(b) remain the same
(c) decrease
(d) none of the options
Answer: (a) increase
In simple words: High pressure squeezes the atoms closer together, which increases the coordination number of the crystal.

Exam Tip: Keep in mind that high pressure increases the coordination number (e.g., from 6 to 8), while high temperature decreases it (e.g., from 8 to 6).

6. The transition metals when exposed to oxygen at low and intermediate temperatures form thin, protective oxide films of up to some thousands of Angstroms in thickness. Transition metal oxides lie between the extremes of ionic and covalent binary compounds formed by elements from the left or right side of the periodic table. They range from metallic to semiconducting and deviate by both large and small degrees from stoichiometry. Since d-electron bonding levels are involved, the cations exist in various valence states and hence give rise to a large number of oxides. The crystal structures are often classified by considering a cubic or hexagonal close-packed lattice of one set of ions with the other set of ions filling the octahedral or tetrahedral interstices. The actual oxide structures, however, generally show departures from such regular arrays due in part to distortions caused by packing of ions of different size and to ligand field effects. These distortions depend not only on the number of d-electrons but also on the valence and the position of the transition metal in a period or group.

Question (i) Assertion: Cations of transition elements occur in various valence states.
Reason: Large number of oxides of transition elements are possible.
Answer: (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
In simple words: Transition metals have many valence states due to their partially filled d-orbitals. While they do form many oxides, this is a result of their multiple valence states, not the cause of them.

Exam Tip: Transition metals display variable oxidation states because their (n-1)d and ns orbital energy levels are very close to each other.

Question (ii) Assertion: Crystal structure of oxides of transition metals often show defects.
Reason: Ligand field effect cause distortions in crystal structures.
Answer: (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
In simple words: Transition metal oxides are highly prone to defects because ligand field effects split the d-orbital energies, which distorts the ideal crystal geometry.

Exam Tip: Ligand field stabilization energy (LFSE) splits d-orbitals, causing spatial distortions like the Jahn-Teller effect in transition metal compounds.

Question (iii) Assertion: Transition metals form protective oxide films.
Reason: Oxides of transition metals are always stoichiometric.
Answer: (c) Assertion is correct statement but reason is wrong statement.
In simple words: Transition metals do form protective oxide layers, but their oxides are frequently non-stoichiometric because transition elements easily exist in different valence states.

Exam Tip: Many transition metal oxides, such as \( \text{Fe}_{0.95}\text{O} \), are non-stoichiometric because of metal deficiency defects.

Question (iv) Assertion: CrO crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by chromium ions.
Reason: Transition metal oxide may be hexagonal close-packed lattice of oxide ions with metal ions filling the octahedral voids.
Answer: (d) Assertion is wrong statement but reason is correct statement.
In simple words: CrO has a 1:1 ratio, meaning all octahedral holes must be occupied. The assertion is wrong because occupying two-thirds of the holes would describe a 2:3 oxide like \( \text{Cr}_2\text{O}_3 \).

Exam Tip: In a 1:1 oxide (like CrO), all of the octahedral voids are occupied; in a 2:3 oxide (like \( \text{Cr}_2\text{O}_3 \)), only two-thirds of those voids are filled.