Chemistry Question Bank : Solid State (1 Mark)

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Section A – Questions of 1 mark Very Short Answers  difficulty level

Q1. Write an example of a pair of isomorphous substances.

  • a) K2SO4 and K2SeO4   and b) NaCl and KCl

Q2. What is a polymorphic form of Sulphur?

  • Sulphur exists in different polymorphic forms, like monoclinic sulphur, rhombic sulphur, and amorphous sulphur.

Q3. Name the semiconductor obtained when Boron is added as an impurity to Silicon.

  • Boron is a trivalent impurity. Hence the semiconductor obtained when Boron is added as an impurity to Silicon is p-type.

Q4. What type of semiconductor is obtained when Silicon is doped with Phosphorus?

  • Phosphorous is a pentavalent impurity. Hence the semiconductor obtained when Silicon is doped with Phosphorus is n-type.

Q.5. Predict the coordination number of Na+ ion in NaCl. rNa+ = 0.95 Å; r Cl- = 1.81 Å.

Lattice Points

Radius ratio = 0.95/1.81 = 0.525

For radius ratio of 0.414 to 0.732, the coordination number is 6

Q 6. The radius of cation and anion is 0.95 A0 and 1.81 A0. Name the structure of the ionic crystal.

Radius ratio = 0.95/1.81 = 0.525

For radius ratio of 0.414 to 0.732, the ionic crystal has octahedral structure.

Q 7.  What is the magnetic property of Cl ion? [Z=17]

Electronic configuration of Cl ion is 1s2 2s2 2p6 3s2 3p6

The electronic configuration of the valence shell is

↓↑ ↓↑ ↓↑ ↓↑
3s2 3px2 3py2 3pz2

All electrons are paired. Hence Cl ion is diamagnetic.

Note: If the electronic configuration of valence shell shows unpaired electron then it is paramagnetic.

Q8.  What is the magnetic property of the Ni atom? [Z= 28]

Electronic configuration of Ni atom is 1s2 2s2 2p6 3s2 3p6 3d4s2.

The electronic configuration of the last two shells is 3s2 3p6 3d4s2

  • Due to the presence of the unpaired electrons in the third orbit, it shows a peculiar magnetic property. There are certain regions in which all the Ni atoms have magnetic moments in the same direction and form domain. Hence Ni shows ferromagnetism.

Q9. Name the method to determine magnetic properties.

  • Guoy’s method is used to determine magnetic properties.

Q10. Atom C and D form FCC structure. Atom C is present at the corners of the cube and D is at the faces of the cube. What is the formula of the compound?

No. of atoms of C from corners of unit cell = 8 x (1/8) = 1
No. of atoms of D from faces of unit cell = 6 x (1/2) = 3
Thus C:D = 1: 3
Thus the formula is CD3.
Q11. Name the type of solid in S8 molecule.
  • In sulphur crystal, lattice points are Smolecules which are held together by means of weak physical forces (van der Waal’s forces). Hence it is molecular solid.
Q12. Name the type of solid formed Present in Ammonium phosphate?
  • Ammonium phosphate is ionic solid.
Q13. Write uses of Fullerene.
  • Fullerenes are used as catalysts and lubricants.
  • They are also used in nanotubes for strengthening materials (for example sports equipment)
  • Sometimes they are used as a way of delivering drugs
Q14. Which type of hybridization is present in the carbon atom in fullerene?
  • In fullerene sp2 hybridization is present.
Q15. What is the force between the layer of carbon atoms in graphite?
  • Different layers in graphite are bonded to each other by van der Walls’forces.
Q16. How the density of solid varies in Schottky defect?
  • Due to a Schottky defect, the observed density of crystal is found to be lower than expected density.
Q17.  How the density of solid varies in Frenkel defect?
  • The presence of a Frenkel defect does not alter the density of the solid.
Q18. How many kg of carbon does 0.1 mole of Buckminster Fullerene contain?
The molecular formula of Fullerine is C60.
Thus 1 mole of Fullerene contains 60 carbon atoms
Thus the mass of 1 mole of Fullerine is 60 x 12 g = 720 g = 0.720 kg
Thus the mass of 0.1 mole of Fullerine is 0.0720 kg
0.1 mole of Buckminster Fullerene contains 0.072 kg of carbon.
Q19. Calculate the number of unit cells in 58.5 g of a compound which forms BCC. [Molar mass of a compound = 58.5]

Number of moles = given mass/molar mass = 58.5 g/58.5 g = 1 mol

By Avogadro’s law, 1 mole of any substance has exactly 6.022 × 10²³ molecules/atoms/unit cells.

Therefore in 58.5 g of a compound, there will be approximately 6.022 × 10²³ unit cells.

Q20. Write an example of an element in which atoms occupy the position of lattice points?
  •  In covalent solids atoms occupy the position of lattice points. e.g. graphite, diamond.

Section A – Questions of 1 Mark VSA (Which can be combined with sections B / C / D)

Q1. Define crystalline solid.

  • A crystalline solid is a substance whose constituent particles possess a regular orderly long-range arrangement.
  • Examples: Sodium chloride, sucrose (cane sugar)

Q2. Define Anisotropy.

  • The ability of crystalline solids to change their physical properties when measured in different directions is called anisotropy.

Q3.  What is a unit cell?

  • A unit cell is the smallest structural repeating unit of crystalline solid (space lattice).

Q4. What is meant by voids in crystal structure?

  • The empty spaces between the particles (sphere) during their closed packing are called voids.

Q5.  Write two examples of metallic solids.

  • Aluminium, copper, iron are examples of metallic solids.

Q6. Write two examples of covalent solids.

  • Diamond, Silicon,silicon carbide (SiC) are examples of covalent solids.

Q7.  Write two examples of molecular solids.

  • Phosphorous, sulphur. chlorine and argon are molecular solids

Q8.  What is meant by the term coordination number?

  • The number of nearest neighbours of a particle is called its coordination number.

Q9.  Define crystal defect.

  • Any deviation from the perfectly ordered arrangement crystal is called a defect or imperfection of crystal.

Q10. Define a line defect.

  • If lattice imperfections extend along the line they are called line defects.

Q11. Write an example of solid in which the forces between the particles having electrostatic force.

  • The force between the particles is an electrostatic force, hence these are ionic solids.
  • Salts like NaCl, BaSO4, potassium bromide, copper nitrate, copper sulphate are ionic solids.

Q12.  What is the relation between the edge lengths ‘a’, ‘b’ and ‘c’ in body centred orthorhombic crystal?

  • No two edges are of the same length. a ≠ b ≠ c.

Q13. What is the relation between the angles ‘α’, ‘β’ and ‘γ ‘ in simple hexagonal crystal?

  •  α = β = 90° and γ = 120°

Q14. Define crystal lattice.

  • Crystal lattice may be defined as an array of points showing how molecules, atoms or ions are arranged in different sites, in a three-dimensional space.

Q15. Which crystal lattice is represented by the following values? a = b = c and α = β = γ ≠ 90°.

  • The crystal lattice is

Q16. Which crystal lattice has the most efficient packing efficiency?

  • Face centred cubic structure has the highest packing efficiency of 74%.

Q17. What is the packing efficiency of the BCC unit cell?

  • The packing efficiency of BCC unit cell is 68%.

Q18. What is the void space of SCC unit cell?

  • The void space of SCC unit cell is 47.6%.

Q19. The coordination number of the cation is 4. What will be the limiting value of the radius ratio?

  • For coordination number 4 he limiting value of the radius ratio is 0.225 to 0.414

Q20. Define Interstitial defect.

  • When cation and anion from ionic solid leave its regular site and moves to occupy a place between the lattice site (interstitial sites) is called an interstitial defect.

Q21. Define Impurity defect.

  • When regular cation of a crystal is replaced by some different cation than the regualr cation, the defect is called the impurity defect.

Q.22. What is a n-type semiconductor?


Q23.  What is a p-type semiconductor?


Q24. Name the two types of void space in close packing.

  • The two types of void space in close packing are tetrahedral voids and octagonal voids.

Q25. How many tetrahedral and octahedral void spaces are there for each sphere in close packing?

  • If the number of close-packed spheres (particles)is equal to ‘N’. Then, the number of octahedral voids formed = N And, the number of tetrahedral voids formed = 2N
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  1. Momin Misbah

    N!ce 4 practice, we like it. We remember it easily.

  2. It’s amazing to have solution so that we don’t waste our time but upload all the chapters solution as soon as possible.
    Plz do the needful
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    So plz upload asap

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