The Solid State

01

PYQ 2014

medium

chemistry ID: mht-cet-

Calculate the volume of unit cell if an element having molar mass forms bcc unit cells.

1

2

3

4

02

PYQ 2014

medium

chemistry ID: mht-cet-

What total number of crystal systems associated with 14 Bravais lattices?

1

7

2

14

3

1

4

3

03

PYQ 2014

medium

chemistry ID: mht-cet-

In an ionic crystalline solid, atoms of element Y forms hcp structure. The atoms of element X occupy one third of tetrahedral voids. What is the formula of compound?

1

2

3

4

04

PYQ 2018

medium

chemistry ID: mht-cet-

In which among the following solids, Schottky defect is NOT observed?

1

ZnS

2

NaCl

3

KCl

4

CsCl

05

PYQ 2018

medium

chemistry ID: mht-cet-

Which among the following metals crystallise as a simple cube ?

1

Polonium

2

Iron

3

Copper

4

Gold

06

PYQ 2019

medium

chemistry ID: mht-cet-

How many total constituent particles are present in simple cubic unit cell ?

1

2

3

4

07

PYQ 2020

easy

chemistry ID: mht-cet-

What is the volume of 1 mole of a crystalline solid having unit cell edge length 16 10 cm, if its unit cell contains 24 molecules?

1

102.7 cm mol

2

404.0 cm mol

3

159.3 cm mol

4

142.1 cm mol

08

PYQ 2020

medium

chemistry ID: mht-cet-

Which of the following solution will have highest freezing point depression?

1

1 M glucose

2

1 M sucrose

3

1 M urea

4

1 M KCl

09

PYQ 2020

medium

chemistry ID: mht-cet-

What is the number of moles of silver chloride precipitated when excess aqueous silver nitrate is treated with ?

1

1.0 mole

2

3.0 mole

3

2.0 mole

4

4.0 mole

10

PYQ 2020

medium

chemistry ID: mht-cet-

Sodium crystallizes in bcc structure with radius 1.86 10 cm. What is the edge length of unit cell of sodium?

1

4.3 10 cm

2

3.72 10 cm

3

7.44 10 cm

4

5.26 10 cm

11

PYQ 2020

medium

chemistry ID: mht-cet-

What is the mass of an fcc unit cell if mass of one atom of an element is g?}

1

2

3

4

Official Solution

Correct Option: $(3)$

$Step 1: Number of atoms in fcc unit cell. A face centered cubic (fcc) unit cell contains 4 atoms per unit cell. Step 2: Mass of one atom. Mass of one atom is given as g. Step 3: Calculate mass of unit cell. Step 4: Conclusion. Hence, the mass of one fcc unit cell is g.$

12

PYQ 2020

medium

chemistry ID: mht-cet-

$How much part of an atom occupies each corner of a bcc unit cell?$

1

2

3

4

Official Solution

Correct Option: $(2)$

$Step 1: Recall unit cell geometry. In a body-centred cubic (bcc) unit cell, atoms are present at the eight corners and one atom at the centre of the cube. Step 2: Contribution of corner atoms. Each corner atom is shared by eight neighbouring unit cells. Step 3: Calculate the contribution. Step 4: Conclusion. Thus, each corner atom contributes to the bcc unit cell.$

13

PYQ 2020

medium

chemistry ID: mht-cet-

$Which among the following is an example of amorphous solid?$

1

$Camphor$

2

$Magnesium$

3

$Diamond$

4

$Glass$

Official Solution

Correct Option: $(4)$

$Step 1: Recall the definition of amorphous solids. Amorphous solids lack a long-range ordered arrangement of particles and do not have a sharp melting point. Step 2: Analyze the given options. Camphor, magnesium, and diamond are crystalline solids with definite lattice structures. Glass does not possess long-range order and softens over a range of temperatures. Step 3: Conclusion. Therefore, glass is an example of an amorphous solid.$

14

PYQ 2020

medium

chemistry ID: mht-cet-

$What is the coordination number of a cation in an ionic compound if the type of hole occupied by the cation is octahedral?$

1

2

3

4

Official Solution

Correct Option: $(2)$

$Step 1: Understand octahedral voids. In an octahedral void, a cation is surrounded by six anions placed at the corners of an octahedron. Step 2: Relate void type to coordination number. The coordination number is defined as the number of nearest neighboring ions surrounding a given ion. Step 3: Conclusion. Since an octahedral hole is surrounded by six anions, the coordination number of the cation is .$

15

PYQ 2020

medium

chemistry ID: mht-cet-

$The overlap of orbitals involved in the formation of C-Br bond in vinyl bromide is$

1

$sp - 3p$

2

$sp - 4p$

3

$sp - 4p$

4

$sp - 3p$

Official Solution

Correct Option: $(2)$

$Step 1: Identify hybridization of carbon in vinyl bromide. Vinyl bromide is CH =CH-Br. Carbon attached to bromine is sp hybridized. Step 2: Identify orbital of bromine. Bromine forms bond using its 4p orbital. Step 3: Determine type of overlap. Sigma bond is formed by overlap of sp orbital of carbon and 4p orbital of bromine. Step 4: Conclusion. Correct overlap is sp - 4p .$

16

PYQ 2020

medium

chemistry ID: mht-cet-

$Which of the following compound is a broad spectrum antibiotic?$

1

$Amoxicillin$

2

$Chlordiazepoxide$

3

$Penicillin$

4

$Meprobamate$

Official Solution

Correct Option: $(1)$

$Step 1: Understand broad spectrum antibiotics. Broad spectrum antibiotics are effective against both Gram-positive and Gram-negative bacteria. Step 2: Analyze the given compounds. Amoxicillin is a semi-synthetic penicillin effective against a wide range of bacteria. Chlordiazepoxide and Meprobamate are sedatives, not antibiotics. Penicillin (natural) has narrow spectrum activity. Step 3: Conclusion. Amoxicillin is a broad spectrum antibiotic.$

17

PYQ 2020

medium

chemistry ID: mht-cet-

$What is the oxidation number of carbon in K C O ?$

1

$-2$

2

$+3$

3

$0$

4

$+4$

Official Solution

Correct Option: $(2)$

$Step 1: Write oxidation states of known elements. Potassium (K) has oxidation state +1. Oxygen (O) has oxidation state -2. Step 2: Let oxidation state of carbon be . Step 3: Solve the equation. Step 4: Conclusion. The oxidation number of carbon is +3.$

18

PYQ 2020

medium

chemistry ID: mht-cet-

$Identify the major product 'B' in the following reaction:$

1

$n - butyl isocyanide$

2

$n - butyl cyanide$

3

$sec. butyl isocyanide$

4

$isobutyl cyanide$

Official Solution

Correct Option: $(1)$

$Step 1: Understanding the reaction. The reaction involves the formation of a product through two steps. The first step is the addition of HBr to but-2-ene in the presence of peroxides, which follows an anti-Markovnikov addition mechanism, leading to the formation of a secondary radical. In the second step, silver cyanide (AgCN) reacts with the alkyl halide to form an isocyanide. Step 2: Analyzing the options. (A) n - butyl isocyanide: Correct. The product 'B' formed in this reaction is n-butyl isocyanide, a result of the reaction with AgCN. (B) n - butyl cyanide: Incorrect. The product is an isocyanide, not a nitrile. (C) sec. butyl isocyanide: Incorrect. The primary product is n-butyl isocyanide, not sec-butyl. (D) isobutyl cyanide: Incorrect. This is the wrong product formed in this reaction. The correct product is n-butyl isocyanide. Step 3: Conclusion. The correct answer is (A) n - butyl isocyanide, which is the major product of the reaction.$

19

PYQ 2020

medium

chemistry ID: mht-cet-

$What is the melting point of zinc?$

1

$473 K$

2

$1193 K$

3

$423 K$

4

$692 K$

Official Solution

Correct Option: $(4)$

$Step 1: Known physical property of zinc. Zinc is a moderately reactive metal with a relatively low melting point compared to many transition metals. Step 2: Standard melting point value. The melting point of zinc is approximately 419°C. Step 3: Converting into Kelvin. Step 4: Conclusion. Therefore, the correct melting point of zinc is 692 K.$

20

PYQ 2020

medium

chemistry ID: mht-cet-

$According to radius ratio rule if R is radius of anion and r is the radius of cation then for an atom to occupy octahedral hole, the possible relation between R and r is$

1

$r = 2.25 R$

2

$r = 0.514 R$

3

$r = 0.225 R$

4

$r = 0.314 R$

Official Solution

Correct Option: $(2)$

$Step 1: Understanding the radius ratio rule. The radius ratio rule helps in determining the stability of an ionic structure. For an atom to occupy an octahedral hole, the radius ratio (r/R) must fall within a certain range. The typical value for this range is to . Based on this, the relation is the correct one. Step 2: Analyzing the options. (A) r = 2.25 R: This is incorrect, as this ratio is far too large for the octahedral hole. (B) r = 0.514 R: This is the correct ratio, as it is within the acceptable range for an atom to fit into the octahedral hole. (C) r = 0.225 R: This ratio is too small for an atom to occupy an octahedral hole. (D) r = 0.314 R: While close, this is still not the ideal ratio for an octahedral hole. Step 3: Conclusion. The correct relation is r = 0.514 R, which corresponds to option (B) .$

21

PYQ 2020

medium

chemistry ID: mht-cet-

$An element crystallizes with hcp structure with atomic radius 17-32 nm. What is the edge length of unit cell?$

1

$29 nm$

2

$33 nm$

3

$64 nm$

4

$40 nm$

Official Solution

Correct Option: $(4)$

$Step 1: Understanding the structure. In hexagonal close-packed (hcp) structure, the relationship between the atomic radius (r) and the edge length (a) is given by: This equation allows us to calculate the edge length of the unit cell using the given atomic radius. Step 2: Analyzing the options. Given that the atomic radius lies between 17 and 32 nm, using the formula will yield the edge length of the unit cell as approximately 40 nm. Step 3: Conclusion. The correct answer is (D) 40 nm, which is the edge length of the unit cell for the given atomic radius.$

22

PYQ 2020

medium

chemistry ID: mht-cet-

$A metallic element has a cubic lattice with edge length of unit cell 2Å. Calculate the number of unit cells in 200 g of the metal, if the density of metal is 2.5 g cm ?$

1

2

3

4

Official Solution

Correct Option: $(3)$

$Step 1: Understanding the relationship. To find the number of unit cells, we first need to calculate the volume of the metal in 200 g. Using the formula: Next, we calculate the volume of one unit cell using the given edge length of 2Å (convert to cm: cm): Step 2: Calculating the number of unit cells. Now, we can calculate the number of unit cells by dividing the total volume by the volume of one unit cell: Step 3: Conclusion. The correct answer is (C), the number of unit cells in 200 g of the metal.$

23

PYQ 2020

medium

chemistry ID: mht-cet-

$Which of the following formulae is used to find edge length of bcc unit cell?$

1

2

3

4

Official Solution

Correct Option: $(3)$

$Step 1: Understanding bcc unit cell. In a body-centred cubic (bcc) unit cell, the relationship between the edge length and the atomic radius is given by: This formula is derived from the geometry of the bcc unit cell. Step 2: Analyzing the options. (A) : This is incorrect; the correct formula has in the numerator, not in the denominator. (B) : This is incorrect, as it does not correspond to the correct formula for bcc unit cells. (C) : This is the correct formula for the edge length of a bcc unit cell. (D) : This is incorrect and does not match the standard formula for bcc. Step 3: Conclusion. The correct formula to find the edge length of a bcc unit cell is (C) .$

24

PYQ 2020

medium

chemistry ID: mht-cet-

$What is the number of atoms in unit cells if an element crystallises in bcc structure?$

1

2

3

4

Official Solution

Correct Option: $(3)$

$Step 1: Recall atoms per unit cell in BCC. In a body-centred cubic (bcc) structure, each unit cell contains 2 atoms. Step 2: Multiply by number of unit cells. Number of unit cells given Step 3: Calculate total atoms. Step 4: Conclusion. The total number of atoms present is .$

25

PYQ 2020

medium

chemistry ID: mht-cet-

$What is the radius of sodium atom if it crystallizes in fcc structure with edge length of unit cell ?$

1

2

3

4

Official Solution

Correct Option: $(3)$

$Step 1: Using the relationship for fcc crystals. For an fcc structure, the edge length is related to the atomic radius by the equation: Given that the edge length, we can solve for : Step 2: Analyzing the options. (A) : Incorrect. This is too large. (B) : Incorrect. This is too small. (C) : Correct — This is the correct atomic radius for sodium in fcc structure. (D) : Incorrect. This is not the correct radius. Step 3: Conclusion. The correct answer is (C) .$

26

PYQ 2020

medium

chemistry ID: mht-cet-

$What are the constituents of Natalite?$

1

$Dimethyl ether and methanol$

2

$Dimethyl ether and ethanol$

3

$Diethyl ether and ethanol$

4

$Diethyl ether and methanol$

Official Solution

Correct Option: $(3)$

$Step 1: Understanding Natalite. Natalite is a mixture of diethyl ether and ethanol. It is used as a solvent and in chemical processes. Step 2: Analyzing the options. (A) Dimethyl ether and methanol: Incorrect. This combination does not form Natalite. (B) Dimethyl ether and ethanol: Incorrect. This combination is not the correct constituent of Natalite. (C) Diethyl ether and ethanol: Correct — This is the correct combination forming Natalite. (D) Diethyl ether and methanol: Incorrect. This is not the correct constituent of Natalite. Step 3: Conclusion. The correct answer is (C) Diethyl ether and ethanol, as these are the constituents of Natalite.$

27

PYQ 2020

medium

chemistry ID: mht-cet-

$Which among the following is NOT a polar molecular solid?$

1

2

3

4

Official Solution

Correct Option: $(1)$

$Step 1: Understand polar molecular solids. Polar molecular solids are composed of polar molecules held together by dipole-dipole interactions. These molecules have a permanent dipole moment. Step 2: Analyse methane (). Methane is a symmetrical tetrahedral molecule. Although the C-H bonds are slightly polar, their dipole moments cancel out due to symmetry. Hence, methane is a non-polar molecule . Step 3: Analyse other options. (B) : Bent molecule with a net dipole moment \to polar. (C) : Polar covalent molecule \to polar. (D) : Bent shape with net dipole moment \to polar. Step 4: Conclusion. Methane does not form a polar molecular solid.$

28

PYQ 2020

medium

chemistry ID: mht-cet-

$Which of the following molecules does not contain oxygen?$

1

$Serotonin$

2

$Phenelzine$

3

$Veronal$

4

$Iproniazid$

Official Solution

Correct Option: $(2)$

$Step 1: Identifying the presence of oxygen. Serotonin, Veronal, and Iproniazid all contain oxygen atoms in their structures. Phenelzine does not contain oxygen. Step 2: Analyzing the options. (A) Serotonin: Contains oxygen. (B) Phenelzine: Correct — Does not contain oxygen. (C) Veronal: Contains oxygen. (D) Iproniazid: Contains oxygen. Step 3: Conclusion. The correct answer is (B) Phenelzine, as it does not contain oxygen.$

29

PYQ 2020

medium

chemistry ID: mht-cet-

$An element crystallizes in a bcc lattice with cell edge of 500 pm. The density of the element is 7.5 g/cm . How many atoms are present in 300 g of metal?$

1

$atoms$

2

$atoms$

3

$atoms$

4

$atoms$

Official Solution

Correct Option: $(2)$

$Step 1: Formula for number of atoms. The number of atoms can be calculated using the formula: Where: - is Avogadro's number. Step 2: Calculation. For a bcc lattice, the number of atoms per unit cell is 2. The volume of the unit cell can be calculated using the formula for the volume of a cube: Where, so: The density is given as 7.5 g/cm, so the mass of the unit cell is: Now, the number of unit cells in 300 g of the metal is: Since each unit cell contains 2 atoms, the total number of atoms is: Step 3: Conclusion. The correct answer is (B) atoms .$

30

PYQ 2020

medium

chemistry ID: mht-cet-

$What is the packing efficiency of fcc crystal structure?$

1

$74.0 %$

2

$68.04 %$

3

$52.4 %$

4

$47.6 %$

Official Solution

Correct Option: $(1)$

$Step 1: Recall packing efficiency. Packing efficiency is the fraction of the total volume of the unit cell that is actually occupied by the constituent particles (atoms/ions). Step 2: fcc crystal characteristics. In a face-centered cubic (fcc) unit cell: Number of atoms per unit cell . Atoms touch along the face diagonal. Step 3: Standard result. For fcc structure, the packing efficiency has a well-known calculated value of . Step 4: Conclusion. The packing efficiency of an fcc crystal structure is .$

31

PYQ 2020

medium

chemistry ID: mht-cet-

$How many unit cells are present in 100 g of an element with fcc crystal structure having density and edge length ?$

1

2

3

4

Official Solution

Correct Option: $(4)$

$Step 1: Convert edge length into cm. Step 2: Calculate volume of one unit cell. Step 3: Find total volume of the element. Step 4: Calculate number of unit cells. Step 5: Conclusion. The number of unit cells present is .$

32

PYQ 2020

medium

chemistry ID: mht-cet-

$What is the value of radius ratio of ionic crystal having coordination number six?$

1

$Greater than 0.732$

2

$In between 0.414 to 0.732$

3

$In between 0.225 to 0.414$

4

$Less than 0.225$

Official Solution

Correct Option: $(2)$

$Step 1: Understanding Radius Ratio for Coordination Number 6. For ionic crystals, the radius ratio defines the relative sizes of the cation and anion. For a coordination number of six, the radius ratio (r /r) should be between 0.414 and 0.732. This range allows for an efficient packing arrangement in the crystal lattice. Step 2: Analyzing the options. (A) Greater than 0.732: Incorrect. This is the range for coordination numbers greater than 6. (B) In between 0.414 to 0.732: Correct. This is the correct range for the radius ratio for ionic crystals with a coordination number of 6. (C) In between 0.225 to 0.414: Incorrect. This range corresponds to crystals with lower coordination numbers. (D) Less than 0.225: Incorrect. This range is not applicable to coordination number 6. Step 3: Conclusion. The correct range for the radius ratio of ionic crystals with a coordination number of six is between 0.414 and 0.732, corresponding to option (B) .$

33

PYQ 2020

medium

chemistry ID: mht-cet-

$Calcite crystals used in Nicol’s prism are formed of:$

1

$CaC$

2

$CaCO$

3

$CaCl$

4

$CaO$

Official Solution

Correct Option: $(2)$

$Step 1: Understanding Nicol’s Prism. Nicol’s prism is made from calcite crystals. These crystals are primarily composed of calcium carbonate (CaCO), which is a mineral form of calcium carbonate. The other compounds listed are not associated with the material used in Nicol’s prism. Step 2: Analyzing the options. (A) CaC : Incorrect. This is calcium carbide, which is not used in Nicol’s prism. (B) CaCO : Correct. Calcite is a naturally occurring form of calcium carbonate, used to make Nicol’s prism. (C) CaCl : Incorrect. This is calcium chloride, which is not used in Nicol’s prism. (D) CaO: Incorrect. This is calcium oxide, which is not used to make Nicol’s prism. Step 3: Conclusion. The correct compound is CaCO, corresponding to option (B) .$

34

PYQ 2020

medium

chemistry ID: mht-cet-

$What is the percentage of void space in bcc type of unit cell?$

1

2

3

4

Official Solution

Correct Option: $(4)$

$Step 1: Packing efficiency of BCC lattice. The packing efficiency of a body-centered cubic (bcc) unit cell is approximately . This represents the fraction of space occupied by atoms. Step 2: Calculation of void space. Void space is the unoccupied volume in the unit cell and is calculated as: Step 3: Substitution of values. Step 4: Conclusion. Hence, the percentage of void space in a bcc unit cell is .$

35

PYQ 2020

medium

chemistry ID: mht-cet-

$Gold crystallises in fcc structure with edge length 396 pm, find atomic radius of gold?$

1

$198 pm$

2

$714 pm$

3

$140 pm$

4

$162 pm$

Official Solution

Correct Option: $(3)$

$Step 1: Relationship in fcc structure. In an fcc (face-centered cubic) structure, the relationship between the edge length and the atomic radius is given by: Given that the edge length, we can calculate the atomic radius as: Step 2: Analyzing the options. (A) 198 pm: Incorrect. This is not the correct atomic radius. (B) 714 pm: Incorrect. This value is too large for the atomic radius. (C) 140 pm: Correct — This is the correct atomic radius based on the given edge length. (D) 162 pm: Incorrect. This is not the correct radius. Step 3: Conclusion. The correct answer is (C) 140 pm, which is the atomic radius of gold in an fcc structure.$

36

PYQ 2020

medium

chemistry ID: mht-cet-

$What is the mass of the unit cell of gold if it crystallizes in fcc structure?$

1

2

3

4

Official Solution

Correct Option: $(4)$

$Step 1: Understanding the fcc structure. In the fcc (face-centered cubic) structure, there are 4 atoms per unit cell. The formula for the mass of the unit cell is: Given that the molar mass of gold is 197 g/mol, the mass of the unit cell is calculated as: Step 2: Analyzing the options. (A) : Incorrect. This is not the correct mass. (B) : Incorrect. This is too small. (C) : Incorrect. This is not the correct mass. (D) : Correct — This is the correct mass of the unit cell of gold in fcc structure. Step 3: Conclusion. The correct answer is (D) .$

37

PYQ 2020

medium

chemistry ID: mht-cet-

$Which among the following lanthanoids shows only +3 oxidation state?$

1

$Terbium$

2

$Gadolinium$

3

$Neodymium$

4

$Cerium$

Official Solution

Correct Option: $(2)$

$Step 1: Understanding the oxidation states of lanthanoids. Lanthanoids typically exhibit a +3 oxidation state, but some elements like cerium can also show +4 oxidation states. Gadolinium, however, exclusively exhibits the +3 oxidation state. Step 2: Analyzing the options. (A) Terbium: Terbium primarily exhibits the +3 oxidation state, though it can also exist in other states. (B) Gadolinium: Gadolinium exclusively shows the +3 oxidation state. (C) Neodymium: Neodymium generally shows +3 oxidation state but can also show +2 in some compounds. (D) Cerium: Cerium can show both +3 and +4 oxidation states. Step 3: Conclusion. The correct answer is (B) Gadolinium, as it only exhibits the +3 oxidation state.$

38

PYQ 2020

medium

chemistry ID: mht-cet-

$How many electrons are involved in the reaction when 0.40 F of electricity is passed through an electrolytic solution?$

1

2

3

4

Official Solution

Correct Option: $(2)$

$Step 1: Using the relation between charge and moles of electrons. The amount of electricity in Faradays (F) is related to the number of moles of electrons by the equation: Where: - (charge passed), - (Faraday constant). Step 2: Calculation. We can calculate the number of electrons as: Step 3: Conclusion. The correct answer is (B) .$

39

PYQ 2020

medium

chemistry ID: mht-cet-

$Which among the following elements when added to silicon forms P-type semiconductor?$

1

$Sb$

2

$Bi$

3

$B$

4

$As$

Official Solution

Correct Option: $(3)$

$Step 1: Understand P-type semiconductors. P-type semiconductors are formed when a trivalent (Group 13) impurity is added to a tetravalent semiconductor like silicon. This creates holes as the majority charge carriers. Step 2: Analyze the given options. (A) Sb: Antimony is pentavalent and forms N-type semiconductor. (B) Bi: Bismuth is also pentavalent and forms N-type semiconductor. (C) B: Boron is trivalent and creates holes, hence forms P-type semiconductor. (D) As: Arsenic is pentavalent and forms N-type semiconductor. Step 3: Conclusion. Boron forms P-type semiconductor when added to silicon.$

40

PYQ 2020

medium

chemistry ID: mht-cet-

$Which among the following crystal structures the edge length of unit cell is equal to twice the radius of one atom?$

1

$End-centred orthorhombic$

2

$Simple cubic$

3

$Face centred cubic$

4

$Body centred cubic$

Official Solution

Correct Option: $(2)$

$Step 1: Recall atom contact in different unit cells. In different crystal structures, atoms touch each other along different directions. Step 2: Simple cubic structure. In a simple cubic unit cell, atoms touch each other along the edge of the cube. Therefore, where is the edge length and is the atomic radius. Step 3: Compare with other structures. In body centred cubic: In face centred cubic: Step 4: Conclusion. The edge length is equal to twice the atomic radius only in the simple cubic structure.$

41

PYQ 2020

medium

chemistry ID: mht-cet-

$Identify the product ‘A’ in the following reaction: \text{Aniline} + (CH CO) O A$

1

$Acetanilide$

2

$Sulphanilic acid$

3

$P-nitroacetanilide$

4

$Benzenediazonium chloride$

Official Solution

Correct Option: $(1)$

$Step 1: Understanding the reaction. In the given reaction, acetic anhydride (Ac O) reacts with aniline in the presence of pyridine, a base, to form acetanilide. This is an example of an acylation reaction. Step 2: Analyzing the options. (A) Acetanilide: This is the correct product formed when aniline reacts with acetic anhydride. (B) Sulphanilic acid: This is an aromatic amine derivative but not the product of this reaction. (C) P-nitroacetanilide: This product requires a nitro group on the aniline ring, which is not introduced in this reaction. (D) Benzenediazonium chloride: This compound is unrelated to the acylation reaction described here. Step 3: Conclusion. The correct answer is (A) Acetanilide .$

42

PYQ 2020

medium

chemistry ID: mht-cet-

$What is the type of hole occupied if limiting value of is in the range of to ?$

1

$Octahedral$

2

$Cubic$

3

$Planar triangular$

4

$Tetrahedral$

Official Solution

Correct Option: $(4)$

$Step 1: Radius ratio rule. The type of interstitial hole occupied by a cation depends on the radius ratio . Step 2: Given range. For tetrahedral voids, the radius ratio lies between and . Step 3: Identification of hole type. Since the given radius ratio falls in this range, the cation occupies a tetrahedral hole. Step 4: Conclusion. Therefore, the correct answer is Tetrahedral .$

43

PYQ 2020

medium

chemistry ID: mht-cet-

$An element crystallizes in bcc type of unit cell, the density and edge length of unit cell is 4 g cm and 500 pm respectively. What is the atomic mass of an element?$

1

$125.5$

2

$100.1$

3

$250.0$

4

$150.0$

Official Solution

Correct Option: $(4)$

$Step 1: Using the formula for density of bcc unit cell. For a bcc unit cell, the formula for density is given by: Where is the number of atoms per unit cell (for bcc,), is the atomic mass, is the edge length, and is Avogadro's number. Step 2: Substituting the given values. Given: Using the formula, we calculate the atomic mass to be 150.0 g/mol. Step 3: Conclusion. The correct answer is (D) 150.0, as this is the atomic mass of the element.$

44

PYQ 2020

medium

chemistry ID: mht-cet-

$Nickel crystallises in a fcc type of unit cell, with edge length 0.3524 nm. Calculate the radius of the nickel atom.$

1

$0.1624 nm$

2

$0.2164 nm$

3

$0.1426 nm$

4

$0.1246 nm$

Official Solution

Correct Option: $(4)$

$Step 1: Understanding the fcc unit cell. In a face-centered cubic (fcc) unit cell, the relation between the edge length and the atomic radius is given by: Given that the edge length, we can solve for . Step 2: Solving for . Step 3: Conclusion. The correct answer is (D) 0.1246 nm, which is the radius of the nickel atom.$

45

PYQ 2020

medium

chemistry ID: mht-cet-

$An element crystallizes in bcc type of unit cell having atomic radius cm, the edge length of unit cell will be$

1

$cm$

2

$cm$

3

$cm$

4

$cm$

Official Solution

Correct Option: $(4)$

$Step 1: Formula for edge length in bcc unit cell. For a bcc unit cell, the relationship between the atomic radius () and the edge length () is given by: Step 2: Substituting the values. Given the atomic radius cm, we can substitute this into the formula: Step 3: Conclusion. The correct answer is (D) cm, as this is the edge length of the unit cell.$

46

PYQ 2020

medium

chemistry ID: mht-cet-

$The edge length of an fcc type unit cell of copper having atomic radius is equal to$

1

2

3

4

Official Solution

Correct Option: $(2)$

$Step 1: Relation for FCC unit cell. In an FCC lattice, atoms touch along the face diagonal. The relation between edge length and atomic radius is: Step 2: Substitute the given value. Step 3: Calculate the edge length. Step 4: Conclusion. Hence, the edge length of the unit cell is .$

47

PYQ 2020

medium

chemistry ID: mht-cet-

$What is the number of atoms present per unit cell of aluminium having edge length ? (Given: Density of Al =, Atomic mass of Al = 27)$

1

$8$

2

$1$

3

$2$

4

$4$

Official Solution

Correct Option: $(4)$

$Step 1: Use density formula for unit cell. Step 2: Substitute given values. Step 3: Solve for . On solving, we get: Step 4: Conclusion. Thus, aluminium unit cell contains 4 atoms, confirming FCC structure.$

48

PYQ 2020

medium

chemistry ID: mht-cet-

$If 2 moles of an ideal gas at 546 K has a volume of 44.8 L, then what will be its pressure?$

1

$1.998 atm$

2

$1.098 atm$

3

$2.408 atm$

4

$3.129 atm$

Official Solution

Correct Option: $(1)$

$Step 1: Use the ideal gas law. The ideal gas law is given by: Where: - is the pressure, - is the volume, - is the number of moles of gas, - is the ideal gas constant (), - is the temperature in Kelvin. Given: -, -, -, - . Substitute the values into the ideal gas law: Step 2: Conclusion. The correct answer is (A) 1.998 atm .$

49

PYQ 2020

medium

chemistry ID: mht-cet-

$Which among the following is an alloy of antimony, tin and copper?$

1

$Babbitt metal$

2

$Spiegeleisen$

3

$Duralumin$

4

$Stainless steel$

Official Solution

Correct Option: $(1)$

$Step 1: Understanding Babbitt metal. Babbitt metal is a type of bearing metal, and it is an alloy of antimony, tin, and copper. It is used in high-speed machinery for bearings due to its ability to reduce friction. Step 2: Analyzing the options. (A) Babbitt metal: Correct — Babbitt metal is an alloy of antimony, tin, and copper. (B) Spiegeleisen: Spiegeleisen is an alloy of iron with high carbon and is used for making steel, not an alloy of antimony, tin, and copper. (C) Duralumin: Duralumin is an alloy of aluminum, copper, and other elements but does not contain antimony or tin. (D) Stainless steel: Stainless steel is an alloy of iron, chromium, and sometimes nickel, not an alloy of antimony, tin, and copper. Step 3: Conclusion. The correct answer is (A) Babbitt metal .$

50

PYQ 2020

medium

chemistry ID: mht-cet-

$Which among the following elements has the highest number of atoms in 1 g each? (at. no. Au = 197, Na = 23, Cu = 63.5, Fe = 56)$

1

$Cu(s)$

2

$Na(s)$

3

$Au(s)$

4

$Fe(s)$

Official Solution

Correct Option: $(2)$

$Step 1: Understanding Avogadro’s number and molar mass. The number of atoms in 1 g of a substance is given by the formula: The substance with the smallest molar mass will have the highest number of atoms in 1 g. Step 2: Calculation. - For Na: atoms. - For Cu: atoms. - For Au: atoms. - For Fe: atoms. Step 3: Conclusion. Sodium (Na) has the highest number of atoms in 1 g. The correct answer is (B) Na(s) .$

51

PYQ 2020

medium

chemistry ID: mht-cet-

$What is the oxidation state of chlorine atom in chloric acid?$

1

$+3$

2

$-1$

3

$+5$

4

$+1$

Official Solution

Correct Option: $(3)$

$Step 1: Understanding the oxidation state of chlorine. In chloric acid (HClO), hydrogen has an oxidation state of +1, and oxygen typically has an oxidation state of -2. The overall charge of the molecule is 0. To find the oxidation state of chlorine, we can use the following equation: Solving for, we get . Thus, the oxidation state of chlorine in HClO is +5. Step 2: Analyzing the options. (A) +3: This is incorrect. The oxidation state of chlorine in HClO is +5, not +3. (B) -1: This is incorrect. Chlorine cannot have an oxidation state of -1 in HClO . (C) +5: Correct — Chlorine has an oxidation state of +5 in chloric acid. (D) +1: This is incorrect. Chlorine cannot have an oxidation state of +1 in HClO . Step 3: Conclusion. The oxidation state of chlorine in chloric acid is (C) +5 .$

52

PYQ 2020

medium

chemistry ID: mht-cet-

$Sodium crystallizes in bcc structure with radius Calculate the edge length of the unit cell?$

1

2

3

4

Official Solution

Correct Option: $(1)$

$Step 1: Understanding the bcc structure. In a body-centered cubic (bcc) structure, the relationship between the atomic radius and the edge length of the unit cell is given by: Thus, we can calculate the edge length using: Step 2: Substituting the known value. Given the atomic radius, we can substitute into the equation: Step 3: Conclusion. The correct edge length of the unit cell is$

53

PYQ 2020

medium

chemistry ID: mht-cet-

$Aluminium crystallises in a face-centred cubic structure, its atomic radius is 125 pm. What is the edge length of unit cell?$

1

$280 pm$

2

$353.5 pm$

3

$335.5 pm$

4

$288.6 pm$

Official Solution

Correct Option: $(2)$

$Step 1: Formula for edge length of unit cell. For a face-centred cubic (FCC) structure, the relationship between the atomic radius and the edge length of the unit cell is given by: Where . Step 2: Calculation. Substitute the given value of : Step 3: Conclusion. The correct answer is (B) 353.5 pm.$

54

PYQ 2020

medium

chemistry ID: mht-cet-

$An element crystallizes in a fcc lattice with cell edge 250 pm. Calculate the density of an element (at mass = 90.3).$

1

$23.12 g cm$

2

$19.20 g cm$

3

$48.40 g cm$

4

$38.40 g cm$

Official Solution

Correct Option: $(4)$

$Step 1: Using the formula for density. Density is given by: Where: - for fcc (face-centered cubic), -, -, - . Step 2: Calculation. Substitute the values: This results in a density of . Step 3: Conclusion. The correct answer is (D) 38.40 g cm .$

55

PYQ 2020

medium

chemistry ID: mht-cet-

$An element with density 2.8 g cm forms FCC unit cell having edge length 4 \times 10 cm. Calculate molar mass of the element.$

1

$33.0 g mol$

2

$22.0 g mol$

3

$27.0 g mol$

4

$36.0 g mol$

Official Solution

Correct Option: $(1)$

$Step 1: Formula for molar mass. Using the formula for molar mass from the unit cell density: Where: - is the density (2.8 g/cm³), - is Avogadro's number, - is the edge length (4 \times 10 cm), - is the number of atoms per unit cell (4 for FCC). Step 2: Calculation. Substituting the values into the formula, we get: This gives the molar mass as 33.0 g/mol. Step 3: Conclusion. The correct answer is (A) 33.0 g mol .}$

56

PYQ 2020

medium

chemistry ID: mht-cet-

$Lithium crystallises into a body centered cubic structure. What is the radius of lithium if the edge length of its unit cell is ?$

1

2

3

4

Official Solution

Correct Option: $(1)$

$Step 1: Relation between edge length and atomic radius in BCC. For a body centered cubic (BCC) lattice, the atoms touch along the body diagonal. The relation is: Step 2: Substitute the given value. Step 3: Calculate the radius. Step 4: Conclusion. Hence, the radius of lithium atom is .$

57

PYQ 2020

medium

chemistry ID: mht-cet-

$An element crystallizes in bcc type crystal structure with edge length of unit cell 300 pm. Calculate radius of element?$

1

$2.299 \times 10^{-8} \, \text{cm}$

2

$1.299 \times 10^{-8} \, \text{cm}$

3

$6.920 \times 10^{-8} \, \text{cm}$

4

$1.440 \times 10^{-8} \, \text{cm}$

Official Solution

Correct Option: $(2)$

$Step 1: Understanding the bcc structure. In a body-centered cubic (bcc) structure, the relation between the edge length and the radius of the atom is given by: From this, we can derive the radius as: Step 2: Calculation of the radius. Given , we can substitute into the equation: Step 3: Conclusion. The correct radius of the element is 1.299 \times 10^{-8} \, \text{cm}.$

58

PYQ 2022

easy

chemistry ID: mht-cet-

$What is the co-ordination number of hcp crystal lattice?$

1

$8$

2

$12$

3

$6$

4

$6$

Official Solution

Correct Option: $(2)$

$Concept: The coordination number in crystallography refers to the number of nearest neighboring atoms or ions that are directly in contact with a given atom or ion in a crystal structure. It is a crucial concept that determines how atoms are arranged and how tightly they are packed in a solid. Hexagonal Close-Packed (hcp) Structure: In a hexagonal close-packed (hcp) crystal lattice, atoms are arranged in a compact and efficient three-layered structure. The stacking follows an ABAB pattern, where: - The first layer (A) is placed at the base. - The second layer (B) is placed in the depressions of the first layer. - The third layer is aligned exactly above the first layer (A). This arrangement allows the atoms to occupy maximum space efficiently. Coordination Number in hcp: In an hcp lattice: - Each atom is in contact with 6 atoms in its own layer. - It also touches 3 atoms in the layer above. - And 3 atoms in the layer below. This gives a total of 12 nearest neighbors, which means the coordination number is 12. This high coordination number is responsible for the excellent packing efficiency of about 74 percent, making hcp one of the most efficient packing structures in solids. Examples of Elements with hcp Structure: - Magnesium (Mg) - Zinc (Zn) - Titanium (Ti) Conclusion: The hexagonal close-packed (hcp) crystal structure has a coordination number of 12. Each atom is closely surrounded by 12 other atoms, resulting in a tightly packed, stable structure. Therefore, the correct option is: (B) 12 .$

59

PYQ 2022

easy

chemistry ID: mht-cet-

$Calculate the number of atoms in 20 gram metal which crystallises to simple cubic structure having unit cell edge length 340 pm. (density of metal = 9.8 gcm -3)$

1

$4.95 x 10 22$

2

$5.81 x10 22$

3

$5.19 x 10 22$

4

$5.42 x 10 22$

Official Solution

Correct Option: $(3)$

$Mass of the metal = 20 g Density of the metal = 9.8 g/cm³ Unit cell edge length (a) = 340 pm = 340 \times 10 -10 m Volume of a cubic unit cell = a 3 Number of unit cells = Convert the mass to kilograms: Mass = 20 g = 20 \times 10 -3 kg Number of unit cells = Now, we need to calculate the number of atoms per unit cell in a simple cubic structure, which is 1. Number of atoms = Number of unit cells \times Number of atoms per unit cell Solving the calculations: Volume = (340 \times 10 -10 m) 3 = 39.304 \times 10 -27 m³ Number of unit cells = Number of unit cells = 5.194 \times 10 22 Number of atoms = (5.194 \times 10 22) \times 1 Therefore, the correct option is (C) 5.19 \times 10 22 atoms.$

60

PYQ 2022

easy

chemistry ID: mht-cet-

$Which of the following compounds is likely to show both Frenkel and Schottky defects in its crystalline form?$

1

$ZnS$

2

$CsCl$

3

$KBr$

4

$AgBr$

Official Solution

Correct Option: $(1)$

61

PYQ 2022

easy

chemistry ID: mht-cet-

$How many atoms are in a bcc unit cell?$

Official Solution

Correct Option: $(1)$

$In a body-centered cubic (BCC) unit cell, there are two atoms. One atom is located at the center of the cube, and the other atom is positioned at each of the eight corners of the cube. Each corner atom is shared between eight adjacent unit cells, so it contributes only th to the total count. Thus, the total number of atoms in a BCC unit cell is: 1 (center atom) + x 8 (corner atoms) = 2 atoms.$

62

PYQ 2023

easy

chemistry ID: mht-cet-

$For a BCC structure, if a = 351 pm, find r. Lithium forms a BCC structure having an edge length of a unit cell 351 pm, then find the atomic radius of lithium.$

Official Solution

Correct Option: $(1)$

$Relationship between the lattice constant (a) and atomic radius (r) for a BCC structure: For a body-centered cubic (BCC) structure, the relationship between the lattice constant and the atomic radius is given by: Step 1: Solve for r: To find the atomic radius, we can rearrange the formula: Step 2: Substitute the given value of : The given lattice constant for lithium is 351 pm (or 3.51 Å). Substituting this value into the equation, we get: Step 3: Simplify the expression: Now, simplify the expression: Next, calculate which equals approximately 5.656, so we get: Final Answer: Therefore, the atomic radius of lithium in its BCC structure is approximately 1.53 Å.$

63

PYQ 2023

medium

chemistry ID: mht-cet-

$What type of bonds are present in molecular crystals?$

Official Solution

Correct Option: $(1)$

64

PYQ 2023

medium

chemistry ID: mht-cet-

$Edge length of a unit cell of a crystal is 288 pm. If its density is 7.2 g/cm3, then determine the type of unit cell assuming mass = 52 g.$

Official Solution

Correct Option: $(1)$

$We can use the formula for the density of a crystal in terms of its unit cell parameters: where: is the density of the crystal, Z is the number of atoms per unit cell, M is the molar mass of the substance, V is the volume of the unit cell, is Avogadro's constant. Step 1: Calculate the volume of the unit cell: To determine the type of unit cell, we first need to calculate the volume of the unit cell using the edge length. For a cubic unit cell, the volume is given by: where is the edge length of the unit cell. Substituting the given values, we get: Step 2: Use the formula for density to solve for Z: Now that we know the volume, we can use the formula for density to solve for, the number of atoms per unit cell: Substituting the given values into the equation: Simplifying the equation to solve for : After evaluating the equation, we get: Step 3: Determine the unit cell type: The value of suggests that the crystal has a face-centered cubic (FCC) unit cell. In an FCC structure, there are 4 atoms per unit cell, with atoms located at the corners and at the center of each face of the cube. Conclusion: Therefore, we can conclude that the crystal has a face-centered cubic (FCC) structure.$

65

PYQ 2024

medium

chemistry ID: mht-cet-

$If the length of the body diagonal of a FCC unit cell is, the distance between two octahedral voids in the cell in is:$

1

2

3

4

Official Solution

Correct Option: $(3)$

$In a face-centered cubic (FCC) unit cell, octahedral voids are located at the body center and edge centers. To calculate the distance between two octahedral voids: Step 1: The length of the body diagonal is: where is the edge length of the unit cell. Step 2: From the body diagonal length, the edge length is: Step 3: The distance between two octahedral voids is along the body diagonal. Since they are located at the body center and edge centers, the distance is: Substitute : Final Answer:$

66

PYQ 2024

medium

chemistry ID: mht-cet-

$The density of a face-centered cubic (FCC) crystal is:$

1

2

3

4

Official Solution

Correct Option: $(2)$

$In a face-centered cubic (FCC) crystal structure, there are 4 atoms per unit cell. The density of a substance is defined as the mass per unit volume. For an FCC structure, the mass of the unit cell can be calculated by multiplying the number of atoms per unit cell (4) by the atomic mass . The volume of the unit cell is, where is the edge length of the unit cell. Thus, the density of an FCC crystal is: This formula represents the density in terms of the atomic mass and the edge length of the unit cell.$

67

PYQ 2024

medium

chemistry ID: mht-cet-

$What is the coordination number in a Hexagonal Close-Packed (hcp) structure?$

1

$8$

2

$10$

3

$12$

4

$14$

Official Solution

Correct Option: $(3)$

$In a Hexagonal Close-Packed (hcp) structure, the coordination number is 12. This is because each atom in the hcp arrangement is surrounded by 12 neighboring atoms. Specifically, 6 atoms are arranged in a plane around a central atom, and 3 atoms are positioned above and 3 atoms are positioned below the central atom, making a total of 12 nearest neighbors. This is a characteristic feature of hcp structures, where the packing is highly efficient, and the atoms are arranged in a way that maximizes the number of nearest neighbors.$

68

PYQ 2024

medium

chemistry ID: mht-cet-

$What is the edge length of a BCC unit cell?$

Official Solution

Correct Option: $(1)$

$For a BCC unit cell, the relation between edge length () and radius () of the atom is: This relation is derived from the geometry of the BCC unit cell. Conclusion: Edge length is .$

69

PYQ 2024

easy

chemistry ID: mht-cet-

$How many unit particles are present in a BCC (Body-Centered Cubic) unit cell?$

1

$2$

2

$1$

3

$4$

4

$3$

Official Solution

Correct Option: $(1)$

$In a BCC unit cell: - There is 1 atom at the center of the cube and 8 atoms at the corners. - Each corner atom contributes, so total corner contributions = . - Adding the center atom, total = . Conclusion: The total number of particles in a BCC unit cell is .$

70

PYQ 2024

medium

chemistry ID: mht-cet-

$The number of nearest neighbours in a BCC unit cell is:$

1

2

3

4

Official Solution

Correct Option: $(2)$

$In a Body-Centered Cubic (BCC) unit cell, the arrangement consists of: - One atom positioned at the center of the cube, - Eight atoms placed at the corners of the cube. The central atom is in contact with the eight corner atoms, making them its nearest neighbors. Therefore, the coordination number of a BCC lattice is . Other options: - A coordination number of corresponds to Face-Centered Cubic (FCC) structures. - A coordination number of applies to Simple Cubic structures. - A coordination number of is not applicable to BCC. Final Answer:$

71

PYQ 2025

medium

chemistry ID: mht-cet-

$Identify the element having highest ionisation enthalpy () from following.$

1

$Ce$

2

$La$

3

$Gd$

4

$Yb$

Official Solution

Correct Option: $(4)$

$Step 1: Concept Ionisation enthalpy generally increases across the lanthanide series due to the increase in effective nuclear charge (Lanthanide Contraction). Step 2: Electronic Configuration - (): - (): - (): - (): Step 3: Analysis Yb has a completely filled orbital () and orbital (), making it exceptionally stable. Additionally, it is located furthest to the right among the given options. Final Answer: (D)$

72

PYQ 2025

medium

chemistry ID: mht-cet-

$Calculate the volume occupied by a particle in fcc unit cell if volume of unit cell is .$

1

2

3

4

Official Solution

Correct Option: $(2)$

$Step 1: Formula Packing Efficiency of FCC = 74% = 0.74. Packing Efficiency Step 2: Volume of 4 atoms Total volume of unit cell . Total volume of 4 particles . Step 3: Volume of 1 particle Since FCC has 4 atoms per unit cell: Volume of 1 particle Volume . Final Answer: (B)$

73

PYQ 2025

medium

chemistry ID: mht-cet-

$Identify product '' in the following sequence of reaction.$

1

$Sodium phenoxide$

2

$Nitrobenzene$

3

$Benzenediazonium chloride$

4

$Phenol$

Official Solution

Correct Option: $(4)$

$Step 1: Diazotization Aniline reacts with at to form Benzenediazonium chloride (Product 'A'). Step 2: Hydrolysis When an aqueous solution of Benzenediazonium chloride is warmed (), it undergoes hydrolysis. Step 3: Conclusion The reaction is: . Product 'B' is Phenol. Final Answer: (D)$

74

PYQ 2025

medium

chemistry ID: mht-cet-

$Which of the following is one of the product of ozonolysis?$

1

$alcohol$

2

$acid$

3

$aldehyde$

4

$ester$

Official Solution

Correct Option: $(3)$

$Concept: Ozonolysis cleaves carbon-carbon double bonds forming carbonyl compounds. Step 1: Reaction mechanism. Step 2: Products formed. Aldehydes Ketones Step 3: Conclusion. Aldehyde is one of the products.$

75

PYQ 2025

medium

chemistry ID: mht-cet-

$Identify the product formed from chlorobenzene on heating with conc. in presence of conc. .$

1

$Only 1-chloro-4-nitrobenzene$

2

$1-chloro-2-nitrobenzene$

3

$Mixture of 1-chloro-4-nitrobenzene and 1-chloro-2-nitrobenzene$

4

$2,4,6-trinitrochlorobenzene$

Official Solution

Correct Option: $(3)$

$Concept: Nitration of aromatic compounds is an electrophilic substitution reaction . Substituents already present on benzene ring influence the position of incoming group. Step 1: Nature of substituent. Chlorine is: Electron withdrawing (effect) But ortho-para directing due to resonance Step 2: Position of substitution. Nitration occurs at: Step 3: Products formed. 1-chloro-2-nitrobenzene (ortho) 1-chloro-4-nitrobenzene (para) Step 4: Conclusion. A mixture of ortho and para products is formed.$

76

PYQ 2025

medium

chemistry ID: mht-cet-

$Which from following polymers needs peroxide as initiator for preparation?$

1

$Nylon 6,6$

2

$Polyacrylonitrile$

3

$Terylene$

4

$Bakelite$

Official Solution

Correct Option: $(2)$

$Concept: Addition polymerization occurs via free radical mechanism and requires initiators like peroxides. Step 1: Classify polymers. Nylon 6,6 â condensation polymer Terylene â condensation polymer Bakelite â condensation polymer Polyacrylonitrile â addition polymer Step 2: Identify initiator requirement. Addition polymers require peroxide initiators. Step 3: Conclusion. Polyacrylonitrile is formed using peroxide.$

77

PYQ 2026

medium

chemistry ID: mht-cet-

$A metal has BCC structure. Atomic radius and molar mass . The density of the metal is:$

1

2

3

4

Official Solution

Correct Option: $(1)$

$Concept: The density of a crystal lattice is given by where = Number of atoms per unit cell = Molar mass = Edge length of the unit cell = Avogadro's number For a Body-Centered Cubic (BCC) structure: Also, the relation between atomic radius and edge length in BCC is Step 1: {Calculate the edge length of the unit cell.} Given Convert to cm: Step 2: {Substitute values into density formula.} Step 3: {Simplify the expression.} Thus, the density of the metal is approximately . Hence, the correct option is (A).$

78

PYQ 2026

medium

chemistry ID: mht-cet-

$How many atoms are present in a unit cell of a body-centered cubic (BCC) lattice?$

1

$1$

2

$2$

3

$3$

4

$4$

Official Solution

Correct Option: $(2)$

$Concept: In a body-centered cubic (BCC) unit cell: • 8 atoms are present at the corners of the cube • 1 atom is present at the center of the cube Each corner atom contributes to the unit cell. Step 1: Contribution from corner atoms. Step 2: Contribution from body-centered atom. The atom at the center belongs completely to the unit cell. Step 3: Total atoms in the unit cell.$

79

PYQ 2026

medium

chemistry ID: mht-cet-

$In a BCC (Body-Centered Cubic) unit cell, what is the coordination number?$

1

2

3

4

Official Solution

Correct Option: $(3)$

$Concept: The coordination number of a crystal structure is the number of nearest neighboring atoms surrounding a given atom. Step 1: Understand BCC structure. In a Body-Centered Cubic (BCC) unit cell: • Atoms are present at the corners of the cube. • One atom is present at the center of the cube. Step 2: Nearest neighbors of the central atom. The atom at the center is directly surrounded by the corner atoms. Step 3: Determine coordination number. Thus, each atom in a BCC structure has 8 nearest neighbors .$

80

PYQ 2026

easy

chemistry ID: mht-cet-

$What is the coordination number of an atom in a Face-Centered Cubic (FCC) crystal structure?$

1

$6$

2

$8$

3

$12$

4

$4$

Official Solution

Correct Option: $(3)$

$Concept: The coordination number of a crystal structure is the number of nearest neighboring atoms surrounding a given atom. In a Face-Centered Cubic (FCC) structure: • Atoms are located at the corners of the cube. • Additional atoms are located at the center of each face. Because of this arrangement, each atom is surrounded by 12 nearest neighboring atoms . Step 1: Understand the FCC arrangement. In an FCC lattice, atoms are present at: • 8 cube corners • 6 face centers Step 2: Determine nearest neighbors. Each atom touches 12 neighboring atoms in the closest packing arrangement. Step 3: State the coordination number.$

81

PYQ 2026

easy

chemistry ID: mht-cet-

$What is the number of unit particles present in a Body-Centered Cubic (BCC) unit cell?$

1

2

3

4

Official Solution

Correct Option: $(2)$

$Concept: A Body-Centered Cubic (BCC) unit cell has atoms located at: • The eight corners of the cube • One atom at the center of the cube However, atoms located at corners are shared among neighboring unit cells. Step 1: Contribution of corner atoms. Each corner atom is shared by unit cells. Contribution of one corner atom: Total contribution from corners: Step 2: Contribution of body-centered atom. The atom at the center belongs completely to the unit cell. Contribution: Step 3: Total number of atoms. Therefore, the number of unit particles in a BCC unit cell is$

High-Yield Trend

Chapter Questions 81 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Chapter Questions
81 MCQs