Class 12 Chemistry MCQ

Chemistry-12 MCQ Q: “Crystalline solids are anisotropic in nature.” What is the meaning of anisotropic in the given statement? a) A regular pattern of arrangement of particles which repeats itself periodically over the entire crystal.b) Different values of some of physical properties are shown when measured along different directions in the same crystals.c) An irregular arrangement of particles over the entire crystal.d) Same values of some of physical properties are shown when measured along different directions in the same crystals.Correct Answer: Different values of some of physical properties are shown when measured along different directions in the same crystals. Class 12 Chemistry MCQs for CBSE, Bihar, UP Board Explanation: What is Anisotropic ? Anisotropic refers to a characteristic of materials where certain physical properties vary depending on the direction in which they are measured. In the context of crystalline solids, this means that when you measure specific properties, such as hardness, electrical conductivity, or refractive index, in different directions within the same crystal, you will obtain different values. Crystalline solids have a structured and orderly arrangement of particles, like atoms or molecules, that repeat in a regular pattern. This regularity can lead to variations in how the material interacts with external forces or fields based on the orientation of those forces relative to the crystal structure. For example, quartz crystals exhibit different hardness along different axes, and graphite conducts electricity well along one direction but poorly in others. Option B accurately describes this behavior by stating that different values of some physical properties are observed when measured along different directions in the same crystal. This directional dependence of properties is what makes a crystalline solid anisotropic.

Chemistry-12 MCQ Q: The Ca2+ and F– are located in CaF2 crystal, respectively at face centred cubic lattice points and in a) tetrahedral voidsb) half of tetrahedral voidsc) octahedral voidsd) half of octahedral voidsCorrect Answer: tetrahedral voids Class 12 Chemistry MCQs for CBSE, Bihar, UP Board Explanation: Calcium Fluoride (CaF₂) Crystal Structure In the calcium fluoride (CaF₂) crystal structure, calcium ions (Ca²⁺) and fluoride ions (F⁻) are arranged based on their sizes and charges. Calcium ions are larger and have a double positive charge (Ca²⁺). They occupy the face-centered cubic (FCC) lattice points, where each unit cell has atoms at the corners and the centers of all the faces of the cube. This arrangement allows for dense packing of the larger calcium ions. Fluoride ions are smaller and carry a single negative charge (F⁻). They fit into the spaces between the calcium ions, specifically in the tetrahedral voids. A tetrahedral void is a space in the crystal lattice surrounded by four atoms arranged in a tetrahedron shape. In the FCC lattice of calcium ions, there are twice as many tetrahedral voids as there are lattice points. For every calcium ion, there are two possible tetrahedral sites, but in CaF₂, each calcium ion is surrounded by eight fluoride ions, effectively filling the tetrahedral positions. This arrangement maximizes the electrostatic forces between the positively charged calcium ions and the negatively charged fluoride ions, leading to a stable and tightly bound crystal structure. In summary, in CaF₂ crystals, the F⁻ ions are located in the tetrahedral voids of the FCC lattice formed by the Ca²⁺ ions.

Chemistry-12 MCQ Q: If Germanium crystallises in the same way as diamond, then which of the following statement is not correct? a) Every atom in the structure is tetrahedrally bonded to 4 atoms.b) Unit cell consists of 8 Ge atoms and co-ordination number is 4.c) All the octahedral voids are occupied.d) All the octahedral voids and 50% tetrahedral voids remain unoccupied.Correct Answer: All the octahedral voids are occupied. Class 12 Chemistry MCQs for CBSE, Bihar, UP Board Explanation: Germanium atom: Diamond Structure In the diamond structure, each germanium atom is arranged in a tetrahedral geometry, meaning every atom is bonded to four neighboring atoms. This arrangement leads to a highly symmetrical and stable crystal lattice. Let’s examine each statement: Option A: Every atom in the structure is tetrahedrally bonded to 4 atoms. – This is correct. In the diamond structure, each germanium atom forms four covalent bonds with neighboring atoms, creating a tetrahedral geometry. Option B: Unit cell consists of 8 Ge atoms and coordination number is 4. – This is also correct. The conventional unit cell for the diamond structure contains eight germanium atoms, and each atom has a coordination number of four, meaning it is connected to four other atoms. Option C: All the octahedral voids are occupied. – This statement is not correct. In the diamond structure, while octahedral voids exist within the lattice, they are not all occupied by atoms. The structure primarily involves tetrahedral bonding, and the octahedral sites remain unfilled or only partially filled depending on the specific structure. Option D: All the octahedral voids and 50% tetrahedral voids remain unoccupied. – This is correct. In the diamond structure, not all the available octahedral voids are occupied by atoms. Additionally, only half of the tetrahedral voids are typically filled, aligning with the tetrahedral bonding nature of the structure. Therefore, the incorrect statement is that all the octahedral voids are occupied in the diamond-like germanium crystal structure.

Chemistry-12 MCQ Q: Doping of silicon (Si) with boron (B) leads to : a) n-type semiconductorb) p-type semiconductorc) metald) insulatorCorrect Answer: p-type semiconductor Class 12 Chemistry MCQs for CBSE, Bihar, UP Board Explanation: Doping silicon with boron Doping silicon with boron introduces impurities into the silicon crystal structure. Silicon is a group IV element, meaning it has four valence electrons. Boron, on the other hand, is a group III element with three valence electrons. When boron atoms are added to the silicon lattice, each boron atom forms three covalent bonds with the surrounding silicon atoms, but there is one bond that remains incomplete because boron has only three valence electrons compared to silicon’s four. This incomplete bond creates a “hole” or a positive charge carrier in the silicon structure. These holes can move through the lattice as neighboring electrons move to fill them, effectively allowing electrical current to flow. Since the majority carriers of charge in this doped silicon are holes (positive charges), the material becomes a p-type semiconductor. The “p” stands for positive, indicating that the charge carriers are positive holes. In contrast, if silicon were doped with a group V element like phosphorus, which has five valence electrons, it would add extra electrons to the silicon lattice, resulting in an n-type semiconductor where electrons are the majority carriers. Therefore, doping silicon with boron specifically creates a p-type semiconductor.

Chemistry-12 MCQ Q: The arrangement ABC ABC ………. is referred to as Octahedral close packingHexagonal close packingTetrahedral close packingCubic close packingCorrect Answer: Cubic close packing Class 12 Chemistry MCQs for CBSE, Bihar, UP Board Explanation:The arrangement ABC ABC is known as cubic close packing. In crystallography, close packing refers to the way layers of atoms are stacked together to achieve the highest possible density. There are two primary types of close-packed structures: hexagonal close packing (HCP) and cubic close packing (CCP), also known as face-centered cubic (FCC) packing. In cubic close packing, the stacking sequence follows an ABCABC pattern. This means that each new layer is positioned in the depressions of the previous two layers, creating a repeating three-layer cycle. This arrangement allows for efficient space utilization and maximizes the packing density of the atoms. On the other hand, hexagonal close packing follows an ABAB pattern, where each layer alternates between two positions. Tetrahedral and octahedral close packing are terms that describe the types of voids created within the packed layers but are not used to describe the overall stacking sequence. Therefore, the correct answer is cubic close packing because the ABCABC arrangement specifically characterizes this type of close-packed structure.

Chemistry-12 MCQ Q: The empty space in the body centred cubic lattice is a) 0.68b) 0.524c) 0.476d) 0.32Correct Answer: d) 0.32 Class 12 Chemistry MCQs for CBSE, Bihar, UP Board Explanation:To understand this, let’s break it down step by step: 1. In a BCC lattice, there are atoms at each corner of the cube and one atom at the center of the cube. 2. Total number of atoms in one unit cell:– Corner atoms: (1/8 × 8) = 1 atom (since each corner atom is shared by 8 unit cells)– Center atom: 1 atom (fully within the unit cell)– Total = 2 atoms per unit cell 3. Volume calculations:– If ‘a’ is the edge length of the cube, the total volume of the unit cell = a³– Each atom has a radius r, where r = (√3a)/4 (this comes from the fact that atoms touch along the body diagonal)– Volume of each atom = (4/3)πr³ 4. The actual space occupied by atoms:– Volume occupied by 2 atoms = 2 × (4/3)πr³– Substituting r = (√3a)/4– Volume occupied = 2 × (4/3)π × ((√3a)/4)³– This works out to about 0.68 × a³ 5. Therefore:– Fraction of space occupied by atoms = 0.68 or 68%– Empty space (void fraction) = 1 – 0.68 = 0.32 or 32% This is why option D (0.32) is the correct answer. The BCC structure has a packing efficiency of 68%, leaving 32% as empty space between the atoms.

Chemistry-12 MCQ Q: An element (atomic mass = 100 g / mol) having bcc structure has unit cell edge 400 pm. Then, density of the element is a) 10.376 g/cm^3b) 5.188 g/cm^3c) 7.289 g/cm^3d) 2.144 g/cm^3 Correct Answer: b) 5.188 g/cm^3 Class 12 Chemistry MCQs for CBSE, Bihar, UP Board Explanation:To determine the density of the element with the given properties, follow these steps: 1. Understand the Given Information: – Atomic Mass (A): 100 g/mol – Unit Cell Structure: Body-Centered Cubic (bcc) – Edge Length of Unit Cell (a): 400 picometers (pm) 2. Convert Edge Length to Centimeters: – 1 pm = 1 × 10^-12 meters – 400 pm = 400 × 10^-12 meters = 4 × 10^-10 meters – Convert meters to centimeters: 4 × 10^-10 meters = 4 × 10^-8 centimeters 3. Calculate Volume of the Unit Cell: – Volume (V) = a³ – V = (4 × 10^-8 cm)³ = 64 × 10^-24 cm³ 4. Determine Number of Atoms per Unit Cell in bcc Structure: – In a bcc lattice, there are 2 atoms per unit cell. 5. Calculate the Mass of Atoms in the Unit Cell: – Molar mass (A) = 100 g/mol – Avogadro’s number (Nₐ) = 6.022 × 10^23 atoms/mol – Mass of one atom = A / Nₐ = 100 / 6.022 × 10^23 ≈ 1.66 × 10^-22 g – Total mass in unit cell = 2 atoms × 1.66 × 10^-22 g ≈ 3.32 × 10^-22 g 6. Compute Density (ρ): – Density formula: ρ = mass/volume – ρ = 3.32 × 10^-22 g / 64 × 10^-24 cm³ ≈ 5.188 g/cm³ Therefore, the density of the element is approximately 5.188 g/cm³, which corresponds to Option B.

Chemistry-12 MCQ Q: Which of the following solids is not an electrical conductor? a) Mg (s)b) TiO (s)c) I2 (s)d) H2O (s) Correct Answer: I2 (s) Class 12 Chemistry MCQs for CBSE, Bihar, UP Board Explanation: Electrical Conductivity Option A: Mg (s) – Magnesium is a metal. Metals generally have free electrons that can move easily through the lattice, allowing them to conduct electricity well. Therefore, magnesium is a good electrical conductor. Option B: TiO (s) – Titanium(II) oxide is an ionic compound. Ionic compounds can conduct electricity when they are melted or dissolved in water because the ions are free to move. However, in the solid state, the ions are fixed in place and cannot move freely, so solid TiO does not conduct electricity. Option C: I2 (s) – Iodine in its solid state consists of molecules held together by intermolecular forces. These molecules do not have free electrons or ions that can move around, which means solid iodine does not conduct electricity. This makes iodine a poor electrical conductor. Option D: H2O (s) – Ice, the solid form of water, is a molecular solid. Pure ice does not conduct electricity because it lacks free ions or electrons. Given these explanations, the correct answer is I2 (s) because solid iodine does not have the free-moving charge carriers necessary to conduct electricity effectively.

Chemistry-12 MCQ Q: Which one of the following forms a molecular solid when solidified? a) Silicon carbideb) Calcium fluoridec) Rock saltd) Methane Correct Answer: Methane Class 12 Chemistry MCQs for CBSE, Bihar, UP Board Explanation: Molecular Solids Methane forms a molecular solid when solidified because it consists of discrete CH₄ molecules held together by weak intermolecular forces known as van der Waals forces. In a molecular solid, the individual molecules are arranged in a regular pattern, but the forces between them are much weaker compared to bonds within the molecules. This allows the solid to have lower melting and boiling points. Let’s examine the other options: Silicon carbide is not a molecular solid. It forms a covalent network solid where each silicon atom is bonded to carbon atoms in a continuous network, resulting in a very hard and high melting material. Calcium fluoride is an ionic solid. It consists of calcium ions (Ca²⁺) and fluoride ions (F⁻) arranged in a repeating lattice structure held together by strong electrostatic forces. Rock salt is also an ionic solid, similar to calcium fluoride. It typically refers to sodium chloride (NaCl), where sodium and chloride ions form a crystalline lattice through strong ionic bonds. Among the given options, only methane consists of discrete molecules held together by weak intermolecular forces, making it a molecular solid when solidified.

Chemistry-12 MCQ Q: Among solids, the highest melting point is exhibited by  a) Covalent solidsb) Ionic solidsc) Pseudo solidsd) Molecular solids Correct Answer: Covalent solids Class 12 Chemistry MCQs for CBSE, Bihar, UP Board Explanation:Covalent solids exhibit the highest melting points among all types of solids. This is because covalent solids consist of a vast network of atoms bonded together by strong covalent bonds. These bonds are continuous throughout the entire structure, creating a rigid and stable lattice that requires a large amount of energy to break. In comparison, ionic solids, while also having strong bonds due to the electrostatic attraction between ions, generally have lower melting points than covalent network solids. This is because the bonds in ionic solids can be overcome more easily under heat. Molecular solids are held together by weaker intermolecular forces such as van der Waals forces or hydrogen bonds. These forces are much easier to break, resulting in significantly lower melting points. The term “pseudo solids” is not a standard classification in chemistry and does not represent a category with high melting points. Therefore, among the options provided, covalent solids have the highest melting points due to their strong and extensive covalent bonding throughout the entire structure.