Class 12 Chemistry MCQ

Chemistry-12 MCQ Q: Which of the following is not a crystalline solid? a) KClb) CsClc) Glassd) Rhombic SCorrect Answer: Glass Class 12 Chemistry MCQs for CBSE, Bihar, UP Board Explanation: Crystalline Solids Crystalline solids are characterized by a highly ordered and repeating arrangement of their constituent particles, such as atoms, ions, or molecules, extending in all three spatial dimensions. This long-range order gives crystalline solids well-defined geometric shapes and distinct melting points. Let’s examine each option: Option A: KCl (Potassium Chloride) – This is an ionic compound that forms a crystalline lattice structure. The potassium and chloride ions are arranged in a regular, repeating pattern, typical of crystalline solids. Option B: CsCl (Cesium Chloride) – Similar to KCl, cesium chloride forms a crystalline structure. The cesium and chloride ions are organized in a specific, orderly lattice arrangement, which is a hallmark of crystalline materials. Option D: Rhombic S (Rhombic Sulfur) – Rhombic sulfur is one of the allotropes of sulfur and has a crystalline structure. Its molecules are arranged in an orderly, repeating pattern, characteristic of crystalline solids. Option C: Glass – Unlike the other options, glass does not have a long-range ordered structure. It is an amorphous solid, meaning its molecules are arranged more randomly without a repeating pattern. This lack of orderly structure means glass does not form crystals and does not exhibit the sharp melting points typical of crystalline solids. Therefore, Glass is not a crystalline solid because it lacks the long-range ordered arrangement of particles that defines crystalline materials.

Chemistry-12 MCQ Q: Graphite cannot be classified as ______. a) conducting solidb) network solidc) covalent solidd) ionic solidCorrect Answer: ionic solid Class 12 Chemistry MCQs for CBSE, Bihar, UP Board Explanation: Graphite’s Structure  Graphite is a form of carbon where each carbon atom is bonded to three other carbon atoms in a flat, two-dimensional hexagonal lattice. These strong covalent bonds within each layer form a network solid, giving graphite its stability and structure. The layers are held together by weaker van der Waals forces, allowing them to slide over one another easily, which is why graphite feels slippery and is used as a lubricant. Graphite can conduct electricity because the electrons in its structure are free to move within the layers. However, it does not consist of positive and negative ions; instead, it is composed entirely of carbon atoms bonded covalently. Ionic solids, on the other hand, are made up of positive and negative ions held together by strong electrostatic forces in a rigid lattice. Since graphite does not have this ionic bonding structure, it cannot be classified as an ionic solid.

Chemistry-12 MCQ Q: Percentages of free space in cubic close packed structure and in body centered packed structure are respectively a) 30% and 26%b) 26% and 32%c) 32% and 48%d) 48% and 26%Correct Answer: 26% and 32% Class 12 Chemistry MCQs for CBSE, Bihar, UP Board Explanation: Cubic Close Packed (CCP) & Body-centered Cubic (BCC) Free Space In crystal structures, the packing efficiency refers to the percentage of space occupied by the atoms, while the remaining percentage represents the free space or voids. For the cubic close packed (CCP) structure, also known as the face-centered cubic (FCC) structure, the packing efficiency is 74%. This means that 26% of the space is unoccupied or free. The CCP structure arranges atoms in a way that minimizes the free space, making it one of the most efficient packing methods. On the other hand, the body centered packed (BCP) structure, commonly referred to as the body-centered cubic (BCC) structure, has a packing efficiency of approximately 68%. Consequently, 32% of the space in the BCP structure is free. The BCP arrangement is less efficient in terms of packing compared to CCP because there is more void space between the atoms. Therefore, the percentages of free space in CCP and BCP structures are 26% and 32% respectively.

Chemistry-12 MCQ Q: In which of the following crystals alternate tetrahedral voids are occupied? a) NaClb) ZnSc) CaF2d) Na2OCorrect Answer: ZnS Class 12 Chemistry MCQs for CBSE, Bihar, UP Board Explanation: In crystal structures, voids or interstices are spaces between the atoms where smaller ions or molecules can fit. These voids come in different shapes and sizes, with tetrahedral voids being one type. A tetrahedral void is a cavity surrounded by four atoms positioned at the corners of a tetrahedron. Different crystal structures occupy these voids in various ways: 1. NaCl (Sodium Chloride): This structure adopts a face-centered cubic (FCC) lattice where each sodium ion is surrounded by six chloride ions and vice versa. The structure primarily involves octahedral coordination rather than tetrahedral. 2. ZnS (Zinc Sulfide): ZnS commonly exists in the zinc blende structure, which is similar to the diamond structure. In this arrangement, each zinc ion is tetrahedrally coordinated by four sulfide ions, and each sulfide ion is similarly coordinated by four zinc ions. This means that the tetrahedral voids in the lattice are alternately occupied by zinc and sulfide ions. 3. CaF2 (Calcium Fluoride): This crystal structure features calcium ions in a cubic arrangement with fluoride ions occupying all the tetrahedral voids. Here, all tetrahedral voids are filled rather than just alternate ones. 4. Na2O (Sodium Oxide): This structure does not typically involve the occupation of alternate tetrahedral voids in the same manner as ZnS. Therefore, among the given options, ZnS is the crystal where alternate tetrahedral voids are occupied.

Chemistry-12 MCQ Q: When molten zinc is converted into solid state, it acquires hcp structure. The number of nearest neighbours of Zn will be  a) 6b) 12c) 8d) 4Correct Answer: 12 Class 12 Chemistry MCQs for CBSE, Bihar, UP Board Explanation: Molten Zinc Solidifies When molten zinc solidifies, it forms a hexagonal close-packed (hcp) crystal structure. In the hcp structure, each zinc atom is surrounded by other zinc atoms in a highly efficient packing arrangement. In the hcp arrangement: – Each atom has 12 nearest neighbors.– These neighbors are positioned as six atoms in the same layer around it, three atoms in the layer above, and three atoms in the layer below.– This configuration allows for maximum packing efficiency, minimizing the space between atoms. Therefore, in the hcp structure of solid zinc, each zinc atom has twelve nearest neighbors. The correct answer is Option B: 12.

Chemistry-12 MCQ Q: Which of the following statements about amorphous solids is incorrect ? a) They melt over a range of temperatureb) They are anisotropicc) There is no orderly arrangement of particlesd) They are rigid and incompressibleCorrect Answer: They are anisotropic Class 12 Chemistry MCQs for CBSE, Bihar, UP Board Explanation: Amorphous Solids Amorphous solids are materials that do not have a long-range ordered atomic structure, unlike crystalline solids which have a highly organized and repeating arrangement of particles. Let’s examine each statement to identify which one is incorrect: Option A: They melt over a range of temperatureThis statement is correct. Amorphous solids do not have a precise melting point. Instead, they soften gradually over a range of temperatures as the kinetic energy of the particles increases. Option B: They are anisotropicThis statement is incorrect. Amorphous solids are generally isotropic, meaning their properties are the same in all directions. This isotropy arises because there is no long-range order or directional dependence in the arrangement of their particles. Option C: There is no orderly arrangement of particlesThis statement is correct. In amorphous solids, the particles are arranged randomly without any long-range periodicity, which distinguishes them from crystalline solids. Option D: They are rigid and incompressibleThis statement is correct. Despite the lack of long-range order, amorphous solids maintain a rigid structure and are typically incompressible because the particles are closely packed together. Therefore, the incorrect statement about amorphous solids is that they are anisotropic.

Chemistry-12 MCQ Q: To get a n- type semiconductor, the impurity to be added to silicon should have which of the following number of valence electrons a) 1b) 2c) 3d) 5Correct Answer: 5 Class 12 Chemistry MCQs for CBSE, Bihar, UP Board Explanation: n-type semiconductor  To create an n-type semiconductor, impurities with five valence electrons are added to silicon. Silicon itself has four valence electrons, which form strong bonds with neighboring silicon atoms to create a stable crystal structure. When a pentavalent impurity, which has five valence electrons, is introduced into the silicon lattice, four of its electrons bond with the surrounding silicon atoms just like silicon does. The fifth electron, however, does not have a bonding partner and becomes free to move within the material. This extra free electron increases the number of negative charge carriers in the semiconductor, enhancing its electrical conductivity. Since the added impurity contributes additional electrons, the semiconductor is termed “n-type,” where “n” stands for negative, indicating the charge carriers are negative electrons. Common pentavalent impurities used for this purpose include phosphorus and arsenic. Therefore, to obtain an n-type semiconductor, an impurity with five valence electrons is required.

Chemistry-12 MCQ Q: Which of the following exists as covalent crystals in the solid state ? a) Iodineb) Siliconc) Sulphurd) PhosphorusCorrect Answer: Silicon Class 12 Chemistry MCQs for CBSE, Bihar, UP Board Explanation: Covalent Crystals Covalent crystals are solids where the atoms are bonded together by strong covalent bonds, forming a continuous network that extends throughout the material. This structure gives covalent crystals their characteristic hardness, high melting points, and stability. Silicon exists as a covalent crystal in the solid state. In its crystalline form, silicon atoms are each bonded to four other silicon atoms in a tetrahedral arrangement, creating a robust and rigid three-dimensional network. This extensive bonding is typical of covalent crystals and is responsible for silicon’s semiconductor properties, making it essential in the electronics industry. The other options do not form covalent crystals in the solid state: – Iodine exists as molecular crystals, where I2 molecules are held together by weaker van der Waals forces rather than strong covalent bonds.– Sulphur typically forms molecular crystals consisting of S8 rings, again held together by weaker intermolecular forces.– Phosphorus can exist in several allotropes, but common forms like white phosphorus consist of molecular structures rather than a covalent network. Therefore, among the given options, silicon is the one that exists as a covalent crystal in the solid state.

Chemistry-12 MCQ Q: With which one of the following elements silicon should be doped so as to give p-type of semiconductor ? a) Germaniumb) Arsenicc) Seleniumd) BoronCorrect Answer: Boron Class 12 Chemistry MCQs for CBSE, Bihar, UP Board Explanation: P-type Semiconductor To create a p-type semiconductor, silicon is doped with an element that has one fewer valence electron than silicon. Silicon has four valence electrons, so an element with three valence electrons is ideal for p-type doping. Boron is a Group III element, which means it has three valence electrons. When boron is introduced into the silicon lattice, it creates “holes” or positive charge carriers because there are not enough electrons to complete the bonds. These holes can move through the silicon structure, allowing electrical current to flow. This process effectively creates a p-type semiconductor, where the majority charge carriers are holes. The other elements listed do not serve this purpose: – Germanium is also a Group IV element, like silicon, and does not introduce holes. – Arsenic and selenium are Group V and Group VI elements, respectively, which have five and six valence electrons. When they are used to dope silicon, they add extra electrons, creating an n-type semiconductor instead of a p-type. Therefore, boron is the appropriate choice for doping silicon to achieve a p-type semiconductor.

Chemistry-12 MCQ Q: Which of the following amorphous solid is used as photovoltaic material for conversion of sunlight into electricity? a) Quartz glassb) Quartzc) Silicond) Both (a) and (b)Correct Answer: Silicon Class 12 Chemistry MCQs for CBSE, Bihar, UP Board Explanation: Photovoltaic Material Silicon is the correct answer because it is widely used as a photovoltaic material to convert sunlight into electricity. Photovoltaic cells, commonly known as solar cells, primarily rely on silicon due to its excellent semiconductor properties. Silicon can efficiently absorb sunlight and convert it into electrical energy through the photovoltaic effect. Amorphous solids lack a long-range ordered structure, and among the options provided: Quartz glass and quartz are forms of silicon dioxide. While they are related to silicon, they do not possess the necessary semiconductor properties required for effective photovoltaic conversion. Silicon, on the other hand, especially in its amorphous form, is utilized in certain types of solar cells. Although amorphous silicon has lower efficiency compared to its crystalline counterpart, it offers advantages like flexibility and lower production costs, making it suitable for specific applications. Therefore, silicon stands out as the appropriate amorphous solid for photovoltaic applications, making option C the correct choice.