Q41.A compound of formula A2B3 has the HCP lattice. Which atom forms the HCP lattice and what fraction of the tetrahedral voids are occupied by the other atoms? (1) HCP lattice - B (2) HCP lattice - A 3 2 tetrahedral voids - A 31 tetrahedral voids - B (3) HCP lattice - B (4) HCP lattice - A 3 1 tetrahedral voids - A 23 tetrahedral voids - B JEE Main 2019 (10 Jan Shift 2) JEE Main Previous Year Paper
What This Question Tests
This question tests the knowledge of HCP lattice properties, specifically the number of atoms and tetrahedral voids, and how these relate to the stoichiometry of a compound formed by occupying a fraction of these voids.
Concepts Tested
📚 NCERT Sections This Tests
14.1 — In An N-Type Silicon, Which Of The Following Statement Is True:
Physics Class 12 · Chapter 14
14.1 In an n-type silicon, which of the following statement is true: (a) Electrons are majority carriers and trivalent atoms are the dopants. (b) Electrons are minority carriers and pentavalent atoms are the dopants. (c) Holes are minority carriers and pentavalent atoms are the dopants. (d) Holes are majority carriers and trivalent atoms are the dopants.
4.2 — Ionic Or Electrovalent Bond Other Factors. The Crystal Structure Of Sodium
Chemistry Class 11 · Chapter 4
4.2 Ionic or Electrovalent Bond other factors. The crystal structure of sodium chloride, NaCl (rock salt), for example isFrom the Kössel and Lewis treatment of the shown below.formation of an ionic bond, it follows that the formation of ionic compounds would primarily depend upon: • The ease of formation of the positive and negative ions from the respective neutral atoms; • The arrangement of the positive and negative ions in the solid, that is, the lattice of the crystalline compound. The formation of a positive ion involves ionization, i.e., removal of electron(s) from the neutral atom and that of the negative ion involves the addition of electron(s) to the Rock salt structure neutral atom. In ionic solids, the sum of the electron gain M(g) → M+(g) + e– ; enthalpy and the ionization enthalpy may be Ionization enthalpy positive but still the crystal structure gets X(g) + e– → X – (g) ; stabilized due to the energy released in the Electron gain enthalpy formation of the crystal lattice. For example: the ionization enthalpy for Na+(g) formation M+(g) + X –(g) → MX(s) from Na(g) is 495.8 kJ mol–1 ; while the electron The electron gain enthalpy, ∆egH, is the gain enthalpy for the change Cl(g) + e–→ enthalpy change (Unit 3), when a gas phase Cl– (g) is, – 348.7 kJ mol–1 only. The sum of the atom in its ground state gains an electron. two, 147.1 kJ mol-1 is more than compensated The electron gain process may be exothermic for by the enthalpy of lattice formation of or endothermic. The ionization, on the other NaCl(s) (–788 kJ mol–1). Therefore, the energy hand, is always endothermic. Electron released in the processes is more than the Reprint 2025-26 Chemical Bonding And Molecular Structure 107 energy absorbed. Thus a qualitative measure of the stability of an ionic compound is provided by its enthalpy of lattice formation and not simply by achieving octet of electrons around the ionic species in gaseous state. Since lattice enthalpy plays a key role in the formation of ionic compounds, it is important that we learn more about it. 4.2.1 Lattice Enthalpy The Lattice Enthalpy of an ionic solid is defined as the energy required to completely separate one mole of a solid ionic compound into gaseous constituent ions. For example, the lattice enthalpy of NaCl is 788 kJ mol–1. This means that 788 Fig. 4.1 The bond length in a covalent kJ of energy is required to separate one mole molecule AB. of solid NaCl into one mole of Na+ (g) and one R = rA + rB (R is the bond length and rA and rB are mole of Cl– (g) to an infinite distance. the covalent radii of atoms A and B respectively) This process involves both the attractive forces between ions of opposite charges in the same molecule. The van der Waals and the repulsive forces between ions of radius represents the overall size of the like charge. The solid crystal being three- atom which includes its valence shell in a dimensional; it is not possible to calculate nonbonded situation. Further, the van der lattice enthalpy directly from the interaction Waals radius is half of the distance between of forces of attraction and repulsion only. two similar atoms in separate molecules in Factors associated with the crystal geometry a solid. Covalent and van der Waals radii of have to be included. chlorine are depicted in Fig. 4.2.
14.3 — Intrinsic Semiconductor
Physics Class 12 · Chapter 14
14.3 INTRINSIC SEMICONDUCTOR We shall take the most common case of Ge and Si whose lattice structure is shown in Fig. 14.3. These structures are called the diamond-like structures. Each atom is surrounded by four nearest neighbours. We know that Si and Ge have four valence electrons. In its crystalline structure, every Si or Ge atom tends to share one of its four valence electrons with each of its four nearest neighbour atoms, and also to take share of one electron from each such neighbour. These shared electron pairs are referred to as forming a covalent bond or simply a valence bond. The two shared electrons can be assumed to shuttle back-and-forth between the associated atoms holding them together strongly. Figure 14.4 schematically shows the 2-dimensional representation of Si or Ge FIGURE 14.3 Three-dimensional dia- structure shown in Fig. 14.3 which overemphasises the mond-like crystal structure for Carbon, covalent bond. It shows an idealised picture in which no Silicon or Germanium with bonds are broken (all bonds are intact). Such a situation respective lattice spacing a equal arises at low temperatures. As the temperature increases, to 3.56, 5.43 and 5.66 Å. more thermal energy becomes available to these electrons and some of these electrons may break–away (becoming free electrons contributing to conduction). The thermal energy effectively ionises only a few atoms in the crystalline lattice and creates a vacancy in the bond as shown in Fig. 14.5(a). The neighbourhood, from which the free electron (with charge –q) has come out leaves a vacancy with an effective charge (+q). This vacancy with the effective positive electronic charge is called a hole. The hole behaves as an apparent free particle with effective positive charge. In intrinsic semiconductors, the number of free electrons, ne is equal to the number of holes, nh. That is ne = nh = ni (14.1) where ni is called intrinsic carrier concentration. Semiconductors posses the unique property in which, apart from electrons, the holes also move. Suppose there is a hole at site 1 as shown 327 Reprint 2025-26 Physics in Fig. 14.5(a). The movement of holes can be visualised as shown in Fig. 14.5(b). An electron from the covalent bond at site 2 may jump to the vacant site 1 (hole). Thus, after such a jump, the hole is at site 2 and the site 1 has now an electron. Therefore, apparently, the hole has moved from site 1 to site 2. Note that the electron originally set free [Fig. 14.5(a)] is not involved in this process of hole motion. The free electron moves completely independently as conduction electron and gives rise to an electron current, Ie under an applied electric field. Remember that the motion of hole is only a convenient way of FIGURE 14.4 Schematic two-dimensional describing the actual motion of bound electrons, representation of Si or Ge structure showing whenever there is an empty bond anywhere in covalent bonds at low temperature the crystal. Under the action of an electric field, (all bonds intact). +4 symbol these holes move towards negative potential indicates inner cores of Si or Ge. giving the hole current, Ih. The total current, I is thus the sum of the electron current Ie and the hole current Ih: I = Ie + Ih (14.2) It may be noted that apart from the process of generation of conduction electrons and holes, a simultaneous process of recombination occurs in which the electrons recombine with the holes. At equilibrium, the rate of generation is equal to the rate of recombination of charge carriers. The recombination occurs due to an electron colliding with a hole. (a) (b) FIGURE 14.5 (a) Schematic model of generation of hole at site 1 and conduction electron due to thermal energy at moderate temperatures. (b) Simplified representation of possible thermal motion of a hole. The electron from the lower left hand covalent bond (site 2) goes to the earlier hole site1, leaving a hole at its site indicating an 328 apparent movement of the hole from site 1 to site 2. Reprint 2025-26 Semiconductor Electronics: Materials, Devices and Simple Circuits An intrinsic semiconductor will behave like an insulator at T = 0 K as shown in Fig. 14.6(a). It is the thermal energy at higher temperatures (T > 0K), which excites some electrons from the valence band to the conduction band. These thermally excited electrons at T > 0 K, partially occupy the conduction band. Therefore, the energy-band diagram of an intrinsic semiconductor will be as shown in Fig. 14.6(b). Here, FIGURE 14.6 (a) An intrinsic semiconductor at T = 0 K some electrons are shown in behaves like insulator. (b) At T > 0 K, four thermally generated the conduction band. These electron-hole pairs. The filled circles ( ) represent electrons have come from the valence and empty circles ( ) represent holes. band leaving equal number of holes there. Example 14.1 C, Si and Ge have same lattice structure. Why is C insulator while Si and Ge intrinsic semiconductors? Solution The 4 bonding electrons of C, Si or Ge lie, respectively, in the second, third and fourth orbit. Hence, energy required to take out an electron from these atoms (i.e., ionisation energy Eg) will be least for Ge, followed by Si and highest for C. Hence, number of free EXAMPLE electrons for conduction in Ge and Si are significant but negligibly small for C. 14.1
📋 Question Details
- Chapter
- Solid State
- Topic
- HCP lattice and voids
- Year
- 2019
- Shift
- 10 Jan Shift 2
- Q Number
- Q41
- Type
- MCQ
- NCERT Ref
- Class 12 Chemistry Ch 1: Solid State
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