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ChemistryEasyMCQ2021 · 27 Jul Shift 1

Q40.The parameters of the unit cell of a substance are a = 2. 5, b = 3. 0, c = 4. 0, α = 90°, β = 120°γ = 90°. The crystal system of the substance is : (1) Hexagonal (2) Orthorhombic (3) Monoclinic (4) Triclinic

What This Question Tests

This question requires identifying the crystal system based on the given unit cell parameters (axial lengths and interfacial angles) by recalling the characteristic features of different crystal systems.

Concepts Tested

Unit cell parametersTypes of crystal systems

📚 NCERT Sections This Tests

3.10In A Reaction Between A And B, The Initial Rate Of Reaction (R0) Was Measured

Chemistry Class 11 · Chapter 3

74% match

3.10 In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below: A/ mol L–1 0.20 0.20 0.40 B/ mol L–1 0.30 0.10 0.05 r0/mol L–1s–1 5.07 × 10–5 5.07 × 10–5 1.43 × 10–4 What is the order of the reaction with respect to A and B? 3.11 The following results have been obtained during the kinetic studies of the reaction: 2A + B ® C + D Experiment [A]/mol L–1 [B]/mol L–1 Initial rate of formation of D/mol L–1 min–1 I 0.1 0.1 6.0 × 10–3 II 0.3 0.2 7.2 × 10–2 III 0.3 0.4 2.88 × 10–1 IV 0.4 0.1 2.40 × 10–2 Determine the rate law and the rate constant for the reaction. 3.12 The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment [A]/ mol L–1 [B]/ mol L–1 Initial rate/ mol L–1 min–1 I 0.1 0.1 2.0 × 10–2 II – 0.2 4.0 × 10–2 III 0.4 0.4 – IV – 0.2 2.0 × 10–2 3.13 Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s–1 (ii) 2 min–1 (iii) 4 years–1 3.14 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample. 3.15 The experimental data for decomposition of N2O5 [2N2O5 ® 4NO2 + O2] in gas phase at 318K are given below: t/s 0 400 800 1200 1600 2000 2400 2800 3200 102 × [N2O5]/ 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 mol L–1 (i) Plot [N2O5] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[N2O5] and t. (iv) What is the rate law ? Chemistry 86 Reprint 2025-26 (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).

5.16Draw Figure To Show The Splitting Of D Orbitals In An Octahedral Crystal Field.

Chemistry Class 11 · Chapter 5

73% match

5.16 Draw figure to show the splitting of d orbitals in an octahedral crystal field.

4.2Ionic Or Electrovalent Bond Other Factors. The Crystal Structure Of Sodium

Chemistry Class 11 · Chapter 4

73% match

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.