![]() Cesium fluoride, therefore, is not Cs 2 +F 2− because the energy cost of forming the doubly charged ions would be greater than the additional lattice energy that would be gained. Not only is an electron being added to an already negatively charged ion, but because the F − ion has a filled 2 p subshell, the added electron would have to occupy an empty high-energy 3 s orbital. Furthermore, forming an F 2− ion is expected to be even more energetically unfavorable than forming an O 2− ion. To form the Cs 2 + ion from Cs +, however, would require removing a 5 p electron from a filled inner shell, which calls for a great deal of energy: I 2 = 2234.4 kJ/mol for Cs. Q- If the formation of ionic lattices containing multiply charged ions is so energetically favorable, why does CsF contain Cs + and F − ions rather than Cs 2 + and F 2− ions?Ī- If we assume that U for a Cs 2 +F 2− salt would be approximately the same as U for BaO, the formation of a lattice containing Cs 2 + and F 2− ions would release 2291 kJ/mol (3048 kJ/mol − 756.9 kJ/mol) more energy than one containing Cs + and F − ions. ![]() Although the internuclear distances are not significantly different for BaO and CsF (275 and 300 pm, respectively), the larger ionic charges in BaO produce a much higher lattice energy. Lattice energies are directly proportional to the product of the charges on the ions and inversely proportional to the internuclear distance. (9.3. If the first four terms in the Born–Haber cycle are all substantially more positive for BaO than for CsF, why does BaO even form?Ī- The answer is the formation of the ionic solid from the gaseous ions. Even though adding one electron to an oxygen atom is exothermic ( EA 1 = −141 kJ/mol), adding a second electron to an O −(g) ion is energetically unfavorable ( EA 2 = +744 kJ/mol)-so much so that the overall cost of forming O 2−(g) from O(g) is energetically prohibitive ( EA 1 + EA 2 = +603 kJ/mol). Q- Forming gaseous oxide (O 2−) ions is energetically unfavorable. The Born–Haber Cycle Illustrating the Enthalpy Changes Involved in the Formation of Solid Cesium Fluoride from Its Elements: Q- Arrange NaCl, MgS, AlN, and KBr in order of increasing lattice energy.Ī- The order of increasing lattice energy is KBr < NaCl < MgS < AlN. Q- Arrange GaP, BaS, CaO, and RbCl in order of increasing lattice energy.Ī- The order of increasing lattice energy is RbCl < BaS < CaO < GaP. Therefore, the angle between bond pairs further decreases.Lattice energy is the most important factor in determining the stability of an ionic compound. H) If there are two bond pairs and two lone pairs of electrons around the nucleus of the central atom in its valence shell, lone pair - lone pair repulsion is greater than lone pair - bond pair repulsion. The remaining three bond pairs come relatively closer as in NH3 molecule. ![]() unshared electron pair, then the lone pair occupies more space around the nucleus of the central atom. 1 Lattice energy and lattice enthalpy Sodium chloride crystal lattice The concept of lattice energy was originally applied to the formation of compounds with structures like rocksalt ( NaCl) and sphalerite ( ZnS) where the ions occupy high-symmetry crystal lattice sites. G) If there are three bond pair and one lone pair i.e. Therefore, the shape of the molecule is trigonal planar.į) If there are four bond pairs in the valence shell of the central atom, the four bond pairs will orient along the four corners of a tetrahedron (three dimensional arrangement) and the bond angle expected is 109º 28'. Thus, the molecule would be linear.Į) If three bond pairs are there in three covalent bonds around the nucleus of the central atom, without any lone pairs they get separated by 120º along three corners of a triangle. * The tendency of losing electrons to form cations (or) gaining electron to form anions depends on the following factors:ĭ) If two bond pairs are present in two covalent bonds around the nucleus of the central atom without any lone pairs in the valence shell, they must be separated by 180º to have minimum repulsion between them. Elements with more electronegative character form anions. This property is called the non-metallic character or electronegativity of the element. Similarly non mentals like Oxygen ( 8O), fluorine ( 9F) and chlorine ( 17C l) acquire electron configuration of element of inert gases by gaining electrons. Elements withmore electropositive character form cations. This property is called as the metallic character or electropositivity. What are the factor that affect the formation of cation and anion? (4 Marks)Ī: * Generally elements of metals have tendency of losing electron to attain the octet in their valence shell. * The electrostatic attractive force that keeps cation and anion (which are formed from metal atoms and non-metal atoms due to transfer of electrons) together to form a new electricallyneutral compounds is called Ionic bond.ġ1. Now, we can define ionic bond as follows * As the valence concept has been explained in terms of electrons, it is also called the electrovalent bond. ![]()
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