For Ca, the second ionization potential requires removing only a lone electron in the exposed outer energy level. In addition, energy is required to unpair two electrons in a full orbital. The second ionization energy for K requires that an electron be removed from a lower energy level, where the attraction is much stronger from the nucleus for the electron. Removal of the 4 s electron in Ca requires more energy than removal of the 4 s electron in K because of the stronger attraction of the nucleus and the extra energy required to break the pairing of the electrons. The +2 charge on calcium pulls the oxygen much closer compared with K, thereby increasing the lattice energy relative to a less charged ion. When two electrons are removed from the valence shell, the Ca radius loses the outermost energy level and reverts to the lower n = 3 level, which is much smaller in radius. The first ionization energy of Mg is 738 kJ/mol and that of Al is 578 kJ/mol. Given these ionization values, explain the difference between Ca and K with regard to their first and second ionization energies. The lattice energy of CaO( s) is –3460 kJ/mol the lattice energy of K 2O is –2240 kJ/mol. The radius of the Ca atom is 197 pm the radius of the Ca 2+ ion is 99 pm. Use principles of atomic structure to answer each of the following: 1Ī. MgO the higher charge on Mg leads to a larger lattice energyĦ.\( \newcommand\).Li 2O the higher charge on O 2– leads to a larger energy additionally, Cl – is larger than O 2– this leads to a larger interionic distance in LiCl and a lower lattice energy. ![]() MgO the higher charges on Mg and O, given the similar radii of the ions, leads to a larger lattice energy.MgO selenium has larger radius than oxygen and, therefore, a larger interionic distance and thus, a larger smaller lattice energy than MgO.The compounds with the larger lattice energy are Since the lattice energy is negative in the Born-Haber cycle, this would lead to a more exothermic reaction.Ĥ. 4008 kJ/mol both ions in MgO have twice the charge of the ions in LiF the bond length is very similar and both have the same structure a quadrupling of the energy is expected based on the equation for lattice energyĥ. The smaller the radius of the anion, the shorter the interionic distance and the greater the lattice energy would be.In the Born-Haber cycle, the more negative the electron affinity, the more exothermic the overall reaction. A higher electron affinity is more negative.The lower it is, the more exothermic the reaction will be. As in part (b), the bond energy is a positive energy.This would make the reaction more exothermic, as a smaller positive value is “more exothermic.” A lower ionization energy is a lower positive energy in the Born-Haber cycle.Since the lattice energy is negative in the Born-Haber cycle, this would lead to a more exothermic reaction. ![]()
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