Question

1. Calculate the amount of energy released in the formation of one mole of BeS bonds (not lattice energy). The radius of the beryllium ion is 0.31 Å, and the radius of the sulfide ion is 1.84 Å. Note that 1Å=10−10m.

2. Arrange the following ionic compounds in order of decreasing amount of energy released in lattice formation: LiF, CaS, GaN, and RbBr.

Answer 1

1. The ennergy releasedin theformati on of onemole of BeScan be calculatedby using Coloumb's law 28 where, kis constant 2.31x10 Jm anda, are charges on theions Ii q,-charge of Beion2 q2 =charge of thesion=-2 distance between thetwoions (0.31+1.84)A-2.15A 2.15x101 m Putting the values 2.31*10 8 Jmx2x2-4.3x101J-4.3 10 kJ 2.15x1010 m Therefore, the energy released per mole 4.3x1021 kJx6.023x103 mol1 = 25.8989x10 kl mol_: 2589.89kJm ol- 1mole 6.023x10 molecules 2. The Coloumb's equation is Therefore, morethe charge of theionsnadless di stance between thei ons larger isthe energy released The given molecules are LiF. CaS, GaN, RbBr The order of distance between theionsfollows Andthe order of chargebetween theionsis Therefore, combining thesetwo effects, the order of energy rel easedis