Exercise 8.122 Trends in First lonization Energy 2372 2081 1312 o 1681 1402 1314

Question

Exercise 8.122 Trends in First lonization Energy 2372 2081 1312 o 1681 1402 1314 P S 1251 As Se 1140 C 1402 B 1086 801 Cl 1521 520 Mg Na 738 Si 1012 1000 8 Al 786 Br 1351 2500 Ge 947 941 Ga 762 558 709 834 8108 1 TI PbBi Po 589 16 | 703 , 812 Te 1008 Sn 834 549 T1 716 703 1500 Rb Sr 1000 403 /549 S 503 1037 0 500-376 1A 2A A 4A 5A 6A 7A 8A Increasing ionization energy Electron Affinities (kJ/mol) -73 2A 3A4A 5A 6A 7A>0 -60 027-122 0141-328 0 -53 | > 0 | |-43-134|-72 |-200|-349| >0 -48230-11978-195-325 >0 Rb Sr In SnSbTe I Xe -47-530-107-103-190-295 0

Explanation / Answer

The reaction Na(g)+Cl(g)Na+(g)+Cl(g) can be written as,

Na(g) ------> Na+ (g) + e-1 , ionization energy = 496 kJ/mol

Cl(g) + e-1 -------> Cl-(g) , electron affinity = -349 kJ/mol

therefore E = I.E + E.A

= 496 + (-349)

= 147 kJ/mol

the reaction is endothermic reaction

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