6. Refer to the table below (Table 1.5 from Shriver et al.) to answer the follow
ID: 717602 • Letter: 6
Question
6. Refer to the table below (Table 1.5 from Shriver et al.) to answer the following questions, where In is t nh ionization energy. In the table, values of n increase from top to bottom; units of In are kJ mol 2373 5250 1086 2352 4619 1402 2855 4577 5300 6050 6122 7297 3386 3353952 1757 2426 11 809 14 844 25 018 CI 495 4562 6911 1000 1251 1520 1903 225 296 2665 3826 3928 737 577 816 2744 7732 3231 2911 3361 1 574 419 3051 4410 579 1979 2963 1391351 2103 3500 1798 3314 2044 2974 1537 2734 3565 1170 558 1821 2704 TI 869 1795 2698 708 1846 3197 064 412 1794 204s 3900 4210 2943 2443 036 375 2420 3400 16 1450 3080 926 1600 2900 610 1800 3619 287B 2700 a. For a given element, why do values of In increase with the value of n? b. Why is Ii of Ne greater than /h of F? c. Why is I of Na less than h of Li? d. Why is / of Al less than I of Mg? e. For Mg, why is Is//2 much greater than h/l?Explanation / Answer
a.
In increases with the value of n because positive charge increases on atom as the electrons get ejected one by one. This increases the effective nuclear charge and further ionization requires more energy.
b.
Neon (Ne) has the following electronic configuration : 1s22s22p6
Fluorine (F) has the following electronic configuration : 1s22s22p5
I1 of Ne costs octet stability of neon (2s22p6 = 8 electrons). Ionization from fully filled p orbital requires more energy than from p orbital when it is not fully filled.
Thus the I1 of Ne > I1 of F.
c.
Sodium (Na) has the following electronic configuration : 1s22s22p63s1
Lithium (Li) has the following electronic configuration : 1s22s1
For Na the outermost electron is in 3s orbital wheras for Li the outermost electron is in 2s orbital. 3s orbital lies further from nucleus than the 2s orbital and this makes ionization of electron from 3s easier. This is the reason why I1 of Na < I1 of Li.
d.
Al has the following electronic configuration : 1s22s22p63s23p1
Mg has the following electronic configuration : 1s22s22p63s2
The outermost electron of Mg lies in fully filled 3s orbital whereas the outermost electron of Al lies in less than half filled 3p orbital. Also the 3s orbital lies closer to nucleus than 3p orbital. This makes the ionization form 3s more energy intensive than the ionization from 3p. Thus I1 of Al < I1 of Mg.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.