6. Refer to the table below (Table 1.5 from Shriver et al.) to answer the follow

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6. Refer to the table below (Table 1.5 from Shriver et al.) to answer the following questions, where I is the nh ionization energy. In the table, values of n increase from top to bottom; units of I, are kJ mol 2373 5259 7297 11 809 1086 1402 1314 681 2080 1757 2426 232 2855 3386 3375 3952 14844 3660 619 477 5300 6122 25 018 ci 495 577 1000 1251520 1577 1903 2251 2296 2665 1011 4562 816 6911 7732 2744 323 1361 3826 3928 11 574 579 1139 135 941 3051 11451979 37 1798 1044 2103 3314 4410 4910 2963 3302 234 2974 3500 36 549 558 2632 1064 900 4210 0 834 869 1008 110 1795 18462045 2943 2443 2698 1 3097 1821 1412 590 04 812 26 1036 716 2420 96 3400 3619 2878 3080 2466 2700 2900 1450 1610 1800 1600 a. For a given element, why do values of In increase with the value of n? b. Why is h of Ne greater than /h of F? c. Why is of Na less than h of Li? d. Why is h of Al less than h of Mg? e. For Mg, why is I/h much greater than b/l?

Explanation / Answer

a. The values of In increases with the value of n because with the removal of each electron there is decrease in the size of the atom and also increase in the net positive charge. Thus the remaining electrons are attracted towards the nucleus with the greater force. So, the energy required to remove further electrons increases.

b.Ne has an atomic number of 10 whereas the atomic number of fluorine is 9 and the two are present in the same period. the ionization energy increases from left to right in the same period because of the increase in the number of protons. so the ionization energy of neon is greater than fluorine.

Also neon being the noble gas has avery stable electronic configuration because of which it has high ionization energy.

c.The ionization energy decreases from top to bottom in a group in the periodic table because of which the I1 of Na less than I1 of Li.

d. The electronic configuration of aluminium being 1s2 2s2 2p6 3s2 3p1 has an unpaired valence electron in the 3p orbital which is readily lost and acquires a stable 3s2 configuration because of this the I1 of aluminium is lower than I1 of Mg.

e. In Mg the second ionization energy is low because Mg readily loses the second electron to attain a stable configuration of 1s2 2s2 2p6 ( configuration of neon) and as discussed before, it is diffcult to remove the electron from this configuration as this is very stable and needs high ionization energy.

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6. Refer to the figure on the right: If a saturated solution of NH4NO3 at 75 °C

6. Refer to the table below (Table 1.5 from Shriver et al.) to answer the follow

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6. Refer to the table below (Table 1.5 from Shriver et al.) to answer the follow

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