The correct values of ionization enthalpies (in kJ mol–1) of Si, P, Cl and S respectively are:a)786, 1012, 999, 1256b)1012, 786, 999, 1256c)786, 1012, 1256, 999d)786, 999, 1012, 1256Correct answer is option 'C'. Can you explain this answer?

Question

The correct values of ionization enthalpies (in kJ mol&ndash;1) of Si, P, Cl and S respectively are:a)786, 1012, 999, 1256b)1012, 786, 999, 1256c)786, 1012, 1256, 999d)786, 999, 1012, 1256Correct answer is option 'C'. Can you explain this answer?

Answer

-1) for the first 20 elements are:

1. Hydrogen - 1312
2. Helium - 2372
3. Lithium - 520
4. Beryllium - 899
5. Boron - 801
6. Carbon - 1086
7. Nitrogen - 1402
8. Oxygen - 1314
9. Fluorine - 1681
10. Neon - 2081
11. Sodium - 496
12. Magnesium - 738
13. Aluminum - 578
14. Silicon - 786
15. Phosphorus - 1012
16. Sulfur - 1000
17. Chlorine - 1251
18. Argon - 1521
19. Potassium - 419
20. Calcium - 590

Answer

All options are wrong.. because we know along the period ionization energy increases.if option C is correct then.it means phosphorous has greater ionization energy than that of sulphur

Answer

C

Answer

Option c
here we know the concept that ionization energy increases along the period.
chlorine has more value of IE than sulphur due to its nature of half field orbitals
thus correct answer is c

Answer

As we move along horizontal in the periodic table, the ionisation enthalpy increases as the internuclear distance decreases, So Cl will be higher ionisation enthalpy.
Out of P and S ,P is half filled p subshell (1s^2 2s^2 2p^6 3s^2 3p^3) where S is not exactly half filled ( 1s^2 2s^2 2p^6 3s^2 3p^ 4) . The more stable electronic configuration is of P. So P will be having higher ionisation enthalpy than S . The lowest ionisation enthalpy is of Si .
So the correct order is :
Cl> P>S>Si .

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