Question

the correct order of decreasing second ionization energy is Li,Be,B,C,Ne with reason ​

Answer 1

Answer:

Maximum will be in li becoz after 1st ionization enthalpy it will attain nobel gas config

Then on boron

Then on neon

Then in carbon

And the least on berilyium

Answer 2

Answer:

The minimum amount of energy required to remove the valence electron from loosely bound electron in a molecule, is ionization energy, denoted by Ei.

Explanation:

The second ionization energy is the energy it takes to remove an electron from a 1+ ion.

There are five factors affecting ionization energy: The effective nuclear charge,  size of an atom, Shielding effect, electronic configuration of an atom and Penetration effect. All of these variables were related to the number of electrons in the valence shell and between the shell and nucleus of the valence.

electron configurations of the ions is given below:

• Li - Li+ (ion) and 1s2 (config)
• Be - Be+(ion)  and 2s (config)
• C - C+ (ion)  and 2s2p (config)
• Ne - Ne+ (ion) and 2s2p5 (config)
• B - B+ (ion) and 2s2 (config)

Li has highest IE2 than because to remove the second electron we must break the stable and B has higer IE2 than C due to the extra stability of the  s2  subshell in the B+ ion.

Hence, The decreasing order of second ionization energy is Li > Ne > B > C > Be.

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