As the nuclear charge is increasing, which holds the electrons more tightly to the nucleus, while the extra electron is also added to a 2p shell so it isn't any further from the nucleus and is at the same energy level. There is no additional electron shielding either, so the increased positive nuclear charge isn't counteracted and the electrons are pulled closer and held more tightly. To then remove the outer electron, much more energy is needed to overcome the electrostatic attraction, so the 1st ionisation energy increases.
Original post by summer044
I'm confused let takes an example :
lets take an example:
nitrogen --> oxygen
the ionization energy will decrease but why?
electronic configuration of nitrogen: 1s2 2s2 2p3
electronic configuration of oxygen: 1s2 2s2 2p4
lets take an example:
nitrogen --> oxygen
the ionization energy will decrease but why?
electronic configuration of nitrogen: 1s2 2s2 2p3
electronic configuration of oxygen: 1s2 2s2 2p4
Spin pair repulsion.
There is a doubly occupied p orbital.
Original post by golgiapparatus31
Spin pair repulsion.
There is a doubly occupied p orbital.
There is a doubly occupied p orbital.
okay so why doesnt the ionsation energy between lithium and be decrease?
lithium config. - 1s2 2s1
be config - 1s2 2s2
the orbital is doubly occupied for the be so shouldnt the IE decrease?
Original post by summer044
okay so why doesnt the ionsation energy between lithium and be decrease?
lithium config. - 1s2 2s1
be config - 1s2 2s2
the orbital is doubly occupied for the be so shouldnt the IE decrease?
lithium config. - 1s2 2s1
be config - 1s2 2s2
the orbital is doubly occupied for the be so shouldnt the IE decrease?
Slater's rules.
The effective nuclear charge is greater in Be, hence more energy is required to ionise it. In other words, first IE is greater for Be
Original post by golgiapparatus31
Slater's rules.
The effective nuclear charge is greater in Be, hence more energy is required to ionise it. In other words, first IE is greater for Be
The effective nuclear charge is greater in Be, hence more energy is required to ionise it. In other words, first IE is greater for Be
so if the effective nuclear charge is greater it cancels out any other factor?
Original post by summer044
so if the effective nuclear charge is greater it cancels out any other factor?
Um.. Yes, but not always
https://www.thoughtco.com/ionization-energy-and-trend-604538
Maybe this can clear your doubt?
There are many factors that affect ionisation energy
1. Effective nuclear charge
2. Distance of electron from nucleus
3. Shielding
etc...
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