Why is 2nd IE of Neon higher than that of Boron.B+ = 1s****22s2 Ne+ = 1s****22s22p5 As far as i know, Full filled and half filled orbitals are more stable and hence require more energy to pull out an electron, Also s orbital is closer to nucleus and thus its more difficult to pull electrons out. In this particular case if we take out one more electron,Boron becomes 2s****1 **and Neon becomes 2p4**, I looked up on google and someone told that Neon is an inert gas, hence stable, But here we arent talking about Neon with full octet, its excited. Please help me out with a detailed explanation as I'm new to this concept. Thanks!!
Basically the discrepancy of half filled orbital stability isn't very large. Boron 2nd exited state having larger exited state is only comparable to elements near it. For example In this video 1st enthalpy of 2nd period be and N only jumped once. My point is that even though Boron will lose it's second exited and half filled it is not going to soo much as to have higher enthalpy than an element that's so much farther away. Hence neon will still have higher enthalpy. The half filled discrepancy and stability is good but it's not huge.it will only be usable against elements beside it, not the ones so far away.hope this helps.
ratta marne ki koi need hai, u don't even tell us that the IE order is s>p>d>f As Be has full 2s orbital than Li and B, and same as Nitrogen haive half filled 2p orbital and it is hard to remove electron from full filled and half filled atom,
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Why is 2nd IE of Neon higher than that of Boron.B+ = 1s****22s2
Ne+ = 1s****22s22p5
As far as i know, Full filled and half filled orbitals are more stable and hence require more energy to pull out an electron, Also s orbital is closer to nucleus and thus its more difficult to pull electrons out.
In this particular case if we take out one more electron,Boron becomes 2s****1 **and Neon becomes 2p4**,
I looked up on google and someone told that Neon is an inert gas, hence stable, But here we arent talking about Neon with full octet, its excited.
Please help me out with a detailed explanation as I'm new to this concept.
Thanks!!
Bhai sabse pehle electron S shell se niklega , neon = 2p⁶ and boron = 2p¹
Basically the discrepancy of half filled orbital stability isn't very large. Boron 2nd exited state having larger exited state is only comparable to elements near it. For example In this video 1st enthalpy of 2nd period be and N only jumped once. My point is that even though Boron will lose it's second exited and half filled it is not going to soo much as to have higher enthalpy than an element that's so much farther away. Hence neon will still have higher enthalpy. The half filled discrepancy and stability is good but it's not huge.it will only be usable against elements beside it, not the ones so far away.hope this helps.
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ratta marne ki koi need hai, u don't even tell us that the IE order is s>p>d>f
As Be has full 2s orbital than Li and B, and same as Nitrogen haive half filled 2p orbital and it is hard to remove electron from full filled and half filled atom,
Penetration power order of these orbital is s>p>d>f. That's why in Be , the 2s orbital present so it's pp is more. That's why it's ie is more
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