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Not considering the electronic spin,the degeneracy of the second excited state of H is 9
What is the degeneracy of the second excited state of H negative
[JEE ADVANCED 2015]
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Not considering the electronic spin if the degeneracy of H atom in 2nd excited state is 9 then what is the degeneracy of H^-1 in second excited state?
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2122
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Abhilasha Choudhary, B.tech Materials Science and Engineering & Metallurgy and Materials Engineering, Punjab Engineering College...
Answered May 2, 2016
JEE ADVANCED 2015; a fairly good question that checks the concepts!
Single electron species don't follow the [ n+l ]rule. [(n+l) rule is the rule to find the energy order of the subshells], thus hydrogen being a single electron specie has all its subshells of equal energy in n=3. thus degeneracy( number of orbitals of equal energy for the electron to be excited in ) is 9 {1 s orbital+3p orbitals +5d orbitals}
however hydride ion has a configuration of 1s(2)
FIRST EXCITED STATE: 1s(1) 2s(1)
SECOND EXCITED STATE:1s(1) 2s(0) 2p(1)
we excite the electron form s subshell to p subshell .but yeah..we have 3 p orbitals of equal energy for the electron to be put in !
and this solves the problem ..the answer is 3
Not considering the electronic spin if the degeneracy of H atom in 2nd excited state is 9 then what is the degeneracy of H^-1 in second excited state?
Answer
2122
Follow
Request
More
You cannot write an answer
You aren't allowed to write answers to questions.
4 ANSWERS
Abhilasha Choudhary, B.tech Materials Science and Engineering & Metallurgy and Materials Engineering, Punjab Engineering College...
Answered May 2, 2016
JEE ADVANCED 2015; a fairly good question that checks the concepts!
Single electron species don't follow the [ n+l ]rule. [(n+l) rule is the rule to find the energy order of the subshells], thus hydrogen being a single electron specie has all its subshells of equal energy in n=3. thus degeneracy( number of orbitals of equal energy for the electron to be excited in ) is 9 {1 s orbital+3p orbitals +5d orbitals}
however hydride ion has a configuration of 1s(2)
FIRST EXCITED STATE: 1s(1) 2s(1)
SECOND EXCITED STATE:1s(1) 2s(0) 2p(1)
we excite the electron form s subshell to p subshell .but yeah..we have 3 p orbitals of equal energy for the electron to be put in !
and this solves the problem ..the answer is 3
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