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Concept of Exchange energy
(Eex)
Dr. Mithil Fal Desai
Shree Mallikarjun and Shri Chetan Manju Desai College
Canacona Goa
Exchange energy (Eex)
𝐸𝑒𝑥 = ෍
𝑁(𝑁 − 1)
2
𝐾
Nitrogen[He] 2s2, 2p3
2px 2py 2pz 2px 2py 2pz
N- number of electrons with parallel spin
K- constant
or
Electrons can exchange their position
with other electrons having parallel spin.
Exchange energy (Eex)
𝐸𝑒𝑥 = ෍
𝑁(𝑁 − 1)
2
𝐾
2px 2py 2pz 2px 2py 2pz
N- number of electrons with parallel spin
K- constant
or
Electrons can exchange their position
with other electrons having parallel spin.
Nitrogen[He] 2s2, 2p3
𝐸𝑒𝑥 =
3(3 − 1)
2
𝐾
= 3K
𝐸𝑒𝑥 =
2(2 − 1)
2
𝐾
= 1K
Exchange energy (Eex)
𝐸𝑒𝑥 = ෍
𝑁(𝑁 − 1)
2
𝐾
N- number of electrons with parallel spin
K- constant
2px 2py 2pz 2px 2py 2pz
or
Nitrogen[He] 2s2, 2p3
𝐸𝑒𝑥 =
5(5 − 1)
2
𝐾 +
5(5 − 1)
2
𝐾
= 20 K
Exchange energy (Eex)
𝐸𝑒𝑥 = ෍
𝑁(𝑁 − 1)
2
𝐾
N- number of electrons with parallel spin
K- constant
Cu [Ar] 3d10, 4s1 Cu [Ar] 3d9, 4s2
𝐸𝑒𝑥 =
5(5 − 1)
2
𝐾 +
4 (4 − 1)
2
𝐾
= 16 K
Higher the exchange energy more will be the stability of the electronic configuration

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Exchange energy (Eex)

  • 1. Concept of Exchange energy (Eex) Dr. Mithil Fal Desai Shree Mallikarjun and Shri Chetan Manju Desai College Canacona Goa
  • 2. Exchange energy (Eex) 𝐸𝑒𝑥 = ෍ 𝑁(𝑁 − 1) 2 𝐾 Nitrogen[He] 2s2, 2p3 2px 2py 2pz 2px 2py 2pz N- number of electrons with parallel spin K- constant or Electrons can exchange their position with other electrons having parallel spin.
  • 3. Exchange energy (Eex) 𝐸𝑒𝑥 = ෍ 𝑁(𝑁 − 1) 2 𝐾 2px 2py 2pz 2px 2py 2pz N- number of electrons with parallel spin K- constant or Electrons can exchange their position with other electrons having parallel spin. Nitrogen[He] 2s2, 2p3
  • 4. 𝐸𝑒𝑥 = 3(3 − 1) 2 𝐾 = 3K 𝐸𝑒𝑥 = 2(2 − 1) 2 𝐾 = 1K Exchange energy (Eex) 𝐸𝑒𝑥 = ෍ 𝑁(𝑁 − 1) 2 𝐾 N- number of electrons with parallel spin K- constant 2px 2py 2pz 2px 2py 2pz or Nitrogen[He] 2s2, 2p3
  • 5. 𝐸𝑒𝑥 = 5(5 − 1) 2 𝐾 + 5(5 − 1) 2 𝐾 = 20 K Exchange energy (Eex) 𝐸𝑒𝑥 = ෍ 𝑁(𝑁 − 1) 2 𝐾 N- number of electrons with parallel spin K- constant Cu [Ar] 3d10, 4s1 Cu [Ar] 3d9, 4s2 𝐸𝑒𝑥 = 5(5 − 1) 2 𝐾 + 4 (4 − 1) 2 𝐾 = 16 K Higher the exchange energy more will be the stability of the electronic configuration