Question

FIND the bond order of O²+​

Answer 1

Answer:

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Answer 2

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The electronic configuration of the O2+ion containing 15 electrons can be written as:

ion containing 15 electrons can be written as:σ1s² σ ∗1s²σ2s² σ*2s σ2pz2l² π2px² π2p y²

π ∗ 2px1

π ∗ 2px1

π ∗ 2px1

π ∗ 2px1 The bond order can be found as:

π ∗ 2px1 The bond order can be found as:B.O = Nb - Na / 2

Nb = Number of electrons in the bonding orbitals = 10

Nb = Number of electrons in the bonding orbitals = 10Na = Number of electron in the anti-bonding orbitals = 5

Nb = Number of electrons in the bonding orbitals = 10Na = Number of electron in the anti-bonding orbitals = 5B.O =(10−5)/2=5/2=2.5

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