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Question

# Pick out the incorrect statement?

A
N2 has greater dissociation energy than N+2
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B
O2 has lower dissociation energy than O+2
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C
Bond length in N+2 is less than N2
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D
Bond length in NO+ is less than in NO
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Solution

## The correct option is C Bond length in N+2 is less than N2(a) The electronic configuration of N2=(σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px)2(π2py)2(σ2pz)2 Thus, electronic configuration of N+2=(σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px)2(π2py)2(σ2pz)2 So bond order of N2=10−42=3 Bond order of N+2=9−52=2.5 Since bond order of N2 is more compared to N+2, thus, thus N2 has greater dissociation energy than N+2 (b) The electronic configuration of O2=(σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(σ2pz)2(π2px)2(π2py)2(π∗2px)1(π∗2py)1 The electronic configuration of O+2=(σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(σ2pz)2(π2px)2(π2py)2(π∗2px)1 So the bond order of O2=10−62=2 Bond order of O+2=10−52=2.5 Since bond order of O2 is less than that of O+2, thus O2 has lower dissociation energy than O+2. (c) Bond order of N2 is more compared to N+2 and again we know, bond length ∝1 bond order . SO bond length of N+2 is more than that of N2. (d) The electronic confiuration of NO=(σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px)2(π2py)2(σ2pz)2(π∗2px)1 The electronic confiuration of NO+=(σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px)2(π2py)2(σ2pz)2 The bond order of NO=10−52=2.5 ​​​​​​​The bond order of NO+=10−42=3 Since bond order of NO+ is more, thus bond length in NO+ is less than in NO

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