Q.
The species having a spin only magnetic moment of 2.82 B.M is
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a
N2
b
N2+
c
N2-
d
N22-
answer is D.
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Detailed Solution
N2-KK'σ2s2σ2s*2π2px2π2py2σ2pz2 ; N2+-KK'σ2s2σ2s*2π2px2π2py2σ2pz1 ; N2--KK'σ2s2σ2s*2π2px2π2py2σ2pz2π2p*1 ; N22--KK'σ2s2σ2s*2π2px2π2py2σ2pz2π2px*1π2py*1 ;Bond order and magnetic moment : μs=nn+2 B. M ; N2-2=10-62=2; Paramagnetic...μs=2.82 B.M ; N2-=10-52=2.5; Paramagnetic...μs=1.732 B.M ; N2=10-42=3; Diamagnetic.....μs=0 ; N2+=9-42=2.5; Paramagnetic..μs=1.732B.M ;Bond order ∝ 1Bond lengthStability ∝Bond order;Stability----- N23 >N2+ 2.5>N2- 2.5>N22-2 ;Higher stability of N2+ over N2- is due to lesser number of ABMO electrons in N2+.
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