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Find the bond order in B2 molecule. (at. no. of B=5).

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Detailed Solution

no. of electrons isB2=(2× at. no. of B)=2×5=10∴ M.O. electron configuration ofB2=σ1s2σ1s∗2σ2s2σ2s∗2(π2px1 =π2py1) ∴ Bond order =Nb−Na2=6−42=1.

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Find the bond order in B2 molecule. (at. no. of B=5).