Place the following molecules and ions in order from largest to smallest bond order. Rank from largest to smallest bond order. To rank items as equivalent, overlap them.

Ne2, F2+, N2+, B2, H2+

The bond order is half of the difference between the number of bonding and number if antibonding electrons.

The bond order is obtained by first writing the molecular orbital configuration of the respective species.

For Ne2, the molecular orbital configuration is written as follows;

σ1s2, σ*1s2, σ2s2, σ*2s2, σ2px2, π2py2, π2pz2, π*2py2, π*2py2, σ*2px2

Bond order = 1/2(Number of bonding - Number of antibonding)electrons

Bond order =  1/2(10 - 10) = 0

For F2+

σ1s2, σ*1s2, σ2s2, σ*2s2, σ2px1

Bond order = 1/2(5 - 4)= 0.5

For N2+

σ1s2, σ*1s2, σ2s2, σ*2s2, π2py2, π2pz2, σ2px1

Bond order = 1/2(9 - 4) = 2.5

For B2

σ1s2, σ*1s2, σ2s2, σ*2s2, π2py2

Bond order = 1/2(6 - 4) = 1

For H2+

σ1s1

Bond order = 1/2(1 - 0) = 0.5

The ranking from largest to smallest bond order is; N2+ > B2 > F2+, H2+ > Ne2

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Ranking from the largest to smallest bond order the given molecules and ions are arranged as

N2^+ (bo = 2.5) > F2^+ (bo =1.5)> B2 (bo= 1) >H2^+ (bo =0.5) > Ne2 (bo = 0)

Explanation:

                       Bond order = 0.5 (Nb – Na)    

Nb = number of bonding electrons or number of electrons in bonding M.O’s

Na = number of anti-bonding electrons

The bond order for the given molecules and ions are calculated as

                        Bond order of N2^+ = 2.5

                        Bond order of F2^+ = 1.5

                        Bond order of B2 = 1

                        Bond order of H2^+ = 0.5

                        Bond order of Ne2 = 0

Ranking from the largest to smallest bond order the given molecules and ions are arranged as N2^+ > F2^+ > B2 > H2^+ > Ne2.

the correct answer is:

a. formation of precipitation

it would be a gas formation