Is $B_{2}$ $B_2$ diamagnetic or paramagnetic?

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Is $B_{2}$ $B_2$ diamagnetic or paramagnetic?

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Answer 1 · 2016-04-05T01:05:19

It depends on the MO diagram.

ORBITAL MIXING EFFECTS

$B_{2}$ $"B"_2$ will experience orbital mixing, just like $N_{2}$ $"N"_2$ and every homonuclear diatomic molecule on period 2 to the left of $N_{2}$ $"N"_2$ , where the $σ_{2 p_{z}}$ $sigma_(2p_z)$ MO increases in energy and the $σ_{2 s}$ $sigma_(2s)$ decreases in energy relative to if there is NOT orbital mixing.

So, for $B_{2}$ $"B"_2$ , the $σ_{2 p_{z}}$ $sigma_(2p_z)$ MO will be higher in energy than the $π_{2 p_{x}}$ $pi_(2p_x)$ and $π_{2 p_{y}}$ $pi_(2p_y)$ MOs, NOT lower!

Inorganic Chemistry, Miessler et al., pg. 126

As predicted, the $σ_{2 p_{z}}$ $sigma_(2p_z)$ MO (labeled $σ_{g} (2 p)$ $sigma_(g)(2p)$ ) is higher in energy than the $π_{2 p_{x}}$ $pi_(2p_x)$ and $π_{2 p_{y}}$ $pi_(2p_y)$ MOs (labeled as $π_{u} (2 p)$ $pi_(u)(2p)$ ).

Notice the trend in how the $σ_{g} (2 p)$ $sigma_(g)(2p)$ MO has been moving upwards as we move from ${Ne}_{2}$ $"Ne"_2$ to ${Li}_{2}$ $"Li"_2$ ; that is indicating that the orbital mixing is greater towards ${Li}_{2}$ $"Li"_2$ , and becomes less than significant for $O_{2}$ $"O"_2$ , $F_{2}$ $"F"_2$ , and ${Ne}_{2}$ $"Ne"_2$ .

RATIONALIZING THE ELECTRON CONFIGURATION OF $B_{2}$ $\mathbf(B_2)$

Since boron contributes three valence electrons, two borons give a total of six valence electrons. So, the $σ$ $sigma$ and $π$ $pi$ MOs generated by the $2 s$ $2s$ and $2 p$ $2p$ atomic orbitals in forming $B_{2}$ $"B"_2$ will be filled with six valence electrons total.

Two fill the $σ_{2 s}$ $sigma_(2s)$ (labeled $σ_{g} (2 s)$ $sigma_(g)(2s)$ ) and two fill the $σ_{2 s}^{*}$ $sigma_(2s)^"*"$ (labeled $σ_{u}^{*} (2 s)$ $sigma_(u)^"*"(2s)$ ). Finally, one electron goes in the $π_{2 p_{x}}$ $pi_(2p_x)$ and the other goes into the $π_{2 p_{y}}$ $pi_(2p_y)$ .

That accounts for all six valence electrons.

CONCLUSION

Therefore, $B_{2}$ $"B"_2$ contains two unpaired electrons and is paramagnetic.

CHALLENGE: Had the $π_{u} (2 p)$ $pi_(u)(2p)$ MOs been higher in energy than the $σ_{g} (2 p)$ $sigma_(g)(2p)$ MO, $B_{2}$ $B_2$ would have been diamagnetic. Why?

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Is $B_{2}$ $B_2$ diamagnetic or paramagnetic?

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Is $B_{2}$ $B_2$ diamagnetic or paramagnetic?