Give electron configuration, magnetic property, bond order and energy diagram for fluorine $\left( {{{\rm{F}}_2}} \right)$ molecule.
Give electron configuration, magnetic property, bond order and energy diagram for fluorine $\left( {{{\rm{F}}_2}} \right)$ molecule.
$\mathrm{F}_{2}(\mathrm{Z}=9) 1 s^{2} 2 s^{2} 2 p^{5} .$ In valence cell 7 electron and bond structure of $\mathrm{F}_{2}=14$ electrons. Electron configuration in MO for $\mathrm{F}_{2}: \mathrm{KK}\left(\sigma_{2 s}\right)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2}=\left(\pi 2 p_{y}\right)^{2}\left(\pi^{*} 2 p_{x}\right)^{2}$ $=\left(\pi^{*} 2 p_{y}\right)^{2}$
Bond order $=\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)=\frac{1}{2}(10-8)=1$
So, $\mathrm{F}$ - $\mathrm{F}$ Single bond
Magnetic property : All electrons are paired, So, diamagnetic. Energy diagram for $\mathrm{F}_{2}$ molecule :
Similar Questions
According to Molecular Orbital Theory,
($A$) $\mathrm{C}_2^{2-}$ is expected to be diamagnetic
($B$) $\mathrm{O}_2{ }^{2+}$ is expected to have a longer bond length than $\mathrm{O}_2$
($C$) $\mathrm{N}_2^{+}$and $\mathrm{N}_2^{-}$have the same bond order
($D$) $\mathrm{He}_2^{+}$has the same energy as two isolated He atoms
According to Molecular Orbital Theory,
($A$) $\mathrm{C}_2^{2-}$ is expected to be diamagnetic
($B$) $\mathrm{O}_2{ }^{2+}$ is expected to have a longer bond length than $\mathrm{O}_2$
($C$) $\mathrm{N}_2^{+}$and $\mathrm{N}_2^{-}$have the same bond order
($D$) $\mathrm{He}_2^{+}$has the same energy as two isolated He atoms
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