Give electron configuration, bond order, magnetic property and energy diagram for berilium $\left( {{\rm{B}}{{\rm{e}}_2}} \right)$ molecule and writ about it existence.
Give electron configuration, bond order, magnetic property and energy diagram for berilium $\left( {{\rm{B}}{{\rm{e}}_2}} \right)$ molecule and writ about it existence.
$Be$ $(\mathrm{Z}=4)$ So, total electron in $\mathrm{Be}_{2}=8$
Electron configuration in $\mathrm{MO}$ for $\mathrm{Be}_{2}$ :
$\mathrm{KK}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}$ OR $(\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}$
Magnetic Property : All electrons are paired. So, diamagnetic
Bond order $\mathrm{BO}=\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)=\frac{1}{2}(2-2)=0 \quad$ OR $\mathrm{BO}=\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)=\frac{1}{2}(4-4)=0$
Bond order is zero, $\mathrm{So}, \mathrm{Be}_{2}$ is unstable and does not exist.
Energy diagram for $\mathrm{Be}_{2}$ molecule is as under.
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- [JEE MAIN 2020]
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$(c)$ $F_2$ is more likely to dissociate into atoms than $OF$
$(d)$ Both have greater number of electrons in $BMO$ than that in $ABMO$
Which of the following statements are true $(T)$ or false $(F)$ respectively
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$(a)$ Bond order for both is one
$(b)$ $OF$ is paramagnetic but $F_2$ is diamagnetic
$(c)$ $F_2$ is more likely to dissociate into atoms than $OF$
$(d)$ Both have greater number of electrons in $BMO$ than that in $ABMO$
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