How many bonds (bond order) does B_2 have? | vedclass

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(B) The $B_2$ molecule is formed by the combination of two Boron atoms.The atomic number of Boron is $5$,and its electronic configuration is $1s^2 \,

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$1$

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(B) The $B_2$ molecule is formed by the combination of two Boron atoms.The atomic number of Boron is $5$,and its electronic configuration is $1s^2 \, 2s^2 \, 2p^1$.According to Molecular Orbital Theory,the molecular orbital configuration for $B_2$ ($10$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1 = \pi 2p_y^1$.Number of bonding electrons $(N_b)$ = $6$.Number of antibonding electrons $(N_a)$ = $4$.Bond order = $\frac{1}{2}(N_b - N_a) = \frac{1}{2}(6 - 4) = \frac{2}{2} = 1$.

How many bonds (bond order) does $B_2$ have?

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