Give electron configuration, bond order,Magnetic property and energy diagram for Baron (B_(2)) Molecule and write about it existence.

Give electron configuration, bond order,Magnetic property and energy d

1.	Give electron configuration, bond order,Magnetic property and energy diagram for Baron (B_(2)) Molecule and write about it existence.
Answer» Solution :`B_(2) (Z = 5) 1s^(2) 2s^(2) 2p^(1) ` So, total electron in `B_(2) = 10 ` Electron configuration in MO for`B_(2)`: KK `(sigma_(2s))^(2) (sigma^(**) 2s)^(2)(pi 2p_(x))^(1) (pi 2p_(y))^(1)("Because " B_(2) pi 2p lt sigma 2p_(z))` BOND order = `(1)/(2) (N_(b) - N_(a)) = (1)/(2) (4 - 2) = 1 ` Two unpaired electron present in `B_(2)` , So it is paramagnetic and Single bond so, it is stable . Bond length B - B ( 159 PM ) and Bond energy `B_(2) ` 290 kj `mol^(-1)` Energy DIAGRAM for `B_(2) ` Molecule is as under (fig).