 # B2 Lewis Structure

## B2 Lewis Structure

How to draw the Lewis Point structure of B2 (two barrows)? ۔

?

Well, you know that the Lewis structure must adhere to the principle with 8 electrons in the outer shell of each atom.

This is an exception, so the two holes are connected by a triple bond (B to K: 2 electrons and L: 3 electrons), with each of the 3 electrons in its outer shell. They cannot hold 8 electrons in their outer shell and this is a fact even with the elements of CI IIA apparatus.

Lew's structure of B2.

The number of formula limits is = N = limit C + limit 2 or atom N = 4 + 2X 6 = 16 (electronic configuration C = (6) = 2.4 4 each or (8 NON atoms) 2, 6 to 6 VALENCE ES V = Stable state limit = 8 + 2x 8 = 24 {In gas or stable order, each atom has 8 limits, except H, in which only 2} S = distribution = VN = 24 16 = 8 (4 bonds are bad) Maybe because each bond is made up of 2 and co2 is shaded in 8 to form 4 bonds; U = hairless = NS = 16 8 = 8 ES; CO2 has only 4 pairs) so STR :: O = C = O :: Note that each atom must complete its byte, meaning that each atom must have 8 ca caes (4 divisions and 4 divisions 2e o) here, but each O has only 4 So the remaining 4 are not paired. 8 are around each one

## B2 Lewis Structure

How to draw the Lewis Point structure of B2 (two bronze)? 3

?

Well, you know that the Lewis structure must comply with the rule of 8 electrons in the outer shell of each atom.

This is an exception, because the two holes are connected by a triple bond (B to K: 2 electrons and L: 3 electrons), with 3 electrons each in its outer shell. Their outer shell cannot contain 8 electrons and this is a fact even with the elements of CI IIA apparatus.

The formula is the number of limits = N = limit C + limit 2 or atom N = 4 + 2X 6 = 16 (electronic configuration C = (6) = 2.4 4 moisture each or (8 non-atoms) 2, 6 to 6 valence e S) V = steady state limit = 8 + 2x 8 = 24 {In gas or steady state each atom has 8 extensions besides H, in which only 2} S = split = VN = 24 16 = 8 (4 bonds) The shape will change because each bond is formed by 2 and co2 is divided into 8, thus 4 bonds are formed; U = no hair = NS = 16 8 = 8 IT; CO2 has only 4 pairs) So STR: : O = C = O:: Please note that each atom must complete its byte, which means that each atom here must have 8 approximately 8s (4 divides a and 4 into 2e o) , But each O has only 4s, so the remaining 4, respectively, have non-paired 8s around each. O) Similarly, you can draw another structure using the same formula. ...