The structure of borazine, \(\mathrm{B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6},\) is a six-membered ring of alternating \(\mathrm{B}\) and \(\mathrm{N}\) atoms. There is one \(\mathrm{H}\) atom bonded to each \(B\) and to each \(\mathrm{N}\) atom. The molecule is planar. (a) Write a Lewis structure for borazine in which the formal charge on every atom is zero. (b) Write a Lewis structure for borazine in which the octet rule is satisfied for every atom. (c) What are the formal charges on the atoms in the Lewis structure from part (b)? Given the electronegativities of \(B\) and \(N,\) do the formal charges seem favorable or unfavorable? (d) Do either of the Lewis structures in parts (a) and (b) have multiple resonance structures? (e) What are the hybridizations at the \(\mathrm{B}\) and \(\mathrm{N}\) atoms in the Lewis structures from parts (a) and (b)? Would you expect the molecule to be planar for both Lewis structures? (f) The six \(\mathrm{B}-\mathrm{N}\) bonds in the borazine molecule are all identical in length at \(144 \mathrm{pm} .\) Typical values for the bond lengths of \(\mathrm{B}-\mathrm{N}\) single and double bonds are \(151 \mathrm{pm}\) and \(131 \mathrm{pm},\) respectively. Does the value of the \(\mathrm{B}-\mathrm{N}\) bond length seem to favor one Lewis structure over the other? (g) How many electrons are in the \(\pi\) system of botazine?

For borazine (\(\mathrm{B}_{3}\mathrm{N}_{3}\mathrm{H}_{6}\)), we have: - 3 B atoms with 3 valence electrons each (total of 9) - 3 N atoms with 5 valence electrons each (total of 15) - 6 H atoms with 1 valence electron each (total of 6) Total valence electrons: 9 + 15 + 6 = 30.

For the structure with a formal charge of zero, we can place single bonds between each alternating B and N atoms and between each B-N pair and their respective H atoms. This structure will consume 12 electrons (2 for each bond): - B-N-B-N-B-N - Each B and N atom has an additional H atom bonded.

We have used 12 electrons in the structure, and we have 30 - 12 = 18 electrons remaining. We can distribute these as lone pairs: - Assign the 3 lone pairs (total of 6 electrons) to each of the N atoms. - Assign the 1 lone pair (total of 2 electrons) to each of the B atoms. - At this point, we have a total of 24 electrons used. No more electrons can be placed on any atom without violating the octet rule. This structure satisfies the condition of having zero formal charges on each atom. (b) Now, we will write a Lewis structure satisfying the octet rule for each atom.

Since the previous Lewis structure already satisfies the octet rule for N, we just need to satisfy it for the B atoms. We can do this by removing one lone pair from each N atom and forming a double bond between N and B atoms. This results in the following structure: - Double bonds between each alternating B-N pair and single bonds between each B-N pair and their H atoms. - One lone pair on B atom and no lone pair on N atom. (c) Determining the formal charges:

To calculate the formal charge, we can use the following formula: Formal charge = (number of valence electrons) - (number of bonding electrons/2) - (number of non-bonding electrons) Applying this formula to the specific atoms: - B: Formal charge = 3 - (6/2) - 2 = 0 - N: Formal charge = 5 - (8/2) - 0 = 1 - H: Formal charge = 1 - (2/2) - 0 = 0 Considering the electronegativities, B (2.04) is less electronegative than N (3.04), so having a positive formal charge on the more electronegative atom (N) is unfavorable, but still possible. (d) Both Lewis structures in parts (a) and (b) have alternate resonance structures. Since the ring contains alternating double and single bonds, we can have different structures by shifting the double bonds. (e) The hybridizations for both Lewis structures are: - B atom: sp² hybridized (3 regions of electron density: one single bond, one double bond, and one lone pair) - N atom: sp² hybridized (3 regions of electron density: one single bond, one double bond, and one lone pair) Both structures should be planar due to the sp² hybridization. (f) Experimental B-N bond length: 144 pm Single bond length: 151 pm Double bond length: 131 pm The experimental bond length is approximately the average of the B-N single and double bond lengths, which favors the resonance structure where the electrons are delocalized over the ring, creating a hybrid bond length between the single and double bond lengths. (g) The π system of borazine has 6 electrons: 3 from the N atoms and 3 gained from the B atoms.