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Chapter 8: Problem 90

Borazine \(\left(\mathrm{B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6}\right)\) has often been called "inorganic" benzene. Write Lewis structures for borazine. Borazine contains a sixmembered ring of alternating boron and nitrogen atoms with one hydrogen bonded to each boron and nitrogen.

Short Answer

Expert verified
The Lewis structure for Borazine (B3N3H6) can be drawn as follows: H \ B - H / \ N - B - H / \ H N \ H In this structure, each boron and nitrogen atom in the six-membered ring is bonded to an adjacent hydrogen atom. Nitrogen atoms have two lone pairs, and all atoms have a formal charge of zero.

Step by step solution

01

1. Count total valence electrons

To draw the Lewis structure, we first need to identify the total number of valence electrons in the molecule. Boron has 3 valence electrons, Nitrogen has 5 valence electrons, and Hydrogen has 1 valence electron. There are equal numbers of boron and nitrogen atoms, and two hydrogen atoms bonded to each boron and nitrogen atom in Borazine. Total valence electrons in Borazine = (3 valence e⁻ for B) × (3 B atoms) + (5 valence e⁻ for N) × (3 N atoms) + (1 valence e⁻ for H) × (6 H atoms) = 9 + 15 + 6 = 30 valence electrons.
02

2. Arrange atoms to form a skeleton structure

Arrange the atoms such that there are alternating boron and nitrogen atoms in a six-membered ring. Attach one hydrogen atom to each boron and nitrogen atom, forming a ring where B and N atoms alternate.
03

3. Distribute the valence electrons as bonds and lone pairs

Distribute the 30 valence electrons among the atoms by setting up single bonds between adjacent atoms in the ring and between each hydrogen atom and the atom it is attached to: Atomic valence e⁻ count: B: 3 valence electrons N: 5 valence electrons H: 1 valence electron Each bond requires 2 electrons. So, we will have 12 electrons to form the single bonds: Bonding: B - N: 2 electrons per bond B - H: 2 electrons per bond N - H: 2 electrons per bond Now there are 18 valence electrons left (30 - 12). We will use these electrons to form lone pairs on nitrogen atoms. Lone pairs: N: 2 lone pairs (6 electrons used) per nitrogen atom. Now, 12 electrons are left unused which will be assigned as lone pairs.
04

4. Check the formal charge and adjust accordingly

Check the formal charge on each atom, and if any atom has a formal charge other than zero, adjust the bonding accordingly. Formal charge for Boron = (3 valence electrons) - (0.5 × 4 bonds) - (0 lone pairs) = 0 Formal charge for Nitrogen = (5 valence electrons) - (0.5 × 4 bonds) - (2 lone pairs) = 0 Formal charge for Hydrogen = (1 valence electron) - (0.5 × 2 bonds) = 0 All formal charges are zero, indicating that this structure is appropriate.
05

5. Draw the Lewis structure for Borazine

Use the skeleton structure, bonds, and lone pairs found to draw the final Lewis structure for borazine. Borazine (B3N3H6) Lewis structure: H \ B - H / \ N - B - H / \ H N \ H With Nitrogen atoms having 2 lone pairs each, and Boron atoms having an empty p-orbital. Each bond between the B and N is single, and the charges are all neutral.

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Most popular questions from this chapter

Consider the following reaction: $$ \mathrm{A}_{2}+\mathrm{B}_{2} \longrightarrow 2 \mathrm{AB} \quad \Delta H=-285 \mathrm{~kJ} $$ The bond energy for \(\mathrm{A}_{2}\) is one-half the amount of the \(\mathrm{AB}\) bond energy. The bond energy of \(\mathrm{B}_{2}=432 \mathrm{~kJ} / \mathrm{mol}\). What is the bond energy of \(\mathrm{A}_{2}\) ?

The alkali metal ions are very important for the proper functioning of biologic systems, such as nerves and muscles, and \(\mathrm{Na}^{+}\) and \(\mathrm{K}^{+}\) ions are present in all body cells and fluids. In human blood plasma, the concentrations are $$ \left[\mathrm{Na}^{+}\right] \approx 0.15 M \text { and }\left[\mathrm{K}^{+}\right] \approx 0.005 M $$ For the fluids inside the cells, the concentrations are reversed: $$ \left[\mathrm{Na}^{+}\right] \approx 0.005 M \text { and }\left[\mathrm{K}^{+}\right] \approx 0.16 M $$ Since the concentrations are so different inside and outside the cells, an elaborate mechanism is needed to transport \(\mathrm{Na}^{+}\) and \(\mathrm{K}^{+}\) ions through the cell membranes. What are the ground-state electron configurations for \(\mathrm{Na}^{+}\) and \(\mathrm{K}^{+}\) ? Which ion is smaller in size? Counterions also must be present in blood plasma and inside intracellular fluid. Assume the counterion present to balance the positive charge of \(\mathrm{Na}^{+}\) and \(\mathrm{K}^{+}\) is \(\mathrm{Cl}^{-}\). What is the ground-state electron configuration for \(\mathrm{Cl}^{-}\) ? Rank these three ions in order of increasing size.

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