Question

1. using your knowledge of ionic transfer of electrons, draw a lewis diagram for the compound bh₃. show all orbital numbers before and after the compound was formed as used in your past studies.

Explanation:

Step1: Analyze Atomic Orbitals and Electrons

• Boron (B): Atomic number 5. Electron configuration: \(1s^2 2s^2 2p^1\). Valence electrons: 3 (in \(2s^2 2p^1\)). Orbital filling before bonding: \(1s^2\) (filled), \(2s^2\) (filled), \(2p^1\) (partially filled).
• Hydrogen (H): Atomic number 1. Electron configuration: \(1s^1\). Valence electron: 1 (in \(1s^1\)). Orbital filling before bonding: \(1s^1\) (partially filled). There are 3 H atoms, so total H valence electrons: \(3\times1 = 3\).

Step2: Determine Bonding (Note: \(BH_3\) is Covalent, Not Ionic)

• Boron has 3 valence electrons, and each H has 1. In \(BH_3\), B forms 3 covalent bonds with H atoms (sharing electrons).
• Orbital Hybridization (for B): B undergoes \(sp^2\) hybridization. The \(2s\) and two \(2p\) orbitals hybridize to form three \(sp^2\) hybrid orbitals (each with 1 electron) and one unhybridized \(2p\) orbital (empty).
• Bond Formation: Each \(sp^2\) hybrid orbital of B overlaps with the \(1s\) orbital of an H atom, sharing an electron pair.

Step3: Draw Lewis Diagram

• Boron (Central Atom): Write "B" in the center. It has 3 valence electrons initially, and after forming 3 bonds, it has 6 electrons (3 shared pairs) around it (since it's a Lewis acid in some cases, but in \(BH_3\) the Lewis structure shows 3 single bonds to H).
• Hydrogen Atoms: Each H is bonded to B. Draw three H atoms around B, each connected by a single line (representing the shared electron pair). Each H has 2 electrons (the shared pair) after bonding, filling its \(1s\) orbital.
• Orbital Numbers (Before Bonding):
• B: \(1s^2\), \(2s^2\), \(2p^1\) (orbitals: \(1s\), \(2s\), \(2p_x\), \(2p_y\), \(2p_z\) – with \(2p_x\) having 1 electron, others in \(2p\) empty).
• H (each): \(1s^1\) (orbital: \(1s\) with 1 electron).
• Orbital Numbers (After Bonding):
• B: \(1s^2\), \(sp^2\) hybrid orbitals (3 orbitals, each with 1 electron from B and 1 from H, so each \(sp^2\) orbital has 2 electrons), and unhybridized \(2p_z\) (empty).
• H (each): \(1s^2\) (filled \(1s\) orbital with 2 electrons from the shared bond).

Answer:

Lewis Diagram of \(BH_3\):

• Central B atom with 3 single bonds to H atoms. Each bond is a pair of electrons. B has 6 valence electrons (3 bonds), and each H has 2 valence electrons (from the bond).

Orbital Details:

• Before Bonding:
• B: \(1s^2\), \(2s^2\), \(2p^1\) (orbitals: \(1s\) (filled), \(2s\) (filled), \(2p_x\) (1 e⁻), \(2p_y\) (0 e⁻), \(2p_z\) (0 e⁻)).
• Each H: \(1s^1\) (orbital: \(1s\) (1 e⁻)).
• After Bonding:
• B: \(1s^2\), \(sp^2\) hybrid orbitals (3 orbitals, each with 2 e⁻ from B - H bonds), unhybridized \(2p_z\) (0 e⁻).
• Each H: \(1s^2\) (orbital: \(1s\) (2 e⁻ from B - H bond)).

(Note: \(BH_3\) is a covalent compound, not ionic, so electron "transfer" is not accurate; it involves electron sharing. The Lewis structure shows 3 B - H single bonds, with B having 6 valence electrons and each H having 2.)