BH4-, B is central atom

BH4-, B is central atom.

Write acceptable Lewis structures for BH4-

The correct answer and explanation is :

The BH₄⁻ ion is composed of one boron (B) atom and four hydrogen (H) atoms, along with a negative charge. The structure of BH₄⁻ follows the basic principles of Lewis structures, where the electrons are arranged to satisfy the octet rule (or a duet rule for hydrogen, which only requires two electrons).

Steps to Draw the Lewis Structure for BH₄⁻:

1. Count the total number of valence electrons:

• Boron (B) is in Group 13 of the periodic table and has 3 valence electrons.
• Each hydrogen (H) atom has 1 valence electron. Since there are 4 hydrogens, this contributes 4 electrons.
• The negative charge on the ion means there is an additional electron. Total valence electrons = (3 from B) + (4 × 1 from H) + (1 additional electron for the negative charge) = 8 valence electrons.

1. Arrange the atoms:

• Boron (B) will be the central atom since it is less electronegative than hydrogen (H).
• Place the 4 hydrogen atoms around the boron atom.

1. Distribute the electrons:

• Each hydrogen atom forms a single bond with the boron atom. A single bond consists of 2 electrons.
• After placing the 4 bonds (which use 8 electrons), all electrons are accounted for.

1. Verify the octet rule:

• Boron is an exception to the octet rule. It can accommodate only 6 electrons in its valence shell, which is satisfied in this case (via 4 single bonds with 4 hydrogens).
• Each hydrogen atom has 2 electrons, satisfying the duet rule.

1. Final Structure:

• The central boron atom has 4 single bonds to 4 hydrogen atoms, and there is a negative charge on the entire ion. The negative charge is usually represented on the boron atom.

Lewis Structure of BH₄⁻:

    H
    |
H—B—H
    |
    H

Explanation:

The structure of BH₄⁻ is a tetrahedral configuration, where the four hydrogen atoms surround the central boron atom. The boron atom has 4 single bonds with hydrogen, and there are no lone pairs of electrons on the boron. The negative charge is distributed across the ion, and it does not affect the bond formation, as the ion’s charge is taken into account in the electron count.

The BH₄⁻ ion is stable because it satisfies the duet rule for hydrogen and the general bonding preferences for boron, which does not require an octet to be stable. This structure is one of the simplest examples of a boron hydride.