Problem 31 - Chemical Principles: The Quest for Insight

Question

Write the Lewis structure of (a) ammonium ion, \(\mathrm{NH}_{4}{\underline{\phantom{xx}}}^{*}\); (b) hypochlorite ion, \(\mathrm{ClO}^{-}\); (c) tetrafluoroborate ion, \(\mathrm{BF}_{4}^{-} .\)

Step-by-Step Solution

Answer

The Lewis structures are: (a) Ammonium ion (\text{NH}_{4}^{+}) has nitrogen in the center with no lone pairs and four single bonds to hydrogen atoms. (b) Hypochlorite ion (\text{ClO}^{-}) has chlorine in the center with three lone pairs, oxygen with two lone pairs, and a double bond between them. (c) Tetrafluoroborate ion (\text{BF}_{4}^{-}) has boron in the center connected to four fluorine atoms each with three lone pairs, and boron itself has no lone pairs.

Step 1: Drawing the Lewis Structure for Ammonium Ion (\text{NH}_{4}^{+})

Begin by considering the valence electrons. Nitrogen has 5 valence electrons, and each hydrogen atom has 1. Since the molecule is positively charged, we remove one electron. Total valence electrons = 5 (from N) + 4 (from H) - 1 (charge) = 8 electrons. Arrange the electrons to form four N-H bonds around the nitrogen atom. The positive charge signifies that nitrogen does not have a lone pair of electrons in this ion.

Step 2: Drawing the Lewis Structure for Hypochlorite Ion (\text{ClO}^{-})

Chlorine has 7 valence electrons, and oxygen has 6. Since the molecule has a negative charge, add an extra electron. Total valence electrons = 7 (from Cl) + 6 (from O) + 1 (charge) = 14 electrons. Place Chlorine at the center and connect it with Oxygen using a single bond (2 electrons). Distribute the remaining 12 electrons to complete the octets of both oxygen and chlorine. Since oxygen still has more open spots after completing the octet, create a double bond with chlorine, which completes both atoms' octets and accounts for the extra electron from the negative charge.

Step 3: Drawing the Lewis Structure for Tetrafluoroborate Ion (\text{BF}_{4}^{-})

Boron has 3 valence electrons, and each fluorine atom has 7. Since the molecule has a negative charge, add an extra electron. Total valence electrons = 3 (from B) + 4 x 7 (from F) + 1 (charge) = 32 electrons. Place Boron at the center and connect it to four fluorine atoms through single bonds (8 electrons). Distribute the remaining 24 electrons to complete the octets of the fluorine atoms. Boron, after forming the single bonds, does not require any additional electrons to complete its octet, which is an exception to the octet rule.

Key Concepts

Valence Electrons

Valence electrons are the outermost electrons of an atom and play a pivotal role in chemical bonding. They are the electrons that are most likely to be involved in interactions with other atoms because they have the most energy and are the furthest away from the nucleus.

In the textbook exercise provided, determining the number of valence electrons is always the first step in drawing Lewis structures. For instance, nitrogen (N) has 5 valence electrons, hydrogen (H) has 1 each, chlorine (Cl) has 7, oxygen (O) has 6, boron (B) has 3, and fluorine (F) has 7. However, when an atom is part of an ion, the charge of the ion alters the number of valence electrons. For positively charged ions (cations), electrons are removed, and for negatively charged ions (anions), electrons are added. This adjustment is crucial for accurately representing the ionic species.

Here are important tips to help students better understand valence electrons: