Lewis Dot Structures: Ions (Simplified): Videos & Practice Problems

In Lewis dot structures, cations have fewer valence electrons due to electron loss, while anions have more from electron gain. For the anion BCl₄^-, calculate total valence electrons: boron (1) + chlorine (4 x 7) + 1 (for the charge) = 32. Place boron at the center, bond it to four chlorines, and ensure each chlorine follows the octet rule. Finally, enclose the structure in brackets with the charge in the top right corner. Understanding these principles is essential for accurately representing ionic compounds.

Lewis Dot Structures: Ions (Simplified) Concept 1

In Lewis dot structures, cations have fewer valence electrons because, remember, cations are positive since they are losing electrons; anions have more valence electrons, and an anion is a negative ion because it's gaining electrons. So using this, let's answer this example question. It says draw the Lewis dot structure for the following anion, BCl₄^-. Alright. So here we have to determine the total number of valence electrons of the structure. And recall that valence electrons equal the group number of the element. So here we have boron, which is in group 3A, and there's one of them. We have chlorine, which is in group 7A, and there are 4 of them. That's 28. And then remember that this minus one means we've gained an electron. So we're going to add 1 electron to this mix, giving us a total of 32 valence electrons. Now, we're going to place the least electronegative element in the center and connect all elements with single bonds. We're going to follow our bonding preferences guide to determine atom connectivity. Now, if you haven't watched my video on bonding preferences, I highly suggest you go back and take a look because it helps you to identify what kind of bonds these elements like to make—2 bonds, 3 bonds, etc. So here we're going to place boron in the center, and we're going to make single bonds to each of the chlorines. Now we're going to add electrons to all surrounding elements until they have 8 electrons because we're trying to follow the octet rule. Now remember, hydrogen doesn't follow the octet rule; it follows the duet rule. It only wants 2 valence electrons around it, so that it can be just like helium. So, I'm adding enough electrons so that each chlorine is following the octet rule. Now place any remaining electrons on the central element or central atom. So we've already used all 32 of our electrons here, so there's none left. Step 5 says that we need to place the ion in brackets and its charge in the top right corner. So that's right. For ions, once you draw the Lewis dot structure, you have to put it in brackets with the charge on the outside. So brackets around it, and you'd put the minus charge here. Here I'm going to bring it down so we can see it better. So here goes boron with the 4 chlorines that it's attached to. So here go our 4 chlorines that we're attached to. Since it's an ion, put it in brackets, and the charge on the outside and top right corner. Now for cations, this is not a cation, but for cations, remove valence electrons from the central element. So just keep that in mind when dealing with cations. So these would be the rule that we use in order to draw either cations or anions when asked to determine the Lewis dot structure.

Draw the Lewis Dot Structure for the following cation:NH₄⁺.

• A.
• B.
• C.
• D.

Answer: (C)

Determine the Lewis Dot Structure for the following ion:O₂^2–.

• A.
• B.
• C.
• D.

Answer: (A)

Determine the Lewis Dot Structure for the following ion:SCl₄²⁺.

• A.
• B.
• C.
• D.

Answer: (C)

Draw the Lewis Dot Structure for the following ion:PCl₄⁺.

• A.
• B.
• C.
• D.

Answer: (B)

Here’s what students ask on this topic:

How do you draw the Lewis dot structure for anions?

To draw the Lewis dot structure for anions, follow these steps: First, determine the total number of valence electrons by adding the group numbers of each element and accounting for the charge (add electrons for negative charge). Place the least electronegative element in the center and connect all elements with single bonds. Add electrons to surrounding elements to satisfy the octet rule. Place any remaining electrons on the central atom. Finally, enclose the structure in brackets and indicate the charge in the top right corner.

What is the significance of the octet rule in Lewis dot structures?

The octet rule is significant in Lewis dot structures because it helps predict the arrangement of electrons around atoms. Atoms tend to form bonds to have eight electrons in their valence shell, achieving a stable electron configuration similar to noble gases. This rule guides the placement of electrons in bonds and lone pairs, ensuring that each atom (except hydrogen, which follows the duet rule) has a complete octet.

How do you determine the total number of valence electrons in a polyatomic ion?

To determine the total number of valence electrons in a polyatomic ion, sum the valence electrons of each atom based on their group number in the periodic table. For anions, add extra electrons equal to the negative charge. For cations, subtract electrons equal to the positive charge. For example, in BCl₄^-, boron (group 3) has 3 electrons, each chlorine (group 7) has 7 electrons, and the negative charge adds 1 electron, totaling 32 valence electrons.

Why do cations have fewer valence electrons in Lewis dot structures?

Cations have fewer valence electrons in Lewis dot structures because they lose electrons to achieve a positive charge. This loss of electrons results in fewer valence electrons around the central atom. For example, if a neutral atom loses one electron to form a cation with a +1 charge, it will have one less valence electron in its Lewis dot structure.

What is the process for placing the charge on a Lewis dot structure for ions?

After drawing the Lewis dot structure for an ion, enclose the entire structure in square brackets. Then, place the charge of the ion in the top right corner outside the brackets. For example, for the anion BCl₄^-, draw the structure, enclose it in brackets, and place the negative charge outside the brackets in the top right corner.