Bonding Preferences - Video Tutorials & Practice Problems

Bonding Preferences Concept 1

Now when it comes to bonding preferences realize that we can predict the number of bonds and non bonding electrons in a molecular compound. Now when it comes to these non bonding electrons, they're just electrons that do not participate in bonding with other elements. And we can say here that a lone pair is just a pair of these non bonding electrons. Now if we take a look here, we have some common bonding preferences. So we have from groups one, a 27 a. And here we have representative elements for each group. Realize here that when it comes to group seven A. To represent a halogen, I can use the variable of X. Which stands in for florine, chlorine, bromine and iodine here. When we talk about bonding preferences, we're going to say here that for rule one, we're gonna say group 18 of four A elements. It's pretty simple, we're gonna say the number of bonded they prefer is equal to their group number. So hydrogen, which is a group, Want a preferably wants to make one bond. And one important thing here is when we start drawing molecular compounds later on, I realized that when drawing it hydrogen never goes in the center of any of those compounds. Beryllium is a group to a So wants to make two bonds bronze and group three. Acer wants to make three. Now, carbon is incredibly important. Carbon wants to make four bonds. Even if that means bonding with itself. Now for groups 58 A seven A. We go by rule two here, we're going to say that the number of bonds equals the number of electrons needed for stable electron configuration or arrangement. So again, groups one a, 2, 4 a. It's based on group number and then we're going to stay here. That nitrogen, which is a group five a. has five valence electrons. So comes in with five. But if it wants to follow the octet rule, it wants to get to eight valence electrons by forming bonds with surrounding elements. It's able to pick up three more. So that's why he wants to make three bonds oxygen's route six A. So it has six valence electrons that it brings to the table and it picks up the two additional ones that it needs by forming to bonds to surrounding elements. And then the halogen as surrounding elements only make a single bond. They come in with seven valence electrons because their in group seven a. So to pick up that eighth electron, they need they form one bond to a surrounding element. Now if we look here in terms of lone pairs, were going to say that the first four have no lone pairs, but then groups 58 A seven A. Here we have one lone pair on nitrogen, Then we have two lone pairs on oxygen, and then we have 1, 2, 3 lone pairs on a halogen when it's a surrounding element. So just remember when it comes to finding preferences. These are the typical ones that we have for elements from groups one A 27 A of the periodic table.

Bonding Preferences Example 1

here it asks how many bonds and lone pairs are typically found around oxygen at Now, we're going to stay here. If we look up above, there'll be a little bit of cheating. So let's just think about this oxygen's and groups six a. Remember from elements from groups 5-7 a. The number of bands make is based on how many electrons they need to get to the idea electron arrangement. Oxygen's group six A. So it has six valence electrons. If it wants to follow the octet rule to get eight valence electrons around it, It wants to make two bonds and thereby pick up to additional electrons. So here we can say the number of bonds and oxygen atom typically makes us too. And then we can say the number of lone pairs around it are one to these lone pairs are not connecting to another element. So here that means, the answer is option B.

How many bonds and nonbonding electrons can be found around Si atoms?

• A. 4, 4
• B. 2, 4
• C. 3, 2
• D. 4, 0

How many bonds and lone pairs can be found around Mg atoms?

a) 2, 1                    b) 2, 0                    c) 3, 1                     d) 3, 0

• A. 2, 1
• B. 2, 0
• C. 3, 1
• D. 3, 0