Represent the following ionic compounds by Lewis structures: (a) barium hydroxide; (b) sodium ni

step 1 So we're going to draw the Lewisot structure for barium hydroxide. And barium hydroxide is an io

Represent the following ionic compounds by Lewis structures:...

Key Concepts

Octet Rule

The octet rule is a chemical guideline stating that atoms tend to form compounds in ways that give them eight electrons in their valence shell, resembling the electron configuration of a noble gas. While there are exceptions, it provides a foundational principle when constructing Lewis structures by dictating how electrons should be shared or transferred in bonds.

Valence Electron Counting

Valence electron counting involves determining the number of electrons available from the outermost shell of each atom for bonding. This is crucial for constructing Lewis structures, as it guides where bonds should be placed and how many electrons will appear as lone pairs, ensuring that the electrons are correctly allocated among atoms.

Polyatomic Ions

Polyatomic ions are groups of atoms bonded together that act as a single charged entity. Understanding their structure, including the arrangement of atoms and the distribution of electrons, is important for drawing accurate Lewis structures and for understanding the behavior of ionic compounds in which they occur.

Ionic Compounds

Ionic compounds consist of positively charged ions (cations) and negatively charged ions (anions) that are held together by electrostatic forces. When representing these compounds, it is important to show the individual ions and to indicate any charges present on them, reflecting the transfer of electrons from one atom to another.

Lewis Structures

Lewis structures are diagrams that represent the bonding between atoms of a molecule and the lone pairs of electrons that may exist. They provide a visual method to depict how valence electrons are arranged in a molecule or ion, which is essential for understanding molecular geometry and chemical reactivity.

Charge Balancing

Charge balancing is the process of ensuring that the total positive and negative charges in an ionic compound cancel out to yield a neutral compound. This conceptual step guides how many ions of each type should be present in the formula and is critical when deriving the Lewis structures for the constituent ions.

Transcript

00:01 So we're going to draw the lewisot structure for barium hydroxide.

00:04 And barium hydroxide is an ionic compound that contains a covalent compound inside of it.

00:12 Right.

00:12 So you have to show both the ionic bonding and the covalent bonding in this molecule.

00:17 So we're going to start with the ionic bonding.

00:20 So we're going to do barium in brackets because it transfer its electrons, two of them, to hydroxide.

00:27 So when you have barium hydroxide, barium hydroxide, barium looks like this, right? and then you have hydroxide, which we're going to draw in a second.

00:34 And the barium gives up one to each hydroxide in order to be stable.

00:40 So it will be oh minus.

00:42 Now we have to draw that oh to show what it looks like.

00:47 So then we're going to do oh.

00:49 So oh count on a number of valence electrons.

00:52 Oxygen has 6 plus hydrogen has 1 plus the one electron.

00:55 Each of them, that should be 2.

00:57 Each of them takes from barium.

00:59 So that's h, connect.

01:02 And then that's 2, 4, 6, 8, and put it in brackets.

01:07 And that is the leuzah structure for hydroxide.

01:09 And we're going to put it together, barium hydroxide.

01:14 So, barium go in the center.

01:17 This is plus 2.

01:19 And then you get hydroxide, oh, h, with your lois dot.

01:25 And then another one, oh, h, and then the church.

01:31 You could also just put this one in here.

01:34 Just don't forget that there's swill them.

01:37 Alright, let's do b.

01:38 For b, we're going to do sodium nitride, which sodium nitride is n -a -no2.

01:52 So then we do the same, draw that ionic bond first.

01:55 So it'll be sodium.

01:56 Sodium has a charge of plus 1, which means that it gives up one electron to nitride.

02:01 And so there's one in one of each, you just need to draw 1.

02:05 So it will be n02 minus.

02:08 And then since this is a covalent bond inside of this, we have to dry it out.

02:12 So it will be n -o -o.

02:15 Count total number of valance electrons, 5 for nitrogen, plus 6 for each oxygen, but we have 2, so it will be 12, plus 1 from the charge, so that will be 6 plus 12, that's 18.

02:29 We connect to 4.

02:31 We have used 4 electrons, remember each of those bonds is 2 electrons, and then we distribute 2 ,000.

02:38 2, 4, 6, 8, 10, 12, and then 14.

02:44 Count that everything has 8 elections around there because none of them can break the opt -to -rule.

02:50 24 -6 -8, 246, 246, that nitrogen is not stable.

02:56 So draw a double bond.

02:58 Now, because this has a resonance, which means that i could have drawn this, you also have to draw resonance for your ionic compound.

03:09 So let me draw the other nitride.

03:20 And then i'm gonna put the sodium around then.

03:24 So we sodium plus and then sodium plus, oh, outside of it.

03:30 And that would be the louisotte structure for sodium nitride, right? this is resonance, so we draw the resonance arrow.

03:38 Usually you put it in the side, but i don't have space on the side.

03:43 All right, let's do c.

03:50 So for c, we're going to have magnesium iodate.

04:08 So this one's a little more complex because the iodide can actually have an expanded octet.

04:13 And i know that because iodine is anything, so anything pasa 3p, any non -metal positive 3p, can have an expanded octet.

04:24 So we have c, magnesium iodate.

04:44 So this is magnesium is mg and then iodate.

04:47 Is i .03 and there's a 2 here right so we're gonna have one magnesium right and then two iodates and then that magnesium is gonna give one electron to each of them so for the lewis s structure to show ionic bonding we do magnesium plus two and then two i .03 minus now we're gonna have to draw those out and do formal charges because the iodine can have an expand let octet based on formal charges.

05:24 So i'm going to do i -o -o -o -o -o -com total number of valence electrons.

05:31 So we have iodine has seven plus each oxygen has six, and there's three of them, plus one from the charge.

05:40 So that will be 18, and then that's 8.

05:43 So plus 8, that will give us 26.

05:46 Connect, 2, 4, 6.

05:49 So now we have 20 electrons, this should be from the outside in, 2, 4, 6, 8, 10, 12, 14, 16, and then 18, and 20.

06:01 Put it in brackets, census charge, and now we're going to do formal charges.

06:07 So we have to do, again, formal charges because this is a charge, not the charge, because iodine can occupy electrons in the d -orbital, right? so we're going to use that.

06:20 So, skycar formal charge for iodine...

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