Write the equation for the reaction, if any, of sodium with the following substances. (a) oxygen

Write the equation for the reaction, if any, of sodium with...

Key Concepts

Formation of Ionic Compounds

The interactions between sodium and nonmetallic elements typically result in the formation of ionic compounds. In these compounds, sodium donates its valence electron to a nonmetal, producing a positively charged sodium ion and a negatively charged anion. This electron transfer leads to the formation of a compound with strong electrostatic attractions.

Redox Reactions

Many reactions involving sodium are redox (reduction-oxidation) reactions, where sodium is oxidized by losing electrons and the other reactant is reduced by gaining electrons. Understanding electron transfer processes is crucial for predicting the nature of the reactants and products formed during these reactions.

Reactions Involving Diatomic Molecules and Water

Reactions of sodium with diatomic molecules (such as oxygen, nitrogen, or chlorine) or with water illustrate different types of chemical behavior. With diatomic molecules, the reaction often involves direct synthesis of compounds, potentially under specific conditions, while with water, sodium reacts vigorously to produce a hydroxide and hydrogen gas, exemplifying an exothermic reaction with significant practical consequences.

Alkali Metal Reactivity

Alkali metals, like sodium, are characterized by their single valence electron which they lose easily, making them highly reactive. This reactivity is the foundation for their vigorous interactions with various nonmetals and compounds, leading to a broad range of reactions including oxidation, formation of ionic compounds, and reactions that release hydrogen gas.

Chemical Equation Balancing

Balancing chemical equations is a fundamental concept in chemistry that ensures the conservation of mass and atoms. When writing reactions, one must account for each atom on both the reactant and product sides of the equation by adjusting the stoichiometric coefficients appropriately.

Transcript

00:01 All right.

00:02 So what we are looking at here is kind of a ionic bonding related problem.

00:09 I want us to write the reaction of sodium with a couple of different elements and compounds.

00:16 Right.

00:17 So the first thing is asking us is to write the reaction for sodium and oxygen if they were combined together.

00:28 Okay.

00:29 We know in the real world that oxygen doesn't just come as a free element.

00:35 Element like this.

00:35 It comes in the form of oxygen gas as o2.

00:40 But what's going to be happening here is that the sodium will be combining with our oxygen to make sodium oxide.

00:50 And what we want to know is what ratio they're going to be combining in.

00:56 Right.

00:56 So we know what this formula over here is going to be.

00:59 To figure that out, we need to look at what the charge of each of these ions is.

01:04 If you look on the periodic table and finds sodium, you'll see that it is in the far left column, meaning that it has one valence electron.

01:12 So when it becomes an ion, it ends up as n -a -plus -1, because it gives that one electron away and has a charge of positive one.

01:23 If it's going to form an ionic bond with oxygen, well, when oxygen ionizes, it needs to gain two electrons, so it becomes o -minus 2.

01:33 All right, so this compound over here we need to make sure is neutral, which means that if if each sodium has a charge of positive one, then each oxygen has the charge of negative 2, we would need two of these sodium to cancel out the charge of the oxygen.

01:49 So our compound over here would be na2o.

01:52 This is sodium oxide.

01:54 And the last thing we've got to do is make sure this equation is balanced.

02:00 Okay.

02:01 On the left side of our reaction here, we actually have two oxygens because oxygen comes, you know, bonded to itself like this.

02:09 And on the right side, we only have one of them.

02:11 So what we'd have to do is put in our coefficient here to make sure we have two total oxygens.

02:18 That also affects the number of sodiums we have.

02:20 Two times two would be four, which means over here we would need a coefficient of four.

02:26 So this would be your final answer for part a.

02:35 Okay, part b is kind of the same idea...

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