BF 3 reacts with F - to give BF 4 - , but AlF 3 reacts with F - to give AlF 6 3- . Explain.

Hi, everyone. Welcome back. Our next question says, how does Inc L three or indium three chloride react with CL minus and produce Inc L six with a three minus charge or indium hex aide. While BC L three or boron trichloride reacts with CL minus and only produces BC L four with a negative one charge or boron tetrachloride. We have four possibilities. I'm going to paraphrase them a little bit because they kind of repeat the entire description of the starting point and the reaction. And if I read them word for word, it will be very long and repetitive. So choice A says or basically says Indium hex alor is produced because indium is much more electron than boron. Hence, when boron ty chloride reacts it with the extra chloride, it only produces BC L four minus choice. B says boron tetrachloride is produced because boron is a non metal. While indium is a metal, which is why it is able to form indium hex chloride. Choice C says boron hex chloride or boron tetrachloride, excuse me is produced because boron is not large enough to be able to hold more than four groups. While indium is large enough to form indium hex alor. And choice of d when indium rea indium tr three chloride reacts, it produces indium hex alor and not indium tetrachloride. Because indium has six valence electrons while boron only has four, hence boron trichloride reacts and only produces boron tetrachloride. Well, to think this through, we need to think about our elements of India and boron to think about why they would react differently in their trichloride form. So born and Indian are both in group three A and thus both of them have three valence electrons. So right off the bat, we can eliminate choice B which says that in India has six valence electrons and B boron only has four since they both have three. So since they're both in the same group, we need to turn to their row to understand the differences between them. So boron is all the way up in row two while India M is down. So quite a bit farther in row five. So boron, therefore, it has a three valence electrons, but it only has five electrons total. So its electron arrangement is just one, S 22, S 22 P one, it has no level three, it has nothing beyond those two P orbitals in India. However, down in row five, we're all the way up to the fifth shell. And that means not only does it have P orbitals available, but it has those four D orbitals available. So it's much bigger, more orbital available. Well, when we think about the ability to make bonds with a central atom, if you only have P orbitals, the biggest or the, you know, highest hybridization you can achieve is an sp three hybridization which only allows four bonds. So boron literally can't take on more than four chlorides. It doesn't have any de orbitals. Indium can follow the expanded octet rule. It has four de orbitals available. It can make different hybridization and it can make more than four bonds. So that takes us to our answer of choice. C that boron is not large enough to be able to hold more than group four groups while in India M is large enough. And that's because boron has no de orbitals. Well, to be thorough, we should look at our other choices just to make sure there's nothing that's a better description choice. A says that Indium is much more electron negative than boron. But if we look at electronegativity trends, electron negativity becomes larger as you go to the right, but they're both equally far to the right given they're in the same group. But electron negativity decreases going down or increase is going up, we can say. So indium being down further than boron. So boron's up high indium's down, lower indium is less electron than boron. So choice A is not going to be our answer. And then finally, we have toys b it does correctly state boron is a non metal indium is a metal. But that isn't why indium forms indium but not our explanation. So that's why choice B isn't our answer. So that could be a little confusing just because it's a correct statement of fact, but not the explanation we need. So once again, the reason that indium three chloride can produce indium hex chloride, while boron trichloride will only produce boron tetrachloride is choice C boron trichloride reacts with chloride ions and only produces boron tetrachloride because boron is not large enough to be able to hold more than four groups. While endi M is large enough to form indium.

Ch.22 - The Main Group Elements

McMurry8th EditionChemistryISBN: 9781292336145Not the one you use?Change textbook

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McMurry 8th Edition

Ch.22 - The Main Group Elements

Chapter 22, Problem 1

BF₃ reacts with F^- to give BF₄^-, but AlF₃ reacts with F^- to give AlF₆^3-. Explain.

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and are both Lewis acids, meaning they can accept electron pairs. However, they differ in their ability to expand their coordination number due to the size and electronic configuration of the central atom.

Boron in has an empty p-orbital and can accept one more pair of electrons from , forming . Boron cannot expand its octet beyond four because it lacks d-orbitals.

Aluminum in can expand its coordination number because it has available d-orbitals. This allows it to accept more than one pair of electrons from ions.

When reacts with , it forms . This is possible because aluminum can accommodate six fluorine atoms by utilizing its d-orbitals, forming an octahedral complex.

The difference in behavior between and is due to the ability of aluminum to expand its coordination number by using d-orbitals, while boron cannot do so.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Lewis Acids and Bases

Lewis acids are substances that can accept an electron pair, while Lewis bases are those that can donate an electron pair. In the reaction between BF3 and F-, BF3 acts as a Lewis acid, accepting an electron pair from the fluoride ion to form BF4-. Understanding this concept is crucial for analyzing how different compounds interact based on their electron pair acceptance or donation.

Coordination Complexes

Coordination complexes consist of a central metal atom bonded to surrounding molecules or ions called ligands. In the case of AlF3 reacting with F-, the aluminum ion can coordinate with multiple fluoride ions, leading to the formation of AlF6^3-. This concept is essential for understanding how metal ions can form complex ions with varying numbers of ligands.

Oxidation States and Coordination Numbers

The oxidation state of an element in a compound indicates its degree of oxidation, while the coordination number refers to the number of ligand atoms bonded to the central atom. In AlF3, aluminum has a +3 oxidation state and can accommodate six fluoride ions, resulting in a coordination number of 6 in AlF6^3-. This concept helps explain the different products formed in reactions involving metal halides.

BF3 reacts with F- to give BF4-, but AlF3 reacts with F- to give AlF63-. Explain.