What is the hybridization of the central atom in (a) PBr 5 ? (b) CH 2 O?

Hello everyone. And this question was asked what is the hybridization of grooming in the following inter halogen compounds for BRF three. We're gonna have seven valence electronic for bro mean Plus seven valence electrons for flooring and there's three flooring And this will give us a total of 28 balance electrons and B. R. Is going to be a central atom because it is less electro negative than flooring. We're gonna have three flooring coming off of it And each bond contains two electrons. So we have 246 eight 10 14 16 18 22 24 26 28. And we can add to long pairs here because bromine can have an expanded octet. So we see that we have one two three four five electron groups around the central atom. And for the habit Ization S has one orbital he has three and so we have three plus one. For a total of four electron groups so far. Look at one in the D orbital. So now we get five electron groups. So this would be sp three D. If a BRF five We have seven valence electrons Plus seven Valence Electrons Times five. And this will give us 42 valence electrons. We have roaming in the center And five flooring coming off of it. We have 2468 10 12 14 18 20 22 26 28 30 34 36 38 40 and 42. So we have one two three four five six electron groups around the central atom. So we're gonna have one and S. RCMP. So now we have four and now we need to add two more, so we have two and d. We get sp three D. Two. Thanks for watching my video, and I hope it was helpful.

Ch.9 - Molecular Geometry and Bonding Theories

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Brown 14th Edition

Ch.9 - Molecular Geometry and Bonding Theories

Problem 52a,b

Chapter 9, Problem 52a,b

What is the hybridization of the central atom in (a) PBr₅? (b) CH₂O?

Verified step by step guidance

Step 1: Draw the Lewis structure of the molecule. For PBr5, the central atom is Phosphorus (P) and it is surrounded by five Bromine (Br) atoms.

Step 2: Count the number of atoms bonded to the central atom. In PBr5, there are five Bromine atoms bonded to the central Phosphorus atom.

Step 3: Count the number of lone pairs on the central atom. In PBr5, Phosphorus does not have any lone pairs.

Step 4: Add the number of atoms bonded to the central atom and the number of lone pairs. In PBr5, this gives us a total of 5 (5 bonded atoms + 0 lone pairs).

Step 5: Determine the hybridization based on this number. If the total is 2, the hybridization is sp. If the total is 3, the hybridization is sp2. If the total is 4, the hybridization is sp3. If the total is 5, the hybridization is sp3d. If the total is 6, the hybridization is sp3d2. In the case of PBr5, the hybridization of the central Phosphorus atom is sp3d.

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Key Concepts

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

Hybridization

Hybridization is a concept in chemistry that describes the mixing of atomic orbitals to form new hybrid orbitals. These hybrid orbitals are used to explain the geometry and bonding properties of molecules. The type of hybridization depends on the number of electron pairs around the central atom, influencing molecular shape and bond angles.

Valence Shell Electron Pair Repulsion (VSEPR) Theory

VSEPR theory is a model used to predict the geometry of individual molecules based on the repulsion between electron pairs in the valence shell of the central atom. According to this theory, electron pairs will arrange themselves as far apart as possible to minimize repulsion, which helps determine the molecular shape and the hybridization of the central atom.

Molecular Geometry of PBr5

PBr5, or phosphorus pentabromide, has a trigonal bipyramidal molecular geometry. In this structure, the phosphorus atom is surrounded by five bromine atoms, leading to a hybridization of sp3d. This hybridization accounts for the five bonding pairs of electrons and the specific arrangement of atoms in three-dimensional space.