Question

The procedure would be whats in portion A " Mechanism of hybridization."

A. Mechanism of Hybridization 

.1. For each of the species in Experiment 19 (use the Lewis Structures drawn), determine the number of electron domains around the central atom. 

2. Write an electron configuration only for the valence electrons of the central atom. 

3. Properly place the valence electrons into the unhybridized orbital diagram using the rules for filling orbital diagrams. Next, illustrate the hybridization of the central atom by placing the valence electrons on the hybridized orbital diagram. Circle which orbitals from the unhybridized diagram were used to hybridize the central atom. 

4. Determine the hybridization of the central atom. NOTE: For C2H2Cl2, both cis and trans have the same hybridization. As well, because both central atoms in each compound have the same electron domains, the hybridization applies to both central atoms. 

B. Identifying the Hybridization of the Central Atom from Lewis Structures 

.1. Drawing orbital diagrams is not necessary to determine the hybridization of the central atom in a Lewis Structure. Since the shape and bond angles of hybrid orbitals parallel that of electron geometry, we can make a correlation between the two. Based on your results from Part A, determine the correlation between electron geometry and the hybridization of the central atom. 

C. Identifying Modes of Orbital Overlap: Sigma and Pi Bonds

 .1. Determine the total number of sigma and pi bonds that are present in the molecule. 

Answer 1

BBr3 The B belongs to group III A. Thus, it contains 3 electons in valence shell No of electron domains : 2 Valence electron configuration : 2s22p Atomic orbitals of Valence electrons - Hybridized orbitals formed sp? Atomic orbitals unhybridized Hybridization

Here, the number of electron domains for BBr3 is 3.

PCi: TheP belongs to group V A. Thus, it contains 5 electons in valence shell No of electron domains : 5 Valence electron configuration :3s 3p%3d Atomic orbitals of Valence electrons Hybridized orbitals formed + A spa A sp A spa A sp Atomic orbitals unhybridized Hybridization

PF5 TheP belongs to group V A. Thus, it contains 5 electons in valence shell No of electron domains : 5 Valence electron configuration :35-3p 3d Atomic orbitals of Valence electrons 1 : *** Hybridized orbitals formed A spd A spåd A spåd A spåd A spåd Atomic orbitals unhybridized 3d Hybridization

POCI; TheP belongs to group V A. Thus, it contains 5 electons in valence shell No of electron domains : 5 Valence electron configuration :3s 3p%3d Atomic orbitals of Valence electrons Hybridized orbitals formed + A spa A sp A spa A sp Atomic orbitals unhybridized Hybridization

Simillarly,

SiCl6(2-):

6, 3s2 3p2 3d, 6 sp3d2(1 e-) hydrid orbitals, 3 d unhybridized orbitals, hybridisation sp3d2.

SCl2:

2, 3s2 3p4 3d, 4 sp3(1 e-) hydrid orbitals, 3 d unhybridized orbitals, hybridisation sp3.

SCl4:

4, 3s2 3p4 3d, 4 sp3(1 e-) hydrid orbitals, 3 d unhybridized orbitals, hybridisation sp3.