Valence Bond Theory Flashcards
Bonds between atoms are e- pairs between nuclei (T/F).
True.
How is potential energy (V) of e-1 & e-2 between nuclei described?
Potential energy comes from e-1 & e-2 being simultaneously attracted to each others nucleus AND repelled by each others negative charge.
What information does the shrodinger equation (ψ) of H2 provide?
Provides information about the potential energy and possible coordinates for e-1 & e-2.
What would the ψ^2 function represent?
The simultanous probability of finding e-1 & e-2 in a volume.
Bonds result when valence orbitals overlap (T/F).
True.
Which kind of orbital overlaps with corresponding bond are possible?
END-ON : overlap of two atomic orbitals of higher e- density centered on the internuclear axis (σ-bond).
SIDE-ON : overlap of two atomic orbitals in two regions straddling the internuclear axis with a nodal plane between them (π-bond).
FACE-ON : overlap of two d atomic orbitals in four regions surrounding the internuclear axis with two nodal planes between them (δ -bond).
What happens when you mix two atomic orbitals?
Atomic orbitals are wavefunctions and when mixed waves interfere & change their shape. Mixing wavefunctions involves sharing valence e-s , therefore molecules adopt e- repulsion minimizing properties.
Propose hybrid atomic orbital sets for common geometries.
- Linear, sp (2x).
- trigonal, sp2 (3x).
- tetrahedral, sp3 (4x).
- trigonal bypiramidal, sp3d (5x)
- octahedral, sp3d2 (6x).
Character of sp3 hybrid orbital.
- 35% s-character and 75% p-character.
Describe the molecular geometry, and hybridization of CH4.
Four hydrogens bound to carbon, no lone pairs, the structure is predicted to be tetrahedral, with sp3 hybridization. Don’t have to show excited valence state e- configuration (promotion step), since e-s would be unpaired in degenerate hybrids.
Describe the molecular geometry, and hybridization of NH3.
Three hydrogens bound to nitrogen, with lone-pair, the structure is tetrahedral basic geometry, with sp3 hybridization. The molecular geometry is trigonal pyramidal. Lone pair requires more space than bond pairs, thus H-N-H angle will be less than the ideal 109.5 °. As a result, the hybrids would be inequivalent, with more s-character associated with the hybrid housing the lone pair, and more p-character in those making the N-H bonds.
Describe the molecular geometry, and hybridization of ClF3.
Three fluorine atoms bound to chlorine, two lone pairs, the basic geometry will be trigonal bipyramidal, with sp3d hybridization. Molecular geometry is T-shape. Here, the empty 3d orbital of Cl is used to make hybrids therefore the excited valence state e- configuration of Cl is needed (promotion step). The lone pairs will make the hybrid orbitals inequivalent and are housed in the equatorial site, because fewer neighboring bond pairs 90 degrees away.
Define sp2 hybrid in relation to unhybridized oribitals.
For sp2 hybrid, one s & two p orbitals were used to make it, one p orbital is left over in unhybridized atomic orbital. This unhybridized p orbital is perpendicular to the plane of the molecule.
Describe the molecular geometry, and hybridization of NOCl.
Nitrogen has three regions of enhanced e- density: the lone pair, double bond to oxygen, and bond to chlorine, with sp2 hybridization. The basic geometry is trigonal planar, and molecular geometry is bent. Unhibridized 2p orbital forms π component of double bond. Bond angles can’t be predicted because the regions of enhanced e- density are all different sizes.
Describe the molecular geometry, and hybridisation of SOCl2.
Sulphur forms four bonds, one to each chlorine and two to oxygen. The predicted basic geometry is tetrahedral, with sp3 hybridization. The molecular geometry is trigonal pyramidal. The unhibridized 3d orbital on Sulphur and unhibridized 2p orbital on oxygen form the π component of the double bond.
