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katen-ka-za [31]
3 years ago
7

Determine if each statement is True or False. [ Select ] Central atoms with four electron groups will be sp3 hybridized. [ Selec

t ] Hybrid orbitals are delocalized over the entire molecule. [ Select ] The number of hybrid orbitals is equal to the number of atomic orbitals that are blended together. [ Select ] Atoms with a single pi bond and an octet are sp2 hybridized. [ Select ] Sometimes oxygen atoms will be sp3d hybridized in organic molecules. [ Select ] All resonance structures must be considered when assigning hybridization
Chemistry
1 answer:
meriva3 years ago
8 0

Answer:

Central atoms with four electron groups will be sp3 hybridized. True

Hybrid orbitals are delocalized over the entire molecule. False

The number of hybrid orbitals is equal to the number of atomic orbitals that are blended together. True

Atoms with a single pi bond and an octet are sp2 hybridized. True

Sometimes oxygen atoms will be sp3d hybridized in organic molecules. False

All resonance structures must be considered when assigning hybridization. False

Explanation:

When a central atom has four electron groups attached to it, then it must be sp3 hybridized. This is the case in ammonia, water, hydrogen sulphide, methane etc.

Hybridization is a valence bond concept while delocalization is a molecular orbital theory concept. Hybridized orbitals are localized on central atoms in a molecule while delocalized orbitals spread across the entire molecule.

According to valence bond theory, the number of atomic orbitals that combined to give hybrid orbitals must be equal to the number of hybrid orbitals formed.

When an atom has a single pi bond and on octet of electrons, then it must be sp2 hybridized. Remember that in ethene for instance, carbon has one pi bond and an octet of electrons.

Oxygen has an empty n=3 level hence it can not have d-orbitals involved in hybridization.

The electron domain geometry of the molecule is considered when assigning hybridization and not the resonance structures. All the resonance structures must have the central atom in the same hybridization state.

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A low electronegativity

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