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What is the hybridization of chromium in chromate and dichromate ions?
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Mako Koiwai
What is the hybridization of chromium in chromate and dichromate ions?
The d orbitals are commonly grouped into two groups because that is how they transform under both octahedral (including a g subscript) and tetrahedral (no g subscript) symmetry:
$\mathrm{e_{(g)}}: \mathrm d_{z^2}$ and $\mathrm d_{x^2-y^2}$
In tetrahedral environments, the p orbitals also transform as $\mathrm t_2$ while in octahedral ones they transform as $\mathrm{t_{1u}}$. By analogy to main group tetrahedral molecules, it would be easy to assume that just the three $\mathrm t_2$ d orbitals take part in the formation of tetrahedral complexes (throw in one s orbital for completion of the four).
In reality, all orbitals that transform as $\mathrm t_2$ will be represented in the final complex in some way or another, so that whatever bonding and antibonding orbitals of $\mathrm t_2$ symmetry you end up with will always include some contribution from the original p orbitals. Furthermore, outside of hydride complexes the ligands will also have available orbitals for π-type bonds which will complicate the picture further. The final MO for chromate will look something like the following:
Figure 1: tetrahedral $\ce{[ML4]}$ complex with ligand π contributions.
The metal orbitals on the left are d, s, p from bottom to top. The ligand orbitals on the right are the oxygens’ p orbitals (all 12 of them).
Don’t worry, these schemes will come to haunt you at one point when studying chemistry.
The d orbitals are commonly grouped into two groups because that is how they transform under both octahedral (including a g subscript) and tetrahedral (no g subscript) symmetry:
$\mathrm{e_{(g)}}: \mathrm d_{z^2}$ and $\mathrm d_{x^2-y^2}$
In tetrahedral environments, the p orbitals also transform as $\mathrm t_2$ while in octahedral ones they transform as $\mathrm{t_{1u}}$. By analogy to main group tetrahedral molecules, it would be easy to assume that just the three $\mathrm t_2$ d orbitals take part in the formation of tetrahedral complexes (throw in one s orbital for completion of the four).
In reality, all orbitals that transform as $\mathrm t_2$ will be represented in the final complex in some way or another, so that whatever bonding and antibonding orbitals of $\mathrm t_2$ symmetry you end up with will always include some contribution from the original p orbitals. Furthermore, outside of hydride complexes the ligands will also have available orbitals for π-type bonds which will complicate the picture further. The final MO for chromate will look something like the following:
Figure 1: tetrahedral $\ce{[ML4]}$ complex with ligand π contributions.
The metal orbitals on the left are d, s, p from bottom to top. The ligand orbitals on the right are the oxygens’ p orbitals (all 12 of them).
Don’t worry, these schemes will come to haunt you at one point when studying chemistry.
The d orbitals are commonly grouped into two groups because that is how they transform under both octahedral (including a g subscript) and tetrahedral (no g subscript) symmetry:
In tetrahedral environments, the p orbitals also transform as $\mathrm t_2$ while in octahedral ones they transform as $\mathrm{t_{1u}}$. By analogy to main group tetrahedral molecules, it would be easy to assume that just the three $\mathrm t_2$ d orbitals take part in the formation of tetrahedral complexes (throw in one s orbital for completion of the four).
In reality, all orbitals that transform as $\mathrm t_2$ will be represented in the final complex in some way or another, so that whatever bonding and antibonding orbitals of $\mathrm t_2$ symmetry you end up with will always include some contribution from the original p orbitals. Furthermore, outside of hydride complexes the ligands will also have available orbitals for π-type bonds which will complicate the picture further. The final MO for chromate will look something like the following:
Figure 1: tetrahedral $\ce{[ML4]}$ complex with ligand π contributions.
The metal orbitals on the left are d, s, p from bottom to top. The ligand orbitals on the right are the oxygens’ p orbitals (all 12 of them).
Don’t worry, these schemes will come to haunt you at one point when studying chemistry.
The d orbitals are commonly grouped into two groups because that is how they transform under both octahedral (including a g subscript) and tetrahedral (no g subscript) symmetry:
In tetrahedral environments, the p orbitals also transform as $\mathrm t_2$ while in octahedral ones they transform as $\mathrm{t_{1u}}$. By analogy to main group tetrahedral molecules, it would be easy to assume that just the three $\mathrm t_2$ d orbitals take part in the formation of tetrahedral complexes (throw in one s orbital for completion of the four).
In reality, all orbitals that transform as $\mathrm t_2$ will be represented in the final complex in some way or another, so that whatever bonding and antibonding orbitals of $\mathrm t_2$ symmetry you end up with will always include some contribution from the original p orbitals. Furthermore, outside of hydride complexes the ligands will also have available orbitals for π-type bonds which will complicate the picture further. The final MO for chromate will look something like the following:
Figure 1: tetrahedral $\ce{[ML4]}$ complex with ligand π contributions.
The metal orbitals on the left are d, s, p from bottom to top. The ligand orbitals on the right are the oxygens’ p orbitals (all 12 of them).
Don’t worry, these schemes will come to haunt you at one point when studying chemistry.
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