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Jan
Jan
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You nailed it, this is almost a chicken-and-egg problem. As long as we don’t know which complex will form, we cannot say which orbitals are populated. This is why it is often and regularly repeated that you need to know the geometry before you can make a call.

In an octahedral complex, the $\mathrm{t_{2g}}$ orbitals will always be populated first as they are lower in energy. (Look up any introduction on crystal field theory or check out the answers on this site.) In a tetrahedral complex, $\mathrm{e}$ — not $\mathrm{e_g}$, as there is no centre of symmetry, meaning the the $\mathrm{u/g}$ designators do not make sense, even though the orbitals are still often called $\mathrm{e_g}$ at introductory level — are the orbitals of choice to begin with. Again, just look at the scheme.

Thankfully, however, we can make an educated guess based on various factors.

  • octahedral complexes are more common especially at introductory levels and for the ‘top middle’ of the d block elements (roughly from titanium to nickel)
  • the LFSE (ligand field stabilisation enthalpy) can be calculated to $\pu{-12Dq}$ in the octahedral case and $\pu{-8Dq}$ in the tetrahedral case which twice prefers octahedral
  • there is no compelling reason (size or charge balance) to consider tetrahedral to be preferable to octahedral.

Thus, the complex you are looking at will most likely be octahedral — and so it is: it is hexaaquachromium(III) $\ce{[Cr(H2O)6]^3+}$, which is octahedral.

You nailed it, this is almost a chicken-and-egg problem. As long as we don’t know which complex will form, we cannot say which orbitals are populated. This is why it is often and regularly repeated that you need to know the geometry before you can make a call.

In an octahedral complex, the $\mathrm{t_{2g}}$ orbitals will always be populated first as they are lower in energy. (Look up any introduction on crystal field theory or check out the answers on this site.) In a tetrahedral complex, $\mathrm{e}$ — not $\mathrm{e_g}$, as there is no centre of symmetry, meaning the the $\mathrm{u/g}$ designators do not make sense, even though the orbitals are still often called $\mathrm{e_g}$ at introductory level — are the orbitals of choice to begin with. Again, just look at the scheme.

Thankfully, however, we can make an educated guess based on various factors.

  • octahedral complexes are more common especially at introductory levels and for the ‘top middle’ of the d block elements (roughly from titanium to nickel)
  • the LFSE (ligand field stabilisation enthalpy) can be calculated to $\pu{-12Dq}$ in the octahedral case and $\pu{-8Dq}$ in the tetrahedral case which twice prefers octahedral
  • there is no compelling reason (size or charge balance) to consider tetrahedral to be preferable to octahedral.

Thus, the complex you are looking at will most likely be octahedral — and so it is: it is hexaaquachromium(III) $\ce{[Cr(H2O)6]^3+}$, which is octahedral.

You nailed it, this is almost a chicken-and-egg problem. As long as we don’t know which complex will form, we cannot say which orbitals are populated. This is why it is often and regularly repeated that you need to know the geometry before you can make a call.

In an octahedral complex, the $\mathrm{t_{2g}}$ orbitals will always be populated first as they are lower in energy. (Look up any introduction on crystal field theory or check out the answers on this site.) In a tetrahedral complex, $\mathrm{e}$ — not $\mathrm{e_g}$, as there is no centre of symmetry, meaning the the $\mathrm{u/g}$ designators do not make sense, even though the orbitals are still often called $\mathrm{e_g}$ at introductory level — are the orbitals of choice to begin with. Again, just look at the scheme.

Thankfully, however, we can make an educated guess based on various factors.

  • octahedral complexes are more common especially at introductory levels and for the ‘top middle’ of the d block elements (roughly from titanium to nickel)
  • the LFSE (ligand field stabilisation enthalpy) can be calculated to $\pu{-12Dq}$ in the octahedral case and $\pu{-8Dq}$ in the tetrahedral case which twice prefers octahedral
  • there is no compelling reason (size or charge balance) to consider tetrahedral to be preferable to octahedral.

Thus, the complex you are looking at will most likely be octahedral — and so it is: it is hexaaquachromium(III) $\ce{[Cr(H2O)6]^3+}$, which is octahedral.

Source Link
Jan
Jan
  • 68.4k
  • 12
  • 202
  • 388

You nailed it, this is almost a chicken-and-egg problem. As long as we don’t know which complex will form, we cannot say which orbitals are populated. This is why it is often and regularly repeated that you need to know the geometry before you can make a call.

In an octahedral complex, the $\mathrm{t_{2g}}$ orbitals will always be populated first as they are lower in energy. (Look up any introduction on crystal field theory or check out the answers on this site.) In a tetrahedral complex, $\mathrm{e}$ — not $\mathrm{e_g}$, as there is no centre of symmetry, meaning the the $\mathrm{u/g}$ designators do not make sense, even though the orbitals are still often called $\mathrm{e_g}$ at introductory level — are the orbitals of choice to begin with. Again, just look at the scheme.

Thankfully, however, we can make an educated guess based on various factors.

  • octahedral complexes are more common especially at introductory levels and for the ‘top middle’ of the d block elements (roughly from titanium to nickel)
  • the LFSE (ligand field stabilisation enthalpy) can be calculated to $\pu{-12Dq}$ in the octahedral case and $\pu{-8Dq}$ in the tetrahedral case which twice prefers octahedral
  • there is no compelling reason (size or charge balance) to consider tetrahedral to be preferable to octahedral.

Thus, the complex you are looking at will most likely be octahedral — and so it is: it is hexaaquachromium(III) $\ce{[Cr(H2O)6]^3+}$, which is octahedral.