Why are low spin tetrahedral complexes so rare?
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As I was going through Concise Inorganic Chemistry by J. D. Lee, I realised that there are simply no low spin tetrahedral complexes mentioned in the book. Is there any specific condition required for the formation of such a complex?
Usually low-spin complexes are in $mathrm{dsp^2}$ electronic configuration. But can this kind of orbital form a tetrahedral geometry? Because usually tetrahedron is usually synonymous to $mathrm{sp^3}$.
inorganic-chemistry coordination-compounds
edited May 9 at 23:37
orthocresol♦
41.1k7124255
asked May 9 at 10:39
evamPUNditevamPUNdit
14112
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add a comment |
$begingroup$
As I was going through Concise Inorganic Chemistry by J. D. Lee, I realised that there are simply no low spin tetrahedral complexes mentioned in the book. Is there any specific condition required for the formation of such a complex?
Usually low-spin complexes are in $mathrm{dsp^2}$ electronic configuration. But can this kind of orbital form a tetrahedral geometry? Because usually tetrahedron is usually synonymous to $mathrm{sp^3}$.
inorganic-chemistry coordination-compounds
edited May 9 at 23:37
orthocresol♦
41.1k7124255
asked May 9 at 10:39
evamPUNditevamPUNdit
14112
$endgroup$
add a comment |
$begingroup$
As I was going through Concise Inorganic Chemistry by J. D. Lee, I realised that there are simply no low spin tetrahedral complexes mentioned in the book. Is there any specific condition required for the formation of such a complex?
Usually low-spin complexes are in $mathrm{dsp^2}$ electronic configuration. But can this kind of orbital form a tetrahedral geometry? Because usually tetrahedron is usually synonymous to $mathrm{sp^3}$.
inorganic-chemistry coordination-compounds
edited May 9 at 23:37
orthocresol♦
41.1k7124255
asked May 9 at 10:39
evamPUNditevamPUNdit
14112
$endgroup$
As I was going through Concise Inorganic Chemistry by J. D. Lee, I realised that there are simply no low spin tetrahedral complexes mentioned in the book. Is there any specific condition required for the formation of such a complex?
Usually low-spin complexes are in $mathrm{dsp^2}$ electronic configuration. But can this kind of orbital form a tetrahedral geometry? Because usually tetrahedron is usually synonymous to $mathrm{sp^3}$.
inorganic-chemistry coordination-compounds
inorganic-chemistry coordination-compounds
edited May 9 at 23:37
orthocresol♦
41.1k7124255
asked May 9 at 10:39
evamPUNditevamPUNdit
14112
edited May 9 at 23:37
orthocresol♦
41.1k7124255
asked May 9 at 10:39
evamPUNditevamPUNdit
14112
edited May 9 at 23:37
orthocresol♦
41.1k7124255
edited May 9 at 23:37
orthocresol♦
41.1k7124255
edited May 9 at 23:37
orthocresol♦
41.1k7124255
41.1k7124255
asked May 9 at 10:39
evamPUNditevamPUNdit
14112
asked May 9 at 10:39
evamPUNditevamPUNdit
14112
asked May 9 at 10:39
evamPUNditevamPUNdit
14112
14112
add a comment |
add a comment |
1 Answer
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Hybridisation won't explain anything in transition metal complexes, so please stop using it, at least to the extent where it is possible to avoid using it. Quite literally everything about transition metal complexes is better rationalised using MO theory, and I am not exaggerating.
The reason why low-spin $T_mathrm d$ complexes are rare is because the splitting parameter, $Delta_t$, is significantly smaller than the corresponding octahedral parameter $Delta_o$. In crystal field theory, there is a complicated derivation which leads to the conclusion that (all things being equal)
$$Delta_t = frac{4}{9}Delta_o$$
For more information, please see: Why do tetrahedral complexes have approximately 4/9 the field split of octahedral complexes? and Why do octahedral metal ligand complexes have greater splitting than tetrahedral complexes?. Of course, this relationship is not exact in the real world, because CFT is a very simplified model; ligands are not point charges. However, it is still true in a qualitative sense.
Since the splitting $Delta_t$ is smaller, it is usually easier to promote an electron to the higher-energy $mathrm t_2$ orbitals, rather than to pair the electrons up in the lower-energy $mathrm e$ orbitals. Consequently, most tetrahedral complexes, especially those of the first-row transition metals, are high-spin. Low-spin ones do exist (e.g. J. Chem. Soc., Chem. Commun. 1986, 1491), but aren't common.
answered May 9 at 11:05
orthocresol♦orthocresol
41.1k7124255
$endgroup$
1
$begingroup$
It's just that we just went through the basics of CFT. I can't think without using hybridisation right now..
$endgroup$
– evamPUNdit
May 9 at 11:10
5
$begingroup$
I can understand that. However, I think it would be a disservice to not at least mention it. It is an absolute travesty that hybridisation for TM complexes is still taught. Also, other people, not just you, will read this too.
$endgroup$
– orthocresol♦
May 9 at 11:22
$begingroup$
CFT contains zero reference to hybridization. It is ridiculous to force into it.
$endgroup$
– Greg
May 10 at 6:28
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hybridisation won't explain anything in transition metal complexes, so please stop using it, at least to the extent where it is possible to avoid using it. Quite literally everything about transition metal complexes is better rationalised using MO theory, and I am not exaggerating.
The reason why low-spin $T_mathrm d$ complexes are rare is because the splitting parameter, $Delta_t$, is significantly smaller than the corresponding octahedral parameter $Delta_o$. In crystal field theory, there is a complicated derivation which leads to the conclusion that (all things being equal)
$$Delta_t = frac{4}{9}Delta_o$$
For more information, please see: Why do tetrahedral complexes have approximately 4/9 the field split of octahedral complexes? and Why do octahedral metal ligand complexes have greater splitting than tetrahedral complexes?. Of course, this relationship is not exact in the real world, because CFT is a very simplified model; ligands are not point charges. However, it is still true in a qualitative sense.
Since the splitting $Delta_t$ is smaller, it is usually easier to promote an electron to the higher-energy $mathrm t_2$ orbitals, rather than to pair the electrons up in the lower-energy $mathrm e$ orbitals. Consequently, most tetrahedral complexes, especially those of the first-row transition metals, are high-spin. Low-spin ones do exist (e.g. J. Chem. Soc., Chem. Commun. 1986, 1491), but aren't common.
answered May 9 at 11:05
orthocresol♦orthocresol
41.1k7124255
$endgroup$
1
$begingroup$
It's just that we just went through the basics of CFT. I can't think without using hybridisation right now..
$endgroup$
– evamPUNdit
May 9 at 11:10
5
$begingroup$
I can understand that. However, I think it would be a disservice to not at least mention it. It is an absolute travesty that hybridisation for TM complexes is still taught. Also, other people, not just you, will read this too.
$endgroup$
– orthocresol♦
May 9 at 11:22
$begingroup$
CFT contains zero reference to hybridization. It is ridiculous to force into it.
$endgroup$
– Greg
May 10 at 6:28
add a comment |
$begingroup$
Hybridisation won't explain anything in transition metal complexes, so please stop using it, at least to the extent where it is possible to avoid using it. Quite literally everything about transition metal complexes is better rationalised using MO theory, and I am not exaggerating.
The reason why low-spin $T_mathrm d$ complexes are rare is because the splitting parameter, $Delta_t$, is significantly smaller than the corresponding octahedral parameter $Delta_o$. In crystal field theory, there is a complicated derivation which leads to the conclusion that (all things being equal)
$$Delta_t = frac{4}{9}Delta_o$$
For more information, please see: Why do tetrahedral complexes have approximately 4/9 the field split of octahedral complexes? and Why do octahedral metal ligand complexes have greater splitting than tetrahedral complexes?. Of course, this relationship is not exact in the real world, because CFT is a very simplified model; ligands are not point charges. However, it is still true in a qualitative sense.
Since the splitting $Delta_t$ is smaller, it is usually easier to promote an electron to the higher-energy $mathrm t_2$ orbitals, rather than to pair the electrons up in the lower-energy $mathrm e$ orbitals. Consequently, most tetrahedral complexes, especially those of the first-row transition metals, are high-spin. Low-spin ones do exist (e.g. J. Chem. Soc., Chem. Commun. 1986, 1491), but aren't common.
answered May 9 at 11:05
orthocresol♦orthocresol
41.1k7124255
$endgroup$
1
$begingroup$
It's just that we just went through the basics of CFT. I can't think without using hybridisation right now..
$endgroup$
– evamPUNdit
May 9 at 11:10
5
$begingroup$
I can understand that. However, I think it would be a disservice to not at least mention it. It is an absolute travesty that hybridisation for TM complexes is still taught. Also, other people, not just you, will read this too.
$endgroup$
– orthocresol♦
May 9 at 11:22
$begingroup$
CFT contains zero reference to hybridization. It is ridiculous to force into it.
$endgroup$
– Greg
May 10 at 6:28
add a comment |
$begingroup$
Hybridisation won't explain anything in transition metal complexes, so please stop using it, at least to the extent where it is possible to avoid using it. Quite literally everything about transition metal complexes is better rationalised using MO theory, and I am not exaggerating.
The reason why low-spin $T_mathrm d$ complexes are rare is because the splitting parameter, $Delta_t$, is significantly smaller than the corresponding octahedral parameter $Delta_o$. In crystal field theory, there is a complicated derivation which leads to the conclusion that (all things being equal)
$$Delta_t = frac{4}{9}Delta_o$$
For more information, please see: Why do tetrahedral complexes have approximately 4/9 the field split of octahedral complexes? and Why do octahedral metal ligand complexes have greater splitting than tetrahedral complexes?. Of course, this relationship is not exact in the real world, because CFT is a very simplified model; ligands are not point charges. However, it is still true in a qualitative sense.
Since the splitting $Delta_t$ is smaller, it is usually easier to promote an electron to the higher-energy $mathrm t_2$ orbitals, rather than to pair the electrons up in the lower-energy $mathrm e$ orbitals. Consequently, most tetrahedral complexes, especially those of the first-row transition metals, are high-spin. Low-spin ones do exist (e.g. J. Chem. Soc., Chem. Commun. 1986, 1491), but aren't common.
answered May 9 at 11:05
orthocresol♦orthocresol
41.1k7124255
$endgroup$
Hybridisation won't explain anything in transition metal complexes, so please stop using it, at least to the extent where it is possible to avoid using it. Quite literally everything about transition metal complexes is better rationalised using MO theory, and I am not exaggerating.
The reason why low-spin $T_mathrm d$ complexes are rare is because the splitting parameter, $Delta_t$, is significantly smaller than the corresponding octahedral parameter $Delta_o$. In crystal field theory, there is a complicated derivation which leads to the conclusion that (all things being equal)
$$Delta_t = frac{4}{9}Delta_o$$
For more information, please see: Why do tetrahedral complexes have approximately 4/9 the field split of octahedral complexes? and Why do octahedral metal ligand complexes have greater splitting than tetrahedral complexes?. Of course, this relationship is not exact in the real world, because CFT is a very simplified model; ligands are not point charges. However, it is still true in a qualitative sense.
Since the splitting $Delta_t$ is smaller, it is usually easier to promote an electron to the higher-energy $mathrm t_2$ orbitals, rather than to pair the electrons up in the lower-energy $mathrm e$ orbitals. Consequently, most tetrahedral complexes, especially those of the first-row transition metals, are high-spin. Low-spin ones do exist (e.g. J. Chem. Soc., Chem. Commun. 1986, 1491), but aren't common.
answered May 9 at 11:05
orthocresol♦orthocresol
41.1k7124255
answered May 9 at 11:05
orthocresol♦orthocresol
41.1k7124255
answered May 9 at 11:05
orthocresol♦orthocresol
41.1k7124255
answered May 9 at 11:05
orthocresol♦orthocresol
41.1k7124255
41.1k7124255
1
$begingroup$
It's just that we just went through the basics of CFT. I can't think without using hybridisation right now..
$endgroup$
– evamPUNdit
May 9 at 11:10
5
$begingroup$
I can understand that. However, I think it would be a disservice to not at least mention it. It is an absolute travesty that hybridisation for TM complexes is still taught. Also, other people, not just you, will read this too.
$endgroup$
– orthocresol♦
May 9 at 11:22
$begingroup$
CFT contains zero reference to hybridization. It is ridiculous to force into it.
$endgroup$
– Greg
May 10 at 6:28
add a comment |
1
$begingroup$
It's just that we just went through the basics of CFT. I can't think without using hybridisation right now..
$endgroup$
– evamPUNdit
May 9 at 11:10
5
$begingroup$
I can understand that. However, I think it would be a disservice to not at least mention it. It is an absolute travesty that hybridisation for TM complexes is still taught. Also, other people, not just you, will read this too.
$endgroup$
– orthocresol♦
May 9 at 11:22
$begingroup$
CFT contains zero reference to hybridization. It is ridiculous to force into it.
$endgroup$
– Greg
May 10 at 6:28
1
1
$begingroup$
It's just that we just went through the basics of CFT. I can't think without using hybridisation right now..
$endgroup$
– evamPUNdit
May 9 at 11:10
$begingroup$
It's just that we just went through the basics of CFT. I can't think without using hybridisation right now..
$endgroup$
– evamPUNdit
May 9 at 11:10
5
5
$begingroup$
I can understand that. However, I think it would be a disservice to not at least mention it. It is an absolute travesty that hybridisation for TM complexes is still taught. Also, other people, not just you, will read this too.
$endgroup$
– orthocresol♦
May 9 at 11:22
$begingroup$
I can understand that. However, I think it would be a disservice to not at least mention it. It is an absolute travesty that hybridisation for TM complexes is still taught. Also, other people, not just you, will read this too.
$endgroup$
– orthocresol♦
May 9 at 11:22
$begingroup$
CFT contains zero reference to hybridization. It is ridiculous to force into it.
$endgroup$
– Greg
May 10 at 6:28
$begingroup$
CFT contains zero reference to hybridization. It is ridiculous to force into it.
$endgroup$
– Greg
May 10 at 6:28
add a comment |
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