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How to determine the greatest d orbital splitting?
How do I determine the crystal field splitting for an arbitrary point group?How to determine peroxy oxygen?Iron chemistry: acetates for ebonizing woodHow can the intense color of potassium permanganate be explained with molecular orbital theory?How to determine the magnetic character of heteroleptic complexes?Why do better π-acceptor ligands cause smaller Δ(T) d-orbital splitting?How to Determine An Element's ColourWhat exactly is the d-orbital splitting and how does this affect the colors for transition metal compounds?Pattern to determine the maximum ionic charge for transition elements?Effect of oxidation state on d-orbital splitting
$begingroup$
This question comes specifically from an IB Chemistry HL Paper 1 in May 2018 TZ1, namely question 8.
Which complex has the greatest d orbital splitting?
It gives 4 Complexes $ce[Fe(H_2O)_6]^2+$, $ce[Fe(H_2O)_6]^3+$, $ce[Co(H_2O)_6]^3+$, $ce[Cr(NH_3)_6]^3+$ and it says that they give the colours green, orange, blue and violet respectively.
Initially I thought that the answer would be $ce[Cr(NH_3)_6]^3+$ because it gives the highest energy light, being violet. However, the answer is given as $ce[Fe(H_2O)_6]^3+$, why is this?
ions transition-metals oxidation-state color
edited Mar 19 at 2:44
Mathew Mahindaratne
1,51313
asked Mar 19 at 1:20
Anthony PAnthony P
172
$endgroup$
add a comment |
$begingroup$
This question comes specifically from an IB Chemistry HL Paper 1 in May 2018 TZ1, namely question 8.
Which complex has the greatest d orbital splitting?
It gives 4 Complexes $ce[Fe(H_2O)_6]^2+$, $ce[Fe(H_2O)_6]^3+$, $ce[Co(H_2O)_6]^3+$, $ce[Cr(NH_3)_6]^3+$ and it says that they give the colours green, orange, blue and violet respectively.
Initially I thought that the answer would be $ce[Cr(NH_3)_6]^3+$ because it gives the highest energy light, being violet. However, the answer is given as $ce[Fe(H_2O)_6]^3+$, why is this?
ions transition-metals oxidation-state color
edited Mar 19 at 2:44
Mathew Mahindaratne
1,51313
asked Mar 19 at 1:20
Anthony PAnthony P
172
$endgroup$
add a comment |
$begingroup$
This question comes specifically from an IB Chemistry HL Paper 1 in May 2018 TZ1, namely question 8.
Which complex has the greatest d orbital splitting?
It gives 4 Complexes $ce[Fe(H_2O)_6]^2+$, $ce[Fe(H_2O)_6]^3+$, $ce[Co(H_2O)_6]^3+$, $ce[Cr(NH_3)_6]^3+$ and it says that they give the colours green, orange, blue and violet respectively.
Initially I thought that the answer would be $ce[Cr(NH_3)_6]^3+$ because it gives the highest energy light, being violet. However, the answer is given as $ce[Fe(H_2O)_6]^3+$, why is this?
ions transition-metals oxidation-state color
edited Mar 19 at 2:44
Mathew Mahindaratne
1,51313
asked Mar 19 at 1:20
Anthony PAnthony P
172
$endgroup$
This question comes specifically from an IB Chemistry HL Paper 1 in May 2018 TZ1, namely question 8.
Which complex has the greatest d orbital splitting?
It gives 4 Complexes $ce[Fe(H_2O)_6]^2+$, $ce[Fe(H_2O)_6]^3+$, $ce[Co(H_2O)_6]^3+$, $ce[Cr(NH_3)_6]^3+$ and it says that they give the colours green, orange, blue and violet respectively.
Initially I thought that the answer would be $ce[Cr(NH_3)_6]^3+$ because it gives the highest energy light, being violet. However, the answer is given as $ce[Fe(H_2O)_6]^3+$, why is this?
ions transition-metals oxidation-state color
ions transition-metals oxidation-state color
edited Mar 19 at 2:44
Mathew Mahindaratne
1,51313
asked Mar 19 at 1:20
Anthony PAnthony P
172
edited Mar 19 at 2:44
Mathew Mahindaratne
1,51313
asked Mar 19 at 1:20
Anthony PAnthony P
172
edited Mar 19 at 2:44
Mathew Mahindaratne
1,51313
edited Mar 19 at 2:44
Mathew Mahindaratne
1,51313
edited Mar 19 at 2:44
Mathew Mahindaratne
1,51313
1,51313
asked Mar 19 at 1:20
Anthony PAnthony P
172
asked Mar 19 at 1:20
Anthony PAnthony P
172
asked Mar 19 at 1:20
Anthony PAnthony P
172
172
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The colour at which the complex absorbs reflects the wavelength of the d–d* electronic transitions. However, this colour is not the same as the transmitted colour (which you see), but is instead complementary to the transmitted colour. Therefore, a complex that appears purple is actually absorbing lower-energy light than a complex that appears red.
answered Mar 19 at 1:30
orthocresol♦orthocresol
39.7k7114243
$endgroup$
$begingroup$
Please explain how is it 'complementary'.
$endgroup$
– Pan
Mar 19 at 8:55
$begingroup$
From what I understand, $ce[Cr(NH_3)_6]^3+$ would be absorbing the complementary colour of violet being yellow, which is low energy. However, how does this make any sense because $ceNH_3$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?
$endgroup$
– Anthony P
2 days ago
$begingroup$
@AnthonyP The splitting of the d orbitals is not only a function of the ligand, but also the metal (the atom type as well as the oxidation state). It therefore isn’t correct to only look at the ligand. Apart from that, NH3 isn’t even particularly strong-field; it’s only marginally higher than H2O in the spectrochemical series.
$endgroup$
– orthocresol♦
2 days ago
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The colour at which the complex absorbs reflects the wavelength of the d–d* electronic transitions. However, this colour is not the same as the transmitted colour (which you see), but is instead complementary to the transmitted colour. Therefore, a complex that appears purple is actually absorbing lower-energy light than a complex that appears red.
answered Mar 19 at 1:30
orthocresol♦orthocresol
39.7k7114243
$endgroup$
$begingroup$
Please explain how is it 'complementary'.
$endgroup$
– Pan
Mar 19 at 8:55
$begingroup$
From what I understand, $ce[Cr(NH_3)_6]^3+$ would be absorbing the complementary colour of violet being yellow, which is low energy. However, how does this make any sense because $ceNH_3$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?
$endgroup$
– Anthony P
2 days ago
$begingroup$
@AnthonyP The splitting of the d orbitals is not only a function of the ligand, but also the metal (the atom type as well as the oxidation state). It therefore isn’t correct to only look at the ligand. Apart from that, NH3 isn’t even particularly strong-field; it’s only marginally higher than H2O in the spectrochemical series.
$endgroup$
– orthocresol♦
2 days ago
add a comment |
$begingroup$
The colour at which the complex absorbs reflects the wavelength of the d–d* electronic transitions. However, this colour is not the same as the transmitted colour (which you see), but is instead complementary to the transmitted colour. Therefore, a complex that appears purple is actually absorbing lower-energy light than a complex that appears red.
answered Mar 19 at 1:30
orthocresol♦orthocresol
39.7k7114243
$endgroup$
$begingroup$
Please explain how is it 'complementary'.
$endgroup$
– Pan
Mar 19 at 8:55
$begingroup$
From what I understand, $ce[Cr(NH_3)_6]^3+$ would be absorbing the complementary colour of violet being yellow, which is low energy. However, how does this make any sense because $ceNH_3$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?
$endgroup$
– Anthony P
2 days ago
$begingroup$
@AnthonyP The splitting of the d orbitals is not only a function of the ligand, but also the metal (the atom type as well as the oxidation state). It therefore isn’t correct to only look at the ligand. Apart from that, NH3 isn’t even particularly strong-field; it’s only marginally higher than H2O in the spectrochemical series.
$endgroup$
– orthocresol♦
2 days ago
add a comment |
$begingroup$
The colour at which the complex absorbs reflects the wavelength of the d–d* electronic transitions. However, this colour is not the same as the transmitted colour (which you see), but is instead complementary to the transmitted colour. Therefore, a complex that appears purple is actually absorbing lower-energy light than a complex that appears red.
answered Mar 19 at 1:30
orthocresol♦orthocresol
39.7k7114243
$endgroup$
The colour at which the complex absorbs reflects the wavelength of the d–d* electronic transitions. However, this colour is not the same as the transmitted colour (which you see), but is instead complementary to the transmitted colour. Therefore, a complex that appears purple is actually absorbing lower-energy light than a complex that appears red.
answered Mar 19 at 1:30
orthocresol♦orthocresol
39.7k7114243
answered Mar 19 at 1:30
orthocresol♦orthocresol
39.7k7114243
answered Mar 19 at 1:30
orthocresol♦orthocresol
39.7k7114243
answered Mar 19 at 1:30
orthocresol♦orthocresol
39.7k7114243
39.7k7114243
$begingroup$
Please explain how is it 'complementary'.
$endgroup$
– Pan
Mar 19 at 8:55
$begingroup$
From what I understand, $ce[Cr(NH_3)_6]^3+$ would be absorbing the complementary colour of violet being yellow, which is low energy. However, how does this make any sense because $ceNH_3$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?
$endgroup$
– Anthony P
2 days ago
$begingroup$
@AnthonyP The splitting of the d orbitals is not only a function of the ligand, but also the metal (the atom type as well as the oxidation state). It therefore isn’t correct to only look at the ligand. Apart from that, NH3 isn’t even particularly strong-field; it’s only marginally higher than H2O in the spectrochemical series.
$endgroup$
– orthocresol♦
2 days ago
add a comment |
$begingroup$
Please explain how is it 'complementary'.
$endgroup$
– Pan
Mar 19 at 8:55
$begingroup$
From what I understand, $ce[Cr(NH_3)_6]^3+$ would be absorbing the complementary colour of violet being yellow, which is low energy. However, how does this make any sense because $ceNH_3$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?
$endgroup$
– Anthony P
2 days ago
$begingroup$
@AnthonyP The splitting of the d orbitals is not only a function of the ligand, but also the metal (the atom type as well as the oxidation state). It therefore isn’t correct to only look at the ligand. Apart from that, NH3 isn’t even particularly strong-field; it’s only marginally higher than H2O in the spectrochemical series.
$endgroup$
– orthocresol♦
2 days ago
$begingroup$
Please explain how is it 'complementary'.
$endgroup$
– Pan
Mar 19 at 8:55
$begingroup$
Please explain how is it 'complementary'.
$endgroup$
– Pan
Mar 19 at 8:55
$begingroup$
From what I understand, $ce[Cr(NH_3)_6]^3+$ would be absorbing the complementary colour of violet being yellow, which is low energy. However, how does this make any sense because $ceNH_3$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?
$endgroup$
– Anthony P
2 days ago
$begingroup$
From what I understand, $ce[Cr(NH_3)_6]^3+$ would be absorbing the complementary colour of violet being yellow, which is low energy. However, how does this make any sense because $ceNH_3$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?
$endgroup$
– Anthony P
2 days ago
$begingroup$
@AnthonyP The splitting of the d orbitals is not only a function of the ligand, but also the metal (the atom type as well as the oxidation state). It therefore isn’t correct to only look at the ligand. Apart from that, NH3 isn’t even particularly strong-field; it’s only marginally higher than H2O in the spectrochemical series.
$endgroup$
– orthocresol♦
2 days ago
$begingroup$
@AnthonyP The splitting of the d orbitals is not only a function of the ligand, but also the metal (the atom type as well as the oxidation state). It therefore isn’t correct to only look at the ligand. Apart from that, NH3 isn’t even particularly strong-field; it’s only marginally higher than H2O in the spectrochemical series.
$endgroup$
– orthocresol♦
2 days ago
add a comment |
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