How to determine the greatest 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 2 days ago
Mathew Mahindaratne
1,50313
asked 2 days ago
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 2 days ago
Mathew Mahindaratne
1,50313
asked 2 days ago
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 2 days ago
Mathew Mahindaratne
1,50313
asked 2 days ago
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 2 days ago
Mathew Mahindaratne
1,50313
asked 2 days ago
Anthony PAnthony P
172
edited 2 days ago
Mathew Mahindaratne
1,50313
asked 2 days ago
Anthony PAnthony P
172
edited 2 days ago
Mathew Mahindaratne
1,50313
edited 2 days ago
Mathew Mahindaratne
1,50313
edited 2 days ago
Mathew Mahindaratne
1,50313
1,50313
asked 2 days ago
Anthony PAnthony P
172
asked 2 days ago
Anthony PAnthony P
172
asked 2 days ago
Anthony PAnthony P
172
172
add a comment |
add a comment |
1 Answer
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$begingroup$
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 2 days ago
orthocresol♦orthocresol
39.7k7114243
$endgroup$
$begingroup$
Please explain how is it 'complementary'.
$endgroup$
– Pan
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 $ce{NH_3}$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?
$endgroup$
– Anthony P
yesterday
$begingroup$
@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♦
yesterday
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
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 2 days ago
orthocresol♦orthocresol
39.7k7114243
$endgroup$
$begingroup$
Please explain how is it 'complementary'.
$endgroup$
– Pan
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 $ce{NH_3}$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?
$endgroup$
– Anthony P
yesterday
$begingroup$
@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♦
yesterday
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 2 days ago
orthocresol♦orthocresol
39.7k7114243
$endgroup$
$begingroup$
Please explain how is it 'complementary'.
$endgroup$
– Pan
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 $ce{NH_3}$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?
$endgroup$
– Anthony P
yesterday
$begingroup$
@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♦
yesterday
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 2 days ago
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 2 days ago
orthocresol♦orthocresol
39.7k7114243
answered 2 days ago
orthocresol♦orthocresol
39.7k7114243
answered 2 days ago
orthocresol♦orthocresol
39.7k7114243
answered 2 days ago
orthocresol♦orthocresol
39.7k7114243
39.7k7114243
$begingroup$
Please explain how is it 'complementary'.
$endgroup$
– Pan
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 $ce{NH_3}$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?
$endgroup$
– Anthony P
yesterday
$begingroup$
@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♦
yesterday
add a comment |
$begingroup$
Please explain how is it 'complementary'.
$endgroup$
– Pan
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 $ce{NH_3}$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?
$endgroup$
– Anthony P
yesterday
$begingroup$
@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♦
yesterday
$begingroup$
Please explain how is it 'complementary'.
$endgroup$
– Pan
2 days ago
$begingroup$
Please explain how is it 'complementary'.
$endgroup$
– Pan
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 $ce{NH_3}$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?
$endgroup$
– Anthony P
yesterday
$begingroup$
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 $ce{NH_3}$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?
$endgroup$
– Anthony P
yesterday
$begingroup$
@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♦
yesterday
$begingroup$
@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♦
yesterday
add a comment |
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