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How to determine the greatest d orbital splitting?
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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
6,004723
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
6,004723
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
6,004723
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
6,004723
asked Mar 19 at 1:20
Anthony PAnthony P
172
edited Mar 19 at 2:44
Mathew Mahindaratne
6,004723
asked Mar 19 at 1:20
Anthony PAnthony P
172
edited Mar 19 at 2:44
Mathew Mahindaratne
6,004723
edited Mar 19 at 2:44
Mathew Mahindaratne
6,004723
edited Mar 19 at 2:44
Mathew Mahindaratne
6,004723
6,004723
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
39.9k7115245
$endgroup$
$begingroup$
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 $ce{NH_3}$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?
$endgroup$
– Anthony P
Mar 20 at 2:06
$begingroup$
@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♦
Mar 20 at 2:40
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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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 Mar 19 at 1:30
orthocresol♦orthocresol
39.9k7115245
$endgroup$
$begingroup$
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 $ce{NH_3}$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?
$endgroup$
– Anthony P
Mar 20 at 2:06
$begingroup$
@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♦
Mar 20 at 2:40
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
39.9k7115245
$endgroup$
$begingroup$
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 $ce{NH_3}$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?
$endgroup$
– Anthony P
Mar 20 at 2:06
$begingroup$
@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♦
Mar 20 at 2:40
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
39.9k7115245
$endgroup$
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
39.9k7115245
answered Mar 19 at 1:30
orthocresol♦orthocresol
39.9k7115245
answered Mar 19 at 1:30
orthocresol♦orthocresol
39.9k7115245
answered Mar 19 at 1:30
orthocresol♦orthocresol
39.9k7115245
39.9k7115245
$begingroup$
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 $ce{NH_3}$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?
$endgroup$
– Anthony P
Mar 20 at 2:06
$begingroup$
@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♦
Mar 20 at 2:40
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 $ce{NH_3}$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?
$endgroup$
– Anthony P
Mar 20 at 2:06
$begingroup$
@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♦
Mar 20 at 2:40
$begingroup$
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 $ce{NH_3}$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?
$endgroup$
– Anthony P
Mar 20 at 2:06
$begingroup$
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
Mar 20 at 2:06
$begingroup$
@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♦
Mar 20 at 2:40
$begingroup$
@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♦
Mar 20 at 2:40
add a comment |
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