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The oxidation state of iodine in iodoxybenzoic acid
How to determine an oxidation number of an atom in the compound, that has at least two elements, which are not hydrogen or oxygen?Electronegativity Considerations in Assigning Oxidation StatesHow to find out oxidation state when compound has multiple electronegativitiesWhy do negative oxidation states not extend to -8?Typesetting oxidation state and charge simultaneouslyOxidation state of nitrogen in triazirine (HN3)How can chlorine have an oxidation number of 3 in chlorite?What are oxidation states used for?Difficulty understanding redox in terms of hydrogen and oxygen transfer?Why are electrons shared equally when calculating formal charge, but unequally when calculating oxidation state?
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I am slightly puzzled by this seemingly trivial question. The idea behind assigning oxidation states is simply to treat every single bond made by the atoms as ionic bonds. That is to say, both the electrons of a covalent bond in a molecule would go to the more electronegative atom. For example, in the case of the $ce O-I$ bond, the bond electrons would go to the oxygen atom, leaving the oxygen atom with a negative charge while the iodine atom incurs a positive charge. Based on this, we would say that the oxygen atom doubly bonded to the $ce I$ atom has an oxidation state of -2.
However, I am slightly confused as to why various sources assign the oxidation state of $ce I$ in IBX as +5, rather than +3. On the Pauling scale of electronegativity, it is shown that $ce I (2.66)$ is indeed slightly more electronegative than $ce C (2.55)$. Based on this premise, we would reasonably assign the $ce I$ atom in IBX with a +3 oxidation state as the pair of electrons in $ce C-I$ bond would go to $ce I$ rather than to $ce C$.
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oxidation-state
asked 10 hours ago
Tan Yong BoonTan Yong Boon
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$begingroup$
I am slightly puzzled by this seemingly trivial question. The idea behind assigning oxidation states is simply to treat every single bond made by the atoms as ionic bonds. That is to say, both the electrons of a covalent bond in a molecule would go to the more electronegative atom. For example, in the case of the $ce O-I$ bond, the bond electrons would go to the oxygen atom, leaving the oxygen atom with a negative charge while the iodine atom incurs a positive charge. Based on this, we would say that the oxygen atom doubly bonded to the $ce I$ atom has an oxidation state of -2.
However, I am slightly confused as to why various sources assign the oxidation state of $ce I$ in IBX as +5, rather than +3. On the Pauling scale of electronegativity, it is shown that $ce I (2.66)$ is indeed slightly more electronegative than $ce C (2.55)$. Based on this premise, we would reasonably assign the $ce I$ atom in IBX with a +3 oxidation state as the pair of electrons in $ce C-I$ bond would go to $ce I$ rather than to $ce C$.
Image taken from here
oxidation-state
asked 10 hours ago
Tan Yong BoonTan Yong Boon
4,7051 gold badge15 silver badges55 bronze badges
$endgroup$
$begingroup$
Yeah, it should be +3.
$endgroup$
– Mithoron
8 hours ago
2
$begingroup$
@Mithoron If that is the case, why are so many authoritative sources claiming otherwise? For example, Zhdankin's Hypervalent Iodine Chemistry also states that it is +5. We can also extend this to another popular reagent, $ce PhI(OAc)2$, why is it +3 and not +1? It seems that the $ce C-I$ bond is treated rather differently by these source.
$endgroup$
– Tan Yong Boon
8 hours ago
add a comment
|
$begingroup$
I am slightly puzzled by this seemingly trivial question. The idea behind assigning oxidation states is simply to treat every single bond made by the atoms as ionic bonds. That is to say, both the electrons of a covalent bond in a molecule would go to the more electronegative atom. For example, in the case of the $ce O-I$ bond, the bond electrons would go to the oxygen atom, leaving the oxygen atom with a negative charge while the iodine atom incurs a positive charge. Based on this, we would say that the oxygen atom doubly bonded to the $ce I$ atom has an oxidation state of -2.
However, I am slightly confused as to why various sources assign the oxidation state of $ce I$ in IBX as +5, rather than +3. On the Pauling scale of electronegativity, it is shown that $ce I (2.66)$ is indeed slightly more electronegative than $ce C (2.55)$. Based on this premise, we would reasonably assign the $ce I$ atom in IBX with a +3 oxidation state as the pair of electrons in $ce C-I$ bond would go to $ce I$ rather than to $ce C$.
Image taken from here
oxidation-state
asked 10 hours ago
Tan Yong BoonTan Yong Boon
4,7051 gold badge15 silver badges55 bronze badges
$endgroup$
I am slightly puzzled by this seemingly trivial question. The idea behind assigning oxidation states is simply to treat every single bond made by the atoms as ionic bonds. That is to say, both the electrons of a covalent bond in a molecule would go to the more electronegative atom. For example, in the case of the $ce O-I$ bond, the bond electrons would go to the oxygen atom, leaving the oxygen atom with a negative charge while the iodine atom incurs a positive charge. Based on this, we would say that the oxygen atom doubly bonded to the $ce I$ atom has an oxidation state of -2.
However, I am slightly confused as to why various sources assign the oxidation state of $ce I$ in IBX as +5, rather than +3. On the Pauling scale of electronegativity, it is shown that $ce I (2.66)$ is indeed slightly more electronegative than $ce C (2.55)$. Based on this premise, we would reasonably assign the $ce I$ atom in IBX with a +3 oxidation state as the pair of electrons in $ce C-I$ bond would go to $ce I$ rather than to $ce C$.
Image taken from here
oxidation-state
oxidation-state
asked 10 hours ago
Tan Yong BoonTan Yong Boon
4,7051 gold badge15 silver badges55 bronze badges
asked 10 hours ago
Tan Yong BoonTan Yong Boon
4,7051 gold badge15 silver badges55 bronze badges
asked 10 hours ago
Tan Yong BoonTan Yong Boon
4,7051 gold badge15 silver badges55 bronze badges
asked 10 hours ago
Tan Yong BoonTan Yong Boon
4,7051 gold badge15 silver badges55 bronze badges
asked 10 hours ago
Tan Yong BoonTan Yong Boon
4,7051 gold badge15 silver badges55 bronze badges
4,7051 gold badge15 silver badges55 bronze badges
$begingroup$
Yeah, it should be +3.
$endgroup$
– Mithoron
8 hours ago
2
$begingroup$
@Mithoron If that is the case, why are so many authoritative sources claiming otherwise? For example, Zhdankin's Hypervalent Iodine Chemistry also states that it is +5. We can also extend this to another popular reagent, $ce PhI(OAc)2$, why is it +3 and not +1? It seems that the $ce C-I$ bond is treated rather differently by these source.
$endgroup$
– Tan Yong Boon
8 hours ago
add a comment
|
$begingroup$
Yeah, it should be +3.
$endgroup$
– Mithoron
8 hours ago
2
$begingroup$
@Mithoron If that is the case, why are so many authoritative sources claiming otherwise? For example, Zhdankin's Hypervalent Iodine Chemistry also states that it is +5. We can also extend this to another popular reagent, $ce PhI(OAc)2$, why is it +3 and not +1? It seems that the $ce C-I$ bond is treated rather differently by these source.
$endgroup$
– Tan Yong Boon
8 hours ago
$begingroup$
Yeah, it should be +3.
$endgroup$
– Mithoron
8 hours ago
$begingroup$
Yeah, it should be +3.
$endgroup$
– Mithoron
8 hours ago
2
2
$begingroup$
@Mithoron If that is the case, why are so many authoritative sources claiming otherwise? For example, Zhdankin's Hypervalent Iodine Chemistry also states that it is +5. We can also extend this to another popular reagent, $ce PhI(OAc)2$, why is it +3 and not +1? It seems that the $ce C-I$ bond is treated rather differently by these source.
$endgroup$
– Tan Yong Boon
8 hours ago
$begingroup$
@Mithoron If that is the case, why are so many authoritative sources claiming otherwise? For example, Zhdankin's Hypervalent Iodine Chemistry also states that it is +5. We can also extend this to another popular reagent, $ce PhI(OAc)2$, why is it +3 and not +1? It seems that the $ce C-I$ bond is treated rather differently by these source.
$endgroup$
– Tan Yong Boon
8 hours ago
add a comment
|
2 Answers
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Whether you're talking about an $ceI-O$ bond or an $ceI=O$ bond, each oxygen bound to the iodine has an oxidation state of -2. Since iodine is more electronegative than carbon, the carbon has an oxidation state of +1. Therefore, the sum of the oxidation states of the atoms connected to iodine is $3(-2) + 1 = -5$. Therefore, iodine's oxidation state is +5 for this neutral compound.
In the case of $cePhI(OAc)2$, the iodine is directly connected to one carbon (+1) and two oxygens (-4, total); therefore the oxidation state of iodine is +3.
edited 2 hours ago
Mathew Mahindaratne
11.3k2 gold badges15 silver badges39 bronze badges
answered 5 hours ago
Greg DwuletGreg Dwulet
3995 bronze badges
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I don't understand that. What about the electrons from hydrogen? If you ignore that, it'll won't come out to 0 total.
$endgroup$
– Martin - マーチン♦
2 hours ago
add a comment
|
$begingroup$
Count the electrons assigned to iodine in the standard oxidation state convention. None of the bonds to the more electronegative oxygen impart any valence electrons to iodine, leaving only the two valence electrons coming from carbon (which, as pointed out in the question, is less electronegative than iodine). So two valence electrons counted to iodine in the molecule versus seven in a neutral iodine atom. Difference is +5.
answered 4 hours ago
Oscar LanziOscar Lanzi
18.8k2 gold badges32 silver badges58 bronze badges
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2 Answers
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2 Answers
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$begingroup$
Whether you're talking about an $ceI-O$ bond or an $ceI=O$ bond, each oxygen bound to the iodine has an oxidation state of -2. Since iodine is more electronegative than carbon, the carbon has an oxidation state of +1. Therefore, the sum of the oxidation states of the atoms connected to iodine is $3(-2) + 1 = -5$. Therefore, iodine's oxidation state is +5 for this neutral compound.
In the case of $cePhI(OAc)2$, the iodine is directly connected to one carbon (+1) and two oxygens (-4, total); therefore the oxidation state of iodine is +3.
edited 2 hours ago
Mathew Mahindaratne
11.3k2 gold badges15 silver badges39 bronze badges
answered 5 hours ago
Greg DwuletGreg Dwulet
3995 bronze badges
$endgroup$
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I don't understand that. What about the electrons from hydrogen? If you ignore that, it'll won't come out to 0 total.
$endgroup$
– Martin - マーチン♦
2 hours ago
add a comment
|
$begingroup$
Whether you're talking about an $ceI-O$ bond or an $ceI=O$ bond, each oxygen bound to the iodine has an oxidation state of -2. Since iodine is more electronegative than carbon, the carbon has an oxidation state of +1. Therefore, the sum of the oxidation states of the atoms connected to iodine is $3(-2) + 1 = -5$. Therefore, iodine's oxidation state is +5 for this neutral compound.
In the case of $cePhI(OAc)2$, the iodine is directly connected to one carbon (+1) and two oxygens (-4, total); therefore the oxidation state of iodine is +3.
edited 2 hours ago
Mathew Mahindaratne
11.3k2 gold badges15 silver badges39 bronze badges
answered 5 hours ago
Greg DwuletGreg Dwulet
3995 bronze badges
$endgroup$
$begingroup$
I don't understand that. What about the electrons from hydrogen? If you ignore that, it'll won't come out to 0 total.
$endgroup$
– Martin - マーチン♦
2 hours ago
add a comment
|
$begingroup$
Whether you're talking about an $ceI-O$ bond or an $ceI=O$ bond, each oxygen bound to the iodine has an oxidation state of -2. Since iodine is more electronegative than carbon, the carbon has an oxidation state of +1. Therefore, the sum of the oxidation states of the atoms connected to iodine is $3(-2) + 1 = -5$. Therefore, iodine's oxidation state is +5 for this neutral compound.
In the case of $cePhI(OAc)2$, the iodine is directly connected to one carbon (+1) and two oxygens (-4, total); therefore the oxidation state of iodine is +3.
edited 2 hours ago
Mathew Mahindaratne
11.3k2 gold badges15 silver badges39 bronze badges
answered 5 hours ago
Greg DwuletGreg Dwulet
3995 bronze badges
$endgroup$
Whether you're talking about an $ceI-O$ bond or an $ceI=O$ bond, each oxygen bound to the iodine has an oxidation state of -2. Since iodine is more electronegative than carbon, the carbon has an oxidation state of +1. Therefore, the sum of the oxidation states of the atoms connected to iodine is $3(-2) + 1 = -5$. Therefore, iodine's oxidation state is +5 for this neutral compound.
In the case of $cePhI(OAc)2$, the iodine is directly connected to one carbon (+1) and two oxygens (-4, total); therefore the oxidation state of iodine is +3.
edited 2 hours ago
Mathew Mahindaratne
11.3k2 gold badges15 silver badges39 bronze badges
answered 5 hours ago
Greg DwuletGreg Dwulet
3995 bronze badges
edited 2 hours ago
Mathew Mahindaratne
11.3k2 gold badges15 silver badges39 bronze badges
edited 2 hours ago
Mathew Mahindaratne
11.3k2 gold badges15 silver badges39 bronze badges
edited 2 hours ago
Mathew Mahindaratne
11.3k2 gold badges15 silver badges39 bronze badges
11.3k2 gold badges15 silver badges39 bronze badges
answered 5 hours ago
Greg DwuletGreg Dwulet
3995 bronze badges
answered 5 hours ago
Greg DwuletGreg Dwulet
3995 bronze badges
answered 5 hours ago
Greg DwuletGreg Dwulet
3995 bronze badges
3995 bronze badges
$begingroup$
I don't understand that. What about the electrons from hydrogen? If you ignore that, it'll won't come out to 0 total.
$endgroup$
– Martin - マーチン♦
2 hours ago
add a comment
|
$begingroup$
I don't understand that. What about the electrons from hydrogen? If you ignore that, it'll won't come out to 0 total.
$endgroup$
– Martin - マーチン♦
2 hours ago
$begingroup$
I don't understand that. What about the electrons from hydrogen? If you ignore that, it'll won't come out to 0 total.
$endgroup$
– Martin - マーチン♦
2 hours ago
$begingroup$
I don't understand that. What about the electrons from hydrogen? If you ignore that, it'll won't come out to 0 total.
$endgroup$
– Martin - マーチン♦
2 hours ago
add a comment
|
$begingroup$
Count the electrons assigned to iodine in the standard oxidation state convention. None of the bonds to the more electronegative oxygen impart any valence electrons to iodine, leaving only the two valence electrons coming from carbon (which, as pointed out in the question, is less electronegative than iodine). So two valence electrons counted to iodine in the molecule versus seven in a neutral iodine atom. Difference is +5.
answered 4 hours ago
Oscar LanziOscar Lanzi
18.8k2 gold badges32 silver badges58 bronze badges
$endgroup$
add a comment
|
$begingroup$
Count the electrons assigned to iodine in the standard oxidation state convention. None of the bonds to the more electronegative oxygen impart any valence electrons to iodine, leaving only the two valence electrons coming from carbon (which, as pointed out in the question, is less electronegative than iodine). So two valence electrons counted to iodine in the molecule versus seven in a neutral iodine atom. Difference is +5.
answered 4 hours ago
Oscar LanziOscar Lanzi
18.8k2 gold badges32 silver badges58 bronze badges
$endgroup$
add a comment
|
$begingroup$
Count the electrons assigned to iodine in the standard oxidation state convention. None of the bonds to the more electronegative oxygen impart any valence electrons to iodine, leaving only the two valence electrons coming from carbon (which, as pointed out in the question, is less electronegative than iodine). So two valence electrons counted to iodine in the molecule versus seven in a neutral iodine atom. Difference is +5.
answered 4 hours ago
Oscar LanziOscar Lanzi
18.8k2 gold badges32 silver badges58 bronze badges
$endgroup$
Count the electrons assigned to iodine in the standard oxidation state convention. None of the bonds to the more electronegative oxygen impart any valence electrons to iodine, leaving only the two valence electrons coming from carbon (which, as pointed out in the question, is less electronegative than iodine). So two valence electrons counted to iodine in the molecule versus seven in a neutral iodine atom. Difference is +5.
answered 4 hours ago
Oscar LanziOscar Lanzi
18.8k2 gold badges32 silver badges58 bronze badges
answered 4 hours ago
Oscar LanziOscar Lanzi
18.8k2 gold badges32 silver badges58 bronze badges
answered 4 hours ago
Oscar LanziOscar Lanzi
18.8k2 gold badges32 silver badges58 bronze badges
answered 4 hours ago
Oscar LanziOscar Lanzi
18.8k2 gold badges32 silver badges58 bronze badges
18.8k2 gold badges32 silver badges58 bronze badges
add a comment
|
add a comment
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$begingroup$
Yeah, it should be +3.
$endgroup$
– Mithoron
8 hours ago
2
$begingroup$
@Mithoron If that is the case, why are so many authoritative sources claiming otherwise? For example, Zhdankin's Hypervalent Iodine Chemistry also states that it is +5. We can also extend this to another popular reagent, $ce PhI(OAc)2$, why is it +3 and not +1? It seems that the $ce C-I$ bond is treated rather differently by these source.
$endgroup$
– Tan Yong Boon
8 hours ago