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How do you determine titanium oxidation state 3+?
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+ Inorganic chemistry
+ Physical chemistry
+ Titanium dioxide
+ Oxidation
+ Titanium
+ Chemistry
+ Oxides
Posted by
Ned Kaufman
How do you determine titanium oxidation state 3+?
Well, it is a fact that the SUM of the oxidation numbers of the elements in a molecule or formula is EQUAL to the CHARGE, neutral, or anionic, or cationic, of that species… For [math]TiCl_{4}[/math], we got [math]4×Cl(-I)[/math], and [math]Ti(+IV)[/math] (and we use Roman numerals to confuse the punters)…. when we “BREAK” the [math]Ti-Cl[/math] bond, the 2 electrons we conceive to comprise that bond are assigned to the MOST electronegative atom, i.e. here the halogen… It should be fairly clear what we got for [math]TiCl_{3}[/math] …. what about for [math]TiCl_{2}[/math]?
Well, it is a fact that the SUM of the oxidation numbers of the elements in a molecule or formula is EQUAL to the CHARGE, neutral, or anionic, or cationic, of that species… For [math]TiCl_{4}[/math], we got [math]4×Cl(-I)[/math], and [math]Ti(+IV)[/math] (and we use Roman numerals to confuse the punters)…. when we “BREAK” the [math]Ti-Cl[/math] bond, the 2 electrons we conceive to comprise that bond are assigned to the MOST electronegative atom, i.e. here the halogen… It should be fairly clear what we got for [math]TiCl_{3}[/math] …. what about for [math]TiCl_{2}[/math]?
Well, it is a fact that the SUM of the oxidation numbers of the elements in a molecule or formula is EQUAL to the CHARGE, neutral, or anionic, or cationic, of that species… For [math]TiCl_{4}[/math], we got [math]4×Cl(-I)[/math], and [math]Ti(+IV)[/math] (and we use Roman numerals to confuse the punters)…. when we “BREAK” the [math]Ti-Cl[/math] bond, the 2 electrons we conceive to comprise that bond are assigned to the MOST electronegative atom, i.e. here the halogen… It should be fairly clear what we got for [math]TiCl_{3}[/math] …. what about for [math]TiCl_{2}[/math]?
Well, it is a fact that the SUM of the oxidation numbers of the elements in a molecule or formula is EQUAL to the CHARGE, neutral, or anionic, or cationic, of that species… For [math]TiCl_{4}[/math], we got [math]4×Cl(-I)[/math], and [math]Ti(+IV)[/math] (and we use Roman numerals to confuse the punters)…. when we “BREAK” the [math]Ti-Cl[/math] bond, the 2 electrons we conceive to comprise that bond are assigned to the MOST electronegative atom, i.e. here the halogen… It should be fairly clear what we got for [math]TiCl_{3}[/math] …. what about for [math]TiCl_{2}[/math]?
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Titanium sometimes has oxidation state +3, e. g. in titanium(III) oxide Ti[math]_2[/math]O[math]_3[/math].
Titanium sometimes has oxidation state +3, e. g. in titanium(III) oxide Ti[math]_2[/math]O[math]_3[/math].
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