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caters

  • 4 months ago

Fe has an electronegativity of 1.83 and O has an electronegativity of 3.44. I calculated the electronegativity difference and it is 1.61. This is very close to the >= 1.7 for an ionic bond. Is Iron oxide ionic? Sc has an electronegativity of 1.36 so Scandium oxide has an electronegativity difference of 2.08. Is this in fact Sc^2+ O^2-(I might be wrong on the positive charge and the negative charge) instead of ScO3?

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  1. caters
    • 4 months ago
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    lots of metal oxides are of this form: A^x+ O^2-(where A is a metal and x is the magnitude of the positive charge). This is because metals often want to form ionic bonds, even with elements that prefer covalent such as oxygen. Lots of metal hydrides are of this form: A^x+ H- and likewise quite a few metal carbides are of this form: A^x+ C^4- and lots of metal halides are of this form: A^x+ X-(the lowercase x is the magnitude of the positive charge whereas the uppercase x represents a halogen) Now except for the halogens which really want to form either extremely polar covalent bonds(like HF) or ionic bonds(like Na+ Cl-) and the metals which really want to form ionic bonds with other elements the rest prefer covalent bonds or no bonds(no bonds as in like neon for example or covalent bonds like H2O or CH4 and occasionally things like XeF6 and HeH(helium hydride)) However because of how metals behave a lot of metal compounds are ionic whether they are carbides, oxides, hydrides, halides, or other compounds.

  2. freethinker
    • 4 months ago
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    start from the very beginning like the definition of a covalent and ionic bonds you will see that Fe is metal and O is non-metal a metal + non-metal = ionic but it is also important to note the role of electronegativity in reactions

  3. freethinker
    • 4 months ago
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    read up on chapter 7 of this book http://finedrafts.com/files/CUNY/chemistry/General/Brown-Holme/Chemistry_Engineering_Students_2nd_txtbk.pdf

  4. caters
    • 4 months ago
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    I know that but some metal oxides are covalent so that why I am asking.

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