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How can the formal charge of the $\ce{B}$ in the lewis structure of $\ce{BF4-}$ ion be minus one ?

I draw the lewis structure of $\ce{BF4-}$ ion such like the following: I put $\ce{B}$ in the central position and make single bonds between $\ce{B}$ and $\ce{F}$s, then completed the octet of $\ce{F}$s.

But in such a structure, the formal charge of $\ce{B}$ is $-1$, and the FC of $\ce{F}$s is 0.However, we know that if someone is going to get a minus FC, it should be the one whose electronegativity is the highest, and in this example this rule is not followed, so what is the reason for that ?

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    $\begingroup$ Formal charges are assigned assuming pure covalent bonds, so it often goes against electronegativity. $\endgroup$ – DHMO Jan 14 '17 at 10:31
  • $\begingroup$ Couple of comments. 1) Please don't use MathJax in titles as far as possible. 2) Please use $\ce{...}$ for chemical formulae and expressions. See here for more info. $\endgroup$ – orthocresol Jan 14 '17 at 11:18
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    $\begingroup$ Formal charge is the charge an atom would have if all the bonds were cleaved homolytically, ignoring electronegativity. $\endgroup$ – DHMO Jan 14 '17 at 11:30
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    $\begingroup$ There's a reason it's a called a "formal" charge and not just charge. $\endgroup$ – Zhe Jan 14 '17 at 13:48
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    $\begingroup$ @Zhe so how does this knowledge, which I have already known, help me with solving the problem ? $\endgroup$ – onurcanbektas Jan 15 '17 at 5:46

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