# How can the formal charge of the B in the Lewis structure of BF4(-) ion be minus one?

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 ?

• Formal charges are assigned assuming pure covalent bonds, so it often goes against electronegativity. – DHMO Jan 14 '17 at 10:31
• 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. – orthocresol Jan 14 '17 at 11:18
• Formal charge is the charge an atom would have if all the bonds were cleaved homolytically, ignoring electronegativity. – DHMO Jan 14 '17 at 11:30
• There's a reason it's a called a "formal" charge and not just charge. – Zhe Jan 14 '17 at 13:48
• @Zhe so how does this knowledge, which I have already known, help me with solving the problem ? – onurcanbektas Jan 15 '17 at 5:46