# How can chlorine have 10 electrons around it in chlorine trifluoride?

I'm relatively new to chemistry, and while doing some exercises about $\ce{ClF_3}$ I tried to draw where the electrons are. I checked Google and got this:

How is this structure possible? From what I've learned from high school every element wants 8 valence electrons. From the figure, $\ce{F}$ has 8, but $\ce{Cl}$ has 10.

• "Every element wants 8 valence electrons" is just a simplified description for people to get started with chemistry. Octet rule has its limits and many exceptions, $\ce{ClF3}$ is a typical example of hypervalent molecules. I think you can find pretty much from the two links above and if there's anything unclear, come back and ask more specific questions regarding your doubts. – Weijun Zhou Jan 28 '18 at 17:34
• @WeijunZhou Ok, thank you! I'll have a look! Should I delete this question? – cansomeonehelpmeout Jan 28 '18 at 17:35
• No need to do so. It can help others who may ask questions similar to yours, and that's how the site works. – Weijun Zhou Jan 28 '18 at 17:36
• There's a total of 4*7 = 28. One of the flourine's shares its lone electron with chlorine's lone electron. The other two flourines get shares of two of the chlorine's remaining six thus allowing covalent bonds to be formed )and leaving two unshared pairs). – A. J. deLange Jan 28 '18 at 17:58
• possible duplicate of What is the favoured geometry of chlorine trifluoride? – ron Jan 28 '18 at 18:14

Interhalogens consist of one $\ce{X-F}$ pair whith a regular two electron bond, and pairs of flourine atoms taking over one free electron pair of X (chlorine, bromine, iodine).
What forms are three-centre four-electron bonds (one pair of electrons from the chlorine, two single electrons from two fluorine atoms). That's why interhalogens always have an uneven number of fluorine atoms (except for the dimer $\ce{Cl_2F_6}$).