My book states that

the total number of electrons present in the central atom of $\ce{I_3^-}$ is equal to 10.

What if I put a negative charge on the central iodine atom such that it gains a noble configuration and then I say that it is coordinatively bonded with other two iodine atoms? Then the number of electrons which the central atom will have would be 8. Wouldn't it?

Am I wrong somewhere in my analogy?

  • $\begingroup$ Maybe you could specify the title of "the book", that could be helpful ; ) $\endgroup$ Commented Dec 4, 2016 at 12:48

1 Answer 1


Looking at the wikipedia article here, your explanation seems to be exactly right:

The ion is linear and symmetrical. According to VSEPR theory, the central iodine atom has three equatorial lone pairs, and the terminal iodines are bonded axially in a linear fashion, due to the three lone pairs bonding to the central iodine-atom. In the molecular orbital model, a common explanation for the hypervalent bonding on the central iodine involves a three-center four-electron bond.

Thus, the central iodine atom has $3\cdot 2$ valence electrons from the lone pairs and 3 three-center four-electron bonds, so that I speculate you end up with $3\cdot 2$ VE + $4/3 \cdot 3$ VE = 6 VE + 4 VE = 10 VE. Which also fits to the reference given by you.


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