Let's do the ICl4- Lewis structure.
Iodine, 7 valence electrons; Chlorine, 7 as
well, we have four Chlorines; and then we
have this up here so we're going to add an
additional valence electron for a total of
36 valence electrons.
Iodine's the least electronegative, we'll
put that at the center, and then we'll put
four Chlorines around it.
We'll form chemical bonds between the I and
the Chlorines.
So we've used 2, 4, 6, 8 valence electrons.
And then we'll complete the octets around
the outer atoms.
So we've used 2, 4, 6, 8, 10, 12, and 32.
So we've completed the octets on the Chlorines,
but we still have four valence electrons left.
Since Iodine is in period 5 of the periodic
table, it can hold more than eight valence
electrons.
So we're going to put those remaining four
valence electrons on the Iodine.
Put a pair here and then a pair over here.
So we've used all of the valence electrons
up.
And if we check our formal charges, you'll
find that the Iodine has a formal charge of
negative one, and the Chlorines all have formal
charges of zero.
That makes sense: negative one, we have a
negative one up here.
So that makes this the best structure for
ICl4-.
One last thing: we need to put brackets around
the structure to show that it is an ion and
it has a negative charge.
And that's the Lewis structure for ICL4-.
This is Dr. B., and thanks for watching.
