Tuesday, August 22, 2017

oxidation state - Iodate ion and Octet rule


'm trying to figure out why the oxidation state of the iodate ion is +5. The iodine atom has 7 electrons in its outermost shell (comprised of s and p sub-shells). Two oxygen atoms receive 2 electrons from iodine to obtain full valence shells (s2 p6) and only one electron goes to the other oxygen atom. This leaves iodine with a full sub-shell s2. Is this a deviation from the Octet rule?


And how is the charge on the iodate ion negative when the third oxygen requires an electron?



Answer



First of all I strongly advise you to go through the following link as it will make my answer more clear to you .


Iodate ion



Two oxygen atoms receive 2 electrons from iodine to obtain full valence shells (s2 p6) and only one electron goes to the other oxygen atom.




The bond between the iodine atom and oxygen atom is a covalent bond . The electrons from the iodine atom are not given but shared .



This leaves iodine with a full sub-shell s2. Is this a deviation from the Octet rule?



The two double bonded oxygens ( refer the structure in my link ) share in all 4 electrons and the 3rd oxygen shares a single electron . Iodine hence has more electrons than octet rule specified 8. This is why it violates the rule .



And how is the charge on the iodate ion negative when the third oxygen requires an electron?



The third oxygen shares a single electron and hence has a negative charge on it which makes the overall molecule an anion with charge -1.



No comments:

Post a Comment

periodic trends - Comparing radii in lithium, beryllium, magnesium, aluminium and sodium ions

Apparently the of last four, $\ce{Mg^2+}$ is closest in radius to $\ce{Li+}$. Is this true, and if so, why would a whole larger shell ($\ce{...