A valid Lewis structure of IF3 cannot be drawn without violating the octet rule.
Answer: IF3 (Iodine Trifluoride)
This is because, I (Iodine) and F (Fluorine) both have odd number of valence electrons (7) which also means that there are too many valence electrons in the valence shell.
Answer: Li is the reducing agentg and O is the oxidizing agent.
Explanation:
1) The oxidizing agent is the one that is reduced and the reducing agent is the one that is oxidized.
2) The given reaction is:
4Li(s) + O₂ (g) → 2 Li₂O(s)
3) Determine the oxidation states of each atom:
Li(s): oxidation state = 0 (since it is alone)
O₂ (g): oxidation state = 0 (since it is alone)
Li in Li₂O (s) +1
O in Li₂O -2
That because 2× (+1) - 2 = 0.
4) Determine the changes:
Li went from 0 to + 1, therefore it got oxidized and it is the reducing agent.
O went from 0 to - 2, therefore it got reduced and it is the oxidizing agent.
Electrons - they are the particles that determine the chemical properties of an element
