Your question isn't quite clear, but if you're wondering if a chemical is polar or non-polar, you simply draw a VSEPR sketch and draw arrows where the bonds are. Only draw arrows between atoms, NOT between an atom and a lone pair of electrons. The arrow should point to the most electronegative atom (you should be given an electronegativity scale). Afterwards, you add up the arrows as vectors, and look at the sum of the vectors. If the sum is zero (CH4 is a good example), the chemical is non-polar. If the sum is a vector, the chemical is polar (H2O, or water, is polar).
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.
Answer:
B) K⁺, Sr²⁺ , O²⁻
Explanation:
Potassium is present in group one. It is alkali metal and have one valance electron.Potassium need to lose its one valance electron and form cation to get complete octet.
That's why it shows K⁺.
Sr is alkaline earth metal. It is present in group two. It has two valance electrons. Strontium needed to lose its two valance electrons and get stable electronic configuration.
When it loses its two valance electrons it shows cation with charge of +2.
Sr²⁺
Oxygen is present in group 16. It has sex valance electrons. It needed two more electrons to complete the octet. That's why oxygen gain two electron and form anion with a charge of -2.
O²⁻
Oh this is extremely hard...i might just die lol jk its the last one measuring cylinder :)
