The instructions are, Write Lewis structures that obey the octet rule for the following species. (Assign atomic charges where appropriate.)
1) XeO4
2) CIO^-4
3) PO4^-3
Would I just draw them in a different way or just the regular Lewis Structure?
I think drawing them the regular way will obey the octet rule.
To write Lewis structures that obey the octet rule, you will follow a specific set of steps. The octet rule states that atoms will gain, lose, or share electrons in order to achieve a stable configuration with eight valence electrons, except for hydrogen, which forms stable configurations with two valence electrons.
Now, let's go through the steps to draw the Lewis structures for the given species:
1) XeO4:
- Start by counting the total number of valence electrons. Xenon (Xe) is in Group 8, so it has 8 valence electrons. Oxygen (O) has 6 valence electrons each.
- Multiply the number of oxygen atoms by the number of valence electrons for each oxygen: 4 x 6 = 24 electrons.
- Add the number of valence electrons for all atoms: 8 (Xe) + 24 (O) = 32 electrons.
- Place the atoms in the structure, with the central atom (Xe) surrounded by the oxygen atoms (O). Connect them with single bonds (one pair of electrons) between Xe and each O.
- Distribute the remaining electrons to satisfy the octet rule. Place lone pairs around the oxygen atoms until they have an octet. If there are remaining electrons, place them on the central atom (Xe).
- Check the structure for each atom's octet: Xe has 8 electrons (2 lone pairs), and each O has 8 electrons (2 lone pairs).
- Assign formal charges where appropriate to ensure the most stable structure. In this case, there are no formal charges.
2) CIO^-4:
- Find the total number of valence electrons. Carbon (C) has 4 valence electrons, Iodine (I) has 7 valence electrons, and the negative charge on the ion indicates an additional electron.
- Add up all the valence electrons: 4 (C) + 7 (I) + 1 (extra electron) = 12 electrons.
- Assign the central atom. Since carbon is the least electronegative atom, it will be the central atom.
- Place the atoms in the structure, with the central atom (C) bonded to the iodine atom (I) by a single bond.
- Distribute the remaining electrons as lone pairs around the atoms to satisfy the octet rule. Place the remaining electrons on the central atom as lone pairs until it has an octet.
- Check the structure for each atom's octet: C has 8 electrons (4 lone pairs), and I has 8 electrons (4 lone pairs).
- Assign formal charges where needed to ensure the most stable structure. In this case, the negative charge on the ion should be located on the iodine atom.
3) PO4^-3:
- Count the total number of valence electrons. Phosphorus (P) has 5 valence electrons, Oxygen (O) has 6 valence electrons each, and the negative charge on the ion indicates an additional three electrons.
- Calculate the total number of valence electrons: 5 (P) + 4 x 6 (O) + 3 (extra electrons) = 32 electrons.
- Assign the central atom. In this case, phosphorus (P) will be the central atom.
- Place the atoms in the structure. Connect each oxygen atom (O) to the central phosphorus (P) atom by a single bond.
- Distribute the remaining electrons as lone pairs around the atoms to satisfy the octet rule. Place the remaining electrons on the central atom as lone pairs until it has an octet.
- Check the structure for each atom's octet: P has 8 electrons (4 lone pairs), and each O has 8 electrons (2 lone pairs).
- Assign formal charges where required for the most stable structure. In this case, there are no formal charges.
Remember, drawing Lewis structures is a visual representation of electron distribution. The steps provided help guide you toward an arrangement that follows the octet rule and maximizes stability.