RbIO2 Draw the Lewis dot structure for RbIO2. Include all hydrogen atoms and nonbonding electrons. Show the formal charges of all atoms. To change the symbol of an atom, double-click on the atom and enter the letter of the new atom.
The correct answer and explanation is:
Lewis Structure of RbIO₂:
To draw the Lewis structure of rubidium iodate (RbIO₂), follow these steps:
- Determine the Total Number of Valence Electrons:
- Rb (rubidium): 1 valence electron.
- I (iodine): 7 valence electrons.
- O (oxygen): 6 valence electrons (since there are two oxygens, we multiply this by 2).
- Choose the Central Atom: Iodine (I) is the least electronegative element (except for hydrogen), so it will be the central atom.
- Connect the Atoms with Single Bonds: Iodine (I) will form single bonds with both oxygen (O) atoms and rubidium (Rb). This uses 3 bonding pairs of electrons, leaving us with 14 electrons.
- Place Nonbonding Electrons (Lone Pairs) on Oxygen: Each oxygen atom requires 2 lone pairs to complete its octet (8 electrons total). Therefore, 12 electrons (2 lone pairs on each oxygen atom) are used, leaving 2 electrons.
- Distribute Remaining Electrons: These remaining 2 electrons are placed as a lone pair on iodine. Now iodine has 8 electrons in its valence shell.
- Formal Charges:
- Rb: Rb donates 1 electron, forming a single bond with iodine. Since Rb does not have any nonbonding electrons and it has no lone pairs, it holds a formal charge of +1.
- I: Iodine has 3 bonds (one to Rb and two to O) and 1 lone pair. Iodine has 7 valence electrons, uses 3 electrons for bonding, and holds 2 electrons as a lone pair, which gives it a formal charge of 0.
- O: Each oxygen atom has 2 lone pairs and 1 bond to iodine, making the oxygen atoms have formal charges of -1.
Structure:
- Rb-I-O with one oxygen having a single bond to iodine and the other oxygen also bonded to iodine.
- Each oxygen has lone pairs of electrons.
- Rb has a +1 formal charge.
In conclusion, the Lewis structure for RbIO₂ shows iodine at the center, bonded to two oxygens and one rubidium atom. Formal charges of the atoms help to identify the most stable configuration, and we confirm that iodine fulfills its octet and that all atoms in the molecule have the appropriate number of electrons.