Draw the Lewis Dot structure for RbIO2.
2 answers:
Rubidium Iodite is an Ionic compound made up of Following two ions,
1) Rb⁺ (Cation)
2) IO₂⁻ (Anion)
Now in order to draw the lewis structure of Iodite ion (poly anion) we will follow following steps,
a) First add up the number of valence electrons of all elements,
So,
Number of valence electrons in I = 7
Number of Valence electrons in O1 = 6
Number of Valence electrons in O2 = 6
Number of negative charges = 1
--------------
20
b) Secondly draw I as the central atom surrounded by two oxygen atoms. Connect I with each O atom via a single bond and subtract 2 electrons per bond, so,
20
- 4
-----------
16
Now, divide these 16 electrons on all elements starting from most electronegative atom. You will find that each oxygen atom will atain 6 electrons in three pairs and I will attain 4 electrons in 2 pairs, after that make a double bond between one O and I so that the formal charge of I gets zero.
Hence, the structure of Iodite will be as shown below,
The RbIO₂ compound has a Lewis structure in IO₂⁻
<h3>Further Explanation </h3>The Lewis formula is used to describe covalent bonds
Lewis structures have a central atom and a terminal atom
The central atom is an atom that is bound to 2 or more other atoms, while a terminal atom is bound to 1 other atom
In describing Lewis's structure the steps that can be taken are:
- 1. Count the number of valence electrons from atoms in a molecule
- 2. Give each bond a pair of electrons
- 3. The remaining electrons are given to the atomic terminal so that an octet is reached
- 4. If the central atom is not yet octet, free electrons are drawn to the central atom to form double bonds
In the RbIO₂ compound, the bond is an ionic bond that is Rb + and IO₂⁻ so that what Lewis can describe is the I and O bonds in IO₂⁻
Steps to draw the Lewis structure
- 1. Place atom I in the middle and O atom on the side
- 2. Draw the electrons that bind each of 2 pairs to form 2 bonds
- 3. Write down the remaining free electrons after subtracting the bonding electrons
- 4. The form of a double bond of one O atom with atom I, because there are electrons that have not been paired
From this it was found that atom I did not follow the octet rule because it binds 10 electrons, this is what is called the octet rule which was developed / expanded, whereas in general period 3 elements such as P, Br, I can have electrons exceeding 8 when they bind
<h3>Learn more </h3>
adding electron dots as needed h3c ss ch3
formal charge
ionic bonding
Keywords: Lewis structure, single-bonded, double-bonded, octet
You might be interested in
<u>Answer:</u> The correct answer is Option D.
<u>Explanation:</u>
To calculate the hybridization of , we use the equation:
where,
V = number of valence electrons present in central atom (S) = 6
N = number of monovalent atoms bonded to central atom = 0
C = charge of cation = 0
A = charge of anion = 0
Putting values in above equation, we get:
The number of electron pair around the central metal atom are 3. This means that the hybridization will be and the electronic geometry of the molecule will be trigonal planar.
Hence, the correct answer is Option D.
