16 valence electrons
Valence electron is the number of electrons that lies in the outermost orbit of an atom. Si lies in group 4 of the periodic table and O lies in group 6 of the periodic table.
In SiO₂ there is one atom of Si and two atoms of O. The valence electron in Si is 4 and the valence electron in O is also 6. The total number of valence electron is
4 + (6 x 2) = 16
Answer : The Lewis-dot structure of is shown below.
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is,
As we know that rubidium has '1' valence electrons, iodine has '7' valence electrons and oxygen has '6' valence electrons.
Therefore, the total number of valence electrons in = 1 + 7 + 2(6) = 20
As we know that is an ionic compound because it is formed by the transfer of electron takes place from metal to non-metal element.
The total valence electrons present in molecule are as follows.
Valency of Si + (Valency of oxygen × number of oxygen atoms)
= 4 + (6 × 2)
= 4 + 12
Therefore, there are total 16 valence electrons in molecule including the ones used to form bonds.
None of these
Valence electrons of oxygen = 6
The total number of the valence electrons = 2(6) = 12
The Lewis structure is drawn in such a way that the octet of each atom and duet for the hydrogen in the molecule is complete. So, The Lewis structure is shown in image below.
Formula for formal charge :
Formal charge for oxygen:
From the structure,
Number of bond pairs = 4 (Double bond)
Number of lone pairs = 4
Hence, the correct option is:- None of these
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,
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
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,
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⁻
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
adding electron dots as needed h3c ss ch3
Keywords: Lewis structure, single-bonded, double-bonded, octet