There are three lone pairs present in chlorine atom
zero
In icl3 central atom is iodine and its valency is 7 out of 7 electrons 3 electrons are in chemical bonding so 2 lone pairs are there. Hybridization = number of sigma bonds + number of lone pairs = 3 sigma bonds + 2 lone pairs = 5 = sp3d ( 1 s + 3 P + 1 d = 5 ).
The answer is not 6 as said here before! The number of valence electrons in O2 (oxygen molecule) is: 12 valence electrons. 6 of them from each oxygen (O) atom. 4 valence electrons make up the double bond between the two oxygen atoms, and the remaining 8 valence electrons form lone pairs (non-bonding pairs) on the oxygen atoms, 2 lone pairs on each. I hope there is some help in this.
ONE!
NO2 only has one lone electron
Two lone pair on the central selenium and three lone pairs on each chlorine. So total of eight lone pairs.
Lone-pair electrons, Bonded pairs of electrons
There will be a total of 10 lone pairs of electrons. In NI3, each I will have 3 lone pairs (total of 9) and the N will also have 1 lone pair, for a grand total of 10 lone pairs.
These pairs of electrons are referred to as lone pairs.
1
Four pairs of electrons. Neon has a full octet.
Such pairs of electrons are called as lone pairs.
Chlorine (nucleus) has 1 lone pair and 3 polar-covalent bonding pairs (the shared pairs with O). Each oxygen (nucleus) has 3 lone pairs and 1 polar-covalent bonding pair (the shared pair with Cl)
There are four electrons, which is two pair.
There are 2 lone pairs in each Oxygen atom. So there are 4 lone pairs in total, which means 8 lone pair electrons.
There is one lone pair of electrons on the nitrogen atom.
If an atom has five valence electrons, it will have one lone pair of electrons.