Question

How many electrons are in the valence shell of $I$ in $I{F_4}^ -$ ?
A. 14
B. 12
C. 10
D. 8

Answer 1

To solve this question, first we have to find the total valence electrons in the given compound. Then according to the Octet Rule, check how many electrons are left. And, finally conclude the bond pairs and lone pairs. Then we will get the valence electrons present in the valence shell of $I$ .

Complete answer:
Firstly, we will find the actual number of valence electrons, i.e. $7 + 4 \times 7 + 1 = 36$ .
If we draw a trial structure in which $I$ is the central atom and
According to the octet rule, every atom has an octet, we will have used 32 electrons.
So, we have 4 extra electrons left going onto the central $I$ atom as two lone pairs.
The $I$ atom in $I{F_4}^ -$ has four bonding pairs and two lone pairs.
So, four bonding pairs means we have 8 electrons and two lone pairs means we have 4electrons in the valence shell.
This makes a total of 12 valence electrons.
Hence, the correct option is (B.) 12.
Additional Information:- In chemistry, Lewis Structure is used to count Valence Electrons present in the valence shell of any compound.
Here, we have few steps to construct the Lewis Electron Structure is as:
Step-1: Determine the total number of valence electrons in the molecule or ion:- Add together the valence electrons from each atom. On the off chance that the species is a polyatomic particle, make sure to add or deduct the quantity of electrons important to give the absolute charge on the particle.
Step-2: Arrange the atoms to show specific connections.
Step-3: Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond.
Step-4: Starting with the terminal atoms, add enough electrons to each atom to give each atom an octet.
Step-5: If any electrons are left over, place them on the central atom.
Step-6: On the off chance that the focal iota has less electrons than an octet, utilize solitary sets from terminal iotas to shape different (twofold or triple) bonds to the focal particle to accomplish an octet.
Step-7: Finally check whether all valence electrons are accounted for or not.

Note: The only exception case for the standard is hydrogen, which is at its most minimal energy when it has two electrons in its valence shell. Helium is comparable in that it, as well, just has space for two electrons in its solitary valence shell. Hydrogen and helium have just a single electron shell.