Answer :
Answer:
To solve for polyatomic molecules and polyatomic ions, particularly in terms of understanding their structure, charge distribution, and formal charge, we can use several steps. Let’s take the example of the polyatomic ion
O
2
2
−
O
2
2−
(known as the peroxide ion).
Steps to Solve for Polyatomic Ions (O₂²⁻)
Determine the Total Number of Valence Electrons:
Each oxygen atom has 6 valence electrons. Since the ion has a
2
−
2− charge, we add 2 additional electrons.
Total valence electrons
=
(
2
×
6
)
+
2
=
14
electrons
Total valence electrons=(2×6)+2=14 electrons
Draw the Lewis Structure:
Place the two oxygen atoms next to each other.
Distribute the electrons to satisfy the octet rule, starting with a single bond and then adding lone pairs.
Since there are 14 electrons and we need to account for each oxygen atom having 8 electrons in its valence shell:
O
−
O
O−O
After placing a single bond between the oxygen atoms, we have used 2 electrons. Distribute the remaining 12 electrons as lone pairs to complete the octet for each oxygen atom:
[O:
−
:O]
2
−
[O:−:O]
2−
Each oxygen now has 8 electrons around it: 4 in lone pairs and 4 shared (2 from the bond and 2 from the charge).
Assign Formal Charges:
Calculate the formal charge on each atom using the formula:
Formal charge
=
Valence electrons
−
(
Non-bonding electrons
+
Bonding electrons
2
)
Formal charge=Valence electrons−(Non-bonding electrons+
2
Bonding electrons
)
For each oxygen atom in
O
2
2
−
O
2
2−
:
Formal charge on each O
=
6
−
(
6
+
2
2
)
=
6
−
7
=
−
1
Formal charge on each O=6−(6+
2
2
)=6−7=−1
So, the formal charge on each oxygen is
−
1
−1, adding up to the overall charge of
2
−
2−.
Verify the Structure:
Ensure that the total number of electrons matches the number calculated in step 1, and that the charges are consistent with the overall charge of the ion.
Summary for
O
2
2
−
O
2
2−
:
The ion has 14 valence electrons.
The Lewis structure is
[O:
−
:O]
2
−
[O:−:O]
2−
.
Each oxygen has a formal charge of
−
1
−1.
General Approach for Other Polyatomic Ions:
Count the total valence electrons, including the extra electrons for negative charges or subtracting for positive charges.
Draw a skeleton structure, typically with the least electronegative atom in the center (not applicable to diatomic ions like
O
2
2
−
O
2
2−
).
Distribute electrons to satisfy the octet rule (or duet rule for hydrogen).
Assign formal charges to each atom to find the most stable structure.
Verify the total charge of the ion and ensure all atoms have their octets satisfied if possible.