Which of the following species exhibits resonance if all the formal charges are minimized?

a. OF2
b. N2
c. PCl5
d. ClO3-
e. BrF3

plz help me out here. its really urgent!!!

Draw the Lewis electron dot structures for each of the five compounds. If there is a double bond or triple bond (as in ClO3^-), the double/triple bond can be moved around and those will be different resonance structures.

To determine which species exhibits resonance, we need to examine the Lewis structures of each species and check if any of them have multiple valid structures. Resonance occurs when a molecule or ion has more than one valid Lewis structure.

Let's analyze each species:

a. OF2:
The Lewis structure for OF2 is:
O - F

There are no multiple valid Lewis structures for OF2, so it does not exhibit resonance.

b. N2:
The Lewis structure for N2 is:
N ≡ N

There are no multiple valid Lewis structures for N2, so it does not exhibit resonance.

c. PCl5:
The Lewis structure for PCl5 is:
Cl - P - Cl
|
Cl

There are no multiple valid Lewis structures for PCl5, so it does not exhibit resonance.

d. ClO3-:
The Lewis structure for ClO3- is:
O - Cl = O
|
O(-)

In this case, the ClO3- ion exhibits resonance because the negative charge can be moved to any of the three oxygen atoms.

e. BrF3:
The Lewis structure for BrF3 is:
F - Br - F
|
F

There are no multiple valid Lewis structures for BrF3, so it does not exhibit resonance.

Therefore, the species that exhibits resonance if all the formal charges are minimized is d. ClO3-.

To determine which species exhibits resonance with minimized formal charges, we need to calculate the formal charges for each of the options.

Let's go through each option one by one:

a. OF2: Start by assigning formal charges to each atom. Oxygen has 6 valence electrons and is bonded to 2 fluorine atoms, which have 7 valence electrons each. Since oxygen is more electronegative than fluorine, it will hold onto the shared electrons more closely. Therefore, one of the fluorine atoms will have a partial positive charge (+1), and the oxygen atom will have a partial negative charge (-1). There are no other possible structures for OF2, so it does not exhibit resonance.

b. N2: Nitrogen has 5 valence electrons and is bonded to another nitrogen atom. Since they have the same electronegativity, the electrons are shared equally between the two atoms. There are no formal charges on the nitrogen atoms, so there are no other possible structures for N2, and it does not exhibit resonance.

c. PCl5: Phosphorous has 5 valence electrons, and each chlorine atom has 7 valence electrons. The phosphorous atom forms bonds with all 5 chlorine atoms, resulting in a full octet for each atom. There are no formal charges on the atoms, so there are no other possible structures for PCl5, and it does not exhibit resonance.

d. ClO3-: Chlorine has 7 valence electrons, and the oxygen atoms have 6 valence electrons each. The chlorine atom forms bonds with all 3 oxygen atoms. To minimize the formal charges, one of the oxygen atoms should have a negative charge, while the chlorine atom should have a positive charge. There are two possible resonance structures for ClO3-:

O-Cl=O : O-Cl=O
/ \
O=Cl-O O=Cl-O

Overall charge: -1 Overall charge: -1

e. BrF3: Bromine has 7 valence electrons, and the fluorine atoms have 7 valence electrons each. The bromine atom forms bonds with 3 fluorine atoms. To minimize the formal charges, one of the fluorine atoms should have a partial negative charge, while the bromine atom should have a partial positive charge. There are no other possible structures for BrF3, so it does not exhibit resonance.

Therefore, the species that exhibits resonance with minimized formal charges is answer choice d. ClO3-.