Question

which of the following molecules contains at least one atom that has an odd - electron count (i.e. is a free radical)?
select all that apply however marks will be deducted for incorrect choices.
select one or more:
o₂n - no₂
no
sf₆
bbr₃
hcl
nh₃
h₂o

Explanation:

To determine which molecules contain a free radical (odd - electron count), we analyze each molecule:

Step 1: Analyze \(O_2N - NO_2\) (dinitrogen tetroxide)

• Nitrogen has 5 valence electrons, and oxygen has 6 valence electrons.
• For \(N_2O_4\) (the formula for \(O_2N - NO_2\)), the total number of valence electrons is \(2\times5 + 4\times6=10 + 24 = 34\), which is an even number. Also, the Lewis structure of \(N_2O_4\) has all atoms with complete octets (or appropriate electron configurations) and no unpaired electrons. So, it is not a free radical.

Step 2: Analyze \(NO\) (nitric oxide)

• Nitrogen has 5 valence electrons and oxygen has 6 valence electrons.
• The total number of valence electrons is \(5 + 6=11\), which is an odd number. In the Lewis structure of \(NO\), there is an unpaired electron, so \(NO\) is a free radical.

Step 3: Analyze \(SF_6\) (sulfur hexafluoride)

• Sulfur has 6 valence electrons and fluorine has 7 valence electrons.
• The total number of valence electrons is \(6+6\times7 = 6 + 42=48\), an even number. The Lewis structure of \(SF_6\) has sulfur with an expanded octet and all fluorine atoms with complete octets, and no unpaired electrons. So, it is not a free radical.

Step 4: Analyze \(BBr_3\) (boron tribromide)

• Boron has 3 valence electrons and bromine has 7 valence electrons.
• The total number of valence electrons is \(3+3\times7=3 + 21 = 24\), an even number. The Lewis structure of \(BBr_3\) has boron with an incomplete octet (but no unpaired electrons in the sense of a free radical) and bromine atoms with complete octets. So, it is not a free radical.

Step 5: Analyze \(HCl\) (hydrogen chloride)

• Hydrogen has 1 valence electron and chlorine has 7 valence electrons.
• The total number of valence electrons is \(1 + 7=8\), an even number. The Lewis structure of \(HCl\) has a single bond between \(H\) and \(Cl\) with \(Cl\) having a complete octet and \(H\) having a duet, and no unpaired electrons. So, it is not a free radical.

Step 6: Analyze \(NH_3\) (ammonia)

• Nitrogen has 5 valence electrons and hydrogen has 1 valence electron.
• The total number of valence electrons is \(5+3\times1 = 8\), an even number. The Lewis structure of \(NH_3\) has a nitrogen atom with a lone pair and three single bonds to hydrogen atoms, with no unpaired electrons. So, it is not a free radical.

Step 7: Analyze \(H_2O\) (water)

• Hydrogen has 1 valence electron and oxygen has 6 valence electrons.
• The total number of valence electrons is \(2\times1+6 = 8\), an even number. The Lewis structure of \(H_2O\) has oxygen with two lone pairs and two single bonds to hydrogen atoms, with no unpaired electrons. So, it is not a free radical.

Answer:

The molecule that contains at least one atom with an odd - electron count (is a free radical) is \(\boldsymbol{NO}\)