Which of the following molecules has an odd number of total valence electrons?  Group of answer choicesSF6BF3NONO2-1

To determine which molecule has an odd number of total valence electrons, we need to look at the valence electrons of each atom in the molecules.

1. SF6: Sulfur (S) has 6 valence electrons and Fluorine (F) has 7. Since there are 6 Fluorine atoms, the total number of valence electrons is 6 (from S) + 6*7 (from F) = 48, which is even.

2. BF3: Boron (B) has 3 valence electrons and Fluorine (F) has 7. Since there are 3 Fluorine atoms, the total number of valence electrons is 3 (from B) + 3*7 (from F) = 24, which is even.

3. NO: Nitrogen (N) has 5 valence electrons and Oxygen (O) has 6. The total number of valence electrons is 5 (from N) + 6 (from O) = 11, which is odd.

4. NO2-1: Nitrogen (N) has 5 valence electrons, Oxygen (O) has 6, and there is an extra electron due to the negative charge. Since there are 2 Oxygen atoms, the total number of valence electrons is 5 (from N) + 2*6 (from O) + 1 (from the charge) = 18, which is even.

Therefore, the molecule with an odd number of total valence electrons is NO.