Question

question 3 (2 points)
using the first steps, a water molecule is shown as h - o - h. next, the number of valence electrons is determined. all bonds are comprised of valence electrons. the goal of each atom in a molecule is to attain happiness. small atoms try to attain the duet representing (1s^{2}). larger atoms try to attain the octet representing (s^{2}p^{6}).
the total number of valence electrons in a water molecule is 1. there are 2 valence electrons from the single o, and 3 valence electron from each of the two h atoms.
when we distribute valence electrons into the molecule we put them first into the 4, then onto the 5, and finally onto the 6. we see that this satisfies the octet rule for o and the duet rule for each h.
a. 0 b. 1 c. 2 d. 3 e. 4 f. 5 g. 6 h. 7 i. 8
j. 9 k. 10 l. central o atom m. outer h atoms n. bonds

Explanation:

• For the total number of valence electrons in \(H_2O\):
• Oxygen (\(O\)) has \(6\) valence electrons (\(O\) is in group \(16\) of the periodic table, and the number of valence electrons for main - group elements is equal to the group number).
• Hydrogen (\(H\)) has \(1\) valence electron (\(H\) is in group \(1\)).
• For \(H_2O\), the total number of valence electrons is \(6+(1\times2)=8\).
• For the valence electrons from \(O\):
• As mentioned above, \(O\) is in group \(16\), so it has \(6\) valence electrons.
• For the valence electrons from each \(H\):
• \(H\) is in group \(1\), so each \(H\) has \(1\) valence electron.
• For the order of distributing valence electrons:
• First, we put valence electrons into the bonds (to form the \(O - H\) covalent bonds). Then we put them onto the outer \(H\) atoms (but \(H\) only needs \(2\) electrons in total, and it gets \(1\) from the bond). The remaining electrons go onto the central \(O\) atom to satisfy its octet.

Answer:

1. I. \(8\)
2. G. \(6\)
3. B. \(1\)
4. N. bonds
5. M. outer \(H\) atoms
6. L. central \(O\) atom