Question

question 5 (2 points)
lets review what we learned about the lewis structure for water.
a water molecule has 1 regions of electron density, reds.
there are 2 reds that represent bonds to h atoms, and 3 reds that represent non - bonding electron pairs.
all 4 reds repel/attract 4 each other.
to get the reds as far apart from one another as possible, water ends up as a tetrahedron with bond angles of 5°.
the x - ray diffractometer is designed to see atoms, and not electron clouds. thus, the x - ray diffractometer see the water molecule as 6.
a. 1 b. 2 c. 3 d. 4 e. 5 f. 6 g. attract h. repel i. 45 j. 90 k. 109.5 l. 120 m. 180 n. diatomic o. bent p. planar trigonal q. tetrahedral r. pyramidal s. linear

Explanation:

• For the number of electron - density regions (REDS) in \(H_2O\):
• The central atom \(O\) has 2 bonding pairs (with \(H\) atoms) and 2 non - bonding pairs. So, the total number of REDs is \(2 + 2=4\).
• For the number of REDs that represent bonds to \(H\) atoms:
• In \(H_2O\), there are 2 \(O - H\) bonds, so there are 2 REDs representing bonds to \(H\) atoms.
• For the number of REDs that represent non - bonding electron pairs:
• The oxygen atom in \(H_2O\) has 2 non - bonding electron pairs.
• For the interaction between REDs:
• Electron - density regions (both bonding and non - bonding) repel each other according to VSEPR (Valence - Shell Electron - Pair Repulsion) theory.
• For the bond angle in the tetrahedral arrangement (predicted by VSEPR for 4 REDs):
• The ideal bond angle for a tetrahedral arrangement of electron - density regions is \(109.5^{\circ}\).
• For the shape of \(H_2O\) as seen by an x - ray diffractometer (which sees atoms):
• \(H_2O\) has a bent shape because the non - bonding electron pairs push the \(O - H\) bonds closer together.

Answer:

1. D. 4
2. B. 2
3. B. 2
4. H. repel
5. K. 109.5
6. O. bent