Question

question 5 (2 points)
bent molecules like h₂o with 1 reds are 2 symmetrical/asymmetrical and 3 nonpolar/polar
bent molecules like ocl₂ with 4 reds are 5 symmetrical/asymmetrical and 6 nonpolar/polar note that ocl₂ has only nonpolar bonds
a. symmetrical b. asymmetrical c. nonpolar d. polar
e. 0 f. 1 g. 2 h. 3 i. 4

Explanation:

• For \(H_2O\):
• The number of regions of electron density (REDS) around the central \(O\) atom is calculated by considering bonding and non - bonding pairs. In \(H_2O\), the central \(O\) atom has 2 bonding pairs (\(O - H\) bonds) and 2 non - bonding pairs. So, the total number of REDS is \(2 + 2=4\).
• Bent molecules like \(H_2O\) are asymmetrical. This is because of the presence of non - bonding electron pairs which distort the symmetrical arrangement (if there were only bonding pairs in a linear or trigonal planar etc. arrangement).
• Since \(H_2O\) is asymmetrical and has polar \(O - H\) bonds (due to the difference in electronegativity between \(O\) and \(H\)), the molecule is polar. The bond dipoles do not cancel out.
• For \(OCl_2\):
• The central \(O\) atom in \(OCl_2\) has 2 bonding pairs (\(O - Cl\) bonds) and 2 non - bonding pairs. So, the number of REDS is \(2+2 = 4\).
• Bent molecules like \(OCl_2\) are asymmetrical. The non - bonding electron pairs cause a distortion from a symmetrical geometry.
• Although \(O - Cl\) bonds are non - polar (assuming a very small electronegativity difference, and the note in the problem states it has non - polar bonds), because of the asymmetrical shape (the non - bonding pairs affect the overall charge distribution), the molecule is non - polar. The bond dipoles (even if they are very small) cancel out due to the symmetrical (in terms of bond dipole cancellation) but asymmetrical (in terms of atom and electron pair arrangement) shape.

Answer:

1. I. 4
2. B. asymmetrical
3. D. polar
4. I. 4
5. B. asymmetrical
6. C. nonpolar