Question

question 4 (2 points)
consider oxygen dichloride, ocl₂.
the en for chlorine is 1.
the en for o is 2.
the δen for the bond between o and cl is 3. this means that the bonding electrons are shared fairly evenly between the o and the cl atoms.
thus, each of the bonds in ocl₂ is a 4 bond.
also, since ocl₂ has only 5 bonds, there is no dipole in the molecule and ocl₂ is a 6 molecule.
a. 0 b. 0.28 c. 1.02 d. 2.96 e. 3 f. 3.16 g. 3.44 h. 4
i. ionic (bond) j. polar covalent (bond) k. nonpolar covalent (bond)
l. polar (molecule) m. nonpolar (molecule)

Explanation:

• Electronegativity (EN) values:
• Chlorine (\(Cl\)) has an EN value of \(3.16\) (from the Pauling scale).
• Oxygen (\(O\)) has an EN value of \(3.44\) (from the Pauling scale).
• \(\Delta EN\) calculation:
• \(\Delta EN=\vert EN_O - EN_{Cl}\vert=\vert3.44 - 3.16\vert = 0.28\). When \(\Delta EN\) is between \(0\) and \(1.7\), the bond is covalent. Since \(0\lt0.28\lt1.7\), and the electrons are shared fairly evenly (as given in the problem statement), it is a non - polar covalent bond.
• Molecular polarity:
• \(OCl_2\) has two \(O - Cl\) bonds. The molecular geometry of \(OCl_2\) (bent shape, similar to \(H_2O\) but with less polarity due to small \(\Delta EN\)) has bond dipoles that cancel out (because of symmetry in the sense that the two \(O - Cl\) bonds are equivalent in a non - linear but symmetric enough way for the overall dipole moment to be zero). So, it is a non - polar molecule.

Answer:

1. F. \(3.16\)
2. G. \(3.44\)
3. B. \(0.28\)
4. K. nonpolar covalent (bond)
5. J. polar covalent (bond)
6. M. nonpolar (molecule)