Question

question 5 (2 points)
consider oxygen difluoride, of₂.
the en for o is 1.
the en for f is 2. the δen for the bond between o and f is 3. this means that the bonding electrons are not shared evenly between the o and the f atoms.
thus, each of the bonds in of₂ is a 4 bond.
also, since of₂ has 5 bonds, there are two dipoles in the molecule.
because of₂ is bent, and not symmetrical, the dipoles dont cancel out and of₂ is a 6 molecule.
a. 0 b. 0.54 c. 1.02 d. 2.96 e. 3 f. 3.44 g. 3.98 h. 4
i. ionic (bond) j. polar covalent (bond) k. nonpolar covalent (bond)
l. polar (molecule) m. nonpolar (molecule)

Explanation:

• Electronegativity (EN) values:
• From the Pauling electronegativity scale, the electronegativity of \(O\) is \(3.44\).
• The electronegativity of \(F\) is \(3.98\).
• \(\Delta EN\) calculation:
• \(\Delta EN=\vert EN_{F}-EN_{O}\vert=\vert3.98 - 3.44\vert=0.54\).
• Bond type:
• When \(0<\Delta EN<1.7\), the bond is polar covalent. Since \(\Delta EN = 0.54\) for the \(O - F\) bond, the bond is polar covalent.
• Number of bonds in \(OF_{2}\):
• The formula \(OF_{2}\) implies there are two \(O - F\) bonds.
• Molecular polarity:
• A bent - shaped molecule (\(OF_{2}\) has a bent geometry similar to \(H_{2}O\)) with polar bonds (\(O - F\)) that do not cancel out (due to non - symmetry) is a polar molecule.

Answer:

1. F. \(3.44\)
2. G. \(3.98\)
3. B. \(0.54\)
4. J. polar covalent (bond)
5. E. \(2\) (Since the formula is \(OF_{2}\), there are two \(O - F\) bonds)
6. L. polar (molecule)