Question

question 2 (2 points)
consider fluorine chloride, fcl.
the en for fluorine is 1.
the en for cl is 2.
the δen for the bond between f and cl is 3. this means that the bonding electrons are not shared evenly between the f and the cl atoms.
thus, the bond in fcl is a 4 bond.
also, since fcl has 5 bonds, there is a dipole in the molecule and fcl is a 6 molecule.
a. 0 b. 0.82 c. 1 d. 2 e. 3 f. 3.16 g. 3.98 h. 4
i. ionic (bond) j. polar covalent (bond) k. nonpolar covalent (bond)
l. polar (molecule) m. nonpolar (molecule)

Explanation:

Step1: Look up electronegativity values

From the Pauling electronegativity scale, the electronegativity (EN) of fluorine (\(F\)) is \(3.98\) (value \(G\)) and the electronegativity of chlorine (\(Cl\)) is \(3.16\) (value \(F\)).

Step2: Calculate \(\Delta EN\)

\(\Delta EN=\vert EN_{F}-EN_{Cl}\vert=\vert3.98 - 3.16\vert= 0.82\) (value \(B\)).

Step3: Determine bond type

Since \(0<\Delta EN<1.7\), the bond is polar covalent (option \(J\)).

Step4: Determine molecular polarity

\(FCl\) has one polar covalent bond. For a di - atomic molecule (\(FCl\) has the formula \(AB\) where \(A\) and \(B\) are different atoms), if the bond is polar, the molecule is polar. So it has one polar bond (answer for blank 5 is \(J\)) and is a polar molecule (option \(L\)).

Answer:

1. G. 3.98
2. F. 3.16
3. B. 0.82
4. J. polar covalent (bond)
5. J. polar covalent (bond)
6. L. polar (molecule)