Question

question 3 (2 points)
consider chlorine bromide, clbr.
the en for cl is 1.
the en for br is 2. the δen for the bond between cl and br is 3. this means that the bonding electrons are shared evenly between the cl and the br atoms.
thus, the bond in clbr is a 4 bond.
also, since clbr has no 5 bonds, there is no dipole in the molecule and clbr is a 6 molecule.
a. 0 b. 0.20 c. 1.02 d. 2.96 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:

• Electronegativity (EN) values:
• Chlorine (\(Cl\)) has an EN value of \(3.16\) (from the Pauling scale).
• Bromine (\(Br\)) has an EN value of \(2.96\) (from the Pauling scale).
• \(\Delta EN\) calculation:
• \(\Delta EN=\vert EN_{Cl}-EN_{Br}\vert=\vert3.16 - 2.96\vert= 0.20\).
• Bond type:
• When \(0<\Delta EN<1.7\), the bond is polar covalent. Here \(\Delta EN = 0.20\), so the bond in \(ClBr\) is polar covalent.
• Molecular polarity:
• Since there is only one polar covalent bond in \(ClBr\), and no non - polar bonds (as there is only one type of bond), the molecule is polar because of the presence of the polar bond which creates a dipole moment.

Answer:

1. F. \(3.16\)
2. D. \(2.96\)
3. B. \(0.20\)
4. J. polar covalent (bond)
5. K. nonpolar covalent (bond)
6. L. polar (molecule)