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Covalent bonds - exercises
\[\require{mhchem}\]
- Draw the Lewis structures for the following compounds:
\(\quad\)a. \(\ce{Cl}_{2}\)
\(\quad\)b. \(\ce{HI}\)
\(\quad\)c. \(\ce{PCl}_{3}\)
- Determine the molecular formulas of the following compounds by completing subscripts \(v,w,x,z\).
\(\quad\)a. \(\ce{H}_{x}\ce{S}\)
\(\quad\)b. \(\ce{CCl}_{z}\)a) \(x=2\). Electron configuration of \(\ce{S}\) is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}\). \(\ce{S}\) has six valence shell electrons. Therefore, \(\ce{S}\) needs two bonds to complete the valence octet. \(\ce{H}\) requires one single bond to acquire the noble gas configuration. Thus, \(\ce{S}\) forms two single bonds with two \(\ce{H}\) atoms (one bond each).
\(\quad\)c. \(\ce{PF}_{w}\)b) \(z=4\). C needs four bonds to acquire the valence shell octet. \(\ce{Cl}\) forms one covalent bond to achieve the valence shell octet. Therefore, one \(\ce{C}\) atom forms four single bonds with four \(\ce{Cl}\) atoms.
\(\quad\)d. \(\ce{HBr}_{v}\)c) \(w=3\). \(\ce{P}\) belongs to main group \(V\) of the periodic table. Therefore, it requires three covalent bonds to achieve the valence shell octet. \(\ce{F}\) is a halide from main group \(VII\) of the periodic table. Hence \(\ce{F}\) needs one covalent bond to acquire the octet. \(\ce{P}\) forms three single bonds with three \(\ce{F}\) atoms.
d) \(v=1\)
- The following compounds contain multiple bonds. Identify the positions of the multiple bonds and draw the structure.
\(\quad\)a. \(\ce{C}_{2}\ce{H}_{4}\)
\(\quad\)b. \(\ce{N}_{2}\)
\(\quad\)c. \(\ce{SO}_{3}\)
\(\quad\)d. \(\ce{HCN}\)
- Use the knowledge of coordinate covalent bonds and explain the formation of the following polyatomic ions
\(\quad\)a. Formation of \(\ce{NH}_{4}^{+}\) by the reaction between \(\ce{NH}_{3}\) and water \(\left(\ce{H}^{+}\right)\)
\(\quad\)b. Formation of \(\ce{H}_{3}\ce{O}^{+}\) by the reaction between \(\ce{H}_{2}\ce{O}\) and \(\ce{H}^{+}\)
\(\quad\)c. Formation of \(\ce{BF}_{4}^{-}\) by the reaction between \(\ce{BF}_{3}\) and \(\ce{F}^{-}\)
In all of these molecules, a coordinate covalent bond is formed between an empty orbital of one atom with a lone pair of another atom. In the ammonium ion (\(\ce{NH}_{4}^{+}\)), the nitrogen atom of ammonia donates a lone pair of electrons to the hydrogen cation, which has an empty \(1s\) orbital. In a hydronium ion, the oxygen atom in the water molecule is donating one of their lone pairs to the hydrogen cation with a vacant \(1s\) orbital. In \(\ce{BF}_{4}^{-}\) ion, the fluoride anion donates one of their lone pairs to an empty \(2p\) orbital of the boron atom of \(\ce{BF}_{3}\) which gives boron four covalent bonds.
- Predict the geometry of the following molecules
\(\quad\)a. \(\ce{PF}_{3}\)
\(\quad\)b. \(\ce{SCl}_{2}\)
\(\quad\)c. \(\ce{SiBr}_{4}\)
a. Step 1 - Draw the Lewis structure of the molecule and identify the central atom.
Step 2 - Sum up the number of bonds and lone pairs in the central atom to get the total number of electron clouds around the central atom
\(\ce{P}\) has three covalent bonds and one lone pair of electrons. Therefore, all electron clouds around the central atom are 4.
Step 3 - Anticipate the shape of the molecule based on the number of electron clouds around the central atom
Total number of electron clouds = 4
So the shape can be tetrahedral, pyramidal or angular. Since one of the electron clouds is a lone pair, the best choice would be pyramidal.
b. Angular/Bent
c. Tetrahedral
- Predict the geometry around each of the carbon atoms in \(\ce{C}_{2}\ce{Cl}_{2}\).
- Predict the geometry of the following polyatomic ions
\(\quad\)a. \(\ce{NH}_{2}^{-}\)
\(\quad\)b. \(\ce{CO}_{3}^{2-}\)
\(\quad\)c. \(\ce{SO}_{4}^{2-}\)
a. Bent/Angular
b. Trigonal planar
c. Tetrahedral
b. Trigonal planar
c. Tetrahedral
- The structure and 3D view of the amino acid Aspartic acid is shown below. Predict the geometry around the indicated carbon atoms.
Trigonal planar and tetrahedral
- Determine whether the following bonds are ionic, polar covalent or covalent
\(\quad\)a. Between \(\ce{S}\) and \(\ce{Cl}\)
\(\quad\)b. Between \(\ce{N}\) and \(\ce{H}\)
\(\quad\)c. Between \(\ce{C}\) and \(\ce{Br}\)
\(\quad\)d. Between \(\ce{Li}\) and \(\ce{F}\)
\(\quad\)a. EN of \(\ce{S}\) \(=2.58\), EN of \(\ce{Cl}\) \(=3.16\). Difference of electronegativities \(=3.16-2.58=0.58\). Based on the following table, \(\ce{S}-\ce{Cl}\) bond is polar covalent.
| Electronegativity difference | Type of bond |
|---|---|
| \(\geqslant2.0\) | Ionic |
| \(0.5-1.9\) | Polar covalent |
| \(0-0.4\) | Covalent |
\(\quad\)b. Polar covalent
\(\quad\)c. Covalent
\(\quad\)d. Ionic
- Predict the polarity (polar or nonpolar) of each of the following molecules.
\(\quad\)a. \(\ce{HCN}\)
\(\quad\)b. \(\ce{NH}_{3}\)
\(\quad\)c. \(\ce{CH}_{3}\ce{Br}\)
First, draw the Lewis structure of the molecule and identify the shape. Next, use the electronegativity differences to find the polarity of the bonds. Finally, decide the net polarity of the molecule by adding each of the polar bonds.
\(\quad\)a. \(\ce{HCN}\) has a linear geometry. The \(\ce{C}-\ce{H}\) bond is a covalent bond (EN difference is \(0.35\), non-polar bond). The \(\ce{C}-\ce{N}\) bond is almost a polar covalent bond (EN difference is \(0.49\)). Therefore, electrons are attracted strongly towards the nitrogen atom in the \(\ce{C}-\ce{N}\) bond, creating a net polarity in the molecule.
\(\quad\)b. Polar molecule \(\quad\)c. Polar molecule
\(\quad\)a. \(\ce{HCN}\) has a linear geometry. The \(\ce{C}-\ce{H}\) bond is a covalent bond (EN difference is \(0.35\), non-polar bond). The \(\ce{C}-\ce{N}\) bond is almost a polar covalent bond (EN difference is \(0.49\)). Therefore, electrons are attracted strongly towards the nitrogen atom in the \(\ce{C}-\ce{N}\) bond, creating a net polarity in the molecule.
\(\quad\)b. Polar molecule \(\quad\)c. Polar molecule