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CHEMISTRY, M1 EQ-Bank 17

The following table gives some information about two covalent molecule substances

\begin{array} {|c|c|c|}
\hline \text{Compound} & \text{Molecular formula} & \text{Boiling Point } (^{\circ}C) \\
\hline \text{Water} & \ce{H2O} & 100 \\
\hline \text{Hydrogen sulfide} & \ce{H2S} & -60 \\
\hline \end{array}

  1. Draw Lewis dot diagrams for each molecule.   (2 marks)
  1. Identify the shape and polarity of each molecule.   (4 marks)
  1. Explain why there is a difference in the boiling points of these molecules.   (3 marks)
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a.   
             

b.   Water: Bent and Polar

Hydrogen Sulfide: Bent and polar 

c.   Even though both molecular are polar:

  • Water has strong hydrogen bonding between molecules because \(\ce{H}\) is bonded to highly electronegative \(\ce{O}\) with lone pairs. These hydrogen bonds require large amounts of energy to break, giving water a high boiling point.
  • Hydrogen sulfide cannot form hydrogen bonds because \(\ce{S}\) is not electronegative enough. The only intermolecular forces are weak dipole–dipole and dispersion forces. Thus less energy is required to overcome the intermolecular forces and so \(\ce{H2S}\) a lower boiling point.
Show Worked Solution

a.   
             

b.   Water: Bent and Polar

Hydrogen Sulfide: Bent and polar
 

c.   Even though both molecular are polar:

  • Water has strong hydrogen bonding between molecules because \(\ce{H}\) is bonded to highly electronegative \(\ce{O}\) with lone pairs. These hydrogen bonds require large amounts of energy to break, giving water a high boiling point.
  • Hydrogen sulfide cannot form hydrogen bonds because \(\ce{S}\) is not electronegative enough. The only intermolecular forces are weak dipole–dipole and dispersion forces. Thus less energy is required to overcome the intermolecular forces and so \(\ce{H2S}\) a lower boiling point.

CHEMISTRY, M1 EQ-Bank 14

Copper and copper \(\text{(II)}\) oxide both conduct electricity in the molten state. However, copper also conducts electricity in the solid state, whereas copper \(\text{(II)}\) oxide does not.

Explain the electrical conductivity of copper and copper \(\text{(II)}\) oxide in terms of their structure and bonding.   (4 marks)

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Copper \(\text{(II)}\):

  • Copper is a metal with a giant metallic lattice structure. In both the solid and molten states, copper atoms are arranged in a lattice surrounded by a “sea of delocalised valence electrons”.
  • These delocalised electrons are free to move and carry charge, so copper conducts electricity in the solid state as well as when molten.

Copper \(\text{(II)}\) oxide :

  • Copper \(\text{(II)}\) oxide is an ionic compound composed of \(\ce{Cu^2+}\) cations and \(\ce{O^2-}\) anions arranged in a giant ionic lattice.
  • In the solid state, the ions are held in fixed positions by strong electrostatic forces and cannot move, so \(\ce{CuO}\) does not conduct electricity.
  • In the molten state, the ionic lattice breaks apart and the ions become mobile, allowing them to act as charge carriers, so molten \(\ce{CuO}\) does conduct electricity.
Show Worked Solution

Copper \(\text{(II)}\):

  • Copper is a metal with a giant metallic lattice structure.
  • In both the solid and molten states, copper atoms are arranged in a lattice surrounded by a “sea of delocalised valence electrons”.
  • These delocalised electrons are free to move and carry charge, so copper conducts electricity in the solid state as well as when molten.

Copper \(\text{(II)}\) oxide :

  • Copper \(\text{(II)}\) oxide is an ionic compound composed of \(\ce{Cu^2+}\) cations and \(\ce{O^2-}\) anions arranged in a giant ionic lattice.
  • In the solid state, the ions are held in fixed positions by strong electrostatic forces and cannot move, so \(\ce{CuO}\) does not conduct electricity.
  • In the molten state, the ionic lattice breaks apart and the ions become mobile, allowing them to act as charge carriers, so molten \(\ce{CuO}\) does conduct electricity.

CHEMISTRY, M1 EQ-Bank 1 MC

The Lewis electron dot diagram of ammonia \(\ce{NH3}\) is shown:
 

Which of the following is correct for this molecule?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Molecule polarity}\rule[-1ex]{0pt}{0pt}& \text{Molecular shape} \\
\hline
\rule{0pt}{2.5ex}\text{Non-polar}\rule[-1ex]{0pt}{0pt}&\text{Trigonal Planar}\\
\hline
\rule{0pt}{2.5ex}\text{Non-polar}\rule[-1ex]{0pt}{0pt}& \text{Tetrahedral}\\
\hline
\rule{0pt}{2.5ex}\text{Polar}\rule[-1ex]{0pt}{0pt}& \text{Tetrahedral} \\
\hline
\rule{0pt}{2.5ex}\text{Polar}\rule[-1ex]{0pt}{0pt}& \text{Pyramidal} \\
\hline
\end{array}
\end{align*}

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\(D\)

Show Worked Solution
  • \(\ce{NH3}\) has 3 bonding pairs and 1 lone pair of electrons around nitrogen.
  • According to VSEPR theory, this gives a pyramidal molecular shape (not planar or tetrahedral as a whole).
  • The bonds between nitrogen and hydrogen are polar covalent bonds. The arrangement causes the whole molecule to be polar.

\(\Rightarrow D\)

CHEMISTRY, M1 2014 HSC 11 MC

Which of the following does NOT represent the formation of a coordinate covalent bond?
 

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`A`

Show Worked Solution
  • A coordinate covalent bond occurs when one atom donates a pair of electrons (both electrons) to form the covalent bond.
  • In \(A\) both atoms donate 1 electron each, therefore it is not a coordinate covalent bond.

`=>A`