Use MO diagrams to place B2+, B2, and B2- in order of (a) decreasing bond energy; (b) decreasing bond length.

1 answer:

3 0

We use the MO diagram for a homonuclear diatomic species (since C and N are neighbours, we treat them as the "same").

The first two electrons contribute to bonding. The next two are anti-bonding.

The next six contribute to bonding, and the following six are anti-bonding.

So, if we start with CN+, which has 4+5-1 (8) valence electrons, we note that the first two electrons contribute to bonding, while the next two cancel this out; the next four contribute to bonding, so the bond order is 4/2 = 2.

If we add one more electron to get CN, there are now 5 bonding electrons, giving bond order 5/2=2.5.

Adding one more to give CN- would give the bond order 6/2 = 3. (If we added more electrons, each one would lower the bond order.)

Given a series of molecules with identical skeletal structures, the one with the highest bond order has the highest bond energy:

CN+ < CN < CN-

Lewis structures will verify that CN- has a triple bond, but they do not work particularly well for CN+ and CN.

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90 POINTSSSS!!!

Katarina [22]

Answer:

$\large \boxed{\text{-2043.96 kJ/mol}}$

Explanation:

Assume the reaction is the combustion of propane.

Word equation: propane plus oxygen produces carbon dioxide and water

Chemical eqn: C₃H₈(g) + O₂(g) ⟶ CO₂(g) + H₂O(g)

Balanced eqn: C₃H₈(g) + 5O₂(g) ⟶ 3CO₂(g) + 4H₂O(g)

(a) Table of enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{C$_{3}$H$_{8}$(g)} & -103.85 \\\text{O}_{2}\text{(g)} & 0 \\\text{CO}_{2}\text{(g)} & -393.51 \\\text{H$_{2}$O(g)} & -241.82\\\end{array}$

(b)Total enthalpies of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)\\= \text{-2147.81 kJ/mol - (-103.85 kJ/mol)}\\= \text{-2147.81 kJ/mol + 103.85 kJ/mol}\\= \textbf{-2043.96 kJ/mol}\\\text{The enthalpy change is $\large \boxed{\textbf{-2043.96 kJ/mol}}$}$