Question

(c) a graph of potential energy versus internuclear distance for two cl atoms is given below. on the same graph, carefully sketch a curve that corresponds to potential energy versus internuclear distance for two br atoms.

Explanation:

1. Understand Atomic Properties: Bromine (Br) atoms are larger than chlorine (Cl) atoms (Br has more electron shells). So, the equilibrium internuclear distance ($r_{eq}$) for $Br_2$ should be greater than for $Cl_2$.
2. Potential Energy Well Depth: The strength of the covalent bond (or the depth of the potential energy well) depends on the overlap of atomic orbitals. Since Br atoms are larger, the overlap of their $p$-orbitals (involved in bonding) is less effective than Cl atoms. Thus, the potential energy minimum (well depth) for $Br_2$ should be less deep (higher potential energy at the minimum) than for $Cl_2$.
3. Sketch the Curve:

• The general shape of the potential energy curve (decreasing to a minimum, then increasing) remains the same for $Br_2$ as for $Cl_2$ (due to similar bonding nature, both diatomic halogens).
• Shift the minimum of the $Br_2$ curve to a larger internuclear distance (right along the x-axis) compared to $Cl_2$.
• The minimum of the $Br_2$ curve should be higher (less deep) than the minimum of the $Cl_2$ curve (since $Br_2$ has a weaker bond, so less stabilization energy).

Answer:

To sketch the curve for $Br_2$:

1. Shape: Maintain the same general curve shape (decreasing to a minimum, then increasing) as the $Cl_2$ curve.
2. Equilibrium Distance: Shift the minimum of the $Br_2$ curve to a larger internuclear distance (right of $Cl_2$’s minimum) because Br atoms are larger.
3. Well Depth: The minimum potential energy for $Br_2$ is less deep (higher on the y-axis) than $Cl_2$ because $Br_2$ has a weaker covalent bond (less orbital overlap).

(Visually, the $Br_2$ curve will be a “wider” and “shallower” version of the $Cl_2$ curve, with its minimum shifted right.)