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As two atoms approach one another in space, the potential energy decreases and then increases as shown in the diagram below.
To what are these two changes attributed?
[Note : ${p^{+}}^{}$ = proton, $e^{-}$ = electron]

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Decrease

(from 0.40 to 0.15 nm)

increase

(from 0.15 to 0 mm)

(c)

decreased

p^{+} p^{+}

repulsion

increased

e^{-} e^{-}

repulsion

Decrease

(from 0.40 to 0.15 nm)

increase

(from 0.15 to 0 mm)

(b)

increased

p^{+} e^{-}

attraction

increased

e^{-} e^{-}

repulsion

Decrease

(from 0.40 to 0.15 nm)

increase

(from 0.15 to 0 mm)

(a)

increased

p^{+} e^{-}

attraction

increased

p^{+} p^{+}

repulsion

Decrease

(from 0.40 to 0.15 nm)

increase

(from 0.15 to 0 mm)

(d)

decreased

p^{+} e^{-}

repulsion

increased

p^{+} p^{+}

repulsion

answer is A.

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Detailed Solution

The strong attraction of each shared electron to both nuclei stabilizes the system, and the potential energy decreases as the bond distance decreases. If the atoms continue to approach each other, the positive charges in the two nuclei begin to repel each other, and the potential energy increases.

Decrease

(from 0.40 to 0.15 nm)

increase

(from 0.15 to 0 mm)

(a)

increased