At 298.2 K the relationship between enthalpy of bond dissociation (in kJ mol$$^{-1}$$) for hydrogen E$$_H$$ and its isotope, deuterium E$$_D$$, is best described by:

We are asked to compare the enthalpy of bond dissociation of the normal hydrogen molecule (that is, the H-H bond) with that of its isotopic analogue deuterium (the D-D bond) at a temperature of 298.2 K.

First, recall the definition of the dissociation enthalpy (sometimes called the bond enthalpy or bond energy). It is the energy required to break one mole of a particular bond in the gaseous state at the specified temperature and convert the atoms to the gaseous state as well:

$$\mathrm{H_2(g)} \longrightarrow 2\,\mathrm{H(g)}, \qquad \Delta H = E_H$$

$$\mathrm{D_2(g)} \longrightarrow 2\,\mathrm{D(g)}, \qquad \Delta H = E_D$$

Now, because deuterium atoms are twice as heavy as protium (ordinary hydrogen) atoms, the two isotopic molecules have different zero-point vibrational energies. Quantum‐mechanically, even at 0 K, a diatomic molecule cannot be perfectly at rest in its bond; it possesses a minimum vibrational energy given by

$$E_{\text{ZP}}=\dfrac{1}{2}h\nu,$$

where $$h$$ is Planck’s constant and $$\nu$$ is the fundamental vibrational frequency of the molecule. For a harmonic oscillator, the frequency is inversely proportional to the square root of the reduced mass $$\mu$$ of the two atoms:

$$\nu \propto \dfrac{1}{\sqrt{\mu}}.$$

Because the reduced mass of D-D is larger than that of H-H (twice as large, in fact), its vibrational frequency is smaller and therefore its zero-point energy is smaller. In simpler words, the “spring” of the chemical bond in D2 jiggles less energetically at the quantum ground state than the same bond in H2.

Let us put that idea into an algebraic statement. Write the observed bond dissociation enthalpy $$E$$ in terms of the intrinsic electronic bond energy $$E_{\text{e}}$$ (the depth of the potential well) and the zero-point contribution:

$$E = E_{\text{e}} - E_{\text{ZP}}.$$

Applying this once for hydrogen and once for deuterium, we have

$$E_H = E_{\text{e}} - E_{\text{ZP}}(\mathrm{H_2}),$$

$$E_D = E_{\text{e}} - E_{\text{ZP}}(\mathrm{D_2}).$$

Subtracting the first equation from the second eliminates the common electronic term $$E_{\text{e}}$$ and gives

$$E_D - E_H = E_{\text{ZP}}(\mathrm{H_2}) - E_{\text{ZP}}(\mathrm{D_2}).$$

Experimental spectroscopy tells us that the numerical value of the zero-point energy difference between H2 and D2 is approximately $$7.5\;\text{kJ mol}^{-1}$$. (One often also quotes a value in the range 7.2-8.0 kJ mol−1; 7.5 kJ mol−1 is the commonly accepted textbook figure.) Substituting this value into the algebraic relation, we obtain

$$E_D - E_H \approx 7.5\;\text{kJ mol}^{-1}.$$

Rearrange the above to express $$E_H$$ explicitly in terms of $$E_D$$:

$$E_H = E_D - 7.5\;\text{kJ mol}^{-1}.$$

This final equation matches exactly the functional form given in Option C.

Hence, the correct answer is Option C.