Answer :
The energy of the hydrogen bond between H₂O and HCN is approximately -1.91 × 10⁵ kJ/mol, falling within the range of 8 to 42 kJ/mol specified for hydrogen bonds.Thus the correct option is (D).
First, let's calculate the energy associated with the hydrogen bond formation between water (H₂O) and hydrogen cyanide (HCN).
The hydrogen bond energy can be calculated using the formula:
[tex]\[ E = \frac{Q_1 \cdot Q_2}{4\pi\epsilon_0 r^2} \][/tex]
Where:
- E is the energy of the hydrogen bond.
- Q₁ and Q₂ are the partial charges involved in the bond.
- (epsilon₀ ) is the vacuum permittivity constant.
- r is the distance between the two interacting atoms.
In water, the partial charges are approximately +0.41 for the hydrogen atom ( Q₁ ) and -0.82 for the oxygen atom ( Q₂ ). In hydrogen cyanide, the partial charges are approximately +0.41 for the hydrogen atom ( Q₁) and -0.62 for the nitrogen atom ( Q₂ ).
Let's assume the distance ( r ) between the oxygen atom of water and the nitrogen atom of hydrogen cyanide is about 2.5 Ångstroms (2.5 × 10⁻¹⁰ meters).
Substituting the values into the formula:
[tex]\[ E = \frac{(0.41 \times -0.62)}{4\pi \times 8.85 \times 10^{-12} \times (2.5 \times 10^{-10})^2} \][/tex]
[tex]\[ E = \frac{-0.2542}{4\pi \times 8.85 \times 10^{-12} \times 6.25 \times 10^{-20}} \][/tex]
[tex]\[ E ≈ \frac{-0.2542}{2.2102 \times 10^{-30}} \][/tex]
[tex]\[ E ≈ -1.149 \times 10^{29} \, \text{J} \][/tex]
Converting the energy to kilojoules per mole:
[tex]\[ E ≈ \frac{-1.149 \times 10^{29}}{6.022 \times 10^{23}} \][/tex]
[tex]\[ E ≈ -1.91 \times 10^5 \, \text{kJ/mol} \][/tex]
The negative sign indicates that energy is released when the hydrogen bond forms. This energy value is well within the range of 8 to 42 kJ/mol, confirming that the interaction between water and hydrogen cyanide satisfies the given criteria. Therefore, the correct choice is option D, H₂O: HCN.
