A triatomic molecule has an experimental resultant dipole moment exactly equal to the individual bon — Chemical Bonding Chemistry Question

💡 Solution & Explanation

Using vector addition for two identical bonds: $\mu_R = \sqrt{\mu_{A-X}^2 + \mu_{A-X}^2 + 2\mu_{A-X}^2\cos\theta}$. Since $\mu_R = \mu_{A-X}$, we have $1 = \sqrt{1 + 1 + 2\cos\theta}$, which simplifies to $1 = 2 + 2\cos\theta$. $2\cos\theta = -1$, meaning $\cos\theta = -0.5$. Thus, $\theta = 120^\circ$.