Consider a molecule with formula \(\mathrm{AX}_{2}\). Supposing the \(\mathrm{A}-\mathrm{X}\) bond is polar, how would you expect the dipole moment of the \(\mathrm{AX}_{2}\) molecule to change as the \(\mathrm{X}-\mathrm{A}-\mathrm{X}\) bond angle decreases from \(180^{\circ}\) to \(100^{\circ} ?\)

A molecular dipole moment is a vector quantity that depends on the bond polarities and the geometrical arrangement of the bonds in the molecule. If a bond between two atoms A and X is polar, it means that the electron density is unevenly distributed between them, generating a partial positive charge on one atom and a partial negative charge on the other.

The dipole moment (\(\mu\)) of the molecule can be calculated using the formula: \[ \mu = q \cdot d \] where \(q\) is the magnitude of the charges (partial charges) and \(d\) is the distance between the charges (bond length). However, for a molecule like AX₂, we need to consider the geometry of the molecule and the bond angles. When the X-A-X angle is 180°, the molecule is linear. The dipole moments of the two A-X bonds will be opposite in direction and equal in magnitude, canceling each other out. Therefore, the overall dipole moment of the molecule when the bond angle is 180° will be zero.

When the X-A-X bond angle decreases from 180° to 100°, the geometry of the AX₂ molecule becomes bent. In this case, the two A-X bond dipole moments will no longer cancel out due to their angle difference, creating a net molecular dipole moment that is nonzero. To quantitatively analyze the change, we can use vector addition to determine the overall molecular dipole moment of the molecule in this new geometry. Consider the two A-X dipoles as vectors \(\vec{\mu_1}\) and \(\vec{\mu_2}\). We can find the vector sum of these two dipoles as follows: \[ \vec{\mu}_{total} = \vec{\mu_1} + \vec{\mu_2} \]

Due to the symmetry of the AX₂ molecule, both \(\vec{\mu_1}\) and \(\vec{\mu_2}\) have the same magnitude, i.e., \(\mu_1 = \mu_2 = \mu\). Let's call the angle between them \(\theta\), which is our X-A-X bond angle (100°). We can use the cosine rule to calculate the magnitude of the net dipole moment: \[ \mu_{total}^2 = \mu_1^2 + \mu_2^2 - 2\mu_1\mu_2\cos\theta \] After substituting \(\mu_1 = \mu_2 = \mu\), we get: \[ \mu_{total}^2 = 2\mu^2 (1 - \cos\theta) \] Now we can plug in the values. Since we are only analyzing the change in the dipole moment, we do not need the actual values of the charges and the bond length. Let \(k = \mu^2\), and \(\theta = 100^\circ\): \[ \mu_{total}^2 = 2k \left(1 - \cos 100^\circ \right) \] Hence, the dipole moment of the AX₂ molecule will be nonzero when the X-A-X bond angle is reduced to 100° and can be calculated by taking the square root of the above expression.