 Solution

$I_{C_4}=4m_B{R}^2(1-\cos \theta )$; the central atom makes no contribution. Let $\rho$ be the radius of the square of outer atoms and $h_A$ the central atom's height above its plane. Then $R^2={\rho }^2+h_A^2$. Adjacent bond vectors have perpendicular in-plane components, soThe centre of mass lies on the fourfold axis, even when $m_A\ne m_B$. Thus the central atom and the shift of origin along the axis contribute no perpendicular lever arm. The four outer atoms each contribute $m_B{\rho }^2$, givingThe result concerns the smallest adjacent bond angle; the angle between opposite outer atoms is a different angle in a pyramid.