Final answer:
Experiments have indeed demonstrated the electric field of polar molecules such as hydrogen fluoride (HF) through electrostatic potential maps, displaying the pronounced dipole due to the large electronegativity difference between hydrogen and fluorine.
Step-by-step explanation:
The question revolves around the polar nature of molecules and specifically addresses the measurement of the electric field generated by polar molecules like hydrogen fluoride (HF). Experiments and computer simulations have indeed illustrated the polarity of HF by visualizing the electron distribution around its covalent bond. These experiments use electrostatic potential maps, which are color-coded to indicate regions of electron density, thus revealing the polarity.
In the case of HF, a significant difference in electronegativity between hydrogen (2.2) and fluorine (4.0) results in a pronounced dipole, with the end closer to fluorine being partially negative and the end closer to hydrogen being partially positive.
An electrostatic potential map for HF shows this polarity with colors varying from blue to red, where blue indicates low electron density (partial positive charge), and red indicates high electron density (partial negative charge). Moreover, the application of an electric field causes polar molecules like HF to align with the dipoles in the direction of the field. The profound electronegativity difference between the hydrogen and fluorine atoms in HF leads to this clear polarization, and HF is thus considered a very polar molecule.