would one consider the molecule NBr3O to be nonpolar or polar? I feel like all of the dipole moments will cancel, when you draw the VSPER structure the Br atom in the plane and out of the plane cancel, however O going up and the Br coming out of the plane, ...I feel like the Br would have a greater pull, but maybe insignificant. I know NBr3 would be polar due to the lone pair on N.

sorry to repost haha I'd hate to get this wrong

To determine if the molecule NBr3O is polar or nonpolar, we need to analyze its molecular geometry and the polarity of its bonds.

First, let's look at the Lewis structure of NBr3O:
N - Br
|
Br - O

From the Lewis structure, we can determine that the central atom, nitrogen (N), has three bonded atoms (Br) and one lone pair of electrons. This gives NBr3O a tetrahedral electron geometry.

Next, let's consider the molecular geometry using the VSEPR (Valence Shell Electron Pair Repulsion) theory. The three bonded Br atoms and the lone pair of electrons will repel each other, resulting in a trigonal pyramidal molecular geometry.

Now, let's consider the polarity of the individual bonds in NBr3O. The N-Br bonds are polar due to the difference in electronegativity between nitrogen and bromine. Bromine is more electronegative than nitrogen, which means it pulls the shared electrons towards itself, creating a partial negative charge on the bromine atom and a partial positive charge on the nitrogen atom.

Now, let's analyze the overall molecular polarity by considering the arrangement of polar bonds in the molecule. In order for a molecule to be nonpolar, all the bond dipoles must cancel each other out. However, in NBr3O, the trigonal pyramidal molecular geometry and the unequal distribution of electron density (due to the polar N-Br bonds) result in a net molecular dipole moment.

The lone pair on nitrogen and the Br-O bond contribute to the overall polarity. The Br atom coming out of the plane has a partial positive charge, and the O atom going up has a partial negative charge. This difference in charges creates an additional dipole moment.

Therefore, considering the presence of the lone pair and the unequal distribution of polar bonds, we can conclude that the molecule NBr3O is polar.