Dipole moments of trans and cis nitrous acid

Question

I have a paper due about the isomers of nitrous acid, using various calculations methods to find molecular properties like their dipole moment. I've found the dipole moment of trans isomer to be 1.987 and the dipole moment of cis to be 1.458. I've also found that ALL bond angles are bigger for cis, and most bond lengths are longer for cis, other than N-O (single bond) which is longer for trans. Also if that helps, the HOMO and LUMO orbitals of trans have a bigger overlap in the N-O bond.

I know that the longer the bond, the bigger the charge separation, thus a bigger dipole moment, which doesn't seem to match my data and the literature, as the bonds are longer in cis (also checked the distance between the oxygens in both isomers). Does it have anything to do with the N-O bond being longer for trans, as the dipole moment for trans is bigger?

I'm matching my data to "Microwave Spectrum, Structure, Dipole Moment and Quadrupole Coupling Constants of cis and trans Nitrous Acids" by A. Peter Cox, Alan H. Brittain and David J. Finnigan from 1970 (The article is quite old but it's the first one I had found, and it is very similar to more recent ones).

Here are the calculated isomers:

Thanks!

Answer 1

The dipole moments are aligned in trans-nitrous acid, adding up yield a higher dipole moment than that of cis-nitrous acid.

Dipole Moment ($\mu$)

Charge and Distance

I know that the longer the bond, the bigger the charge separation, thus a bigger dipole moment.

$\mu$ constitutes two terms, charge and distance:

$$ \mu = q\times d $$

You need to account for both factors, even though the bond may be longer, the charge separation may be lesser reducing $\mu$. As an example, consider hydrogen halides:

$\ce{HX}$	Bond Length ($\pu{pm}$)	$\mu$	Source
$\ce{HF}$	94	$ \pu {1.51 D}$	1
$\ce{HCl}$	127	$ \pu {1.76 D}$	2
$\ce{HBr}$	147	$ \pu {1.73 D}$	3
$\ce{HI}$	168	$ \pu {1.55 D}$	4

Notice that even though the bond lengths are increasing, $\mu$s are not following the same trend, this is because of increasing covalent character of the bond (lesser charge separation) which is becoming prominent going down the group.

Vectorial Nature

Also note that $\mu$ is actually a vector quantity. Thus, the correct formula would be:

$$ \vec{\mu} = q\vec{d} $$

Although this is a simplification, see note at the end, if you have multiple bonds with individual $\mu_i$s:

$$ \vec{\mu} = \vec{\mu_1} + \vec{\mu_2} + \vec{\mu_3} + \cdots \tag{1} $$

When these vectors are aligned, the dipole moments add up vectorially. A classic example would be the $\ce{CF4}$, where all bonds are polar, yet $\mu$ is close to zero because of the symmetry of the molecule.

In case of cis- and trans-nitrous acid, $\mu$s are aligned for the trans isomer, adding up vectorially, and vice-versa for the cis isomer.

Structure	Polarity	$\mu$	Source
cis-nitrous acid		$\pu{1.46 D}$	5
trans-nitrous acid		$\pu{2.56 D}$	6

Note

In a molecule, charges are not localized on the atoms or within bonds, and thus, Equation 1 is a simplification. You can read more about the quantum mechanical nature of $\mu$ for a better understanding.