# H2Te dipole and electronegativities of H and Te (high school level chemistry)

I saw somewhere that $\ce{H2Te}$ is a polar molecule, but as far as I know the only time dipole-dipole forces arise (giving molecules the possibility of becoming polar) are when there is a difference in electronegativity. I know $\ce{H2Te}$ is not a linear molecule but if there is almost no (less than 0.4) difference in electronegativity values the shape shouldn't matter as electrons wont be more drawn to either atom (or 'side' of the molecule). In other words, it seems to me the intermolecular forces should be induced-dipole induced-dipole (making the molecule non-polar), not dipole dipole.

I should clarify, I first saw this statement in the answers to a high school chemistry exam. I then googled it, and got the same answer (i.e that the molecule is polar).

• I think you mean $\ce{H2Te}$. Electronegativity of $\ce{H}$ is 2.2. Electronegativity of $\ce{Te}$ is 2.1. You can have Van-der- Waals forces between non-polar molecules, such as $\ce{I2}$ – Yomen Atassi Nov 27 '15 at 6:50
• sorry yes meant H2Te. Yes van der Waals forces do exist, but shouldnt they be London forces? not dipole-dipole, making the molecule polar – Rani Nov 27 '15 at 8:34
• It would be educating for you to learn about ozone dipole moment. I guess, that case of azulene would be interesting too. – permeakra Nov 27 '15 at 12:57
• BTW, hydrogen-deiterium molecule has non-zero dipole (about 0.001 D) =)/ – permeakra Nov 27 '15 at 17:36
• @Rani - what's the electronegativity of a lone pair? – Dissenter Nov 27 '15 at 18:44

Information in this question is all over the place and often out of context. Let me get a few things straight:

• Van der Waals forces, also known as London interactions, occur between every pair of molecules regardless of polarity. They are the only intermolecular force observed for non-polar molecules such as $\ce{I2}$. They are explained by induced and spontaneous dipoles at a molecular level and femtosecond timescale.

• Dipole interactions are present between macroscopic dipoles and account for the boiling point difference between e.g. tetrahydrofurane and cyclopentane ($\Delta \vartheta_\mathrm{b} \approx 15~\mathrm{^\circ C}$). For those, you need a notable difference in electronegativity resulting in an overall dipole (so $\ce{CO2}$ is not a dipole because the dipole moments of both $\ce{C=O}$ bonds cancel each other out). For strong dipoles, these interactions are orders of magnitude larger than the underlying van der Waals forces.

• Electronegativities are, of course, atomic properties (and also depend on things like oxidation state if one is truly accurate). For no two different elements is the electronegativity difference ever zero. Every $\ce{X-Z}$ bond is always polarised ever so slightly in one of the two directions. Sometimes, as in the case of $\ce{NCl3}$, this is best determined experimentally. ($\ce{N-Cl}$ bonds in $\ce{NCl3}$ are polarised towards nitrogen.)

• However, only bonds which are polarised to a certain extent (usually a cutoff value of $\approx 0.4$) are called polar on a high school level as per convention.

• $\ce{H2Te}$ has non-polar bonds as per convention of the fourth bullet point. Therefore, you should consider it a non-dipole and hence only consider van der Waals forces/London dispersion forces. (DavePhD’s source is likely correct, but the dipole moment is neglegible at high school level.)

• I would love to see a link to the pages that say $\ce{H2Te}$ be notably polar/display dipole dipole interactions. (Compare the $0.2~\mathrm{D}$ to the $1.78~\mathrm{D}$ dipole moment of ethyl acetate.) Calling van der Waals forces dipole interactions is usually considered wrong (although they build up upon the same principles: electrostatic interaction).

I tried to stick to a high-school level as well as possible. Hence a number of simplifications in this answer. At research levels, every molecule without improper rotation (of which inversion and symmetry planes are a special case) should be considered to have a non-zero dipole moment as DavePhD’s answer states. Thus, every molecule of that type will display dipole-dipole interactions however weak they may be.

• "An ab initio Calculation of the Dipole Moment Surfaces and the Vibrational Transition Moments of the H2Te Molecule" sciencedirect.com/science/article/pii/S0022285296971453 is the latest I can find, they have 0.298 D calculated and experimentally "certainly larger than 0.19 Debye and very probably smaller than 0.26 Debye" – DavePhD Nov 27 '15 at 16:59

You can indeed have induced dipole-dipole between non-polar molecules, especially between molecules/particles with a large and "easily polarizable" electron cloud.

The electron cloud is not stationary in time, but dynamic. This means that the electron cloud is subject to fluctuations in its distribution. So, at some point in time, the electron cloud will be shifted in one direction. Hence, the molecule has become slightly polar. These fluctuations take place on a very, very, very short timescale, and basically the "non-polar" molecule is not really non-polar (although it may be close).

When this slightly polar molecule approaches a non-polar one, the small dipole moment induces a shift in the other molecule's electron distribution (due to electron-electron repulsion). The slightly negative part of molecule A will be attracted to the slightly positive part of molecule B, and a lower-energy state is achieved.

This is called induced dipole-dipole interactions, and is a simple explanation of how certain melting-point trends can be observed for groups in the periodic table. $\ce{He2}$ is another example of a molecule held together by induced dipole-dipole interactions.

Here is a (not very good) picture depicting the process in forming $\ce{He2}$.

See this youtube video for an introduction to various dipole-dipole interactions (skip to 4:35 to start on the induced dipole-dipole interactions)

More generally, the symmetry of the molecule should be first considered to decide if the molecule has a dipole moment.

If the symmetry group is $C_1$ (only $E$ meaning identity symmetry operation), $C_\mathrm{s}$ (only a single plane of symmetry and $E$), $C_n$ or $C_{n\mathrm{v}}$, then think that the molecule has a dipole moment regardless of what particular atoms are involved. Someone could argue that these could still hypothetically have no dipole moment at some particular state, such as a specific rotational and vibrational state, but realistically these molecules will have a dipole moment. Then the specific elements and electronegativities could be considered to quantify the dipole moment.