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I am confused about positive, negative and zero overlaps.

Do they represent the extent of overlap or are they related to the bonding and anti-bonding orbitals as described in Molecular Orbital Theory?

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  • $\begingroup$ Where do you get these? Mention the resource or the book where you got that & elaborate it. $\endgroup$ – user5764 Oct 14 '15 at 16:27
  • $\begingroup$ Page 7 @ chem.tamu.edu/rgroup/dunbar/Teaching/CHEM362/Lecture_7.pdf $\endgroup$ – the_random_guy42 Oct 14 '15 at 16:29
  • $\begingroup$ I think by positive overlap, constructive interference between the atomic orbitals may've been referred; negative overlap to anti-bonding that is, destructive interference & zero-overlap to zero-bonding`. $\endgroup$ – user5764 Oct 14 '15 at 16:34
  • $\begingroup$ Related: What are overlap integrals? $\endgroup$ – orthocresol Dec 1 '16 at 21:55
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The terms in the lecture slides you refer to describe the situation in a certain, very specific subset of molecular orbital theory: the idea of linear combining single atomic orbitals to form bonding and antibonding orbitals. While one can arrive at the same bonding/antibonding orbitals with elaborate molecular orbital theory through the localisation of orbitals approach, the simplified view these lecture slides (and many others) present is not consistent with molecular orbital theory. (However, a subset of the result can be obtained by a proper symmetry analysis.)

Let’s consider a rather simple case such as $\ce{HCN}$. It is obvious that the carbon atom’s atomic orbitals will in some way interact with the atomic orbitals of both hydrogen and nitrogen to form bonds. In the method presented in the lecture slides, you would look at certain pairs of orbitals and then determine what type of interaction would be between. I will only present a few representative examples. For consistency, I will define the molecule axis to be my coordinate system’s $z$ axis.

  1. The $2\mathrm p_z$ orbital of carbon and the $2\mathrm p_z$ orbital of nitrogen.

    These two p-type orbitals are aligned in the same way. Without taking their phases into account, we can see that the $-z$ lobe of the nitrogen orbital and the $+z$ lobe of the carbon orbital can interact. Depending on whether the phases match or do not match, we would arrive at a bonding and an antibonding orbital, respectively. The resulting bonding orbitals would be a $\sigma$ and $\sigma^*$ orbital.

  2. The $2\mathrm p_x$ orbital of nitrogen and the $2\mathrm p_x$ orbital of carbon.

    These orbitals are parallel and still well aligned. However, at any one time both lobes of one will interact with both lobes of the other. There is a nodal plane that completely contains the bond axis. If the $+x$ lobe of both orbitals feature the same phase, the interaction is bonding. Otherwise it is again antibonding. The bonding orbitals are $\pi$ and $\pi^*$, respectively.

  3. The $2\mathrm p_x$ orbital of carbon and the $1\mathrm s$ orbital of hydrogen.

    Now, we are trying to mix an orbital that only has one phase (the s orbital) and one that has two phases (the p orbital). This would be okay if only one of the two phases of the p orbital were pointing towards the s orbital; however, because the p orbital is perpendicular to the bond axis, both its phases would have to interact at the same time. Thus, we have one interaction on the $+x$ half of the coordinate system which is bonding and at the same time an interaction on the $-x$ half which is antibonding. Mathematically, the bonding and antibonding contributions cancel out perfectly. Thus, there is no interaction between the two orbitals. This is zero overlap.

  4. The $2\mathrm p_x$ orbital of carbon and the $2\mathrm p_y$ orbital of nitrogen.

    Now, we are attempting to mix two different p orbitals. However, they are not oriented in the same direction. The $+x$ direction of one orbital and the $+y$ direction of the other have the same phase while the $-y$ and $-x$ directions have the other. If we start at $+x$ and go around in a circle, we will start in an area of no interaction (the $\mathrm p_y$ orbital has a nodal plane in the $xz$ plane). Then, turning towards $+y$ we reach an area of bonding overlap. After crossing $45°$, the bonding overlap decreases until we reach zero again at $+y$. From there to $-x$, we experience an increasing and decreasing antibonding interaction. We then rinse and repeat what we just saw for the remaining $180°$.

    It would probably not be impossible to consider this interaction a curly bond that rotates around the central axis by $90°$ between the two atoms. However, traditional analysis would note that at the same time there is something like an antibonding second part displaced by $90°$ to this. Therefore, the analysis would consider these two orbitals nonbonding with zero overlap.

In a proper MO or LCAO approach, molecular orbitals will stretch across the entire molecule. Thus, the analysis of whether an orbital is bonding or antibonding with respect to a certain bond reduces itself to the question of ‘does this orbital feature a nodal plane along the bond axis or does it not?’. From the $+,-,0$ picture we had above, we arrive at a black and white picture. This is because nonbonding interactions are typically separated out before the molecular orbitals are even calculated by a symmetry analysis. If you do that, you would realise that the $2\mathrm px$ orbital and the $1\mathrm s$ mentioned in case three are representative of two different irreducible representations. (Strictly speaking, $2\mathrm p_x$ and $2\mathrm p_y$ both belong to the same degenerate irreducible representation.) Only orbitals that belong to the same irreducible representation can be combined in bonding or antibonding orbitals.

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The atoms combine by colliding with each other. But what does this mean on atomic level. This situation refers to the process in which the two atoms comes so close to each other that they penetrate each other’s orbital and form a new hybridized orbital where the bonding pair of electrons reside. This hybridized orbital has lower energy than the atomic orbital and hence are stable. It is in the minimum energy state. This partial penetration of orbital is known as orbital overlap.

The extent of overlap depends on the two participating atoms, their size and the valence electrons. In general, greater the overlap, stronger is the bond formed between the two atoms. Thus, according to the orbital overlap concept, atoms combine together by overlapping their orbital and thus forming a lower energy state where their valence electrons with opposite spin, pair up to form covalent bond.

The importance of orbital overlap was emphasized by Linus Pauling while explaining the molecular bond angles observed through experimentation and is the basis for the concept of orbital hybridization.

Directional Properties of bond

The molecular bond angles were explained through the directional properties of bond. The molecule of hydrogen is formed by overlap of 1s orbital in head on collision.

Overlapping of Atomic Orbital

When two atoms come in contact with each other to form a bond, their overlap can be positive, negative or even zero depending upon the phase and sign of the two interacting orbital.

Positive Overlapping of Atomic Orbital – When the phase of two interacting orbital is same, then the overlap is positive and in this case, the bond is formed. The phase of the two interacting orbital (+ or -) comes from the sign of orbital wave function and is not related to the charge in any sense.

Negative Overlapping of Atomic Orbital – When the phase of two interacting atomic orbital is opposite, then the overlap is negative and in this case, the bond is not formed.

Zero Overlapping of Atomic Orbital – When the orientation of two interacting atomic orbital is such that there is no overlapping of the orbital, that is known as zero overlapping.

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    $\begingroup$ That is very simplistic and also part wrong. It is not true that more overlap means stronger bonding. There is always a careful balance, as electron election repulsion is also increasing. $\endgroup$ – Martin - マーチン Dec 23 '17 at 11:37

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