Boron orbital diagram

To show how orbital diagrams are obtained from electron configurations, consider the boron atom (Z = 5). Its electron configuration is ls22s22p1. The pair of electrons in the Is orbital must have opposed spins (+j, or f j). The same is true of the two electrons in the 2s orbital. There are three orbitals in the 2p sublevel. The single 2p electron in boron could be in any one of these orbitals. Its spin could be either up or down. The orbital diagram is ordinarily written... 

Following this principle, the orbital diagrams for the elements boron through neon are shown in Figure 6.10. Notice that—... 

It is readily apparent that the three a bonds are capable of holding the six bonding electrons in the a t and e molecular orbitals. The possibility of some 7r-bonding is seen in the molecular orbital diagram as a result of the availability of the a2" orbital, and in fact there is some experimental evidence for this type of interaction. The sum of the covalent radii of boron and fluorine atoms is about 152 pm (1.52 A), but the experimental B-F bond distance in BF3 is about 129.5 pm (1.295 A). Part of this "bond shortening" may be due to partial double bonds resulting from the 7r-bonding. A way to show this is by means of the three resonance structures of the valence bond type that can be shown as... 

The combination of boron and hydrogen orbitals in the three-center bond can be shown in a molecular orbital diagram as in Figure 13.2. Using this approach to bonding, the structures of some of the... 

Let s look at the ground state electron configuration and orbital diagram of Boron (5B) which is the first element of group 3A. 

Boron s fifth electron must go into the 2p energy sublevel. Since 1=1, mi may be -1, 0, or +1. The fifth electron can go into any of these orbitals, because they all have the same energy. When you draw orbital diagrams, it is customary to place the electron in the first available box, from left to right. 

A molecular orbital diagram for a three-center B-H-B bond in diborane. In this diagram, represents an atomic orbital and and represent the bonding and antibonding three-center wave functions, respectively. Note the relative energies of the atomic orbitals on boron and hydrogen atoms (ionization potentials 8.3 eV and 13.6 eV, respectively). 

Describe the bonding of the chlorine bridges in this dimer in molecular orbital terms. 8-6 BF can be obtained by reaction of BF3 with boron at 1850° C and low pressure BF is highly reactive but can be preserved at liquid nitrogen temperature (77 K). Prepare a molecular orbital diagram of BF. How would the molecular orbitals of BF differ from CO, with which BF is isoelectronic ... 

Next we will look at the BF3 (boron trifluoride) molecule, known to have planar geometry based on VSEPR. Considering only the valence electrons, the orbital diagram of... 

Fill in the following orbital diagram for a boron atom in an excited state. 

Boron, atomic number 5, occurs naturally as two isotopes, and B, with natural abundances of 19.9% and 80.1%, respectively, (a) In what ways do the two isotopes differ from each other Does the electronic configuration of differ from that of B (b) Draw the orbital diagram for an atom of B. Which electrons are the valence electrons (c) Indicate three major ways in which the Is electrons in boron differ from its 2s electrons, (d) Elemental boron reacts with fluorine to form BF3, a gas. Write a balanced chemical equation for the reaction of solid boron with fluorine gas. (e) AHf for Bp3(g) is —1135.6 kl/mol Calculate the standard enthalpy change in the reaction of boron with fluorine, (f) When BCI3, also a gas at room temperature, comes into contact with water, the two react to form hydrochloric acid and boric add, H3BO3, a very weak acid in water. Write a balanced net ionic equation for this reaction. 

Carbon s group 13 neighbor, boron, has four orbitals but only three electrons in its valence shell. For most boron compounds, the appropriate hybridization scheme combines the 2s and two 2p orbitals into three sp hybrid orbitals and leaves one p orbital unhybridized. Valence-shell orbital diagrams for this hybridization scheme for boron are shown here, and fhe scheme is further outlined in Figure 11-10. 

In the MO treatment of the D3h BF3 molecule the 2p orbital of the boron atom transforms as an a/" representation, as does one of the linear combinations of fluorine 2p, group orbitals (all F-F n bonding). The MO diagram is shown in Figure 6.6. 

We can now go on to apply the same ideas to some other simple molecules. In boron trifluoride, for example, we start with the boron atom, which has three outer-shell electrons in its normal or ground state, and three fluorine atoms, each with seven outer electrons. As shown in the upper diagram, one of the three boron electrons is unpaired in the ground state. In order to explain the trivalent bonding of boron, we postulate that the atomic s- and p orbitals in the outer shell of boron mix to form three equivalent hybrid orbitals. These particular orbitals are called sp2 hybrids, meaning that this set of orbitals is derived from one s-orbital and two p-orbitals of the free atom. 

Answer. There are a total of 30 orbitals and 26 electrons to be utilized in bonding. The external B-H bonds utihze 12 orbitals and 12 electrons leaving 18 orbitals and 14 electrons for cluster bonding. Four three-center B-B-B and three two-center B-B bonds utilize 12 + 6=18 orbitals and 8 + 6 = 14 electrons. They may be placed on the framework as shown in the diagram above where the top and the bottom of the octahedron are shown separately. By counting you can find that B(l) and B(2) are associated with three three-center-two-electron bonds (and a B-H bond), B(3) and B(4) with two three-center and one two-center bonds (and a B-H), and B(5) and B(6) with one three-center and two two-center bonds (and a B-H), i.e., eight electrons around each B. Notice that one would need to draw a considerable number of resonance structures to give all the boron atoms the same electronic environment. 

The expected MO energy-level diagram for the combination of the 2p orbitals on two boron atoms. 

In the atom of beryllium the fourth electron completes the as orbital and thus with boron the fifth electron must enter a ap orbital. The additional electrons of carbon and nitrogen each occupy a ap orbital as shown in the diagrams above. Once the three ap orbitals contain an electron each, the introduction of further electrons, as in the atoms of oxygen, fluorine and neon, bring about the completion of these orbitak. With neon the electronic shell n = 2 is completed since all the possible eight electron positions are filled and in accordance with this, the second period of the periodic classification contains eight elements. 

FIGURE 6.19 Correlation diagram for heteronuclear diatomic molecules, AB. The atomic orbitals for the more electronegative atom (B) are displaced downward because they have lower energies than those for A. The orbital filling shown is that for (boron monoxide) BO. 

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