Hybrid Atomic Orbitals Quiz

PDF page 277: Quantum-mechanical calculations suggest why the observed bond angles in H2O differ from those predicted by the overlap of the 1s orbital of the hydrogen atoms with the 2p orbitals of the oxygen atom. The mathematical expression known as the wave function, ψ, contains information about each orbital and the wavelike properties of electrons in an isolated atom. When atoms are bound together in a molecule, the wave functions combine to produce new mathematical descriptions that have different shapes. This process of combining the wave functions for atomic orbitals is called hybridization and is mathematically accomplished by the linear combination of atomic orbitals, LCAO, (a technique that we will encounter again later). The new orbitals that result are called hybrid orbitals. The valence orbitals in an isolated oxygen atom are a 2s orbital and three 2p orbitals. The valence orbitals in an

PDF page 278: 3. A set of hybrid orbitals is generated by combining atomic orbitals. The number of hybrid orbitals in a set is

PDF page 278: equal to the number of atomic orbitals that were combined to produce the set.

PDF page 277: two hydrogen atoms. Oxygen has the electron configuration 1s 2s 2p , with two unpaired electrons (one in each of the two 2p orbitals). Valence bond theory would predict that the two O–H bonds form from the overlap of these two 2p orbitals with the 1s orbitals of the hydrogen atoms. If this were the case, the bond angle would be 90°, as shown in Figure 5.6, because p orbitals are perpendicular to each other. Experimental evidence shows that the bond angle is 104.5°, not 90°. The prediction of the valence bond theory model does not match the real-world observations of a water molecule; a different model is needed.

PDF page 277: two hydrogen atoms. Oxygen has the electron configuration 1s 2s 2p , with two unpaired electrons (one in each of the two 2p orbitals). Valence bond theory would predict that the two O–H bonds form from the overlap of these two 2p orbitals with the 1s orbitals of the hydrogen atoms. If this were the case, the bond angle would be 90°, as shown in Figure 5.6, because p orbitals are perpendicular to each other. Experimental evidence shows that the bond angle is 104.5°, not 90°. The prediction of the valence bond theory model does not match the real-world observations of a water molecule; a different model is needed.

PDF page 278: oxygen atom in a water molecule differ; they consist of four equivalent hybrid orbitals that point approximately toward the corners of a tetrahedron (Figure 5.7). Consequently, the overlap of the O and H orbitals should result in a tetrahedral bond angle (109.5°). The observed angle of 104.5° is experimental evidence for which quantummechanical calculations give a useful explanation: Valence bond theory must include a hybridization component to give accurate predictions.

PDF page 278: oxygen atom in a water molecule differ; they consist of four equivalent hybrid orbitals that point approximately toward the corners of a tetrahedron (Figure 5.7). Consequently, the overlap of the O and H orbitals should result in a tetrahedral bond angle (109.5°). The observed angle of 104.5° is experimental evidence for which quantummechanical calculations give a useful explanation: Valence bond theory must include a hybridization component to give accurate predictions.

PDF page 278: The beryllium atom in a gaseous BeCl2 molecule is an example of a central atom with no lone pairs of electrons in a linear arrangement of three atoms. There are two regions of valence electron density in the BeCl2 molecule that correspond to the two covalent Be–Cl bonds. To accommodate these two electron domains, two of the Be atom’s four valence orbitals will mix to yield two hybrid orbitals. This hybridization process involves mixing of the valence s orbital with one of the valence p orbitals to yield two equivalent sp hybrid orbitals that are oriented in a linear geometry (Figure 5.8). In this figure, the set of sp orbitals appears similar in shape to the original p orbital, but there is an important difference. The number of atomic orbitals combined always equals the number of hybrid orbitals formed. The p orbital is one orbital that can hold up to two electrons. The sp set is two equivalent orbitals that point 180° from each other. The two electrons that were originally in the s orbital are now distributed to the two sp orbitals, which are half filled. In gaseous BeCl2, these half-filled hybrid orbitals will overlap with orbitals from the chlorine atoms to form two identical σ bonds.

PDF page 278: The beryllium atom in a gaseous BeCl2 molecule is an example of a central atom with no lone pairs of electrons in a linear arrangement of three atoms. There are two regions of valence electron density in the BeCl2 molecule that correspond to the two covalent Be–Cl bonds. To accommodate these two electron domains, two of the Be atom’s four valence orbitals will mix to yield two hybrid orbitals. This hybridization process involves mixing of the valence s orbital with one of the valence p orbitals to yield two equivalent sp hybrid orbitals that are oriented in a linear geometry (Figure 5.8). In this figure, the set of sp orbitals appears similar in shape to the original p orbital, but there is an important difference. The number of atomic orbitals combined always equals the number of hybrid orbitals formed. The p orbital is one orbital that can hold up to two electrons. The sp set is two equivalent orbitals that point 180° from each other. The two electrons that were originally in the s orbital are now distributed to the two sp orbitals, which are half filled. In gaseous BeCl2, these half-filled hybrid orbitals will overlap with orbitals from the chlorine atoms to form two identical σ bonds.

PDF page 279: (yellow). Each hybrid orbital is oriented primarily in just one direction. Note that each sp orbital contains one lobe that is significantly larger than the other. The set of two sp orbitals are oriented at 180°, which is consistent with the geometry for two domains.

PDF page 279: When atomic orbitals hybridize, the valence electrons occupy the newly created orbitals. The Be atom had two valence electrons, so each of the sp orbitals gets one of these electrons. Each of these electrons pairs up with the unpaired electron on a chlorine atom when a hybrid orbital and a chlorine orbital overlap during the formation of the Be–Cl bonds. Any central atom surrounded by just two regions of valence electron density in a molecule will exhibit sp

PDF page 280: The valence orbitals of a central atom surrounded by three regions of electron density consist of a set of three sp

PDF page 280: hybrid orbitals and one unhybridized p orbital. This arrangement results from sp hybridization, the mixing of one s orbital and two p orbitals to produce three identical hybrid orbitals oriented in a trigonal planar geometry (Figure 5.10).

PDF page 280: hybrid orbitals and one unhybridized p orbital. This arrangement results from sp hybridization, the mixing of one s orbital and two p orbitals to produce three identical hybrid orbitals oriented in a trigonal planar geometry (Figure 5.10).

PDF page 280: Figure 5.10 The hybridization of an s orbital (blue) and two p orbitals (red) produces three equivalent sp hybridized

PDF page 280: orbitals (yellow) oriented at 120° with respect to each other. The remaining unhybridized p orbital is not shown here, but is located along the z axis.

PDF page 281: The observed structure of the borane molecule, BH3, suggests sp hybridization for boron in this compound. The molecule is trigonal planar, and the boron atom is involved in three bonds to hydrogen atoms (Figure 5.12). We can illustrate the comparison of orbitals and electron distribution in an isolated boron atom and in the bonded atom in BH3 as shown in the orbital energy level diagram in Figure 5.13. We redistribute the three valence electrons of the