Question

Which of the following statements about hybrid orbitals is or are true? (i) After an atom undergoes sp hybridization there is one unhybridized \(p\) orbital on the atom, (ii) Under \(s p^{2}\) hybridization, the large lobes point to the vertices of an equilateral triangle, and (iii) The angle between the large lobes of \(s p^{3}\) hybrids is \(109.5^{\circ}\).

Short answer

Statement (i) is false, as there are actually two remaining unhybridized p orbitals after sp hybridization. Statements (ii) and (iii) are true, as the large lobes of sp2 hybrid orbitals form the vertices of an equilateral triangle, and the angle between the large lobes of sp3 hybrid orbitals is approximately \(109.5^{\circ}\).

Step-by-step solution

Statement (i)

The statement says that after an atom undergoes sp hybridization, there is one unhybridized p orbital on the atom. In sp hybridization, one s and one p orbital combine to form two sp hybrid orbitals. Since an atom has a total of three p orbitals, after one combines with the s orbital, there are indeed two remaining unhybridized p orbitals. Therefore, statement (i) is false.

Statement (ii)

The statement says that under sp2 hybridization, the large lobes point to the vertices of an equilateral triangle. In sp2 hybridization, one s and two p orbitals combine to form three sp2 hybrid orbitals. These orbitals arrange themselves in a trigonal planar geometry with bond angles of 120° between them, which indeed corresponds to the vertices of an equilateral triangle. Therefore, statement (ii) is true.

Statement (iii)

The statement says that the angle between the large lobes of sp3 hybrids is 109.5°. In sp3 hybridization, one s and three p orbitals combine to form four sp3 hybrid orbitals. These orbitals arrange themselves in a tetrahedral geometry with bond angles of approximately 109.5° between them. Therefore, statement (iii) is true. To summarize, statement (i) is false, while statements (ii) and (iii) are true.

Key concepts

sp Hybridization

Understanding the concept of sp hybridization is critical for grasping the structure of molecules with a linear shape. In sp hybridization, one s orbital and one p orbital from the same shell of an atom mix to form two sp hybrid orbitals. These orbitals are oriented in a straight line, 180 degrees apart. Since only one p orbital is involved, two remaining p orbitals do not participate in the hybridization and are thus available for other bonding interactions such as pi bonding.
An example of a molecule with sp hybridized orbitals is acetylene (C2H2). Each carbon atom in acetylene has two sp hybrid orbitals. One orbital forms a sigma bond with the other carbon atom, while the other forms a sigma bond with a hydrogen atom. The two remaining unhybridized p orbitals on each carbon overlap to form two pi bonds between the carbon atoms, resulting in a triple bond overall.

sp2 Hybridization

Moving along to sp2 hybridization, it's essential to note that this involves the mixing of one s orbital with two p orbitals, leading to the formation of three equivalent sp2 hybrid orbitals. These orbitals are laid out in a trigonal planar arrangement, with each orbital 120 degrees apart from the others. The unhybridized p orbital remains perpendicular to the plane of the sp2 orbitals and can engage in pi bonding.
Ethylene (C2H4) serves as an illustrative case where each carbon atom is sp2 hybridized. The three sp2 orbitals form sigma bonds with two hydrogen atoms and the other carbon atom, while the unhybridized p orbitals overlap to create a pi bond between the carbon atoms, giving rise to the double bond characteristic of alkenes.

sp3 Hybridization

The sp3 hybridization occurs when one s orbital and all three p orbitals from the same shell mix together. This blending creates four sp3 hybrid orbitals that are equivalent in energy. The spatial arrangement of these orbitals is tetrahedral, with each orbital positioned at roughly 109.5 degrees to the others. This particular geometry leads to the formation of tetrahedral molecular shapes in compounds like methane (CH4).
In methane, each sp3 hybrid orbital of the carbon atom forms a sigma bond with a hydrogen atom, which leads to a stable, tetrahedral shape. These bond angles are critical for the three-dimensional structure of molecules and strongly influence their physical and chemical properties.

Molecular Geometry

The molecular geometry of a substance dictates the three-dimensional arrangement of atoms in a molecule. It's influenced by several factors, including the number of bonds an atom forms, lone pairs of electrons, and the electronic repulsions between these electron pairs. Hybrid orbitals play a decisive role in determining the shape of molecules.
The key geometries that emerge from hybridization are linear (resulting from sp hybridization), trigonal planar (from sp2 hybridization), and tetrahedral (from sp3 hybridization). Understanding molecular geometry is critical as it affects the molecule's polarity, reactivity, color, magnetism, and biological activity.

Bond Angles

Bond angles are the angles between adjacent lines representing bonds emanating from an atom. They are pivotal in determining the shape of a molecule. These angles result from the repulsion between electrons in the bonds and lone pairs according to the VSEPR (Valence Shell Electron Pair Repulsion) theory. For example, 180-degree bond angles are characteristic of linear sp-hybridized molecules, while 120-degree angles are found in trigonal planar sp2-hybridized molecules, and 109.5-degree angles are associated with tetrahedral sp3-hybridized molecules.
Deviations from ideal bond angles can occur due to factors such as lone pair-lone pair, lone pair-bond pair, and bond pair-bond pair repulsions, as well as differences in the size of the atoms involved. Nonetheless, predicting and understanding these angles helps with visualizing molecular structures and is invaluable in explaining a molecule's geometry and properties.