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

Expert verified
Statement (i) is false as there are two unhybridized p orbitals after sp hybridization. Statements (ii) and (iii) are both true, as the large lobes in \( s p^{2} \) hybridization point to the vertices of an equilateral triangle, and the angle between the large lobes of \( s p^{3} \) hybrids is \( 109.5^{\circ} \).

Step by step solution

01

Statement (i) Evaluation

The given statement is: After an atom undergoes sp hybridization, there is one unhybridized p orbital on the atom. In sp hybridization, one s and one p orbital from the atom combine to form two new sp hybrid orbitals. Since we initially had three p orbitals and only one of them combined with the s orbital, there will still be two unhybridized p orbitals on the atom. So, the given statement is incorrect.
02

Statement (ii) Evaluation

The given statement is: Under \( s p^{2} \) hybridization, the large lobes point to the vertices of an equilateral triangle. In \( s p^{2} \) hybridization, one s orbital and two p orbitals from the atom combine to form three new sp2 hybrid orbitals. These hybrid orbitals will have a trigonal planar geometry, with angles of 120° between them. The large lobes will indeed point to the vertices of an equilateral triangle. So, the given statement is correct.
03

Statement (iii) Evaluation

The given statement is: The angle between the large lobes of \( s p^{3} \) hybrids is \( 109.5^{\circ} \). In \( s p^{3} \) hybridization, one s orbital and three p orbitals from the atom combine to form four new sp3 hybrid orbitals. These hybrid orbitals will have a tetrahedral geometry, with angles of \( 109.5^{\circ} \) between them. The large lobes will point to the vertices of a tetrahedron. So, the given statement is correct. In conclusion, statement (i) is false, while statements (ii) and (iii) are true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

sp Hybridization
When we delve into the concept of sp hybridization, it unveils a fascinating aspect of chemical bonding where one s and one p orbital merge to form two equivalent orbitals known as sp hybrids. This process is akin to reshuffling a deck of cards to create a new hand, with the goal to allow atoms to form stronger covalent bonds.

Visualize the sp hybrid orbitals as two vast lobes extending in opposite directions, forming a linear geometry with a bond angle of 180°. It's essential to highlight a common misunderstanding — after undergoing sp hybridization, an atom will possess not one, but two unhybridized p orbitals. These remaining p orbitals are available for further bonding such as pi bonds in multiple-bonded systems like ethyne, commonly referred to as acetylene.
sp2 Hybridization
Now, let us turn to sp2 hybridization. This is where the dance of electrons comes into play with one s and two p orbitals getting into the groove. By assimilating and rebalancing themselves, they form three equivalent sp2 hybrid orbitals.

The result is a molecule flaunting a flat, triangular layout, also known as trigonal planar, with each bond making a 120° angle with its neighbors. The geometry is as symmetrical as the points of an equilateral triangle, optimizing the distribution of electron clouds and thus stabilizing the molecule. This form of hybridization is a key player in structures like ethene (ethylene), showcasing a double bond structure with a sigma and pi bond.
sp3 Hybridization
Moving over to sp3 hybridization, we witness an atom extending its bonding reach even further. By combining one s and three p orbitals, it gives rise to four sp3 hybrid orbitals, each equally adept at forming covalent bonds.

The orbited molecules take on a tetrahedral structure, resembling a pyramid with a triangular base. Within this geometric arrangement, the bond angles are precisely 109.5°, striking a perfect balance that minimizes electron repulsion. This tetrahedral model forms the backbone of many organic molecules, including ethane and methane, reinforcing the foundations of organic chemistry.
Orbital Geometry
The term orbital geometry refers to the spatial arrangement of the hybrid orbitals around a central atom. It's the spatial consideration of where these newly mixed 'hands' of electrons are placed for maximum efficiency and minimal repulsion.

The directional attributes of hybrid orbitals give us clues to how atoms bond and the shape molecules adopt. Whether it's the 180° apart hands of sp hybridization, the 120° spaced arms of sp2 hybridization, or the four corners of a tetrahedron in sp3 hybridization, the blueprint of each tells a tale of molecular architecture that dictates the physical and chemical properties of the substance.
Molecular Geometry
Molecular geometry, unlike orbital geometry, is concerned with the arrangement of the actual atoms in a molecule. Picture it as the framework that results when the bones (orbitals) come together, determined by the vertices where the atoms reside, and the flesh (electron pairs) fill out the structure.

Atoms bonded through sp3 hybridization form a tetrahedral molecular geometry, yet if there are lone pairs involved, the geometry transforms — creating bent or trigonal pyramidal shapes instead. These forms are central to understanding why water is a polar molecule and why ammonia has a distinct odor. They're the sculptors of molecules, carving out the final form that defines function.

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Most popular questions from this chapter

Molecules that are brightly colored have a small energy gap between filled and empty electronic states (the HOMO-LUMO gap; see Exercise 9.104 ). Sometimes you can visually tell which HOMO-LUMO gap is larger for one molecule than another. Suppose you have samples of two crystalline powders-one is white, and one is green. Which one has the larger HOMO-LUMO gap?

Antibonding molecular orbitals can be used to make bonds to other atoms in a molecule. For example, metal atoms can use appropriate \(d\) orbitals to overlap with the \(\pi_{2 p}^{\star}\) orbitals of the carbon monoxide molecule. This is called \(d-\pi\) backbonding. (a) Draw a coordinate axis system in which the \(y\) -axis is vertical in the plane of the paper and the \(x\) -axis horizontal. Write \(^{4} \mathrm{M}^{\prime \prime}\) at the origin to denote a metal atom. (b) Now, on the \(x\) -axis to the right of M, draw the Lewis structure of a CO molecule, with the carbon nearest the M. The CO bond axis should be on the \(x\) -axis. (c) Draw the CO \(\pi_{2 p}^{*}\) orbital, with phases (see the "Closer Look" box on phases) in the plane of the paper. Two lobes should be pointing toward M. (d) Now draw the \(d_{x y}\) orbital of \(\mathrm{M},\) with phases. Can you see how they will overlap with the \(\pi_{2 p}^{\star}\) orbital of CO? (e) What kind of bond is being made with the orbitals between \(\mathrm{M}\) and \(\mathrm{C}, \sigma\) or \(\pi ?(\mathrm{f})\) Predict what will happen to the strength of the CO bond in a metal-CO complex compared to CO alone.

(a) If the valence atomic orbitals of an atom are sp hybridized, how many unhybridized \(p\) orbitals remain in the valence shell? How many \(\pi\) bonds can the atom form? (b) Imagine that you could hold two atoms that are bonded together, twist them, and not change the bond length. Would it be easier to twist (rotate) around a single \(\sigma\) bond or around a double \((\sigma\) plust (rotate) around a single \(\sigma\) bond same?

In which of these molecules or ions does the presence of nonbonding electron pairs produce an effect on molecular shape? (a) \(\operatorname{SiH}_{4,}(\mathbf{b}) \mathrm{PF}_{3},(\mathbf{c}) \mathrm{HBr},(\mathbf{d}) \mathrm{HCN},(\mathbf{e}) \mathrm{SO}_{2}\)

Describe the bond angles to be found in each of the following molecular structures: (a) trigonal planar, (b) tetrahedral, (c) octahedral, (d) linear.

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