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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

The lactic acid molecule, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COH},\) gives sour milk its unpleasant, sour taste. (a) Draw the Lewis structure for the molecule, assuming that carbon always forms four bonds in its stable compounds. (b) How many \(\pi\) and how many \(\sigma\) bonds are in the molecule? (c) Which CO bond is shortest in the molecule? (d) What is the hybridization of atomic orbitals around the carbon atom associated with that short bond? (e) What are the approximate bond angles around each carbon atom in the molecule?

(a) What is the physical basis for the VSEPR model? (b) When applying the VSEPR model, we count a double or triple bond as a single electron domain. Why is this justified?

From their Lewis structures, determine the number of \(\sigma\) and \(\pi\) bonds in each of the following molecules or ions: (a) \(\mathrm{CO}_{2} ;\) (b) cyanogen,\((\mathrm{CN})_{2} ;(\mathbf{c})\) formaldehyde, \(\mathrm{H}_{2} \mathrm{CO}\) (d) formic acid, HCOOH, which has one H and two O atoms attached to \(\mathrm{C}\) .

For each statement, indicate whether it is true or false. (a) In order to make a covalent bond, the orbitals on each atom in the bond must overlap. (b) A p orbital on one atom cannot make a bond to an s orbital on another atom. (c) Lone pairs of electrons on an atom in a molecule influence the shape of a molecule. (d) The 1 s orbital has a nodal plane. (e) The \(2p\) orbital has a nodal plane.

Ethyl acetate, \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2},\) is a fragrant substance used both as a solvent and as an aroma enhancer. Its Lewis structure is (a) What is the hybridization at each of the carbon atoms of the molecule? (b) What is the total number of valence electrons in ethyl acetate? (c) How many of the valence electrons are used to make \(\sigma\) bonds in the molecule? (d) How many valence electrons are used to make \(\pi\) bonds? (e) How many valence electrons remain in nonbonding pairs in the molecule?

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