Problem 13
Question
Indicate which of the following molecules and ions are linear, which are planar, and which are neither. Then propose hybridization schemes for the central atoms. (a) \(\mathrm{Cl}_{2} \mathrm{C}=\mathrm{CCl}_{2} ;(\mathrm{b}) \mathrm{N} \equiv \mathrm{C}-\mathrm{C} \equiv \mathrm{N} ;(\mathrm{c}) \mathrm{F}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{N}\) (d) \([\mathrm{S}-\mathrm{C} \equiv \mathrm{N}]^{-}\)
Step-by-Step Solution
Verified Answer
(a) Planar, sp2 hybridized central atom. (b) Linear, sp hybridized central atom. (c) Neither, sp3 hybridized for the first central atom and sp for the second. (d) Linear, sp hybridized central atom.
1Step 1: Determine Geometry of Molecules
Analyze the structure of given molecules or ions in each part of the problem. (a) \(\mathrm{Cl}_{2} \mathrm{C}=\mathrm{CCl}_{2}\): The central atom is a carbon represented by \(C=\). This molecule is planar due to the sp2 hybridization. (b) \(\mathrm{N} \equiv \mathrm{C}-\mathrm{C} \equiv \mathrm{N}\): Here, the molecule is linear with the central atom being carbon, due to sp hybridization. (c) \(\mathrm{F}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{N}\): We have two central atoms here the first being an sp3 hybridized carbon making the molecule 'neither' planar nor linear. The second carbon atom is sp hybridized and forms a linear portion of the molecule. (d) \([\mathrm{S}-\mathrm{C} \equiv \mathrm{N}]^{-}\): The molecule is linear with sp hybridization.
2Step 2: Propose Hybridization Scheme
Determine the hybridization type for each central atom. (a) The central carbon facilitating a double bond with another carbon and single bonds with chlorine atoms is sp2 hybridized. (b) The central carbon atom linked with two nitrogen atoms with three bond pairs is sp hybridized. (c) The first central carbon atom linked with three fluorine atoms and another carbon atom is sp3 hybridized. The second carbon atom linked to the first carbon and nitrogen atom with three bond pairs is sp hybridized. (d) The central carbon linked to sulfur and nitrogen with three bond pairs is sp hybridized.
Key Concepts
HybridizationLinear MoleculesPlanar MoleculesGeometry Determination
Hybridization
Hybridization is a concept in chemistry that helps explain how atomic orbitals mix to form new hybrid orbitals. These hybrid orbitals determine the geometry of a molecule. For example, in the molecule \( \mathrm{Cl}_{2} \mathrm{C}=\mathrm{CCl}_{2} \), the carbon atoms are \( sp^2 \) hybridized. This means that one s orbital and two p orbitals combine to form three equivalent \( sp^2 \) orbitals.
The hybridization type dictates the molecular shape.
The hybridization type dictates the molecular shape.
- \( sp \) hybridization typically results in a linear shape.
- \( sp^2 \) hybridization leads to a trigonal planar shape.
- \( sp^3 \) hybridization forms a tetrahedral shape.
Linear Molecules
Linear molecules are those in which the atoms are arranged in a straight line. This often results from \( sp \) hybridization. For example, in \( \mathrm{N} \equiv \mathrm{C}-\mathrm{C} \equiv \mathrm{N} \), both carbons are \( sp \) hybridized, making the molecule linear.
A molecule is linear if the angle between adjacent bonds around the central atom is 180 degrees. This straight arrangement can involve:
A molecule is linear if the angle between adjacent bonds around the central atom is 180 degrees. This straight arrangement can involve:
- Triple bonds, as seen in acetylene (\( \mathrm{C}_2 \mathrm{H}_2 \))
- A mix of single and triple bonds
Planar Molecules
Planar molecules have all their atoms in a single plane, often due to \( sp^2 \) hybridization. The molecule \( \mathrm{Cl}_{2} \mathrm{C}=\mathrm{CCl}_{2} \) is a good example of this. Its geometry is described as trigonal planar because the \( sp^2 \) hybridization forms angles of 120 degrees.
Such planar arrangements affect the molecule’s reactivity and physical properties.
Such planar arrangements affect the molecule’s reactivity and physical properties.
- Common in benzene rings or other aromatic systems
- Provide stability through conjugated systems
Geometry Determination
Determining the geometry of a molecule involves analyzing the arrangement of atoms and the types of bonds present. By understanding the hybridization of the central atom, one can predict the molecular shape. For example:
- Identify the hybridization of the central atom, such as \( sp, sp^2, \) or \( sp^3 \).
- Use the hybridization to predict the geometric arrangement.
- Consider resonance and multiple bonds for accurate geometry.
Other exercises in this chapter
Problem 11
Match each of the following species with one of these hybridization schemes: \(s p, s p^{2}, s p^{3}, s p^{3} d, s p^{3} d^{2} .\) (a) \(\mathrm{PF}_{6}^{-}\) (
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Propose a hybridization scheme to account for bonds formed by the central carbon atom in each of the following molecules: (a) hydrogen cyanide, HCN; (b) methyl
View solution Problem 15
Write Lewis structures for the following molecules, and then label each \(\sigma\) and \(\pi\) bond. (a) \(\mathrm{HCN} ;\) (b) \(\mathrm{C}_{2} \mathrm{N}_{2}\
View solution Problem 16
Represent bonding in the carbon dioxide molecule, \(\mathrm{CO}_{2},\) by \((\mathrm{a})\) a Lewis structure and \((\mathrm{b})\) the valencebond method. Identi
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