Chapter 9: Problem 8
Specify the electron-pair and molecular geometry for each underlined atom in the following list. Describe the hybrid orbital set used by this atom in each molecule or ion. $$\begin{aligned} &\text { (a) } \underline{\mathrm{CSe}_{2}}\\\ &\text { (b) } \underline{\mathbf{S O}_{2}} \end{aligned}$$ (c) \(\underline{\mathrm{CH}_{2} \mathrm{O}}\) (d) \(\underline{\mathrm{NH}}_{4}^{+}\)
Short Answer
Step by step solution
Identify the central atom in CSe2
Determine the electron-pair geometry for CSe2
Assign the hybridization for C in CSe2
Identify the central atom in SO2
Determine the electron-pair geometry for SO2
Determine the molecular geometry for SO2
Assign the hybridization for S in SO2
Identify the central atom in CH2O
Determine the electron-pair geometry for CH2O
Determine the molecular geometry for CH2O
Assign the hybridization for C in CH2O
Identify the central atom in NH4+
Determine the electron-pair geometry for NH4+
Determine the molecular geometry for NH4+
Assign the hybridization for N in NH4+
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Electron-Pair Geometry
To determine electron-pair geometry, we focus on the number of electron groups surrounding a central atom. Each single, double, or triple bond counts as one electron group, as does each lone pair.
- In carbon diselenide (CSe\(_2\)), the carbon atom has two single bonds and no lone pairs, resulting in a linear electron-pair geometry.
- In sulfur dioxide (SO\(_2\)), sulfur has two double bonds with the oxygens and one lone pair, which leads to a trigonal planar electron-pair geometry overall.
- For formaldehyde (CH\(_2\)O), carbon has three bonding groups (two single bonds with hydrogens and one double bond with oxygen), producing a trigonal planar electron-pair geometry.
- In ammonium ion (NH\(_4^+\)), nitrogen has four single bonds to hydrogens, giving a tetrahedral electron-pair geometry.
Hybridization
Each type of hybridization corresponds to a different arrangement of electron pairs around a central atom. This helps in understanding the geometrical structure of a molecule.
- In CSe\(_2\), the central carbon atom is sp-hybridized, allowing it to maintain a linear shape with two electron domains.
- Sulfur in SO\(_2\) is sp\(^2\)-hybridized. This accounts for the trigonal planar electron-pair geometry due to three regions of electron density.
- In CH\(_2\)O, the carbon exhibits sp\(^2\) hybridization, with three groups of electron density keeping the trigonal planar shape.
- For NH\(_4^+\), nitrogen is sp\(^3\)-hybridized, suitable for four single bonds surrounding the central atom with a tetrahedral electron-pair geometry.
Central Atom Identification
In the given molecules:
- In CSe\(_2\), the central atom is carbon (C) because it connects equally with the two selenium atoms.
- In the example of SO\(_2\), sulfur (S) is the central atom, forming bonds with two oxygen atoms and holding a lone pair.
- For CH\(_2\)O, the central atom is carbon (C) due to its ability to form multiple bonds, including a double bond with oxygen and single bonds with hydrogen atoms.
- In NH\(_4^+\), nitrogen (N) serves as the central atom, connecting to four hydrogens in the tetrahedral ion.
Chemical Bonding
For our discussion about molecular geometries:
- CSe\(_2\) involves covalent bonding, with carbon sharing electrons equally with selenium, forming linear bonds due to its electron arrangement.
- In SO\(_2\), sulfur forms covalent bonds with oxygen. The molecule bends due to the lone pair effect, shifting the angle from a perfect trigonal planar.
- CH\(_2\)O's carbon forms covalent bonds with hydrogen and a double bond with oxygen. Its trigonal planar shape aligns with the sp\(^2\) hybridization.
- NH\(_4^+\) sees nitrogen form four covalent bonds with hydrogen, resulting in a symmetrical tetrahedral ion with no lone pairs.