Chapter 4: Problem 42
Which of the following statement is true? (a) Hybridization of the central atom in \(\mathrm{NH}_{3}\) and \(\mathrm{CH}_{4}\) is \(\mathrm{sp}^{2}\) (b) \(\mathrm{BeCl}_{2}\) has "V" shape while \(\mathrm{SO}_{2}\) is linear (c) \(\mathrm{SF}_{6}\) is octahedral and \(\mathrm{F}-\mathrm{S}-\mathrm{F}\) bond angle is \(90^{\circ}\) (d) \(\mathrm{CO}_{2}\) has dipole moment
Short Answer
Step by step solution
Understanding the Options
Analyze Option (a)
Analyze Option (b)
Analyze Option (c)
Analyze Option (d)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Hybridization
- For example, in ammonia (\(\mathrm{NH}_3\)), the central nitrogen atom has three bonding pairs and one lone pair, resulting in \(\mathrm{sp}^3\) hybridization.
- This means four orbitals mix to form four equivalent hybrids that explain the molecule's 3D shape.
- Similarly, methane (\(\mathrm{CH}_4\)) has four bonding pairs and no lone pairs, yielding \(\mathrm{sp}^3\) hybridization, where four equivalent \(\mathrm{sp}^3\) orbitals align in a tetrahedral shape.
Molecular Shape
- For instance, beryllium chloride (\(\mathrm{BeCl}_2\)) is linear because beryllium forms only two bonds, with no lone pairs on the central atom, allowing a 180-degree separation.
- Sulfur dioxide (\(\mathrm{SO}_2\)), in contrast, exhibits a "V" shape or bent geometry due to the presence of lone pairs on sulfur that repel bonding pairs, compressing the bond angle.
- Molecular shapes such as linear, bent, tetrahedral, trigonal planar, and octahedral are among the basics you'll encounter when exploring molecular geometry.
Dipole Moment
- For example, carbon dioxide (\(\mathrm{CO}_2\)) has polar bonds; however, because of its linear shape, the dipoles from the two \(\mathrm{C=O}\) bonds cancel each other out, resulting in no overall dipole moment.
- This is why \(\mathrm{CO}_2\) is considered a non-polar molecule despite having polar bonds.
Bond Angles
- In sulfur hexafluoride (\(\mathrm{SF}_6\)), the octahedral geometry creates bond angles of \(90^{\circ}\) between the \(\mathrm{F}-\mathrm{S}-\mathrm{F}\) bonds.
- This arrangement ensures the six fluorine atoms are as far apart as possible, minimizing repulsion.
- On the other hand, a \(\mathrm{CH}_4\) molecule with \(\mathrm{sp}^3\) hybridization has bond angles of \(109.5^{\circ}\).