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Chapter 10: Problem 50

Because both tin and carbon are members of Group \(4 \mathrm{~A}(14),\) they form structurally similar compounds. But tin exhibits a greater variety of structures because it forms several ionic species. Predict the shapes and ideal bond angles, including any deviations: (a) \(\operatorname{Sn}\left(\mathrm{CH}_{3}\right)_{2}\) (b) \(\mathrm{SnCl}_{3}^{-}\) (c) \(\operatorname{Sn}\left(\mathrm{CH}_{3}\right)_{4}\) (d) \(\mathrm{SnF}_{5}^{-}\) (e) \(\mathrm{SnF}_{6}^{2-}\)

Short Answer

Expert verified
(a) Bent, <120º (b) Trigonal pyramidal, ~107º (c) Tetrahedral, 109.5º (d) Square pyramidal, ~90º (e) Octahedral, 90º

Step by step solution

01

Understand VSEPR theory

The Valence Shell Electron Pair Repulsion (VSEPR) theory helps predict the shapes and bond angles of molecules based on electron pairs repelling each other.
02

Electron domain count for each molecule

Determine the number of bonding and lone pairs of electrons around the central tin (Sn) atom.
03

Shape and bond angles of \(\text{Sn}(\text{CH}_3)_2\)

Sn has 2 bonding pairs and 1 lone pair. VSEPR yields a bent shape with bond angles slightly less than 120º due to the lone pair.
04

Shape and bond angles of \(\text{SnCl}_3^-\)

Sn has 3 bonding pairs and 1 lone pair. VSEPR yields a trigonal pyramidal shape with bond angles around 107º due to the lone pair repulsion.
05

Shape and bond angles of \(\text{Sn}(\text{CH}_3)_4\)

Sn has 4 bonding pairs and no lone pairs. VSEPR yields a tetrahedral shape with ideal bond angles of 109.5º.
06

Shape and bond angles of \(\text{SnF}_5^-\)

Sn has 5 bonding pairs and 1 lone pair. VSEPR yields a square pyramidal shape with bond angles of approximately 90º.
07

Shape and bond angles of \(\text{SnF}_6^{2-}\)

Sn has 6 bonding pairs and no lone pairs. VSEPR yields an octahedral shape with ideal bond angles of 90º.

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Key Concepts

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

Molecular Geometry
Molecular Geometry is the three-dimensional arrangement of atoms within a molecule. It fundamentally affects the molecule’s properties and reactivity. Using the Valence Shell Electron Pair Repulsion (VSEPR) theory, we can predict the molecular geometry by considering the repulsions between electron pairs around a central atom. For example,
  1. Sn(CH₃)₂ has a bent geometry due to 2 bonding pairs and 1 lone pair.
  2. SnCl₃⁻ adopts a trigonal pyramidal structure because of 3 bonding pairs and 1 lone pair.
  3. Sn(CH₃)₄ forms a tetrahedral geometry with its 4 bonding pairs.
  4. SnF₅⁻ takes on a square pyramidal shape with 5 bonding pairs and 1 lone pair.
  5. SnF₆²⁻ results in an octahedral structure with 6 bonding pairs and no lone pairs.
Using this systematic approach, predicting molecular geometry becomes intuitive.
Bond Angles
Bond Angles are the angles between adjacent bonds at an atom within a molecule. They are key to determining the shape of a molecule. VSEPR theory helps us understand these angles by highlighting how electron pairs repel each other to adopt orientations that minimize repulsion.
For instance:
  • In Sn(CH₃)₂, the bond angle is slightly less than 120º due to the lone pair repulsion causing a bent shape.
  • SnCl₃⁻ features bond angles around 107º, typical of a trigonal pyramidal configuration with lone pair influence.
  • Sn(CH₃)₄ showcases ideal bond angles of 109.5º in a tetrahedral geometry.
  • SnF₅⁻ has bond angles of approximately 90º due to its square pyramidal shape.
  • SnF₆²⁻ demonstrates ideal bond angles of 90º in an octahedral arrangement.
Bond angles are crucial as they directly affect the molecule’s spatial distribution and physical properties.
Electron Pair Repulsion
Electron Pair Repulsion is the core idea behind VSEPR theory. It involves the repulsive force between electron pairs around the central atom, which dictates the spatial arrangement of the molecule to minimize this repulsion. This theory considers both bonding pairs and lone pairs.
Step-by-step analysis:
  • Sn(CH₃)₂ has 2 bonding pairs and 1 lone pair, causing a bent shape.
  • SnCl₃⁻ contains 3 bonding pairs and 1 lone pair, leading to a trigonal pyramidal shape.
  • Sn(CH₃)₄ possesses 4 bonding pairs with no lone pairs, forming a tetrahedral shape.
  • SnF₅⁻ includes 5 bonding pairs and 1 lone pair, resulting in square pyramidal geometry.
  • SnF₆²⁻ has 6 bonding pairs without lone pairs, forming an octahedral shape.
By understanding electron pair repulsion, we can easily predict the overall shape and bond angles of a molecule. Knowing this, students can solve VSEPR-related problems more effectively.

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Most popular questions from this chapter

Draw a Lewis structure for a resonance form of each ion with the lowest possible formal charges, show the charges, and give oxidation numbers of the atoms: (a) \(\mathrm{BrO}_{3}^{-} ;\) (b) \(\mathrm{SO}_{3}^{2-}\)

Cryolite \(\left(\mathrm{Na}_{3} \mathrm{AlF}_{6}\right)\) is an indispensable component in the electrochemical production of aluminum. Draw a Lewis structure for the \(\mathrm{AlF}_{6}^{3-}\) ion.

Perchlorates are powerful oxidizing agents used in fireworks, flares, and the booster rockets of space shuttles. Lewis structures for the perchlorate ion \(\left(\mathrm{ClO}_{4}^{-}\right)\) can be drawn with all single bonds or with one, two, or three double bonds. Draw each of these possible resonance forms, use formal charges to determine the more important structure, and calculate its average bond order.

These species do not obey the octet rule. Draw a Lewis structure for each, and state the type of octet-rule exception: (a) \(\mathrm{BrF}_{3}\) (b) \(\mathrm{ICl}_{2}^{-}\) (c) \(\mathrm{BeF}_{2}\)

How can a molecule with polar covalent bonds not be polar? Give an example.
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