4 Chemical Bonding — Molecular Geometry and Hybridization: Identify Central Atom Hybridization

Question 

Identify the hybridization of the central atom in each of the following molecules and ions that contain multiple bonds:

  1. CS2
  2. Cl2CO (C is the central atom)
  3. Cl2SO (S is the central atom)
  4. SO2F2 (S is the central atom)
  5. XeO2F2 (Xe is the central atom)

 

Show/Hide Answer
  1. sp
  2. sp2
  3. sp3
  4. sp3
  5. sp3d

Refer to Section 5.3 Hybrid Atomic Orbitals (1).

Strategy Map

Do you need a little help to get started?

Check out the strategy map.

Show/Hide Strategy Map
Table 1: Strategy Map
Strategy Map Steps
1. Create a Lewis structure of the molecule.

Show/Hide Hint

Refer to Section 4.4: Lewis Symblos and Structures (2).

2. Determine the electron-pair geometry of the molecule.

Show/Hide Hint

Recall using VSEPR to determine electron-pair geometry.

Refer to Section 5.1: Molecular Structure Polarity (3).

3. Identify the hybridization of the centre atom based on the molecules’ electron-pair geometry.

Show/Hide Hint

The hybridization of the centre atom will be determined by the electron-pair geometry, not the atom itself. For example, in a linear molecule, the centre atoms hybridization will always be sp.

Solution

Do you want to see the steps to reach the answer?

Check out this solution.

Show/Hide Solution

1. CS2

16 valence electrons, C central atom

 

 

This molecule has a linear electron-pair geometry: sp hybridized

(s + p)

2. Cl2CO (C is the central atom)

24 valence electrons

 

 

This molecule has a trigonal planar electron-pair geometry: sp2 hybridized

(s + p + p)

3.Cl2SO (S is the central atom)

26 valence electrons

 

 

This molecule has a tetrahedral electron-pair geometry: sp3 hybridized

(s + p + p + p)

4. SO2F2 (S is the central atom)

32 valence electrons

 

 

This molecule has a tetrahedral electron-pair geometry: sp3 hybridized

(s + p + p + p)

5. XeO2F2 (Xe is the central atom)

34 valence electrons

 

 

This molecule has a trigonal bipyramidal electron-pair geometry: sp3d hybridized

(s + p + p + p + d)

Guided Solution

Do you want more help?

The guided solution below will give you the reasoning for each step to get your answer, with reminders and hints.

Show/Hide Guided Solution
Table 2: Guided Solution
Guided Solution Ideas
This question is a theory problem where you use your knowledge of molecular bonding and identify atomic orbital hybridization.

Show/Hide Resource

Refer to Section 5.3 Hybrid Atomic Orbitals (1).

What is the pattern associated with atomic orbital hybridization?

Show/Hide Think About This!

The hybridization of the centre atom will be determined by the electron-pair geometry, not the atom itself. For example, the centre atoms hybridization will always be sp in a linear molecule.

Number of electron domains (as determined by the number of bonding directions and lone pairs in the best Lewis structure) = Number of atomic orbitals required for hybridization.

Number of hybrid orbitals = Number of orbitals mixed.

Hybrid orbitals are named by type and number of orbitals mixed.

Show/Hide Don’t Forget!

Recall the atomic orbitals: for n=2 there is 1 ‘s’ orbital (l=0) and 3 degenerate ‘p’ orbitals (l=1).

Note that ‘d’ orbitals (l=2) can only exist when n is greater than or equal to 3.

  • 2 domains = 2 atomic orbitals must be hybridized = s + p = sp hybridization = linear electron domain geometry (EDG)
  • 3 domains = 3 atomic orbitals must be hybridized = s + p + p = sp2 hybridization = trigonal planar EDG
  • 4 domains = 4 atomic orbitals must be hybridized = s + p + p + p = sp3 hybridization = tetrahedral EDG
  • 5 domains = 5 atomic orbitals must be hybridized = s + p + p + p + d = sp3d hybridization = trigonal bipyramidal EDG
  • 6 domains = 6 atomic orbitals must be hybridized = s + p + p + p + d + d = sp3d2 hybridization = octahedral EDG
Table 3: Complete Solution
Complete Solution
CS2

16 valence electrons, C central atom

 

This molecule has a linear electron-pair geometry: sp hybridized

2. Cl2CO (C is the central atom)

24 valence electrons

 

 

This molecule has a trigonal planar electron-pair geometry : sp2 hybridized

3. Cl2SO (S is the central atom)

26 valence electrons

 

 

This molecule has a tetrahedral electron-pair geometry : sp3 hybridized

4. SO2F2 (S is the central atom)

32 valence electrons

 

 

This molecule has a tetrahedral electron-pair geometry : sp3 hybridized

5. XeO2F2 (Xe is the central atom)

34 valence electrons

 

 

This molecule has a trigonal bipyramidal electron-pair geometry: sp3d hybridized

Check Your Work

Verify that:

  • Your Lewis structure has all the valence electrons.
  • You have counted the central atom’s electron domains correctly.

The name of our electron pair geometry and number of electron domains leads us to the correct hybridization.

Does your answer make chemical sense?

Show/Hide Answer

Hybrid orbitals form due to the covalent bonding within molecules. When 2 or more atoms create a covalent bond, their atomic orbitals will overlap to form these hybrid orbitals.

In this case, we are looking at the hybridization of the centre atom to see what kind of hybridization is happening to overlap with all the atoms. When molecules grow and involve more atoms, more bonds form, and thus, more orbitals become involved. This is why a linear molecule only uses two orbitals (s + p), while an octahedral molecule uses six (s + p + p + p + d + d).

PASS Attribution

References

1. OpenStax. 5.3: Hybrid Atomic Orbitals. In CHEM1500: Chemical Bonding and Organic Chemistry. LibreTexts, 2022. https://chem.libretexts.org/Courses/Thompson_Rivers_University/CHEM1500:_Chemical_Bonding_and_Organic_Chemistry/05:_Chemical_Bonding_II-_Molecular_Geometry_and_Hybridization_of_Atomic_Orbitals/5.03:_Hybrid_Atomic_Orbitals.

2. OpenStax. 4.4: Lewis Symbols and Structures. In General Chemistry 1. LibreTexts, 2020. https://chem.libretexts.org/Courses/Nassau_Community_College/General_Chemistry_1/04%3A_Chemical_Bonding_and_Molecular_Geometry/4.04%3A_Lewis_Symbols_and_Structures.

3. OpenStax. 5.1: Molecular Structure and Polarity. In CHEM1500: Chemical Bonding and Organic Chemistry. LibreTexts, 2022. https://chem.libretexts.org/Courses/Thompson_Rivers_University/CHEM1500%3A_Chemical_Bonding_and_Organic_Chemistry/05%3A_Chemical_Bonding_II-_Molecular_Geometry_and_Hybridization_of_Atomic_Orbitals/5.01%3A_Molecular_Structure_and_Polarity.

4. Blackstock, L.; Brewer, S.; Jensen, A. PASS Chemistry Book CHEM 1510/1520; LibreTexts, 2023. https://chem.libretexts.org/Courses/Thompson_Rivers_University/PASS_Chemistry_Book_CHEM_1510%2F%2F1520.

5. Blackstock, L.; Brewer, S.; Jensen, A. 5.3: Question 5.E.59 PASS – Identify Central Atom Hybridization. In PASS Chemistry Book CHEM 1510/1520. LibreTexts, 2023. https://chem.libretexts.org/Courses/Thompson_Rivers_University/PASS_Chemistry_Book_CHEM_1500/05%3A_Chemical_Bonding_II_-_Molecular_Geometry_and_Hybridization_of_Atomic_Orbitals/5.03%3A_Question_5.E.59_PASS_-_identify_central_atom_hybridization.

6. OpenStax. 8.E: Advanced Theories of Covalent Bonding (Exercises). In Chemistry 1e (OpenSTAX). LibreTexts, 2022. https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/08%3A_Advanced_Theories_of_Covalent_Bonding/8.E%3A_Advanced_Theories_of_Covalent_Bonding_(Exercises).

7. Flowers, P.; Robinson, W. R.; Langley, R.; Theopold, K. Ch. 8 Exercises. In Chemistry. OpenStax, 2015. https://openstax.org/books/chemistry/pages/8-exercises.

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

PASSchem Copyright © by Sharon Brewer, Lindsay Blackstock, TRU Open Press is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Share This Book