In the previous post, we talked about the hybridization of atomic orbitals, so this practice set is to master to concepts of determining whether an atom is sp, sp2, sp3, sp3d, or sp3d2 hybridized.

#### Practice

Determine the **hybridization**, the **electron- and molecular geometry** of the following molecules.

(a) BF_{3} |
(b) CH_{2}O |
(c) HCN | (d) BeCl_{2} |
(e) CH_{2}Cl_{2} |

(f) SOCl_{2} |
(g) SO_{2} |
(h) PCl_{5} |
(i) XeO_{4} |
(j) NCl_{3} |

(k) SiCl_{4} |
(l) SF_{2} |
(m) H_{2}S |
(n) SO_{3} |
(o) COCl_{2} |

(p) PCl_{3} |
(q) OF_{2} |
(r) BrF_{5} |
(s) N_{2}O |
(t) SF_{6} |

(u) POCl_{3} |
(x) XeF_{2} |
(y) XeF_{4} |
(z) C_{2}H_{2} |

**Follow these steps to determine the electron and molecular geometries:**

- Determine the central atom and draw the Lewis structure for the molecule.
- Count the number of
*atoms*and lone pairs of electrons on the central atom (*steric number*) - Arrange them in a way that minimizes repulsion (as far apart as possible).
- Determine the name of the electron and molecular geometry.

_{ }

**To draw the Lewis structures:**

- Write the correct skeletal structure for the molecule.

* Hydrogen atoms are always terminal (only one bond)

* Put more electronegative elements in terminal positions

- Sum the valence electrons from all the atoms.
- Use a pair of electrons to form a bond between each pair of bound atoms.
- Add the remaining electrons to satisfy the octet for a more electronegative atom first.
- If any atoms lack an octet, make a double or triple bond to give them an octet.

**(a)** BF_{3 }– trigonal planar, trigonal planar

Boron is the central atom, so we can draw the skeletal structure as:

There is a total of 3 + 3 x 7 = 24 electrons, and halogens on terminal positions are always going to have 3 lone pairs of electrons. Therefore, 3 x 6 = 18 electrons on the fluorenes, and the remaining 6 are used to make the three covalent bonds:

The central atom has three bonds and no lone pairs, therefore, **both** the electron and molecular geometries are **trigonal planar**. The bond angle between the fluorine atoms is 120^{o}.

The steric number (the sum of the number of the atoms and lone pairs) of B is 3 which corresponds to *sp*^{2}-hybridization.

**(b)** CH_{2}O – trigonal planar, trigonal planar

In formaldehyde, the carbon is the central atom, so we can draw a skeletal structure:

There is a total of 4 + 2×1 + 6 = 12 electrons, and 6 are already used for making the bond. The remaining 6 go to oxygen as it is the most electronegative atom.

Because the carbon lacks an octet, we use one lone pair from the oxygen to make a π bond with the carbon, and thus, a double bond is formed.

The central atom has three bonds and no lone pairs, therefore, both the **electron and molecular geometries are trigonal planar**.

For a trigonal planar geometry, the expected bond angles are 120^{o}, however, because the three groups are not identical, the bond angles will not be as in idealized geometry.

In general, remember that oxygens on terminal positions are always going to have 2 bonds and 2 lone pairs of electrons unless there is a formal charge on them.

Knowing these patterns for bonding, lone pairs, and formal charges will make these types of exercises a lot easier.

The steric number (the sum of the number of the atoms and lone pairs) of the carbon is 3 which corresponds to *sp*^{2}-hybridization.

**(c)** HCN – linear, linear

Carbon is the central atom, so we can draw the skeletal structure:

There is a total of 1 + 4 + 5 = 10 valence electrons, and we use four of them to make the bonds.

The remaining 6 go on the nitrogen as lone pairs:

Because the carbon lacks an octet, we use two lone pairs from the nitrogen to make two additional bonds:

The steric number is 2, and there are no lone pairs on the central atom. Therefore, **both the electron and molecular geometries are linear**.

Steric number 2 corresponds to *sp*-hybridization where the bond angles are 180^{o}.

**(d)** BeCl_{2 }– linear, linear

Be is the central atom, so we can draw the skeletal structure:

There is a total of 2 + 2×7 = 16 valence electrons, and we use four of them to make the bonds. The remaining 12 go on the Cl atoms as lone pairs:

The steric number is 2, and there are no lone pairs on the central atom. Therefore, **both the electron and molecular geometries are linear**.

Steric number 2 corresponds to *sp*-hybridization where the bond angles are 180^{o}.

**(e)** CH_{2}Cl_{2 }– tetrahedral, tetrahedral

Carbon is the central atom, so we can draw a skeletal structure first:

There are 4 + 2 + 2×7 = 20 electrons, and 8 have been used to make four bonds. The remaining 12 go on the two chlorines as lone pairs:

The central atom has four atoms and no lone pair, therefore, both the **electron and molecular geometries are tetrahedral**:

_{}

Steric number 4 corresponds to *sp ^{3}*-hybridization where the idealized bond angles are 109.5

^{o}.

**(f)** SOCl_{2 }-tetrahedral, trigonal pyramidal

Sulfur is the central atom, so we can draw the skeletal structure:

_{}

There are 6 + 6 + 2×7 = 26 electrons, and 6 are used to make 3 bonds. Two chlorines and the oxygen take 3 lone pairs, and the remaining one goes to the sulfur:

All the atoms have an octet, however, there are three lone pairs on the oxygen which gives it a -1 formal charge. Therefore, one of the lone pairs on the oxygen is used to make another bond with the sulfur, and even if it exceeds the octet, that is acceptable for elements in the third row and below because they have d orbitals.

The central atom has 3 atoms and a lone pair (SN = 4). Therefore, the **electron geometry is tetrahedral** while the **molecular geometry is trigonal pyramidal**:

Areic number 4 indicates an *sp*^{3}-hybridization with idealized angles of 109.5^{o}. However, because the groups on the central atom are not identical, the angles are not going to be exactly 109.5^{o}.

**(g)** SO_{2 }– trigonal planar, bent

Sulfur is the central atom, so we can draw the skeletal structure:

There are 3×6 = 18 electrons, and 4 of them are used to make 2 bonds. The two oxygens take 6 lone pairs, and the remaining one goes to the sulfur:

As it is drawn, the problems with this structure are that the sulfur lacks an octet and the oxygens have only one bond and three lone pairs. Remember, the normal valency of oxygens is having two bonds and two lone pairs otherwise a formal charge needs to be assigned.

Therefore, one lone pair from each oxygen is used to make an additional bond with the sulfur:

The central atom has a steric number of 3 – two atoms and one lone pair. The **electron geometry,** therefore, is **trigonal planar**, and the **molecular geometry is bent**.

The steric number (the sum of the number of the atoms and lone pairs) of the sulfur is 3 which corresponds to *sp*^{2}-hybridization.

Phosphorus is the central atom, so we can draw a preliminary skeletal structure:

There are 5×7 + 5 = 40 electrons, out of which, 10 are used to make 5 covalent bonds. All the remaining 30 are divided between the five chlorine atoms, each taking 6 electrons as 3 lone pairs:

There are 5 atoms and no lone pairs on the central atom, therefore, **both geometries are trigonal bipyramidal**:

_{}

The steric number is 5, and therefore, there should be 5 orbitals accommodating the bonding electrons. This is achieved through *sp*^{3}*d* hybridization, and even though *there is a constant debate about whether the d orbitals are involved in these types of hybridizations, this answer is usually accepted in undergraduate general chemistry courses. Ultimately, it is on your professor, as they are the authority in the class.*

Xe is the central atom, so we can draw a preliminary skeletal structure:

There are 4×6 + 8 = 32 electrons and this time, instead of putting three lone pairs on the oxygen, we are going to directly add double bonds to leave two lone pairs for each oxygen:

All the electrons are used: 4 double bonds = 16, 8 lone pairs = 16.

The central atom has four atoms and no lone pair, therefore, both the **electron and molecular geometries are tetrahedral**.

Steric number 4 corresponds to *sp ^{3}*-hybridization where the idealized bond angles are 109.5

^{o}.

N is the central atom and there are 5 + 3×7 = 26 electrons. Each chlorine taking three lone pairs leaves the nitrogen with one lone pair:

The central atom has 3 atoms and a lone pair (SN = 4). Therefore, the **electron geometry is tetrahedral** while the **molecular geometry is trigonal pyramidal**:

Steric number 4 corresponds to *sp ^{3}*-hybridization where the idealized bond angles are 109.5

^{o}.

Silicon is the central atom, so we can draw a preliminary skeletal structure:

There are 4 + 4×7 = 32 electrons, and 8 have been used to make four bonds. The remaining 24 go on the two chlorines as lone pairs:

The central atom has four atoms and no lone pair, therefore, both the **electron and molecular geometries are tetrahedral**:

_{ }

*sp ^{3}*-hybridization where the idealized bond angles are 109.5

^{o}.

Sulfur is the central atom. There are 6 + 2×7 = 20 electrons, and 4 of them are used to make 2 bonds. The two fluorines take 6 lone pairs, and the remaining four electrons go to the sulfur:

The central atom has a steric number of 4 – two atoms and two lone pairs. The **electron geometry, **therefore**, is tetrahedral**, and the **molecular geometry is bent**.

_{ }

Steric number 4 corresponds to *sp ^{3}*-hybridization where the idealized bond angles are 109.5

^{o}. Because the groups on the central atom are not identical, the bond angles differ from the idealized values.

There is a debate that the sulfur is *sp*^{2}-hybridized and the lone pairs are in p orbitals and angles are close to 90^{o} (92^{o}). This contradicts the VSEPR theory, and your final reference should be your instructor.

Sulfur is the central atom:

There are 6 + 2 = 8 electrons, and 4 of them are used to make 2 bonds. The remaining four electrons go to the sulfur:

The central atom has a steric number of 4 – two atoms and two lone pairs. The **electron geometry,** therefore, is **tetrahedral**, and the **molecular geometry is bent**.

_{}

Steric number 4 corresponds to *sp ^{3}*-hybridization where the idealized bond angles are 109.5

^{o}. Because the groups on the central atom are not identical, the bond angles differ from the idealized values.

There is a debate that the sulfur is *sp*^{2}-hybridized and the lone pairs are in p orbitals and angles are close to 90^{o} (92^{o}). This contradicts the VSEPR theory, and your final reference should be your instructor.

Sulfur is the central atom:

There are 3×6 + 6 = 24 electrons, and 6 of them are used to make 3 bonds. Three oxygens take 6 lone pairs and make an additional bond with the sulfur.

All the electrons have been used, and the steric number of the central atom is 3 with no lone pairs. Therefore, ** both geometries are trigonal planar**.

Steric number 3 corresponds to *sp** ^{2}*-hybridization where the idealized bond angles are 120

^{o}.

The carbon is the central atom, so we can draw a preliminary skeletal structure:

There is a total of 4 + 2×7 + 6 = 24 electrons, and 6 are already used for making the bond. The remaining 18 go to oxygen and the chlorine atoms as lone pairs. Because the carbon lacks an octet, we use one lone pair from the oxygen to make a π bond with the carbon, and thus, a double bond is formed.

The central atom has three atoms and no lone pairs, therefore, both the **electron and molecular geometries are trigonal planar**.

The steric number (the sum of the number of the atoms and lone pairs) of the carbon is 3 which corresponds to *sp*^{2}-hybridization.

P is the central atom and there are 5 + 3×7 = 26 electrons. Each chlorine taking three lone pairs leaves the nitrogen with one lone pair:

The central atom has 3 atoms and a lone pair (SN = 4). Therefore, the **electron geometry is tetrahedral** while the **molecular geometry is trigonal pyramidal**:

Steric number 4 corresponds to *sp ^{3}*-hybridization where the idealized bond angles are 109.5

^{o}. Because the groups on the central atom are not identical, the bond angles differ from the idealized values.

Oxygen is the central atom, so we can draw a preliminary skeletal structure:

There are 6 + 2×7 = 20 electrons, and 4 of them are used to make 2 bonds. The two fluorines take 6 lone pairs, and the remaining four electrons go to the sulfur:

The central atom has a steric number of 4 – two atoms and two lone pairs. The **electron geometry, **therefore,** is tetrahedral**, and the **molecular geometry is bent**.

*sp ^{3}*-hybridization where the idealized bond angles are 109.5

^{o}. Because the groups on the central atom are not identical, the bond angles differ from the idealized values.

Br is the central atom, so we can draw a preliminary skeletal structure:

There are 5×7 + 7 = 42 electrons, out of which, 10 are used to make 5 covalent bonds. The remaining 30 are divided between the five fluorine atoms, each taking 6 electrons as 3 lone pairs, and Br takes the last pair of electrons:

There are 5 atoms and one lone pair on the central atom, therefore, the **electron geometry is octahedral**, while the **molecular geometry is square pyramidal**:

There are 6 units (atoms and lone pairs) on the central atom, and to accommodate them, it needs 6 orbitals which is achieved through *sp*^{3}*d*^{2} hybridization.

Oxygen is more electronegative, therefore, it goes in a terminal position:

There are 2×5 + 6 = 16 valence electrons and four are already taken to make two bonds. The remaining 12 are distributed between the oxygen and nitrogen:

Because the middle nitrogen lacks an octet, two lone pairs from the nitrogen are used to make two pi bonds with the middle nitrogen and the Lewis structure of this compound can be given as follows:

There are formal charges here because the middle nitrogen and the oxygen are not in their common bonding pattern:

This is one resonance structure and the other would be:

There are formal charges here because the middle nitrogen and the oxygen are not in their common bonding pattern:

This is one resonance structure and the other would be:

In both cases, the central atom has two atoms and no lone pairs, therefore, **both geometries are linear**.

Steric number 2 corresponds to *sp*-hybridization where the bond angles are 180^{o}.

Sulfur is the central atom:

_{ }

There are 6 + 6×7 = 48 valence electrons, and 12 are taken for making 6 covalent bonds. Each fluorine takes three lone pairs, so there are no electrons left: 48 – (12 + 6×6) = 0.

The central atom has 6 atoms connected to it, and no lone pairs, therefore, **both geometries are octahedral**:

There are 6 units (atoms and lone pairs) on the central atom, and to accommodate them, it needs 6 orbitals which is achieved through *sp*^{3}*d*^{2} hybridization.

P is the central atom, so we can draw a preliminary skeletal structure:

There are 5 + 3×7 + 6 = 32 valence electrons out of which, 8 are already taken to make four covalent bonds. From the remaining 24, 18 go to the Cl atoms as three lone pairs per each, three 6 to the oxygen as three lone pairs. This leaves the central atom with 4 atoms connected to it and no lone pairs. Therefore, **both geometries are tetrahedral**.

*sp ^{3}*-hybridization where the idealized bond angles are 109.5

^{o}. Because the groups on the central atom are not identical, the bond angles differ from the idealized values.

Xe is the central atom, so we can draw a preliminary skeletal structure:

There are 2×7 + 8 = 22 electrons and 4 are taken to make 2 covalent bonds. Each fluorine takes 3 lone pairs, so there are 22 – (4+2×6) = 6 electrons left which go to Xe as 3 lone pairs:

There are 2 atoms and 3 lone pairs on the central atom, so the steric number is 5, and therefore, the **electron geometry is trigonal bipyramidal** and the **molecular geometry is liner**:

There are 5 units (atoms and lone pairs) on the central atom, and to accommodate them, it needs 5 orbitals which is achieved through *sp*^{3}*d* hybridization.

Xe is the central atom:

There are 4×7 + 8 = 36 electrons and 8 are taken to make 4 covalent bonds. Each fluorine takes 3 lone pairs, so there are 36 – (8+4×6) = 4 electrons left which go to Xe as 2 lone pairs:

There are 4 atoms and 2 lone pairs on the central atom, therefore, the steric number is 6 and the **electron geometry is octahedral** and the **molecular geometry is square planar**:

_{ }

There are 6 units (atoms and lone pairs) on the central atom, and to accommodate them, it needs 6 orbitals which is achieved through *sp*^{3}*d*^{2} hybridization.

Because the hydrogens are always in terminal positions, the carbons must be connected, and therefore, we can draw a preliminary skeletal structure to start with:

_{ }

There are 2×4 + 2×1 = 10 valence, and 6 have been used to make 3 covalent bonds. Therefore, the remaining 4 go to the carbon atoms as a lone pair on each:

The carbons atoms are lacking octet, so we use the two lone pairs for making 2 additional bonds between them:

Even though there is no single central atom, both carbons are linear since their steric number is 2 which makes the molecule **linear** as well.

Additionally, if you have covered the hybridization theory, you should recognize that both carbons are *sp*-hybridized and thus the bond angle is 180^{o} which makes the molecule **linear**.

In sp hybridization, the s orbital of the excited state carbon is mixed with only one out of the three 2p orbitals. It is called ** sp** hybridization because two orbitals (one s and one p) are mixed:

The resulting two ** sp** hybrid orbitals are then arranged in a linear geometry (180

^{o}) and the two unhybridized 2p orbitals are placed at 90

^{o}:

Let’s see how this happens in acetylene- C2H2. The two carbon atoms make a sigma bond by overlapping the *sp* orbitals.

One hydrogen bonds to each carbon atom by overlapping its s orbital with the other ** sp** orbital.

The two p orbitals of each carbon overlap to make two **π** bonds.

The key parameters about the ** sp **hybridization and triple bond:

* All the atoms have linear geometry.

* The angle between atoms is 180^{o}.

* In a triple bond there is one **σ **(sigma) and two **π** (pi) bonds.

For each marked atom, add any missing lone pairs of electrons to determine the steric number, electron and molecular geometry, approximate bond angles, and the hybridization state.

You can also download the questions as a PDF worksheet to print and work on here.

Determine the hybridization state of each carbon and heteroatom (any atom except C and H) in the following compounds.

**Hint:** Remember to add any missing lone pairs of electrons where necessary.

**Check Also**

- Lewis Dot Symbols
- The Ionic Bond
- The Covalent Bond
- Sigma and Pi Bonds
- Electronegativity and Bond Polarity
- The Octet Rule
- Formal Charges
- Lewis Structures and the Octet Rule
**Lewis Structures Practice Problems**- Resonance Structures
- The VSEPR Model
**VSEPR Theory Practice Problems**- Hybridization of Atomic Orbitals

for 1n, could SO3’s electron geometry possibly be trigonal planar, with the most stable configuration being 3 S-O double bonds? that way there is no lone pair on Sulfur, and the charges on two oxygens could be eliminated as well. But then SO3’s shape would be trigonal planar and sp3

**trig planar; sp2

Hello,

Yes, it definitely is trigonal planar as there is no lone pair on the sulfur. Please see the updated information. The reason for the mistake was I saw it as SO32- ion which has more electrons. For this, you can check the VSEPR examples here. Thanks for letting me know.