**Introduction to the Hybridization**

Let’s start first by answering this question: Why do we need the hybridization theory?

Here is one answer to this. It is confirmed experimentally that the carbon atom in methane (CH_{4}) and other alkanes has a tetrahedral geometry. A reminder that in tetrahedral geometry, all the angles are 109.5^{o} and the bonds have identical lengths. Remember also that covalent bonds form as a result of orbital overlapping and sharing two electrons between the atoms. So, in order to predict the valency and geometry of the carbon atom, we are going to look at its electron configuration and orbitals.

**C – 1s ^{2}2s^{2}2p^{2}**

The valence electrons are the ones in the 2s and 2p orbitals and these are the ones that participate in bonding and chemical reactions.

You can see from the electron configuration that it is impossible to make four, **identical** in bond length, energy, and everything else (degenerate) bonds because one of the orbitals is a spherical **s**, and the other three are **p** orbitals. And this is where we get into the need of a theory that can help us explain the known geometry and valency of the carbon atom in many organic molecules. **Hybridization is a theory that is used to explain certain molecular geometries that would have not been possible otherwise.**

** **

**The*** sp*^{3} hybridization

*sp*hybridization

^{3}Now, let’s see how that happens by looking at methane, CH_{4} as an example. In the first step, one electron jumps from the 2s to the 2p orbital. This leads to the **excited state** of the carbon:

Pay attention that the electron goes uphill as the p subshell is higher in energy than the s subshell and this is not energetically favorable, but we will see how it is compensated in the next step when orbitals are mixed (hybridized).

So, in the next step, the **s** and **p** orbitals of the excited state carbon are hybridized to form **four identical** in size, shape, and energy orbitals.

**The number of the hybrid orbitals is always the same as the number of orbitals that are mixed.** So, four orbitals (one 2s + three 2p) are mixed and the result is four ** sp^{3} **orbitals. These are hybrid orbitals and look somewhat like the s and p orbitals. And again, we call them

**because they are formed from one s orbital and three p orbitals.**

*sp*^{3}The formation of these degenerate hybrid orbitals compensates for the energy uphill of the s-p transition as they have lower energy than the p orbitals.

The four ** sp^{3}**-hybridized orbitals arrange in a tetrahedral geometry and make bonds by overlapping with the s orbitals of four hydrogens: This explains the symmetrical geometry of methane (CH

_{4}) where all the bonds have the same length and bond angle.

All four C – H bonds in methane are single bonds that are formed by **head-on (or end-on)** overlapping of ** sp^{3}** orbitals of the carbon and s orbital of each hydrogen.

The bonds that form by the head-on overlap of orbitals are called **σ (sigma)** bonds because the electron density is concentrated on the axis connecting the C and H atoms.

**Ethane, CH**_{3}-CH_{3} and other alkanes

_{3}-CH

_{3}and other alkanes

If instead of one hydrogen, we connect another sp3-hybridized carbon, we will get ethane:

And consequently, in all the alkanes, there is a sigma bond between the carbon atoms and the carbon-hydrogen atoms, and the carbons are *sp ^{3}*

^{ }hybridized with tetrahedral geometry:

To generalize this, any atom with four groups (either an atom or a lone pair) is ** sp^{3}** hybridized. And the way to look at this is, in order for the four groups to be as far away from each other as possible like we learned in the VSEPR theory, the groups need to be in identical four orbitals which is only possible in the

**hybridization.**

*sp*^{3}

*sp*^{3}** Hybridization with A Lone Pair(s)**

*sp*

^{3}Let’s consider an example where the atom is *sp*^{3}-hybridized but instead of four atoms, it has three atoms and a lone pair. A typical example is ammonia, NH_{3}. Nitrogen is the next element after carbon, with an electron configuration of **1s ^{2}2s^{2}2p^{3}.**

Now, the principle is the same as we have just seen for carbon, however, when the three p orbitals are mixed with one s orbital, four *sp*^{3} orbitals are formed where one of them is filled with two electrons:

** **

This one hybrid orbital accommodates the lone pair of the nitrogen, while the other three overlap with the s orbitals of hydrogen thus making three sigma N-H bonds.

The central atom has 3 atoms and a lone pair, so the steric number is equal to 4 (SN = 4). Remember, the steric number is the sum of the atoms and lone pairs on an atom, and it dictates the hybridization and the geometry of that atom. Therefore, the **electron geometry is tetrahedral** while the **molecular geometry ****is ****trigonal pyramidal**:

A practice example: what is the hybridization of the oxygen in the following molecule?

The oxygen is connected to two atoms and has two lone pairs. In total – four groups, and that and therefore, it is ** sp^{3}** hybridized. The difference with nitrogen in ammonia is that it has two lone pairs, which makes a tetrahedral electron geometry and bent molecular geometry. It is very similar to what we have water:

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**.

**The*** sp*^{2} **hybridization**

*sp*

^{2}

When the excited state carbon atom is formed, the *sp ^{3}* hybridization is not the only option of mixing the orbitals. The

*sp*hybridization occurs when the s orbital is mixed with only two p orbitals as opposed to the three p orbitals in the

^{2}**hybridization. So, three orbitals are mixed, and the outcome is three hybrid orbitals which are called**

*sp*^{3}**.**

*sp*hybrid orbitals^{2}For example, the central atom in BF_{3} is ** sp^{2} – hybridized**. One s orbital of boron is mixed with two p orbitals forming thee

*sp*hybrid orbital and one empty p orbital:

^{2}

The thee *sp*^{2 }orbitals overlap with the p orbitals of fluorine making three covalent bonds. There is an argument that the fluorine atoms are also hybridized, and more specifically, they are *sp*^{3}-hybridized which would mean that instead of the p orbitals, they use one of the *sp*^{3 }orbitals to make these bonds and the other three accommodate the lone pairs. I accept both answers from my students, so check with your instructor for their view on this.

As for the geometry, 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*** sp*^{2} **Hybridization of Carbon Atoms**

*sp*

^{2}

Moving a little bit towards organic molecules, we see that with the same mechanism, the carbon atom makes *sp*^{2} hybridization:

The 2s orbital is mixed with two 2p orbitals and the resulting 3 ** sp^{2}** orbitals are then arranged in a trigonal planar geometry (120

^{o}). An important difference here, compared to the

**hybridization, is the left-out (unhybridized)**

*sp*^{3}**p**orbital that did not participate in the hybridization. This orbital is placed at 90

^{o}to the plane of the trigonal planar arrangement of the three

**orbitals:**

*sp*^{2}

**Carbon hybridization in Ethylene – C**_{2}H_{4}

_{2}H

_{4}

Two ** sp^{2}** hybridized carbon atoms can make a sigma bond by overlapping one of the three

**orbitals and thus bond with two hydrogens each and two hydrogens make sigma bonds with each carbon by overlapping their**

*sp*^{2}**s**orbitals with the other two

**orbitals.**

*sp*^{2}

This makes three bonds for each carbon and one p orbital left. Remember, the standard valency of carbon is four and it likes to have four bonds.

This fourth bond is formed by the side-by-side overlap of the two 2p orbitals on each carbon. This type of bonding by a side-by-side overlap of the orbitals is called a **π bond**.

So, the two carbons in ethylene, which is the first member of the alkene family, are double-bonded. In each double bond, there is one **sigma** and one **π bond**.

Here are some key parameters about the ** sp^{2}** hybridization and double bonds that you need to know:

* All the atoms on the double bond are in one plane.

* The angle between atoms is 120^{o}.

* The angle between the plane and p orbitals is 90^{o}.

**The ***sp* hybridization

*sp*hybridization

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- C_{2}H_{2}. 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 of 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.

So far, we have talked about geometries that are adopted when the central atom has a steric number 1-4. That is, the sum of atoms and lone pairs on it does not exceed four. Now, for molecules with shapes trigonal bipyramidal or octahedral, the steric number is 5 and 6 respectively, and therefore, 5 or 6 orbitals are needed to make these bonds and accommodate any lone pairs. For example, PCl_{5} has a trigonal bipyramidal geometry and SF_{6} has an octahedral geometry. So, let’s start with the trigonal bipyramidal geometry which is achieved when the central atom adopts an *sp ^{3}d* hybridization.

**The*** sp*^{3}d hybridization

*sp*hybridization

^{3}dThe first thing to mention here is that the elements capable of adopting trigonal bipyramidal or octahedral geometry must be in period 3 or higher, because *d *orbitals are needed to produce five or more hybrid orbitals. For example, the electron configuration of phosphorous is:

P – 1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}

And in addition to the 3s and 3p orbitals, it also has the five 3d orbitals available for hybridization. Let’s draw these orbitals and see how the *sp ^{3}d* hybridization is achieved:

The five half-filled *sp ^{3}d *orbitals are formed by mixing one s (the 3s orbital), three p (the 3p orbitals), and one d (one 3d orbital). Four 3

*d*orbitals are left unhybridized and empty. Each of the

*sp*orbitals makes a bond with the p orbital of chlorine forming a trigonal bipyramidal geometry.

^{3}d

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.*

Let’s also consider the structure of BrF_{3} where the central atom is sp^{3}d hybridized but only has 3 atoms connected.

The electron configuration of Br is:

Br – 1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}

Just like phosphorus, in addition to the 3s and 3p orbitals, it also has the five 3d orbitals available and one of them is involved in the *sp ^{3}d* hybridization:

Notice that unlike the phosphorus in PCl_{5}, only three of the five hybrid orbitals are half-filled while the two are accommodating the lone pairs.

We can also see this by drawing the Lewis structure of BrF_{3}:

The central atom has 3 atoms and 2 lone pairs, therefore, the **electron geometry is trigonal bipyramidal**, while the **molecular geometry is T-shaped**:

*Notice that the lone pair does not go in axial positions (up or down).*

**The*** sp*^{3}d^{2}^{ }hybridization

*sp*

^{3}d^{2}^{ }hybridization

Let’s consider the structure of sulfur hexafluoride, SF_{6. }The central atom has six atoms connected, and therefore, it needs six equivalent orbitals, which is achieved by the *sp ^{3}d^{2}*

^{ }hybridization.

The electron configuration of sulfur is:

S – 1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}

You may look at it and say, it is still similar to the configuration of P and Br that adopted *sp ^{3}d *hybridization, so how is the sulfur going to be

*sp*And the difference is that there are two d orbitals mixed with the s and p orbitals thus making six hybrid orbitals each hosting one electron:

^{3}d^{2 }and make six bonds?

** **

There are six *sp ^{3}d^{2}* half-filled orbitals of the S atom and each overlaps with a half-filled 2

*p*orbital of an F atom.

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

To summarize the key concepts about the hybridization of atomic orbitals, remember that the hybridization theory helps explain the geometry of molecules. The type and the shape of hybrid orbitals formed depend on the types of atomic orbitals mixed. What is true for all the hybridizations is that **the number of the hybrid orbitals is always equal to the number of orbitals that are mixed.**

For those of you taking organic chemistry, you will not need to deal with the hybridizations and geometries for steric number > 4 as carbon can only make four bonds at most. Check the shorter version of this article (and hundreds of other) tailored of the structure of organic molecules.

#### 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 + 2 = 26 electrons, and 6 of them are used to make 3 bonds. Three oxygens take 3 lone pairs each and one goes to the sulfur:

All the electrons have been used; however, one oxygen needs to share a pair of electrons with the sulfur otherwise the ionic would be -3:

The central atom has 3 atoms and a lone pair and 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.

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**:

*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.