VSEPR Theory

Here is a discussion about VSEPR (Valence Shell Electron Pair Repulsion) Theory and its applications. Questions are asked and it can be downloaded from the link given at the end of this discussion.

In the year 1957 Gillespie developed a VSEPR theory  to explain molecular shapes and bond angles more accurately. The main postulates of this theory are:

1. The shape of a molecule can be predicted from the number and type of valence shell electron pairs around the central atom

2. Electron pairs around a central atom must stay as far as possible to minimize repulsion.

3. Lone pairs occupy more space around the central atom than the bonded electron pairs.

  The order of repulsion between different types of electron pairs is as follows:

  Lone pair - Lone pair > Lone Pair - Bond pair > Bond pair - Bond pair 

4. The magnitude of repulsion between bond pair's of electrons depend on the electro negativity difference between central atom and bonded atoms.

5. The repulsion order in relation to the bonds are as follows:

     double bond-double bond > double bond-single bond > single bond-single bond. 

Applications of VSEPR theory to simple molecules:

 If there are two electron pairs around the central atom, the only way to keep them as far apart as possible is to arrange them at an angle of 180° to each other. The molecule in such a case will adopt linear geometry. Similarly, the molecule forms trigonal planar geometry for three electron pairs around the central atom, and for four electron pairs around the central atom, the molecule adopts tetrahedral geometry.

Molecules with only bond pairs adopt the above geometry depending upon the number of bonded atoms present, since the above arrangement gives least repulsion among electron pairs and maximum stability.

Distortions from a molecular arrangement arises only when lone pairs are present in a molecule.

Let us take two examples NH3 and H2O to explain the effect of lone pairs in molecular arrangement and compare them with methane.

Shape of NH3 molecule:

We Know that ammonia contains  3 bp and 1 lp ( calculation of bond pair and lone pair of molecule is discussed below)

As lone pair-bond pair repulsion is more than bond pair-bond pair repulsion, the presence of lone pair in ammonia pushes the bond pair from regular tetrahedral geometry. ( ie,lone pair occupy more space) Hence the bond angle decreases from 109.5° to 107°. The geometry of ammonia molecule is also considered as pyramidal.

Shape of H2O molecule:

We Know that water contains  2 bp and 2 lp ( calculation of bond pair and lone pair of molecule is discussed below)

As the repulsive force between lone pair-lone pair is greater lone pair-bond pair repulsion,  the presence of two lone pairs in water pushes the bond pair more strongly from regular tetrahedral geometry. ( ie,lone pair occupy more space) Hence the bond angle decreases from 109.5° to 104°. The geometry of water molecule is also considered as bent or angular.

Shape of CH4 molecule:

We Know that methane contains 4 bp. These four electron pairs, trying to remain as far apart as possible, adopt tetrahedral structure. In this geometry, all the H-C-H bond angles are of 109.5°.

Here is a simple calculation used to predict the number of electron pairs both bonded and lone pairs.

Ex. 1: CH4

Step -1

Number of valence electron of carbon (central atom)                     =  4

Number of bonded atoms (four hydrogen)                                      =  4

Total                                                                                                 =  8

Step - 2

divide this by 2 (8/2)                                                                       =  4

so there must be 4 electron pairs in this molecule and all are bonded with four hydrogen. (ie, no lone pair is present).

Hence, this molecule posses a regular geometry (no distortion) tetrahedral.

Ex. 2: NH3

Step -1

Number of valence electron of nitrogen (central atom)                   =  5

Number of bonded atoms (three hydrogen)                                     =  3

Total                                                                                                 =  8

Step - 2

divide this by 2 (8/2)                                                                       =  4

so there must be 4 electron pairs in this molecule and three electron pairs are bonded with three hydrogen. (ie, one lone pair is present). 3 bp and 1 lp.

Hence, this molecule posses an irregular geometry (distortion from tetrahedral) Pyramidal.

Ex. 3: H2O

Step -1

Number of valence electron of Oxygen (central atom)                   =  6

Number of bonded atoms (two hydrogen)                                      =  2

Total                                                                                                 =  8

Step - 2

divide this by 2 (8/2)                                                                       =  4

so there must be 4 electron pairs in this molecule and two electron pairs are bonded with two hydrogen. (ie, two lone pairs are present). 2 bp and 2 lp.

Hence, this molecule posses an irregular geometry (distortion from tetrahedral) bent shaped.

Ex. 4: BrF5

Step -1

Number of valence electron of bromine (central atom)                  =  7

Number of bonded atoms (five fluorine)                                        =  5

Total                                                                                                = 12

Step - 2

divide this by 2 (12/2)                                                                       =  6

so there must be 6 electron pairs in this molecule, but only five electron pairs are bonded with five fluorine atoms. (ie, one lone pair is present). 5 bp and 1 lp.

Hence, this molecule is distorted from octahedral geometry (irregular geometry) and is square pyramidal in shape.

We can use the above calculation for predicting the shape of ions also.

Ex -5: ‎NH⁺₄ (ammonium ion)

Step -1

Number of valence electron of nitrogen (central atom)                   =  5

Number of bonded atoms (three hydrogen)                                     =  4

Number of positive charge (one) subtract 1 (loss of e-)                  = -1

Total                                                                                                 =  8

Step - 2

divide this by 2 (8/2)                                                                       =  4

so there must be 4 electron pairs in this molecule and four electron pairs are bonded with four hydrogen. (ie, no lone pair is present). 4 bp and 0 lp.

Hence, this molecule posses an regular geometry  tetrahedral. (Compare this with ammonia)

For acquiring a positive charge one electron must be removed. Hence, in the above example we subtract 1(one). Similarly for negative ions we have to add electrons in calculation.

For molecules which have oxygen atom doubly bonded with central atom (ex. CO2, SO3) the following formula gives the expected geometry.

Number of electron pairs            =    1/2{V+(n x v)}-(3 x n)

V - valance electrons of central atom

n -  number of bonded atoms

v - valance electrons of bonded atom

EX - 6: CO2

Number of electron pairs            =    1/2{4+(2x6)}-(3 x 2)

                                                    =    1/2{4 + 12} - (6)

                                                    =    1/2{16} - 6

                                                    =     8 -6  

                                                    =     2                  

There are 2 electron pairs are present and both are bonded with oxygen hence it is linear.

Download questions here.

ARTIFICIAL HEART

Artificial hearts have  been in clinical use for more than 35 years to help people who are suffering from heart failures.  — here’s what you should know. 

The following Presentation is prepared and presented in our class by a vibrant and aspiring student Ms. Sowmiya of III B.Sc., Chemistry, St. John's College. The references at the end of the presentation tells about her efforts in preparing this ppt. All credits to her only. A connection game is also included in the ppt for fun. If You find the answer please comment. Thank You.
                                                                                        - D. JIM LIVINGSTON.

Artificial Heart by on Scribd

POLYMER PROCESSING - II (MOLDING)

This post  gives you more about polymer processing techniques. Videos and animations are given at the end. Hope you can understand better.

Polymer Processing - II from Jim Livingston

INJECTION MOLDING ANIMATION

EXTRUSION MOLDING ANIMATION

VIDEO SHOWING BLOW MOLDING PROCESS

ANIMATIONS OF  INJECTION, EXTRUSION AND STRETCH BLOWING PROCESS

HYBRIDIZATION IN AMMONIA, WATER,ETHYLENE AND ACETYLENE

A short discussion about the hybridization in the remaining  molecules in our syllabus. Hybridization in ethylene and acetylene (not in our syllabus) are posted as video.

Hybridization in Ammonia 

E.C.of ammonia 1s²,2s²,2p³

The four sp³ hybrid orbitals in  ammonia is formed by the overlapping of three half filled orbitals and one filled s- orbital of Nitrogen atom. This is shown in the following diagram.

Geometry of  ammonia is pyramidal

Hybridization in Water

 Electronic configuration of O 1s²,2s²,2p⁴

The four sp³ hybrid orbitals in  water is formed by the overlapping of two half filled p orbitals. one filled p - orbital and one filled s- orbital of oxygen atom. This is shown in the following diagram.

 Geometry of  water  is V-shaped 

The bond angle in water and ammonia is depicted in the following figure.

Hybridization in Ethylene

Hybridization in Acetylene

THEORIES OF COVALENT BONDING (VB Theory) Cont....

The remaining types of hybridization are discussed in this post with clear pictures and animated videos. Hope this will make you understand better.

sp³d hybridization :

The mixing of one s ,three p and one d-atomic orbitals to form five sp³d hybrid orbitals of equal energy is called sp³d hybridization

Example:  PCl₅

The ground state electronic configuration of phosphorus atom is: 1s² 2s²2p⁶ 3s²3p_x¹3p_y¹3p_z¹.

 Phosphorus atom undergoes excitation to promote one electron from 3s orbital to one of empty 3d orbital.

Thus the electronic configuration of 'P' in the excited state is 1s² 2s²2p⁶ 3s¹3p_x¹3p_y¹3p_z¹ 3d¹.

Intermixing of half - filled one 3s, three 3p and one 3dz orbitals to give five  sp³d hybrid orbitals. The shape of the the orbitals participating in sp³d hybridization are shown below.

The sp³d hybrid orbitals which are arranged in trigonal bipyramidal symmetry are shown in the diagram below.

IMPORTANT:

 The five orbitals in sp³d hybridizationare not equivalent. These are divided into two sets :

a) Equatorial hybrid orbitals :

       Three hybrid orbitals are directed towards the corners of an equilateral triangle are called equatorial  hybrid orbitals. These are planar and bond angle is 120⁰.

b) Axial hybrid orbitals :

Two hybrid orbitals are perpendicular to the plane of equatorial hybrid orbital .These are called axial hybrid orbital . They make an angle of 90⁰ with equatorial hybrid orbitals.

Diagram showing equatorial and axial bonds in PCl5.

Geometry :  trigonal bipyramidal. 

Bond angle is 90⁰ and 120⁰.

sp³d² hybridization :

          The mixing of one s, three p and  two d- atomic orbitals to form six equivalent sp³d² hybrid orbitals of equal energy. This hybridization is known as sp³d² hybridization.

The electronic configuration of 'S' in ground state is 1s² 2s²2p⁶ 3s²3p_x²3p_y¹3p_z¹.

  Promotion of two electrons (one from 3s and one from 3p_x)into two of the 3d orbitals gives the excited state configuration of 1s² 2s²2p⁶ 3s¹3p_x¹3p_y¹3p_z¹3d².

In the  excited state, sulfur under goes sp³d² hybridization by mixing a 3s, three 3p and two 3d orbitals. Thus formed six half filled sp³d² hybrid orbitals are arranged in octahedral symmetry. 

Diagram showing the shape of orbitals involved in sp³d² hybridization

In SF6, four out of six hybrid orbitals are lying in one plane while remaining two are directed above and below the plane containing four hybrid orbitals  perpendicularly.

Geometry:  octahedral 

 Bond angle:  90⁰.

Sp³d³ hybridization :

        The mixing of one s, three p and  three d- atomic orbitals to form seven equivalent sp³d³hybrid orbitals of equal energy. This hybridization is known as sp³d³ hybridization.

Example :  IF₇

The electronic configuration of Iodine atom in the ground state is: [Kr]4d¹⁰5s²5p⁵. Since the formation of IF₇ , the iodine atom promotes three of its electrons (one from 5s orbital and two from 5p sublevel) into empty 5d orbitals. 

The electronic configuration of Iodine in the  excited state can be written as: [Kr]4d¹⁰5s¹5p³5d³.

 

In the excited state, iodine atom undergoes sp³d³ hybridization to give 7 half filled sp³d³ hybrid orbitals in pentagonal bipyramidal symmetry. These will form 7 σ_sp3_d3_-p bonds with fluorine atom. Diagram below shows the orbitals involved in the formation of IF7.

Geometry:  pentagonal bipyramidal  

Bond angle:  72⁰ and 90⁰.

Courtesy:

https://chemistryonline.guru/hybridization.

https://www.adichemistry.com/general/chemicalbond/vbt/hybridization-illustrations.html

POLYMER PROCESSING - I

In this post some of the techniques used in the processing of polymers are discussed. The remaining processing techniques (in our univ. syllabus) will be discussed in the next post. Review questions are also asked at the end of this post you can download the questions and try to find the answers.

Polymer Processing by Jim Livingston on Scribd

Video of die-casting

A short video of rotational casting

Questions

Here you can download the review questions.

THEORIES OF COVALENT BONDING (Cont...)

Before we are going to discuss about other examples of hybridization, the first part of this post  deals with how to present your answers in our university examination regarding the questions on hybridization concept .

Q: What is sp² hybridization? Explain the formation of BF₃ molecule.

Ans:
          Mixing of one pure 's' and two pure 'p' orbitals to form three new 'sp² ' hybrid orbitals of nearly equal energy is known as sp²  hybridization. This hybridization leads to trigonal planar geometry.

Characteristics:

1. 1. All three hybrid orbitals are equivalent in shape and energy.
2. 2. Three hybrid orbitals lie in the same plane and are directed towards three corners of an        equilateral triangle.
3. 3. Bond angle is 120⁰.

Formation of BF₃ molecule

The central boron atom has the following electronic configuration.

B: 1s²,2s²,2p¹

In the formation of BF₃, one of the electron from completely filled 2s orbital is promoted to 2p orbital to give the following excited electronic state configuration.

B*: 1s²,2s¹,2px¹2py¹

Now the three singly filled orbitals hybridize to (inter mixing) give three sp² hybrid orbitals which are identical in shape and in energy. These three sp²  hybrid orbitals are pointed towards the three corners of an equilateral triangle. The three sp²  hybrid orbitals overlaps with the singly filled p orbitals of three fluorine atoms t o give BF₃ molecule.

NOTE: 

The above format may be followed to explain other types of hybridization.

SP² and SP³ hybridization

sp² hybridization

   Intermixing of one 's' and two 'p' orbitals of almost equal energy to give three identical                          and degenerate hybrid orbitals is known as sp2 hybridization.  The three sp2 hybrid orbitals are oriented in trigonal planar symmetry at an angle of120^O to each other. The sp2 hybrid orbitals have 33.3% 's' character and 66.6%'p'character.

EXAMPLE :
                      BF₃


A view of sp2 hybrid orbital.

Diagram showing the overlapping of sp2 hybrid orbitals of B with 2P orbital of F to form  BF3 molecule

SP³ Hybridization

Intermixing of one 's'and three 'p'orbitals of almost equal energy to form four identical and degenerate sp3 hybrid orbitals is called sp3 hybridiztion. The four sp3 hybrid orbitals are oriented in tetrahedral symmetry at an angle of 109^o.28'. It has 25% s character and 75% p character.
EXAMPLE:
                      CH₄



Diagram showing the sp3 hybridization in methane.