Lewis Acids

 Hello, welcome to today’s lesson. I believe that this series is making a lot of sense to you. Yesterday, we showed that in order to have a substance behave as a Lewis acid, then the substance would have to possess a central atom that has an empty orbital into which it can be able to



receive an electron pair. We can never stop repeating this fact because it is the most fundamental fact that you need to know about Lewis acids.


We also showed that BF3 and AlCl3 could all act as Lewis acids because we established the fact that there is a free 2pz and 3pz orbital on the central boron and aluminum atoms respectively. This is the orbital into which the electron pair is received when the BF3 or AlCl3 acts as a Lewis acid.


Here we have shown the reaction of ammonia with BF3 and AlCl3. In either case, we can see that what is going on actually the formation of a dative or coordinate covalent bond (Refer to the earlier lesson on the formation of H3O^+ ion). The ammonia molecule has a lone pair of electrons localized on the nitrogen atom. BF3 and AlCl3 have central atoms that contains an empty orbital into which the lone pairs could be donated to form a coordinate covalent bond.


Now let me tell you the common thread that links all the central atoms in compounds that acts as Lewis acids; they all possess an OPEN SEXTET. The term open sextet implies that the atom requires two electrons to complete its octet and this is readily done by accepting an electron pair as shown in the images attached.


Transition metal ions readily act as Lewis acids and this is the foundation for the formation of the kind of compounds called coordination compounds or “complexes”. We shall deal with this kind of compounds in  detail later in the series.


Hope you enjoyed our lesson today?

Follow  our page  and share!

Do you have assignments, term papers, projects and difficult homework questions? Contact us for help now! We also offer home and remote tutoring services to students at all levels in science subjects.



Comments

Post a Comment

Popular posts from this blog

Definition of acids 2 - Introduction to Lewis Acids

Image
 Hello fam! Welcome to today’s lesson. We started to look at the concept of acid definition in yesterday’s lesson. We saw yesterday that acids have been defined by Arrhenius and BrØnsted – Lowry. The Arrhenius definition is limited in the sense that it does not account for behavior of acids in nonaqueous media. Did you try the assignment yesterday? The answers are attached to this post. I believe that the concept of proton donation is now clear to you after looking at the images attached. Reach out to me if you have any challenge. Now let us talk about the Lewis definition of acids. This definition of acids was put out by G. N Lewis in 1923. There is a foundation that we need in order to have an understanding of the idea that was put forward by Lewis.  There is something called an orbital in an atom. It is a region in space where there is a high probability of finding the electron. Now let us assume that your office is at No.38 Bronx street. I may not find you everywhere in the compoun
Read more

Calculations on pKa and pH

Image
 Hello! I heartily welcome you all to today’s lesson. We are going to look at the last two calculations that pertain to pH, acids and bases today and tomorrow we would commence our discussion on salts to pave way for a discussion on buffers. I hope all the topics so far are making sense? Your feedback would be highly appreciated. Example 8; What is the pH of 0.2 M solution of sodium acetate. The pKa of acetic acid is 1.8 * 10^-5. Solution; We have that the hydrolysis of sodium acetate occurs as follows; CH3CH2COONa(aq) <-> CH3CH2COO^-(aq) + Na^+(aq)    CH3CH2COO^-(aq) + H2O(l) <-> CH3CH2COOH + OH^-(aq) We can see that that the acetate ion acted as a base hence it accepted a proton. We would have to obtain the Kb Kb = 1 * 10^-14/Ka Kb = 1 * 10^-14/1.8 * 10^-5 Kb = 5.6 * 10^-10 Setting up the ICE TABLE, we have that;          CH3CH2COO^-(aq) + H2O(l) <-> CH3CH2COOH + OH^-(aq) I            0.2                                                      0                       
Read more

Bonding in Complex Salts

Image
 Hello! yesterday, we started looking at the complex salt which we called tetramine copper II chloride with the formula; [Cu(NH3)4]Cl2. We did say that this is a complex salt because of the presence to the complex cation [Cu(NH3)4]^2+. I want to state that if we had a complex anion, the salt would also be a complex salt. A typical example of this is; K4 [Fe(CN)6]. The complex ion is  [Fe(CN)6]4- which is anionic and the counter ions in this case are potassium cations. I told you that we will look at the bonding in the complex salt [Cu(NH3)4]Cl2 and why it does not produce simple ions in water. I will try to explain the basics of the concept today as we would still look at coordination chemistry in more detail sometime later.  Now, the complex salt [Cu(NH3)4]Cl2 has both ionic and covalent bonds. The four ammonia molecules are attached to the central copper II ion by a coordinate covalent bond(see the page timeline for an explanation of what a coordinate covalent bond is). We showed the
Read more