Coordination Compounds

Full chapter of coordination compounds. Firstly, let me teach you the basic introduction of coordination compounds. Consider the d-block elements like iron, copper, zinc, etc. These all are metals. We know that metals love to lose electrons. But here, the case is opposite. I mean, these metal elements love to accept a pair of electrons. This iron loves to accept a pair of electrons. This copper loves to accept a pair of electrons and this zinc loves to accept a pair of electrons. Here if you ask me why these metal love to accept a pair of electrons, I would say it is due to two reasons. Firstly, these metals have empty shell. Secondly, they complete their octet rule to get stability. That's why they love to accept a pair of electrons. So these metals accept a pair of electrons. Now let me teach you that how metals accept a pair of electrons. Well, consider a metal atom like copper and a neutral molecule like NH3. This metal atom of copper has empty shell. I mean copper can easily accommodate to upcoming electrons. On the other hand, there is one lone pair of electrons present on nitrogen. So, NH3 will donate this lone pair of electrons to copper and copper will accept this one pair of electrons. Hence, coordinate bond is formed between NH3 and copper. We call this compound as coordinate compound because there exists coordination bound between them. Therefore, we define coordination compounds as the complex compound in which metal atom is bonded to a number of ions or a molecule is called coordination compound. Let me repeat it. The complex compound in which metal atom is bonded to a number of ions or a molecule is called coordination compound. Remember that this NH3 donate a pair of electrons to a metal atom. We call it ligand. So we define ligand as Any ion or molecule attached to a central metal atom by donating a pair of electron is called ligand. Thus remember that any coordination compound is made up of two parts. One part is a metal atom like copper and the second part is a ligand like NH3. Thus noted down all these important points. Now let me teach you some basic terms of coordination compounds. Let's consider this coordination compound. There are three species in this coordination compound. Chromium, NH3 and Chlorine. Now listen carefully. This metal atom of chromium is known as central atom. We know that NH3 donate a pair of electrons. So it is elegant. This 6 means 6 molecules of NH3. One molecule of NH3 donate one pair of electrons. 6 molecules of NH3 donate 6 pair of electrons. We call this 6 as coordination number. Remember that coordination number shows the number of pair of electrons donated. This chromium, NH3 and the square bracket is called coordination sphere or coordination entity. This chlorine outside the bracket is called counter ion. Secondly, Consider this complex compound. Here cobalt is a metal atom and we call it central atom. This chlorine is ligand, this 2 is the coordination number of chlorine, this NH3 is another ligand and this 4 is its coordination number. This whole thing and the square bracket is known as coordination sphere or coordination entity. This chlorine outside the square bracket or coordination sphere is called counter ion. Thus noted down all these basic terms of coordination compounds. Now let me teach you Warner's theory. Well, Warner is the first scientist who successfully studied coordination compounds. He took cobalt chloride and he react it with NH3. After reaction, he got cobalt chloride. .6NH3. Secondly, he added silver nitrate to cobalt.6NH3. After reaction, he got 3 moles of HCl3 or silver chloride. in the form of white precipitate or white ppt. Now listen carefully, Warner baba thought that 3 chlorine are weakly or indirectly bonded to cobalt and 6 NH3 strongly or directly bonded to cobalt that's why chlorine love to react with silver. So he rearranged this compound. He writes the central metal atom of cobalt then he takes NH3 and there are six molecules of NH3. He puts square bracket around these two species. Finally, he writes chlorine and there are three of them. So, Werner Baba therefore say that six NH3 are directly bonded to cobalt and three chlorine are indirectly bonded to cobalt. He called this compound as coordination compound. Remember that This is the first coordination compound in chemistry. Now what is the conclusion of Warner's theory? Well, Warner-Baba states that every metal atom of coordination compounds has two types of valencies. Primary valency and secondary valency. Primary valency is satisfied by negative ion. For example, here in this coordination compound, chlorine is a negative ion. It satisfies the primary valency of cobalt. While secondary valency is satisfied by positive ion or neutral molecule. For example, here in this compound, the NH3 is a neutral molecule. It satisfies the secondary valency of cobalt. Secondly, primary valency represents the oxidation state of a central metal atom while secondary valency represents the coordination number. Thirdly, primary valency is denoted by dots and secondary valency is denoted by line. Fourthly, primary valency is ionizable. For example, chlorine is ionizable and aqueous solution. While secondary valency is not ionizable, for example, NH3 is directly bonded to cobalt, it is not ionized. Thus noted down all these important points. Now how can we calculate primary and secondary valencies of a central metal atom? Well, consider these coordination compounds. We have already learned that primary valency is the oxidation number of a central atom and secondary valency is coordination number. Now in case of this molecule, the central atom is cobalt. I am going to find its oxidation state. I write cobalt plus 6 into NH3 plus 3 chlorine equals 0. We know that NH3 is a neutral molecule so its oxidation state is 0 and that of chlorine is positive 1. So I write cobalt plus 6 into 0 plus 3 into negative 1 equals 0. After calculation I get positive 3. The oxidation state of cobalt is positive 3. So, the primary valency of cobalt is 3. Here, we can see that there are 6 ligands of NH3. So, the secondary valency of cobalt is 6. Secondly, in case of this molecule, I find the oxidation state of iron. I write iron plus 6 into Cn equals negative 4. Because the overall charge on this coordination compound is negative 4. We know that The oxidation state of Cn is negative 1. I write iron plus 6 into negative 1 equals negative 4. After calculation, I get positive 2. The oxidation state of iron is positive 2. So, the primary valency of iron is positive 2. Here, there are 6 ligands of Cn. The coordination number is 6. So, the secondary valency of iron is 6. Therefore, Using this trick, we can easily calculate the primary and secondary valences of a central metal atom. Now let me teach you the structure of Warner's theory. Well, consider these four coordination compounds. I write cobalt and I draw six lines as the octahedral structure. Similarly, in case of these three compounds, I write cobalt and I draw six lines as a octahedral structure. We know that primary valency is denoted by dots and secondary valency is denoted by line. Now listen carefully. These are 6 molecules of NH3. I draw these 6 molecules of NH3 around the central atom cobalt. Now I take these 3 ions of chlorine. They satisfy the primary valency. So I randomly draw 3 dot lines. and I put three chlorine ions around cobalt. Remember that you can draw this dotted line in any direction around the central atom. Secondly, in this coordination compound, there are five molecules of NH3. I write them around the central atom. Now one secondary valency is left behind. So I place one chlorine here. Remember that This chlorine can fulfill the secondary valency and primary valency of the central metal atom. I draw two dotted lines and I place two chlorine. Similarly, I write 4 NH3 around the central atom. I write here one chlorine and one chlorine. Here, these two chlorine act as a primary valency and secondary valency. I draw one dotted line and I write chlorine. Lastly, I write 3 NH3 around the central atom. I write chlorine, chlorine and chlorine. Here, these three chlorine act as a primary valency and secondary valency. Now, Warner-Baba added silver nitrate to each of these coordination compound. After chemical reaction, he got different amount of products. In case of this molecule, these three dotted lines mean 3 chlorine are out of coordination sphere. Let me repeat it. These 3 dotted lines mean 3 chlorine are out of coordination sphere. Thus here 3 moles of silver chloride is formed. In this case, 2 chlorine are out of coordination sphere. So, 2 moles of silver chloride are formed. In this case, only 1 chlorine is out of coordination sphere. So, 1 mole of silver chloride is formed. In case of this coordination compound, no chlorine is out of coordination sphere so no silver chloride is formed. Therefore, the structure of this compound is cobalt NH3 6 square bracket Cl 3. The structure of this compound is cobalt NH3 5 chlorine square bracket Cl 2. The structure of this compound is cobalt NH3 4 Cl 2. square bracket Cl and the structure of this compound is cobalt NH3 3 Cl 3 square bracket. So these are the different structures of coordination compounds. Hence noted down all these structures. Now we will learn about different types of ligands. To learn the types of ligands we must learn density of a ligand. It is defined as the number of long pair of electrons Donated by a ligand is called Density of a ligand. For example, water can donate one pair of electrons, so its Density is 1. Now there are 4 types of ligands, unidentate or monodentate, bidentate, ambidentate and polydentate. Unidentate are those ligands which can donate one pair of electrons. For example, chlorine ion. Bromine iron, cyanide iron, carbon monoxide etc. Secondly, bidentate are those ligands which can donate two pair of electrons. For example, En. Remember that En stands for Ethan 1,2-diamine. We know that Ethan is CH2CH2. Di means 2 and Amines mean NH2. 1,2 means at first carbon. and at second carbon. So this is the structure of En which we often use in coordination compounds. Hence note it down. The another one is oxylate COO-bond COO-. They can donate two pair of electrons. Thirdly, ambidentate are those ligands which has two different atoms to donate one pair of electron. For example, Consider SCN. Here, sulfur can donate one pair of electrons. So, it is known as Ti-Sin-I2S. Also, N has the ability to donate one pair of electrons, so it is known as thiocyanide to N. Hence remember that in this molecule, both sulfur and nitrogen have the ability to donate one pair of electrons. Fourthly, polydentate are those ligands which can donate many long pair of electrons like EDTA. EDTA stands for polycyclic acid. ethylene diamine tetra acetate. It can donate six pair of electrons so it is a polydentate. Thus noted down these four types of ligands. Now what is homoleptic complex and heteroleptic complex? Well, the complex in which central metal atom is bonded to only one type of ligand is called homoleptic complex. For example, consider this complex compound. Here the central metal atom is iron, the ligand is Cn, and this K is counter ion. Here there is only one type of ligand Cn, so it is a homoleptic complex. On the other hand, the complex in which central metal atom is bonded to more than one type of ligands is called a heteroleptic complex. For example, Consider this complex compound. Here the central atom is cobalt, NH3 is one type of ligand and chlorine is another type of ligand. There are two types of ligands so it is a hydroleptic complex. Thus note it down this difference between homoleptic complex and heteroleptic complex. Now let me teach you the nomenclature of complex compounds. Firstly, Let me teach you the types of ligands based on charts and their names. Well, on the basis of charts, ligands are neutral, anionic and cationic. Neutral ligands are NH3 ammonia, nitrocyl, H2 aqua, CO carbonyl, CS thio carbonyl. Secondly, in case of anionic ligands, we use the suffix O. like ClCl, BrBr, H-H, SO-2-Sulfato, NH2-1-Amido, NH2-2-Amido, N-3-Azido, O-2-Oxo, etc. Thirdly, in case of cationic ligands, we use the suffix IUM, like NO-Nitrosonium, NO2-Nitronium, NH2-NH3-Hydrazinium. Here let me teach you one important MCQs. Remember that NH4 or ammonium ion is not a ligand because it has no lone pair of electrons. Sometime it is asked an MCQs. So note down the names of all these ligands. Now it is time to learn nomenclature of coordination compounds. I use these rules to name any coordination compound. I write the name of coordination compound from left to right. Secondly, if there are more than one ligand, I follow alphabetic order. Thirdly, if the coordination sphere is negative, I put ATE with central metal atom. Now consider this coordination compound. Firstly, I find the oxidation state of central atom. I write silver plus 2NH3 plus Cl equals 0. The oxidation state of NH3 is 0 and that of chlorine is negative 1. I write silver plus 2 into 0 plus negative 1 equals 0. I get silver equals positive 1. Secondly, I find the charge on coordination sphere. We know that This chlorine is a negative ion and it carries negative one charge. So, the charge on coordination sphere is positive one. Hence, the charge on coordination sphere is positive. Now, listen carefully. Firstly, I write the ligand and it is ammonia. There are two molecules of ammonia. I write di. Secondly, I write the central metal atom. It is silver. and the oxidation state of silver is 1. So, I write here 1. Thirdly, the counter ion is chloride. So, I get di ammonia silver chloride. Remember that I capitalized the first letter. There is no gap between ligand and central metal atom. But there is a gap between coordination sphere name and counter ion name. Secondly, consider this coordination compound. As usual, firstly, I find the oxidation number of the central atom. I write 2K plus nickel. plus 4 chlorine equals 0. R2 into positive 1 plus nickel plus 4 into negative 1 equals 0. After calculation, I get nickel equals 2. Secondly, I find the charge on the coordination sphere. We know that potassium is positive 1. Hence, the charge on the coordination sphere is negative. Now, according to the first rule, we go from left to right. Here, I write potassium. In case of coordination sphere, I write ligand name which is Chloro. There are four of them. I write Tetra. Secondly, the central metal atom is Nickel. According to the third rule, if the charge on coordination sphere is negative, we put Ate with central metal atom. So I put Ate with Nickel. The oxidation state of Nickel is positive too. I write here too. This I gate. Potassium tetra chloronicolate 2. Thirdly, consider this coordination compound. Firstly, I calculate the oxidation state of a central metal atom. I write iron plus 6 fluorine equals negative 4 because the charge on this coordination compound is negative 4. I write iron plus 6 into negative 1 equals negative 4. I get iron is equal to 2. Secondly, I find the charge on coordination sphere. We can see that the charge on coordination sphere is already given and it is negative 4. Now, as usual, I write the name of ligand. Fluoro, there are 6 of them. I write Hexa. For iron Fe, I write Fe. The coordination sphere has negative charge. I write Ate. The oxidation number of iron is 2. I write Fe. And lastly, I write iron because it has negative 4 charge. Thus, the name of this compound is hexafluoroferrate 2 iron iron. Fourthly, consider this coordination compound. I find the oxidation state of a central metal atom. I write chromium plus 4H2O plus 2 chlorine plus NO3 equals 0. We know that H2O is a neutral molecule, the oxidation number of chlorine is negative 1 and that of nitrate ion is also negative 1. I write chromium plus 4 into 0 plus 2 into negative 1 plus negative 1 equals 0. I get chromium equals positive 3. Secondly, I find the charge on coordination sphere. We know that nitrate ion carry negative charge. So the charge on coordination sphere is positive. Now there are two types of ligands, water and chlorine. I write aqua and chloro. And the first position, I write aqua. Because a comes first then c in alphabetic order. Now there are four water molecules. I write tetra. There are two chlorine ions. I write di. The central atom is chromium. and its oxidation state is 3. The counter ion is nitrate. Thus the name of this coordination compound is tetra-aqua-dichlorochromium-3-nitrate. Lastly, let me teach you one advance question. Name this coordination compound. Well, the oxidation state of cobalt is positive 3 and the charge on coordination sphere is positive. Now here, there are two types of ligand. Chlorine and En. I write Chloro. We have already learned that En is Ethane 1,2-Diamine. We can see that there are two chlorine ions. I write Di. Secondly, there are two En molecule. I write here Base. Because Di is already there and Diamine, I cannot write two times Di and a same molecule. Let me repeat it. I cannot write 2 times die and a same molecule. So if there are 3 molecules of EN, I will write trace. We can see that central atom is cobalt and its oxidation number is 3. The counter ion is chloride. So the name of this compound is dichlorobisethan-1,2-diamine-cobalt-3-chloride. Therefore, using this simple trick, we can easily name any coordination compound. Now let me teach you the most difficult topic in the most easy way. What is crystal field theory? Well, consider a metal atom like iron. We know that iron is a d-block element. I mean iron has 5 degenerate d-orbitals. Now listen carefully. When a ligand approaches near to the iron, the 5 d-orbitals of the iron break down. Let me repeat it. When a ligand approaches near to the iron, the 5 d-orbitals of the iron break down. Now here if I ask you that how these 5d orbitals break down and what happens to them after breaking? Your answer is simple. Crystal field theory explains the breaking or splitting of the 5d orbitals when ligand approaches towards them. Let me repeat it. Crystal field theory explains the breaking or splitting of the 5d orbitals when ligand approaches towards them. So remember that crystal field theory explains the splitting of 5d orbitals after interacting with ligand. Now let me teach you my personal trick of learning this topic. Firstly, I learned CFD and chips of d orbitals. We know that d has 5 degenerate orbitals dxy, dyz, dxz, dx square minus 5 square and dz square. Now try to understand to draw the structure of these five orbitals using this simple trick. In case of d x y, I draw x axis and y axis. I draw y axis and z axis. Then I draw x axis and z axis. Then I draw x axis and y axis. Lastly, I draw only z axis. Now we know that the shape of the d orbital is double-double. I draw double-double shape between these axes like this. Now I just draw double-double shape between these axes. In case of dx square minus y squared and dz square, this square always tells me to draw me along the x axes. So I draw one loop along x axes and another one along y axes. In case of dz square. I draw one loop along z-axis and I put a ring in the center. Now we can see that the lobes of the orbitals are between axes in these three orbitals. While the lobes of the orbitals are along axes in these two orbitals. Here many students do not understand CFD because they do not learn these three points. So you must learn these three points. In case of octahedral complex, ligand approaches or attack along axes. For example, only these two orbitals have lobes along axes. So, ligand will attack on this side or on this side along axes. Secondly, in case of tetrahedral complex, ligand approaches or attack between axes. For example, these three orbitals have lobes between axes so ligand will attack either from this side, this side or this side. Thirdly, when ligand attacks on lobes of orbital, there is repulsion between them and energy increases. Let me repeat this important point. When ligand attack on lobes of orbitals, there is repulsion between them and energy increases. Therefore, we conclude that, affligants attack along x-axis, energy of only dx squared minus y squared and dz squared increases. We know that it only happens in octahedral complex. Secondly, affligant attacks between x-axis, energy of dxy, dyz and dxz increases. We know that, It only happens in tetrahedral complex. To summarize this whole concept, we say that in case of octahedral complex, the energy of these two orbital increases and the energy of these three orbital decreases. While in case of tetrahedral, the situation is opposite. The energy of these three orbitals increases and the energy of these two orbitals decreases. Thus noted down all these important points. Now let me teach you the trick to draw the structures of CFT for octahedral. Well, consider these five degenerate d orbitals. The average energy of these five orbitals increases. when ligand approach near the d orbitals. A lot of students are confused that why energy of d orbital increases. I teach them a very simple example. I always say when you are in danger, your adrenaline level pumps up and you get ready. Similarly, when ligand approach toward these orbitals, they get excited and their average energy increases. Remember that I use the term average energy. After interaction, the 5d orbitals split into two sets. We have already learned that in case of octahedral complex, the ligand approach along axes and the energy of dx square minus y square and dz square increases. While that of dxy, dyz and dz, dxz decreases. And we call these three orbitals as T2g. This is the energy gap between them which we call CFS, crystal field splitting. Here we can see that the energy of these two orbital increases from the average energy level. So I write positive 0.6 delta. Secondly, the energy of these three orbital decreases. So I write... negative 0.4 delta. So the value of EG equals positive 0.60 and that of T2G equals negative 0.40. Thus noted down this diagram of octahedral. Now let me teach you crystal field theory for tetrahedral. Well consider these five degenerate d orbitals. The average energy of these orbitals increases. when a ligand approaches nearer them. After interaction, the 5d orbital splits into two sets. We have already learned that in case of tetrahedral complex, ligand approaches between axes. So the energy of dxy, dyz, dxz increases and that of dx square minus y square and dz square decreases. Remember that in case of tetrahedral This is T2 set and this is E set of orbitals. We do not try G with them. Secondly, this is the energy gap or CFS between these two sets of orbitals. Thirdly, the energy of these three orbital increases from average level. It is positive 0.4 delta and the energy of these two orbital decreases from average level. So its value is negative. Thus the value of T2 equals positive 0.40 delta and that of E equals negative 0.60 delta. Therefore using this trick we can easily write CFD for octahedral and tetrahedral complex. Finally let me teach you my personal trick of VBT and hybridization of coordination compounds which I have already explained in my previous video. Now consider this coordination compound. Firstly, I find the oxidation state of a central atom. I write cobalt plus 6 NH3 equals positive 3 or cobalt plus 6 N2O equals positive 3. I get cobalt equals positive 3. So the oxidation state of cobalt is positive 3. Secondly, I find the number of electrons of cobalt. We know that the atomic number of cobalt is 27. Now listen carefully. I always subtract oxidation state of central atom minus 18 from the atomic number of central atom. Let me repeat it. I always subtract oxidation state of a central atom minus 18 from the atomic number of central atom. Here, the oxidation state of cobalt is positive 3. So I write cobalt equals 27 minus 3 minus 18. I get 6 electrons. Now we can see that here the ligand is NH3 and it is a strong ligand. Remember that in case of a strong ligand the electrons of the central atom get paired. While in case of weak ligand the electrons of the central atom do not get paired. Now I use the police trick. I say dsp doctor. We know that d has 5 orbitals S has one orbital, P has three orbitals and D has five orbitals. Now listen carefully. I take the six electrons of central atom and I place it in this D orbital. 1, 2, 3, 4, 5, 6. I have paired these electrons because the ligand is strong. We can see that there are six ligands of NH3 so I select six completely empty orbitals. 1, 2, 3, 4, 5, 6. We know that each ligand can donate one pair of electrons. There are 6 ligands. So, 6 ligand will donate 6 pair of electrons. Thus, I fill these 6 orbitals. Now, these 2 d orbitals, this 1 s orbital and these 3 p orbitals are hybridized. So, the hybridization of this compound is D2sp3. and it is inner orbital complex because inner d orbital is hybridized. Secondly, the geometry of this compound is octahedral. Thirdly, we can see that all the electrons are paired, so the magnetic nature of this compound is dimagnetic. Remember that if electrons are paired, it is low spin complex. If electrons are unpaired, it is a high spin complex. Fourthly, What about magnetic moment? Well, we know that magnetic moment mu equals under root n n to n plus 2. Here n is the number of unpaired electrons. We can see that there is no unpaired electron present. So, n equals zero. Thus I get mu equals zero Bohr magneton. Fifthly, what about spin multiplicity? We know that Spin multiplicity is 2s plus 1. Here s is the spin quantum number. For every unpaired electron s equals 1 upon 2. Here we can see that there is no unpaired electron. So s equals 0. I put s equals 0 in this equation. After calculation I get 1. So the spin multiplicity of this coordination compound is 1. If you want to learn more about this trick, Watch our video and its link is given in the description. I hope that you have learned the chapter of coordination compounds.

Full chapter of coordination compounds. Firstly, let me teach you the basic introduction of coordination compounds. Consider the d-block elements like iron, copper, zinc, etc. These all are metals.

We know that metals love to lose electrons. But here, the case is opposite. I mean, these metal elements love to accept a pair of electrons. This iron loves to accept a pair of electrons. This copper loves to accept a pair of electrons and this zinc loves to accept a pair of electrons.

Here if you ask me why these metal love to accept a pair of electrons, I would say it is due to two reasons. Firstly, these metals have empty shell. Secondly, they complete their octet rule to get stability. That's why they love to accept a pair of electrons.

So these metals accept a pair of electrons. Now let me teach you that how metals accept a pair of electrons. Well, consider a metal atom like copper and a neutral molecule like NH3.

This metal atom of copper has empty shell. I mean copper can easily accommodate to upcoming electrons. On the other hand, there is one lone pair of electrons present on nitrogen. So, NH3 will donate this lone pair of electrons to copper and copper will accept this one pair of electrons.

Hence, coordinate bond is formed between NH3 and copper. We call this compound as coordinate compound because there exists coordination bound between them. Therefore, we define coordination compounds as the complex compound in which metal atom is bonded to a number of ions or a molecule is called coordination compound.

Let me repeat it. The complex compound in which metal atom is bonded to a number of ions or a molecule is called coordination compound. Remember that this NH3 donate a pair of electrons to a metal atom. We call it ligand. So we define ligand as Any ion or molecule attached to a central metal atom by donating a pair of electron is called ligand.

Thus remember that any coordination compound is made up of two parts. One part is a metal atom like copper and the second part is a ligand like NH3. Thus noted down all these important points. Now let me teach you some basic terms of coordination compounds.

Let's consider this coordination compound. There are three species in this coordination compound. Chromium, NH3 and Chlorine.

Now listen carefully. This metal atom of chromium is known as central atom. We know that NH3 donate a pair of electrons.

So it is elegant. This 6 means 6 molecules of NH3. One molecule of NH3 donate one pair of electrons.

6 molecules of NH3 donate 6 pair of electrons. We call this 6 as coordination number. Remember that coordination number shows the number of pair of electrons donated.

This chromium, NH3 and the square bracket is called coordination sphere or coordination entity. This chlorine outside the bracket is called counter ion. Secondly, Consider this complex compound.

Here cobalt is a metal atom and we call it central atom. This chlorine is ligand, this 2 is the coordination number of chlorine, this NH3 is another ligand and this 4 is its coordination number. This whole thing and the square bracket is known as coordination sphere or coordination entity. This chlorine outside the square bracket or coordination sphere is called counter ion.

Thus noted down all these basic terms of coordination compounds. Now let me teach you Warner's theory. Well, Warner is the first scientist who successfully studied coordination compounds.

He took cobalt chloride and he react it with NH3. After reaction, he got cobalt chloride. .6NH3. Secondly, he added silver nitrate to cobalt.6NH3.

After reaction, he got 3 moles of HCl3 or silver chloride. in the form of white precipitate or white ppt. Now listen carefully, Warner baba thought that 3 chlorine are weakly or indirectly bonded to cobalt and 6 NH3 strongly or directly bonded to cobalt that's why chlorine love to react with silver.

So he rearranged this compound. He writes the central metal atom of cobalt then he takes NH3 and there are six molecules of NH3. He puts square bracket around these two species. Finally, he writes chlorine and there are three of them.

So, Werner Baba therefore say that six NH3 are directly bonded to cobalt and three chlorine are indirectly bonded to cobalt. He called this compound as coordination compound. Remember that This is the first coordination compound in chemistry.

Now what is the conclusion of Warner's theory? Well, Warner-Baba states that every metal atom of coordination compounds has two types of valencies. Primary valency and secondary valency.

Primary valency is satisfied by negative ion. For example, here in this coordination compound, chlorine is a negative ion. It satisfies the primary valency of cobalt.

While secondary valency is satisfied by positive ion or neutral molecule. For example, here in this compound, the NH3 is a neutral molecule. It satisfies the secondary valency of cobalt.

Secondly, primary valency represents the oxidation state of a central metal atom while secondary valency represents the coordination number. Thirdly, primary valency is denoted by dots and secondary valency is denoted by line. Fourthly, primary valency is ionizable.

For example, chlorine is ionizable and aqueous solution. While secondary valency is not ionizable, for example, NH3 is directly bonded to cobalt, it is not ionized. Thus noted down all these important points. Now how can we calculate primary and secondary valencies of a central metal atom? Well, consider these coordination compounds.

We have already learned that primary valency is the oxidation number of a central atom and secondary valency is coordination number. Now in case of this molecule, the central atom is cobalt. I am going to find its oxidation state. I write cobalt plus 6 into NH3 plus 3 chlorine equals 0. We know that NH3 is a neutral molecule so its oxidation state is 0 and that of chlorine is positive 1. So I write cobalt plus 6 into 0 plus 3 into negative 1 equals 0. After calculation I get positive 3. The oxidation state of cobalt is positive 3. So, the primary valency of cobalt is 3. Here, we can see that there are 6 ligands of NH3. So, the secondary valency of cobalt is 6. Secondly, in case of this molecule, I find the oxidation state of iron.

I write iron plus 6 into Cn equals negative 4. Because the overall charge on this coordination compound is negative 4. We know that The oxidation state of Cn is negative 1. I write iron plus 6 into negative 1 equals negative 4. After calculation, I get positive 2. The oxidation state of iron is positive 2. So, the primary valency of iron is positive 2. Here, there are 6 ligands of Cn. The coordination number is 6. So, the secondary valency of iron is 6. Therefore, Using this trick, we can easily calculate the primary and secondary valences of a central metal atom. Now let me teach you the structure of Warner's theory.

Well, consider these four coordination compounds. I write cobalt and I draw six lines as the octahedral structure. Similarly, in case of these three compounds, I write cobalt and I draw six lines as a octahedral structure. We know that primary valency is denoted by dots and secondary valency is denoted by line. Now listen carefully.

These are 6 molecules of NH3. I draw these 6 molecules of NH3 around the central atom cobalt. Now I take these 3 ions of chlorine.

They satisfy the primary valency. So I randomly draw 3 dot lines. and I put three chlorine ions around cobalt.

Remember that you can draw this dotted line in any direction around the central atom. Secondly, in this coordination compound, there are five molecules of NH3. I write them around the central atom. Now one secondary valency is left behind. So I place one chlorine here.

Remember that This chlorine can fulfill the secondary valency and primary valency of the central metal atom. I draw two dotted lines and I place two chlorine. Similarly, I write 4 NH3 around the central atom. I write here one chlorine and one chlorine. Here, these two chlorine act as a primary valency and secondary valency.

I draw one dotted line and I write chlorine. Lastly, I write 3 NH3 around the central atom. I write chlorine, chlorine and chlorine. Here, these three chlorine act as a primary valency and secondary valency. Now, Warner-Baba added silver nitrate to each of these coordination compound.

After chemical reaction, he got different amount of products. In case of this molecule, these three dotted lines mean 3 chlorine are out of coordination sphere. Let me repeat it. These 3 dotted lines mean 3 chlorine are out of coordination sphere.

Thus here 3 moles of silver chloride is formed. In this case, 2 chlorine are out of coordination sphere. So, 2 moles of silver chloride are formed.

In this case, only 1 chlorine is out of coordination sphere. So, 1 mole of silver chloride is formed. In case of this coordination compound, no chlorine is out of coordination sphere so no silver chloride is formed. Therefore, the structure of this compound is cobalt NH3 6 square bracket Cl 3. The structure of this compound is cobalt NH3 5 chlorine square bracket Cl 2. The structure of this compound is cobalt NH3 4 Cl 2. square bracket Cl and the structure of this compound is cobalt NH3 3 Cl 3 square bracket. So these are the different structures of coordination compounds.

Hence noted down all these structures. Now we will learn about different types of ligands. To learn the types of ligands we must learn density of a ligand. It is defined as the number of long pair of electrons Donated by a ligand is called Density of a ligand.

For example, water can donate one pair of electrons, so its Density is 1. Now there are 4 types of ligands, unidentate or monodentate, bidentate, ambidentate and polydentate. Unidentate are those ligands which can donate one pair of electrons. For example, chlorine ion. Bromine iron, cyanide iron, carbon monoxide etc.

Secondly, bidentate are those ligands which can donate two pair of electrons. For example, En. Remember that En stands for Ethan 1,2-diamine.

We know that Ethan is CH2CH2. Di means 2 and Amines mean NH2. 1,2 means at first carbon. and at second carbon. So this is the structure of En which we often use in coordination compounds.

Hence note it down. The another one is oxylate COO-bond COO-. They can donate two pair of electrons. Thirdly, ambidentate are those ligands which has two different atoms to donate one pair of electron.

For example, Consider SCN. Here, sulfur can donate one pair of electrons. So, it is known as Ti-Sin-I2S. Also, N has the ability to donate one pair of electrons, so it is known as thiocyanide to N. Hence remember that in this molecule, both sulfur and nitrogen have the ability to donate one pair of electrons.

Fourthly, polydentate are those ligands which can donate many long pair of electrons like EDTA. EDTA stands for polycyclic acid. ethylene diamine tetra acetate. It can donate six pair of electrons so it is a polydentate.

Thus noted down these four types of ligands. Now what is homoleptic complex and heteroleptic complex? Well, the complex in which central metal atom is bonded to only one type of ligand is called homoleptic complex.

For example, consider this complex compound. Here the central metal atom is iron, the ligand is Cn, and this K is counter ion. Here there is only one type of ligand Cn, so it is a homoleptic complex.

On the other hand, the complex in which central metal atom is bonded to more than one type of ligands is called a heteroleptic complex. For example, Consider this complex compound. Here the central atom is cobalt, NH3 is one type of ligand and chlorine is another type of ligand. There are two types of ligands so it is a hydroleptic complex.

Thus note it down this difference between homoleptic complex and heteroleptic complex. Now let me teach you the nomenclature of complex compounds. Firstly, Let me teach you the types of ligands based on charts and their names.

Well, on the basis of charts, ligands are neutral, anionic and cationic. Neutral ligands are NH3 ammonia, nitrocyl, H2 aqua, CO carbonyl, CS thio carbonyl. Secondly, in case of anionic ligands, we use the suffix O. like ClCl, BrBr, H-H, SO-2-Sulfato, NH2-1-Amido, NH2-2-Amido, N-3-Azido, O-2-Oxo, etc.

Thirdly, in case of cationic ligands, we use the suffix IUM, like NO-Nitrosonium, NO2-Nitronium, NH2-NH3-Hydrazinium. Here let me teach you one important MCQs. Remember that NH4 or ammonium ion is not a ligand because it has no lone pair of electrons.

Sometime it is asked an MCQs. So note down the names of all these ligands. Now it is time to learn nomenclature of coordination compounds. I use these rules to name any coordination compound.

I write the name of coordination compound from left to right. Secondly, if there are more than one ligand, I follow alphabetic order. Thirdly, if the coordination sphere is negative, I put ATE with central metal atom.

Now consider this coordination compound. Firstly, I find the oxidation state of central atom. I write silver plus 2NH3 plus Cl equals 0. The oxidation state of NH3 is 0 and that of chlorine is negative 1. I write silver plus 2 into 0 plus negative 1 equals 0. I get silver equals positive 1. Secondly, I find the charge on coordination sphere.

We know that This chlorine is a negative ion and it carries negative one charge. So, the charge on coordination sphere is positive one. Hence, the charge on coordination sphere is positive. Now, listen carefully.

Firstly, I write the ligand and it is ammonia. There are two molecules of ammonia. I write di. Secondly, I write the central metal atom. It is silver.

and the oxidation state of silver is 1. So, I write here 1. Thirdly, the counter ion is chloride. So, I get di ammonia silver chloride. Remember that I capitalized the first letter.

There is no gap between ligand and central metal atom. But there is a gap between coordination sphere name and counter ion name. Secondly, consider this coordination compound. As usual, firstly, I find the oxidation number of the central atom.

I write 2K plus nickel. plus 4 chlorine equals 0. R2 into positive 1 plus nickel plus 4 into negative 1 equals 0. After calculation, I get nickel equals 2. Secondly, I find the charge on the coordination sphere. We know that potassium is positive 1. Hence, the charge on the coordination sphere is negative.

Now, according to the first rule, we go from left to right. Here, I write potassium. In case of coordination sphere, I write ligand name which is Chloro.

There are four of them. I write Tetra. Secondly, the central metal atom is Nickel. According to the third rule, if the charge on coordination sphere is negative, we put Ate with central metal atom.

So I put Ate with Nickel. The oxidation state of Nickel is positive too. I write here too.

This I gate. Potassium tetra chloronicolate 2. Thirdly, consider this coordination compound. Firstly, I calculate the oxidation state of a central metal atom. I write iron plus 6 fluorine equals negative 4 because the charge on this coordination compound is negative 4. I write iron plus 6 into negative 1 equals negative 4. I get iron is equal to 2. Secondly, I find the charge on coordination sphere. We can see that the charge on coordination sphere is already given and it is negative 4. Now, as usual, I write the name of ligand.

Fluoro, there are 6 of them. I write Hexa. For iron Fe, I write Fe. The coordination sphere has negative charge.

I write Ate. The oxidation number of iron is 2. I write Fe. And lastly, I write iron because it has negative 4 charge. Thus, the name of this compound is hexafluoroferrate 2 iron iron.

Fourthly, consider this coordination compound. I find the oxidation state of a central metal atom. I write chromium plus 4H2O plus 2 chlorine plus NO3 equals 0. We know that H2O is a neutral molecule, the oxidation number of chlorine is negative 1 and that of nitrate ion is also negative 1. I write chromium plus 4 into 0 plus 2 into negative 1 plus negative 1 equals 0. I get chromium equals positive 3. Secondly, I find the charge on coordination sphere.

We know that nitrate ion carry negative charge. So the charge on coordination sphere is positive. Now there are two types of ligands, water and chlorine.

I write aqua and chloro. And the first position, I write aqua. Because a comes first then c in alphabetic order.

Now there are four water molecules. I write tetra. There are two chlorine ions. I write di. The central atom is chromium.

and its oxidation state is 3. The counter ion is nitrate. Thus the name of this coordination compound is tetra-aqua-dichlorochromium-3-nitrate. Lastly, let me teach you one advance question. Name this coordination compound. Well, the oxidation state of cobalt is positive 3 and the charge on coordination sphere is positive.

Now here, there are two types of ligand. Chlorine and En. I write Chloro. We have already learned that En is Ethane 1,2-Diamine. We can see that there are two chlorine ions.

I write Di. Secondly, there are two En molecule. I write here Base. Because Di is already there and Diamine, I cannot write two times Di and a same molecule. Let me repeat it.

I cannot write 2 times die and a same molecule. So if there are 3 molecules of EN, I will write trace. We can see that central atom is cobalt and its oxidation number is 3. The counter ion is chloride.

So the name of this compound is dichlorobisethan-1,2-diamine-cobalt-3-chloride. Therefore, using this simple trick, we can easily name any coordination compound. Now let me teach you the most difficult topic in the most easy way. What is crystal field theory?

Well, consider a metal atom like iron. We know that iron is a d-block element. I mean iron has 5 degenerate d-orbitals.

Now listen carefully. When a ligand approaches near to the iron, the 5 d-orbitals of the iron break down. Let me repeat it.

When a ligand approaches near to the iron, the 5 d-orbitals of the iron break down. Now here if I ask you that how these 5d orbitals break down and what happens to them after breaking? Your answer is simple.

Crystal field theory explains the breaking or splitting of the 5d orbitals when ligand approaches towards them. Let me repeat it. Crystal field theory explains the breaking or splitting of the 5d orbitals when ligand approaches towards them. So remember that crystal field theory explains the splitting of 5d orbitals after interacting with ligand.

Now let me teach you my personal trick of learning this topic. Firstly, I learned CFD and chips of d orbitals. We know that d has 5 degenerate orbitals dxy, dyz, dxz, dx square minus 5 square and dz square. Now try to understand to draw the structure of these five orbitals using this simple trick. In case of d x y, I draw x axis and y axis.

I draw y axis and z axis. Then I draw x axis and z axis. Then I draw x axis and y axis.

Lastly, I draw only z axis. Now we know that the shape of the d orbital is double-double. I draw double-double shape between these axes like this. Now I just draw double-double shape between these axes.

In case of dx square minus y squared and dz square, this square always tells me to draw me along the x axes. So I draw one loop along x axes and another one along y axes. In case of dz square.

I draw one loop along z-axis and I put a ring in the center. Now we can see that the lobes of the orbitals are between axes in these three orbitals. While the lobes of the orbitals are along axes in these two orbitals. Here many students do not understand CFD because they do not learn these three points. So you must learn these three points.

In case of octahedral complex, ligand approaches or attack along axes. For example, only these two orbitals have lobes along axes. So, ligand will attack on this side or on this side along axes.

Secondly, in case of tetrahedral complex, ligand approaches or attack between axes. For example, these three orbitals have lobes between axes so ligand will attack either from this side, this side or this side. Thirdly, when ligand attacks on lobes of orbital, there is repulsion between them and energy increases.

Let me repeat this important point. When ligand attack on lobes of orbitals, there is repulsion between them and energy increases. Therefore, we conclude that, affligants attack along x-axis, energy of only dx squared minus y squared and dz squared increases.

We know that it only happens in octahedral complex. Secondly, affligant attacks between x-axis, energy of dxy, dyz and dxz increases. We know that, It only happens in tetrahedral complex.

To summarize this whole concept, we say that in case of octahedral complex, the energy of these two orbital increases and the energy of these three orbital decreases. While in case of tetrahedral, the situation is opposite. The energy of these three orbitals increases and the energy of these two orbitals decreases. Thus noted down all these important points.

Now let me teach you the trick to draw the structures of CFT for octahedral. Well, consider these five degenerate d orbitals. The average energy of these five orbitals increases.

when ligand approach near the d orbitals. A lot of students are confused that why energy of d orbital increases. I teach them a very simple example.

I always say when you are in danger, your adrenaline level pumps up and you get ready. Similarly, when ligand approach toward these orbitals, they get excited and their average energy increases. Remember that I use the term average energy.

After interaction, the 5d orbitals split into two sets. We have already learned that in case of octahedral complex, the ligand approach along axes and the energy of dx square minus y square and dz square increases. While that of dxy, dyz and dz, dxz decreases. And we call these three orbitals as T2g.

This is the energy gap between them which we call CFS, crystal field splitting. Here we can see that the energy of these two orbital increases from the average energy level. So I write positive 0.6 delta. Secondly, the energy of these three orbital decreases.

So I write... negative 0.4 delta. So the value of EG equals positive 0.60 and that of T2G equals negative 0.40.

Thus noted down this diagram of octahedral. Now let me teach you crystal field theory for tetrahedral. Well consider these five degenerate d orbitals. The average energy of these orbitals increases.

when a ligand approaches nearer them. After interaction, the 5d orbital splits into two sets. We have already learned that in case of tetrahedral complex, ligand approaches between axes. So the energy of dxy, dyz, dxz increases and that of dx square minus y square and dz square decreases.

Remember that in case of tetrahedral This is T2 set and this is E set of orbitals. We do not try G with them. Secondly, this is the energy gap or CFS between these two sets of orbitals. Thirdly, the energy of these three orbital increases from average level. It is positive 0.4 delta and the energy of these two orbital decreases from average level.

So its value is negative. Thus the value of T2 equals positive 0.40 delta and that of E equals negative 0.60 delta. Therefore using this trick we can easily write CFD for octahedral and tetrahedral complex. Finally let me teach you my personal trick of VBT and hybridization of coordination compounds which I have already explained in my previous video.

Now consider this coordination compound. Firstly, I find the oxidation state of a central atom. I write cobalt plus 6 NH3 equals positive 3 or cobalt plus 6 N2O equals positive 3. I get cobalt equals positive 3. So the oxidation state of cobalt is positive 3. Secondly, I find the number of electrons of cobalt.

We know that the atomic number of cobalt is 27. Now listen carefully. I always subtract oxidation state of central atom minus 18 from the atomic number of central atom. Let me repeat it. I always subtract oxidation state of a central atom minus 18 from the atomic number of central atom.

Here, the oxidation state of cobalt is positive 3. So I write cobalt equals 27 minus 3 minus 18. I get 6 electrons. Now we can see that here the ligand is NH3 and it is a strong ligand. Remember that in case of a strong ligand the electrons of the central atom get paired. While in case of weak ligand the electrons of the central atom do not get paired.

Now I use the police trick. I say dsp doctor. We know that d has 5 orbitals S has one orbital, P has three orbitals and D has five orbitals. Now listen carefully. I take the six electrons of central atom and I place it in this D orbital.

1, 2, 3, 4, 5, 6. I have paired these electrons because the ligand is strong. We can see that there are six ligands of NH3 so I select six completely empty orbitals. 1, 2, 3, 4, 5, 6. We know that each ligand can donate one pair of electrons.

There are 6 ligands. So, 6 ligand will donate 6 pair of electrons. Thus, I fill these 6 orbitals. Now, these 2 d orbitals, this 1 s orbital and these 3 p orbitals are hybridized. So, the hybridization of this compound is D2sp3.

and it is inner orbital complex because inner d orbital is hybridized. Secondly, the geometry of this compound is octahedral. Thirdly, we can see that all the electrons are paired, so the magnetic nature of this compound is dimagnetic.

Remember that if electrons are paired, it is low spin complex. If electrons are unpaired, it is a high spin complex. Fourthly, What about magnetic moment?

Well, we know that magnetic moment mu equals under root n n to n plus 2. Here n is the number of unpaired electrons. We can see that there is no unpaired electron present. So, n equals zero. Thus I get mu equals zero Bohr magneton. Fifthly, what about spin multiplicity?

We know that Spin multiplicity is 2s plus 1. Here s is the spin quantum number. For every unpaired electron s equals 1 upon 2. Here we can see that there is no unpaired electron. So s equals 0. I put s equals 0 in this equation. After calculation I get 1. So the spin multiplicity of this coordination compound is 1. If you want to learn more about this trick, Watch our video and its link is given in the description.

I hope that you have learned the chapter of coordination compounds.

Transcript for:Coordination Compounds

Transcript for:
Coordination Compounds