Important Books of Inorganic Chemistry for Preparation of CSIR-NET | GATE | IIT-JAM | TIFR [with PDF links]

Important Books of Inorganic Chemistry for Preparation of CSIR-NET | GATE | IIT-JAM | TIFR [with PDF links]

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The Post includes all important books for Inorganic Chemistry preparation for various M.Sc level examinations, including CSIR-UGC NET, GATE, IIT-JAM, TIFR, BARC etc. The books and links provided below are to make these books accessible to all and develop book reading culture among aspirants. 

The Books contain links to Buy Hard Copy (from online websites like Amazon.in) and also Free PDF download. There can be differences in the edition of book provided as hard copy link and PDF file.

INORGANIC CHEMISTRY:

1. Coordination Chemistry by Ajai Kumar 

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Jahn Tellar Distortion/Theorem/Effect/Problems

Jahn Tellar Distortion/Theorem/Effect/Problems

By   06:24   No comments:

The Jahn-Teller effect explains why certain 6 coordinated complexes undergo distortion to assume distorted octahedral (tetragonal) geometry. "This is a geometric distortion of a non-linear molecular system that reduces its symmetry and energy". This distortion is typically observed among octahedral complexes where the two axial bonds can be shorter or longer than those of the equatorial bonds. This effect can also be observed in tetrahedral compounds. This effect is dependent on the electronic state of the system.

Introduction

In 1937, Hermann Jahn and Edward Teller postulated a theorem stating that "stability and degeneracy are not possible simultaneously unless the molecule is a linear one," in regards to its electronic state.This leads to a break in degeneracy which stabilizes the molecule and by consequence, reduces its symmetry. Since 1937, the theorem has been revised which House-croft and Sharpe have eloquently phrased as "any non-linear molecular system in a degenerate electronic state will be unstable and will undergo distortion to form a system of lower symmetry and lower energy, thereby removing the degeneracy." This is most commonly observed with transition metal octahedral complexes, however, it can be observed in tetrahedral compounds as well.

For a given octahedral complex, the five d atomic orbitals are split into two degenerate sets when constructing a molecular orbital diagram. These are represented by the sets' symmetry labels: t_2g (d_xz, d_yz, d_xy) and e_g (d_z² and d_x²−y²). When a molecule possesses a degenerate electronic ground state, it will distort (Jahn-Teller effect) to remove the degeneracy and form a lower energy (and by consequence, lower symmetry) system. The octahedral complex will either elongate or compress the z ligand bonds as shown in Figure 1 below:

Figure 1: Jahn-Teller distortions for an octahedral complex.

When an octahedral complex exhibits elongation, the axial bonds are longer than the equatorial bonds. For a compression, it is the reverse; the equatorial bonds are longer than the axial bonds. Elongation and compression effects are dictated by the amount of overlap between the metal and ligand orbitals. Thus, this distortion varies greatly depending on the type of metal and ligands. In general, the stronger the metal-ligand orbital interactions are, the greater the chance for a Jahn-Teller effect to be observed.

Elongation (Z-Out distortion)

Elongation Jahn-Teller distortions occur when the degeneracy is broken by the stabilization (lowering in energy) of the d orbitals with a z component, while the orbitals without a z component are destabilized (higher in energy) as shown in Figure 2 below:

Figure 2: Illustration of tetragonal distortion (elongation) for an octahedral complex.

This is due to the dxy and dx2−y2 orbitals having greater overlap with the ligand orbitals, resulting in the orbitals being higher in energy. Since the dx2−y2orbital is antibonding, it is expected to increase in energy due to elongation. The dxy orbital is still nonbonding, but is destabilized due to the interactions. 

Jahn-Teller elongations are well-documented for copper(II) octahedral compounds. A classic example is that of copper(II) fluoride as shown in Figure 3.

Figure 3: Structure of octahedral copper(II) fluoride.

Notice that the two axial bonds are both elongated and the four shorter equatorial bonds are the same length as each other. According the theorem, the orbital degeneracy is eliminated by distortion, making the molecule more stable based on the model presented in Figure 2.

Compression (Z-In distortion)

Compression Jahn-Teller distortions occur when the degeneracy is broken by the stabilization (lowering in energy) of the d orbitals without a z component, while the orbitals with a z component are destabilized (higher in energy) as shown in Figure 4 below:

Figure 4: Illustration of tetragonal distortion (compression) for an octahedral complex.

This is due to the z-component d orbitals having greater overlap with the ligand orbitals, resulting in the orbitals being higher in energy. Since the d_z² orbital is anti-bonding, it is expected to increase in energy due to compression. The d_xz and d_yz orbitals are still non-bonding, but are destabilized due to the interactions.

Electronic Configurations and JT Distortions:

For Jahn-Teller effects to occur in transition metals there must be degeneracy in either the t_2g or e_g orbitals. The electronic states of octahedral complexes are classified as either low spin or high spin. The spin of the system is dictated by the chemical environment. This includes the characteristics of the metal center and the types of ligands.

Distortions are more pronounced if the degeneracy occurs in an eg orbital

Low Spin

Figure 5 (below) shows the various electronic configurations for low spin octahedral complexes, The Red Colored electronic configurations does not show any distortions whereas the Black Colored electronic configurations does:

Figure 5: Low spin octahedral coordination diagram (red indicates no degeneracies possible, thus no Jahn-Teller effects).

The figure illustrates that low spin complexes with d³, d⁵, d⁸, and d¹⁰ electrons that do no exhibit Jahn-Teller distortions. These electronic configurations correspond to a variety of transition metals. Many common examples include Cr³⁺, Co³⁺, and Ni²⁺.

High Spin

Figure 6 (below) shows the various electronic configurations for high spin octahedral complexes, The Red Colored electronic configurations does not show any distortions whereas the Black Colored electronic configurations does:

Figure 6: High spin octahedral coordination diagram (red indicates no degeneracies possible, thus no Jahn-Teller effects)

The figure illustrates that low spin complexes with d³, d⁵, d⁸, and d¹⁰ electrons cannot have Jahn-Teller distortions. In general, degenerate electronic states occupying the eg orbital set tend to show stronger Jahn-Teller effects. This is primarily caused by the occupation of these high energy orbitals. Since the system is more stable with a lower energy configuration, the degeneracy of the eg set is broken, the symmetry is reduced, and occupations at lower energy orbitals occur.

Spectroscopic Effects of JT Distortion

Jahn-Teller distortions can be observed using a variety of spectroscopic techniques. In UV-VIS absorption spectroscopy, distortion causes splitting of bands in the spectrum due to a reduction in symmetry (O_h to D_4h). Consider a hypothetical molecule with octahedral symmetry showing a single absorption band. If the molecule were to undergo Jahn-Teller distortion, the number of bands would increase as shown in Figure 7 below:

Figure 7: Hypothetical absorption spectra of an octahedral molecule (left) and the same molecule with Jahn-Teller elongation (right). The red arrows indicate electronic transitions.

A similar phenomenon can be seen with IR and Raman vibrational spectroscopy. The number of vibrational modes for a molecule can be calculated using the 3n - 6 rule (or 3n - 5 for linear geometry) rule. If a molecule exhibits an O_h symmetry point group, it will have fewer bands than that of a Jahn-Teller distorted molecule with D_4h symmetry. Thus, one could observe Jahn-Teller effects through either IR or Raman techniques. This effect can also be observed in EPR experiments as long as there is at least one unpaired electron.

Table 1: Examples of Jahn-Teller distorted complexes

CuBr2	4 Br at 240 pm 2 Br at 318 pm
CuCl2	4 Cl at 230 pm 2 Cl at 295 pm
CuCl 2.2H 2O	2 O at 193 pm 2 Cl at 228 pm 2 Cl at 295 pm
CsCuCl3	4 Cl at 230 pm 2 Cl at 265 pm
CuF2	4 F at 193 pm 2 F at 227 pm
CuSO4.4NH3.H2O	4 N at 205 pm 1 O at 259 pm 1 O at 337 pm
K2CuF4	4 F at 191 pm 2 F at 237 pm
KCuAlF6	2 F at 188 pm 4 F at 220 pm
CrF2	4 F at 200 pm 2 F at 243 pm
KCrF3	4 F at 214 pm 2 F at 200 pm
MnF3	2 F at 209 pm 2 F at 191 pm 2 F at 179 pm

The Jahn-Teller Theorem predicts that distortions should occur for any degenerate state, including degeneracy of the t_2g level, however distortions in bond lengths are much more distinctive when the degenerate electrons are in the e_g level.

Summary

We can summarize all the above contents, the type of distortion and the strength of distortion for different type of complexes in the below table:

Change in geometry and arrangement of orbitals

Change in geometry and arrangement of orbitals in different Distortions

Consequences of JT Distortion

Jahn Tellar Distortion leads to some amazing consesquences, some of which are given below:

1. Stability of Cu²⁺ ion

As given by the Irwing-William Series, the relative stabilities of complexes in (+2) oxidation states of different metals is as follows:

Ba²⁺ < Sr²⁺ < Ca²⁺ < Mg²⁺ < Mn²⁺ < Fe²⁺ < Co²⁺ < Ni²⁺ < Cu²⁺ > Zn²⁺

The Cu²⁺ complexes are more stable than Zn²⁺ complexes due to the Jahn-Tellar distortion observed in Cu²⁺ complexes which is absent in the later.

2. Instability of Au²⁺ ion

Au²⁺ disproportionate into Au³⁺ and Au¹⁺ because in Au²⁺ one electron is present in very high energy,  which easily gets removed.

Orbital Splitting in ion Au²⁺ion (The 9th electron is in very high energy state)

3. Instability of Chelating Complexes

[Cu(en)₃]²⁺ is Not Stable. Why?

The above complex is a chelating complex of Cu²⁺ .This shows Jahn Tellar distortion. Cu-N bonds at axial positions try to elongate (due to JT Effect) but this elongation causes strain in the molecule. Hence the complex becomes unstable.

Practice Questions

1. Why do d³ complexes not show Jahn-Teller distortions?
2. Does the spin system (high spin v. low spin) of a molecule play a role in Jahn-Teller effects?
3. What spectroscopic method would one utilize in order to observe Jahn-Teller distortions in a diamagnetic molecule?
4. What spectroscopic method(s) would one utilize in order to observe Jahn-Teller distortions in a paramagnetic molecule?
5. Why are Jahn-Teller effects most prevalent in inorganic (transition metal) compounds?

Answers

1. Complexes with d³ electron configurations do not show Jahn-Teller distortions because there is no ground state degeneracy.
2. Yes. Examining the d⁵ electron configuration, one finds that the high spin scenario contains all singly occupied d orbitals (no degeneracy). However, the low spin d⁵ electron configuration shows degeneracy, which then leads to possible Jahn-Teller effects.
3. UV-VIS absorption spectroscopy is one of the most common techniques for observing these effects. In general, it is independent of magnetism (diamagnetic v. paramagnetic). Thus, one would see the effect in the spectrum of UV-VIS absorption analysis. Note that EPR requires at least one unpaired electron, and therefore not EPR active.
4. Inorganic, specifically transition metal, complexes are most prevalent in showing Jahn-Teller distortions due to the availability of d orbitals. The most common geometry that the Jahn-Teller effect is observed is in octahedral complexes (see Figures 2, 4, 5 and 6 above) due to the splitting of d orbitals into two degenerate sets. Due to stabilization, the degeneracies are removed, making a lower symmetry and lower energy molecule.
5. In addition to UV-VIS absorption, one can also employ EPR spectroscopy if a molecule possesses and unpaired electron.

References

1. Jahn, H. A.; Teller, E. Proc. R. Soc. London A, 1937, 161, 220-235. DOI: 10.1098/rspa.1937.0142
2. Housecroft, C.; Sharpe, A. G. Inorganic Chemistry. Prentice Hall, 3rd Ed., 2008, p. 644. ISBN: 978-0-13-175553-6
3. Billy, C.; Haendler, H. A. J. Am. Chem. Soc., 1957, 79, 1049–1051. DOI: 10.1021/ja01562a011
4. Chemistry Libretexts (https://chem.libretexts.org/)
5. Wikipedia (https://en.wikipedia.org/wiki/Main_Page)
6. Google (for images)

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Simplest Method to find Bond Order

Simplest Method to find Bond Order

By   00:34   No comments:

Bond Order questions are very common in CSIR NET and other entrance exams. In this post, we will discuss about this topic and at the end I will share a very useful trick that helps to solve questions based on bond order very quickly. 

Let us start with some introduction. 

Bond Order 

Bond order is a number that tells us how strong the bond between two atoms forming a molecule is. If the bond order is high, the bond is strong and so the molecule is stable. If the bond order is low, the molecule is unstable.

How do we calculate Bond Order? (Usual Method)

There are many methods to finding bond order. One of the most famous and most used formulas states:

where, N_b = No. of electrons in Bonding M.O.
and N_a = No. of electrons in Anti-Bonding M.O.

This is USUAL method for finding bond order, but this method is NOT good for solving problems. Why?

1. You need to properly find the number of electrons in bonding molecular orbitals and antibonding molecular orbitals.
2. It takes a lot of time.

 Also See: Hybridisation In Five Simple Steps

Simple Method of Calculating Bond Order: 

With a little memorization and practice, you can find out the bond order of a species just by looking at it. The method that we will be using IS NOT applicable everywhere, but it works in 90% of  CSIR NET, IIT JAM, and Other Entrance Examination's problems. So, you see the worth in learning it.

 The Basic Principle Behind this Trick is: N₂ has 14 electrons and its bond order is 3. Every electron added or subtracted to 14, reduces the bond order by 0.5.

For example, if the number of electrons is 15, then the bond order is 3 – 0.5 = 2.5. Done!


Here's a Table which shows this relation easily:

You can follow these steps:

1. Just count the total number of electrons. 
2. Add number of Negative Charges and Subtract number of Positive Charges on molecule. 
3. Find the difference of total number of electrons from 14. Call it ‘n’.
4. Bond order = 3  – 0.5n

 Few Examples:

Q.1. Find the Bond Order of CO molecule?

Total number of electrons = 6 + 8 = 14.
 B.O. = 3. 
 Done! simple.



Q2. Find the B.O. of O₂^2-.
Total number of electrons = 8 x 2 + 2(for the – charge) = 18.
B.O. = 3 – 0.5 x 4 = 1



Q3. Which of the following has highest bond order?
         a) O₂²⁺ 
         b) O₂^2- 
         c) F₂^2- 
         d) O₂⁺

Just count the total number of electrons for each.
O₂²⁺ has 14 electrons. B.O = 3
O₂^2- has 18 electrons. B.O = 1.
You can find out the B.O. for the other two, and conclude that the answer is (a).

Q4. Which of the following orders regarding the bond order is correct?

1. O₂^– > O₂ > O₂⁺
2. O₂^– < O₂ < O₂⁺
3. O₂^– > O₂ < O₂⁺
4. O₂^– < O₂ >O₂⁺

Once again, we count the number of electrons for each of these species.
For O₂^– no. of electrons = 17. B.O. = 1.5
For O₂, no. of electrons = 16. B.O. = 2
For O₂⁺, no. of electrons = 15. B.O. = 2.5

So, the answer is (b).

NOTE: This method will work for any species that have between 8 and 20 electrons. Most problems asked in Entrance Exams have electrons between 8 and 20, therefore it is very handy.

Application Of Bond Order:

With Calculation of Bond Order you can also predict Bond Length, Bond Strength, Stability of Molecule etc.
As:

1. Bond Order is directly proportional to Bond Strength, means a molecule having larger value of Bond Order will be more strongly held and will require more energy to break it's bonds.
2. Bond Order is Inversely proportional to Bond Length, means a molecule having large value of Bond Order will have smalller bond length than a molecule having less value of Bond Order.
3. Molecules Having Bond Order value of ZERO have no existance, e.g. He₂ molecule have no significance becoz its bond order is zero.

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