Calculate cfse for d6 tetrahedral complex. show the number of unpaired electrons also.

calculate cfse for d6 tetrahedral complex. show the number of unpaired electrons also.

Calculate cfse for d6 tetrahedral complex. show the number of unpaired electrons also.


Predict the number of unpaired electrons in a tetrahedral d ion and in square planar d ion.
Note: [If answer is 1 and 2, then represent as 12.]
Verified by Toppr
Correct option is A)
The number of unpaired electrons in a tetrahedral d ion is 4 .
The number of unpaired electrons  in square planar d ion is 1.
Electronic distribution in a tetrahedral d ion is .
Number of unpaired electrons is 4
Electronic distribution in square planar d ion .
Number of unpaired electrons is 1


Crystal Field Stabilization Energy, Pairing, and Hund’s Rule

The splitting of the d-orbitals into different energy levels in transition metal complexes has important consequences for their stability, reactivity, and magnetic properties. Let us first consider the simple case of the octahedral complexes [M(H2O)6]3+, where M = Ti, V, Cr. Because the complexes are octahedral, they all have the same energy level diagram:

Octahedral energy level diagrams of titanium 3+ with 1 electron, Vanadium 3+ with 2 electrons, Chromium 3+ with 3 electrons, and Chromium 2+ with 4 electrons. Each diagram has two energy levels with the bottom, T 2 g, having three orbitals and the top, E g, having two orbitals. Chromium 2+ has four electrons so it has a high spin and low spin diagram. High spin has the fourth electron in an e g orbital. Low spin has the fourth electron share a t 2 g orbital with another electron.

The Ti3+, V3+, and Cr3+ complexes have one, two and three d-electrons respectively, which fill the degenerate t2g orbitals singly. The spins align parallel according to Hund’s rule, which states that the lowest energy state has the highest spin angular momentum.

For each of these complexes we can calculate a crystal field stabilization energy, CFSE, which is the energy difference between the complex in its ground state and in a hypothetical state in which all five d-orbitals are at the energy barycenter.

For Ti3+, there is one electron stabilized by 2/5 ΔO, so CFSE=(1)(25)(ΔO)=25ΔOCFSE=−(1)(25)(ΔO)=−25ΔO

Similarly, CFSE = -4/5 ΔO and -6/5 ΔO for V3+ and Cr3+, respectively.

For Cr2+ complexes, which have four d-electrons, the situation is more complicated. Now we can have a high spin configuration (t2g)3(eg)1, or a low spin configuration (t2g)4(eg)0 in which two of the electrons are paired. What are the energies of these two states?

High spin: CFSE=(3)(25)ΔO+(1)(35)ΔO=35ΔOCFSE=(−3)(25)ΔO+(1)(35)ΔO=−35ΔO

Low spin: CFSE=(4)(25)ΔO+P=85ΔO+PCFSE=(−4)(25)ΔO+P=−85ΔO+P, where P is the pairing energy

Energy difference = O + P

The pairing energy P is the energy penalty for putting two electrons in the same orbital, resulting from the electrostatic repulsion between electrons. For 3d elements, a typical value of P is about 15,000 cm-1.

The important result here is that a complex will be low spin if ΔO > P, and high spin if ΔO < P.

Because ΔO depends on both the metals and the ligands, it determines the spin state of the complex.

Rules of thumb:

3d complexes are high spin with weak field ligands and low spin with strong field ligands.

High valent 3d complexes (e.g., Co3+ complexes) tend to be low spin (large ΔO)

4d and 5d complexes are always low spin (large ΔO)

Note that high and low spin states occur only for 3d metal complexes with between 4 and 7 d-electrons. Complexes with 1 to 3 d-electrons can accommodate all electrons in individual orbitals in the t2g set. Complexes with 8, 9, or 10 d-electrons will always have completely filled t2g orbitals and 2-4 electrons in the eg set.

d-orbital energy diagrams for high spin and low spin Cobalt 2+ complexes. Both have seven electrons occupying orbitals. Low spin has all six T 2 g orbitals filled with one electron in an E g orbital. High spin has two electrons in E g orbitals and one half empty T 2 g orbital.

d-orbital energy diagrams for high and low spin Co2+ complexes, d7



Examples of high and low spin complexes:

[Co(H2O)62+] contains a d7 metal ion with a weak field ligand. This complex is known to be high spin from magnetic susceptibility measurements, which detect three unpaired electrons per molecule. Its orbital occupancy is (t2g)5(eg)2.

We can calculate the CFSE as (5)(25)ΔO+(2)(35)ΔO=45ΔO−(5)(25)ΔO+(2)(35)ΔO=−45ΔO

[Co(CN)64-] is also an octahedral d7 complex but it contains CN, a strong field ligand. Its orbital occupancy is (t2g)6(eg)1 and it therefore has one unpaired electron.

In this case the CFSE is (6)(25)ΔO+(1)(35)ΔO+P=95ΔO+P.−(6)(25)ΔO+(1)(35)ΔO+P=−95ΔO+P.

Magnetism of transition metal complexes

Compounds with unpaired electrons have an inherent magnetic moment that arises from the electron spin. Such compounds interact strongly with applied magnetic fields. Their magnetic susceptibility provides a simple way to measure the number of unpaired electrons in a transition metal complex.

If a transition metal complex has no unpaired electrons, it is diamagnetic and is weakly repelled from the high field region of an inhomogeneous magnetic field. Complexes with unpaired electrons are typically paramagnetic. The spins in paramagnets align independently in an applied magnetic field but do not align spontaneously in the absence of a field. Such compounds are attracted to a magnet, i.e., they are drawn into the high field region of an inhomogeneous field. The attractive force, which can be measured with a Guoy balance or a SQUID magnetometer, is proportional to the magnetic susceptibility (χ) of the complex.

The effective magnetic moment of an ion (µeff), in the absence of spin-orbit coupling, is given by the sum of its spin and orbital moments:


In octahedral 3d metal complexes, the orbital angular momentum is largely “quenched” by symmetry, so we can approximate:


We can calculate µs from the number of unpaired electrons (n) using:


Here µB is the Bohr magneton (= eh/4πme) = 9.3 x 10-24 J/T. This spin-only formula is a good approximation for first-row transition metal complexes, especially high spin complexes. The table below compares calculated and experimentally measured values of µeff for octahedral complexes with 1-5 unpaired electrons.

Ion Number of
moment /μB
moment /μB
Ti3+ 1 1.73 1.73
V4+ 1 1.68–1.78
Cu2+ 1 1.70–2.20
V3+ 2 2.83 2.75–2.85
Ni2+ 2 2.8–3.5
V2+ 3 3.87 3.80–3.90
Cr3+ 3 3.70–3.90
Co2+ 3 4.3–5.0
Mn4+ 3 3.80–4.0
Cr2+ 4 4.90 4.75–4.90
Fe2+ 4 5.1–5.7
Mn2+ 5 5.92 5.65–6.10
Fe3+ 5 5.7–6.0

The small deviations from the spin-only formula for these octahedral complexes can result from the neglect of orbital angular momentum or of spin-orbit coupling. Tetrahedral d3, d4, d8 and d9 complexes tend to show larger deviations from the spin-only formula than octahedral complexes of the same ion because quenching of the orbital contribution is less effective in the tetrahedral case.

Summary of rules for high and low spin complexes:

3d complexes: Can be high or low spin, depending on the ligand (d4, d5, d6, d7)

4d and 5d complexes: Always low spin, because ΔO is large

Maximum CFSE is for d3 and d8 cases (e.g., Cr3+, Ni2+) with weak field ligands (H2O, O2-, F,…) and for d3-d6 with strong field ligands (Fe2+, Ru2+, Os2+, Co3+, Rh3+, Ir3+,…)

Irving-Williams series. For M2+ complexes, the stability of the complex follows the order Mg2+ < Mn2+ < Fe2+ < Co2+ < Ni2+ < Cu2+ > Zn2+. This trend represents increasing Lewis acidity as the ions become smaller (going left to right in the periodic table) as well as the trend in CFSE. This same trend is reflected in the hydration enthalpy of gas-phase M2+ ions, as illustrated in the graph at the right. Note that Ca2+, Mn2+, and Zn2+, which are d0, d5(high spin), and d10 aquo ions, respectively, all have zero CFSE and fall on the same line. Ions that deviate the most from the line such as Ni2+ (octahedral d8) have the highest CFSE.

Graph with various elements on the x-axis and change in enthalpy on the y-axis. From left to right, the elements on the y-axis are: calcium, scandium, titanium, vanadium, chromium, magnesium, iron, cobalt, nickle, copper, and zinc. Pink line with a positive slope showing increasing change in enthalpy, or decreasing stability from Calcium to Zinc. Blue line starting at Magnesium showing deviations from the pink line. Nickle has the highest deviation.

The hues and spectra of compounds containing transition metals

Because the energy of their d-d transitions may be in the visible region of the spectrum, transition metal complexes can exhibit stunning coloration. This was covered in the previous section. As a result of breaking the Laporte selection criterion, the colors associated with octahedral complexes are quite muted, and the transitions between hues are very subtle. In centrosymmetric complexes, this rule states that transitions from g to g and from u to u are not allowed to take place. d-orbitals have g (gerade) symmetry, hence d-d transitions are Laporte-forbidden. Octahedral complexes, on the other hand, have the ability to absorb light if, while the molecule is vibrating, they briefly deviate from their centrosymmetrical shape. Due to the fact that the spin selection rule prohibits spin flips during optical transitions, the excited state will always have the same spin multiplicity as the ground state.

Because there are many different ways in which the d-electrons might occupy the t2g and eg orbitals, the spectra of even the most basic transition metal complexes are fairly complex. This is due to the fact that the t2g and eg orbitals are filled by d-orbitals. For instance, if we think about a d2 complex like V3+(aq), we know that the two electrons may be located in any one of the five d-orbitals, and they can have either a spin-up or spin-down orientation. This is true even if we take the complex as a whole. For a d2 complex, there are really 45 distinct possible configurations of this kind, which are referred to as microstates, that do not violate the Pauli exclusion principle. In most cases, we are only interested with the six states that have the lowest energy. These are states in which both electrons occupy individual orbitals in the t2g set, and all of their spins are oriented either clockwise or anticlockwise.

Four test tubes. From left to right, lilac liquid, green liquid, blue liquid, yellow liquid.

From left: [V(H2O)6]2+ (lilac), [V(H2O)6]3+ (green), [VO(H2O)5]2+ (blue) and [VO(H2O)5]3+ (yellow).

We can see how these microstates play a role in electronic spectra when we consider the d-d transitions of the [Cr(NH3)6]3+ ion. This ion is d3, so each of the three t2g orbitals contains one unpaired electron. We expect to see a transition when one of the three electrons in the t2g orbitals is excited to an empty eg orbital. Interestingly, we find not one but two transitions in the visible.

The reason that we see two transitions is that the electron can come from any one of the t2g orbitals and end up in either of the eg orbitals. Let us assume for the sake of argument that the electron is initially in the dxy orbital. It can be excited to either the dz2 or the dx2-y2 orbital:

dxydz2dxy→dz2 (higher energy)

dxydx2y2dxy→dx2−y2 (lower energy)

The first transition is at higher energy (shorter wavelength) because in the excited state the configuration is (dyz1dxz1dz21). All three of the excited state orbitals have some z-component, so the d-electron density is “piled up” along the z-axis. The energy of this transition is thus increased by electron-electron repulsion. In the second case, the excited state configuration is (dyz1dxz1dx2-y21), and the d-electrons are more symmetrically distributed around the metal. This effect is responsible for a splitting of the d-d bands by about 8,000 cm-1. We can show that all other possible transitions are equivalent to one of these two by symmetry, and hence we see only two visible absorption bands for Cr3+ complexes.

F.A.Q calculate cfse for d6 tetrahedral complex. show the number of unpaired electrons also.

1. How many electrons are found in the d7 tetrahedral complexes that are not paired?

The square planar d7 ion has one electron that is not coupled with another electron in its structure.

2. How many electrons are found in d6 low spin that are not coupled with another electron?

Answer. Because electron pairing takes place, there are no unpaired electrons in a low spin octahedral complex of d6 configuration. This is because there are no unpaired electrons.

3. How many electrons that aren’t linked up do you think you’ll discover in the tetrahedral structure?

There are four electrons that are not coupled with another.

4. In the fire tetrahedral complex, how many electrons are found that are not coupled with another electron?

There is one electron that is not coupled with another.

See more articles in category: Wiki