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Lanthanide Contraction Q&A on Magnetism The magnetic susceptibility plot of a manganese compound (1) is shown below. 3 -1 The room temperature MT value is 4.36 cm mol K. Assuming an isotropic g- value (a) Calculate the number of unpaired electron in 1 and what is the oxidation state of Mn atom? so number of unpaired electrons = 5 = 2.828 T = n(n + 2) so so Oxidation state of Mn is + 2 assuming n = 5 so = 2.828 4.36 so = 5.9 so = 5.91 b) The susceptibility plot presented above is not corrected for temperature independent paramagnetism. How is this expected to affect your answers given in a) and b). No TIP as there are no excited state for Mn2+. 2 Magnetic properties of complexes with A and E Ground Terms Temperature independence is often the case, however the fit with the calculated spin only moment is not always good… second-order spin-orbit coupling can mix in some orbital contribution from higher T terms of the same multiplicity as the ground term 2+ 6 Why is it such a good fit for Mn ? A1g ground term 2+ 5 6 The ground state for Mn d high spin is A1g - a sextet. There are no higher sextet T terms, and so no mixing of excited and ground terms is possible Q2. [Fe(dtc)3] complex shows µ= 2.2 B.M at low T which enhances to 5.92 B.M. at higher T indicating spin-state isomerism. State the % equilibrium mixture? (At room temperature µobs= 4.3 B.M.) Ans. µ= 2.2 B.M for the low spin isomer (low T) µ= 5.92 B.M for the high spin isomer (high T) µ= 4.3 B.M in room T (observed one) LS HS ; Keq= / (1-) ; = mole fraction of the high spin state 2 2 2 (µobs) = (µhs) + (1-) (µls) Hence, (4.3)2 = (5.92)2 + (1-) (2.2)2 So, = 0.45 giving rise to 45% of the high spin isomer and (1- ) = 0.55 leading to 55% of the low spin isomer Q3. Justify the following statement : Bis(salicylaldoximato)NiII is colourless and diamagnetic, but when recrystallized from a coordinating solvent like pyridine, green coloured paramagnetic crystals are obtained. Square planar octahedral d-Block metal chemistry: the second and third row metals Contextually, considerable part of the Periodic Table Understanding electronic configuration trend • The electronic configurations of the ground state M(g) atoms in second/third row transition series change rather irregularly with increasing atomic number more so than for the first row metals • the nd and (n+1)s atomic orbitals are closer in energy for n = 4 or 5 than for n = 3 Hence, for ions of the first row metals, the electronic configuration is generally dn and this gives a certain amount of order to discussions of the properties of M2+ and M3+ ions. Simple Mn+ cations of the heavier metals are rare, and it is not possible to discuss their chemistry in terms of simple redox couples (e.g. M3+/M2+) as applicable for the most of the first row metals. Outer electronic configurations Ground electronic Metallic radii configuration Encapsulated 4d/5d elements exhibiting anomalous electronic configuration Ref: https://doi.org/10.1590/S0100-40422013000600027 Rich Considerations: • each subshell energy and level is split into two Suter levels, and , related Diagram to the spin of the electrons i..e spin up and spin down respectively Ref: https://doi.org/10.1590/S0100-40422013000600027 ❖ Noteworthy that, irregularities of electronic configuration among transition elements primitively owes to huge stability of half/full filled d orbitals. However, cumulative contributions from following factors rationalise the anomalous electronic configurations : • (i) nuclear electronic attraction •(ii) shielding of one electron by several other electrons •(iii) inter-electronic repulsion •(iv) the exchange-energy forces etc. • Besides, Rich-Suter diagram plays remedial role ❖ In the second-row transition metals, electron–electron repulsions within the 4d subshell cause additional irregularities in electron configurations that are not easily predicted. ❖ For example, Nb and Tc, with atomic numbers 41 and 43, both have a half-filled 5s subshell, with 5s14d4 and 5s14d6 valence electron configurations, respectively. ❖ Further complications occur among the third-row transition metals, in which the 4f, 5d, and 6s orbitals are extremely close in energy. ❖ Although La has a 6s25d1 valence electron configuration, the valence electron configuration of the next element—Ce—is 6s25d04f2. ✓From this point through element 71, added electrons enter the 4f subshell, giving rise to the 14 elements known as the lanthanides. ✓ After the 4f subshell is filled, the 5d subshell is populated, producing the third row of the transition metals. ❖In accordance with maximum spin multiplicity fulfilment norms, levels will be progressively filled with spin up electrons first followed by the spin down electron filling. As known, s and s can have maximum one electrons respectively ❖ As known, d and d can have maximum five electrons each respectively ❖ Nb : The levels sequence from previous R-S diagram : 5s- < 4d- < 5s- < 4d- This explains that, one electron should occupy 5s followed by remaining four electrons occupation in 4d leading to s1d4 electronic configuration for Nb ground state ❖ Mo : The levels sequence from previous R-S diagram : 5s- < 4d- < 5s- < 4d- leading to s1d5 electronic configuration for Mo ground state. Moreover, half filled stability of d subshell rationalises the observation ❖Pd : The levels sequence from previous R-S diagram : 4d- < 4d- < 5s- < 5s- This explains that, considering the level sequencing, each of tow sets of d orbitals will occupy 5 electrons and Pd having in total 10 electrons led to occupation of all ten electrons in 4d level with 5s remaining vacant. ❖ Pt: The levels sequence from previous R-S diagram : 5d- < 6s- < 5d- < 6s- leading to s1d9 electronic configuration for Mo ground state. Moreover, half filled stability of d subshell rationalises the observation Assess yourself : 1. Is Tungsten ground state associated with unique electronic configuration: Justify your answer. 2. What is the electronic configuration of Re? Any anomaly do you observe? Rationalise? 3. Which ion is more stable Ag+ or Ag2+ and why? Understanding Atomic Radii trend ✓As we saw in the s-block and p-block elements, the size of neutral atoms of the d-block elements gradually decreases from left to right across a row, due to an increase in the effective nuclear charge (Zeff) with increasing atomic number. ✓ In addition, the atomic radius increases down a group, just as it does in the s and p blocks. Because of the Lanthanide contraction, however, the increase in size between the 3d and 4d metals is much greater than between the 4d and 5d metals The 4d and 5d Transition Series • There is a decrease in size going from left to right for each of the series. There is an increase in radius in going from the 3d to the 4d metals but 4d and 5d metals are remarkably similar in size. This phenomenon is the result of lanthanide contraction. • In the lanthanide series electrons are filling the 4f orbitals. The 4f orbitals are buried in the interior of these atoms, the increasing nuclear charge causes the radii of the lanthanide elements to decrease significantly going from left to right. Lanthanide contraction It is observed that in lanthanide series, there is a progressive decrease in the atomic and ionic radii with increasing atomic number . This regular decrease with increase in atomic number is called lanthanide contraction. This is due to the weak shielding of f orbitals. These f orbitals are unable to counter balance the effect of increasing nuclear charge because of which the size keeps on decreasing with increase in atomic umber. Genesis : As we move along the period from left to right in lanthanide series, the atomic number increases i.e. number of protons keeps on increasing.For every proton added in the nucleus the extra electron goes to the same 4f orbital The 4f orbital shows poor shielding effect because of which there is a gradual increase in the effective nuclear charge experienced by the outer electrons. Thus, the attraction of the nucleus for the electrons in the outermost shell increases in atomic number. Effects of the lanthanoid contraction pairs of metals in a triad (Zr and Hf, Nb and Ta etc.) are of similar radii. This is due to the lanthanoid contraction: the steady decrease in size along the series of lanthanoid metals Ce–Lu which lie between La and Hf in the third row of the d-block. This observation is due to the presence of a filled 4f level the shielding of one 4f electron by another is less than for one d electron by another, and as the nuclear charge increases from La to Lu, there is a fairly regular decrease in the size of the 4f n sub-shell. Atomic and ionic sizes: The Lanthanide Contraction As the atomic number increases, each succeeding element contains one more electron in the 4f orbital and one proton in the nucleus. The 4f electrons are ineffective in screening the outer electrons from the nucleus causing imperfect shielding. As a result, there is a gradual increase in the nucleus attraction for the outer electrons. Consequently gradual decrease in size occur. This is called lanthanide contraction. Relativistic effects If Einstein’s theory of relativity is combined with quantum mechanics, in which case they are attributed to relativistic effects. According to the theory of relativity, the mass m of a particle increases from its rest mass m0 when its velocity v approaches the speed of light, c, and m is then given by the equation: 28 For a one-electron system, the Bohr model of the atom leads to the velocity of the electron being expressed by the equation: where Z = atomic number, e = charge on the electron, 0 = permittivity of a vacuum, h = Planck constant For n = 1 and Z = 1, v/c is only (1/137) but for Z = 80, v/c = 0.58 leading to m = 1.2m0.
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