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. Since the radius of the Bohr orbit is given by the equation

29 the increase in m results in an approximately 20% contraction of the radius of the 1s (n = 1) orbital; this is called relativistic contraction.

Other s orbitals are affected in a similar way and as a consequence, when Z is high, s orbitals have diminished overlap with orbitals of other atoms. p orbitals (which have a low electron density near to the nucleus) are less affected. d orbitals (which are more effectively screened from the nuclear charge by the contracted s and p orbitals) undergo a relativistic expansion; a similar argument applies to f orbitals.

30 The relativistic contraction of the s orbitals means that for an atom of high atomic number, there is an extra energy of attraction between s electrons and the nucleus.

31

•Similarity of 2nd and 3rd transition series i.e. 4d and 5d series:

•The atomic sizes of second row transition elements and third row transition elements are almost similar. This is also an effect of lanthanide contraction. As we move down the from form 4d to 5d series, the size must increase but it remains almost same due to the fact that the 4f electrons present in the 5d elements show poor shielding effect. Atomic radii section contd. Radii difference (slight noted) Radii similar resulting identical rendering gradual reduction of chemical behaviour pertinent resemblance in chemical behaviour • The atomic radii of 4d/5d elements tend to decrease starting from Y/La upto middle transition element which again tends to enhance beyond that upto the last elements •Notably, the radii differences among Mo-Pd or W-Pt is minuscule i.e. reduction of radii is negligibly small compared to the diminution of radii trend among Y-Nb. This is because of the increased screening effect of the 4d/5d electrons which more or less counter balance the nuclear pull exerted on the 5s/6s electrons. Then the screening effect becomes more and more pronounced thereby decreasing the attractive force between the nucleus and the outer electrons. As a result, atomic radii of Ag/Au and Cd/Hg are increased. • Rad of 3d ions < 4d ions < 5d ions due to the change in the value of n from 3 to 4 to 5 for the three series of elements. Hunt for the factors behind disparity in chemical behaviour among heavier congeners?

• Size relationship is overcome by other predominant factors i.e. • delicate penetration effect dictating ionization energy •Crystal field stabilization energy procurement via involvement of d orbitals • diversity in spatial extension of the orbitals • relativistic effect reduces binding on 5d electron and favours better participation in bonding Assess yourself : Explain the cause and two consequences of lanthanoid contraction.

Solution: The poor shielding effect of f-electrons is cause of lanthanoid contraction. Consequences There is close resemblance between 4d and 5d transition series. Ionization energy of 5d transition series is higher than 3d and 4d transition series. Difficulty in separation of lanthanides Assess yourself :

Why Zr and Hf have almost similar atomic radii? Solution Zr and Hf have almost similar atomic radii as a consequence of lanthanide contraction due to which their properties becomes similar. Understanding oxidation state trend Like 3d elements, the elements of this series also exist in various oxidation states in their compounds. This is because of the availability of several electrons in 4d/5d and 5s/6s subshells whose energies are is proximity. Hence, under various experimental conditions different number of electrons can be used from both the subshells for bonding. O.S. without brackets are stable while those within brackets are unstable

• Contrary to 3d series, for 4d/5d series elements, higher O.S progressively becomes more stable and prominent . • Following upto the previous table, we can state that, for Fe stable O.S.= +2,+3 while unstable O.S. are +4,+5,+6 • The first element Y (+3) and the last element Cd (+2) exhibit sole oxidation state due to the stable valence shell configurations Y3+ [Kr]4d0 5s0 and Cd2+ [Kr]4d105s0 • All other elements show variable oxidation states, both stable and unstable, the variability being the maximum towards the middle of the series as happens in case of elements of 3d series. • In both 4d/5d series, starting from the scratch, elements lying in the middle of series shows maximum number of O.S. i.e. Ru/Os shows max. no. of O.S. compared to its congeners • Ru/Os also show highest possible O.S. +8 Analysis:

• Oxidation state increases with the probability of available electrons to lose/share • with paramagnetic materials more tendency to diversity • with pairing up of the electrons simultaneous reduced diversity • trend : first and last elements less oxidation state with generally maxima achieved in the centre of the periodic table Analysis:

Improved overlap by d orbitals , increased penetrating by 4d and 5d orbitals ns and (n=1)d orbital gap lowered with Z increase; Ease of gaining higher valence state

• For heavy TMs Higher with less frequent O.S. lower O.S. preferred Common oxidation states: Scrutinising 4d and 5d:

+3 for Y to +8 for Ru/Os

Due to loss of ns and (n-1)d electrons

Upon moving right along row, maximum oxidation state decreases

+2 O.S. achieved for Cd, Hg owing to filled (n-1)d subshell Examining 4d and 5d:

Down a group, higher O.S. progressively becomes more stable 2- 2- CrO4 is strong oxidant but WO4 is stable

4d and 5d with lower O.S. can easily attain higher O.S.

CrIII are highly stable but MoIII/WIII are highly reactive

Higher O.S. becomes less stable/frequent along the row

Higher O.S. becomes more stable/frequent along the column Examining 4d and 5d:

Lower O.S. stabilization in 4d and 5d unconventional Only achieved by  type ligands i.e. CO

higher O.S. achieved by electronegative elements

Higher O.S. of these metals are principally fluorides and oxides

Oxides of lower O.S have ionic character and tend to be basic

Oxides of higher O.S have covalent character and tend to be acidic Highest O.S. becomes more stable on descending group from 3d to 4d to 5d :

Gr. 4 5 6 7 8 9 10 11

TiI4 VF5 CrF5 MnF4 FeBr3 CoF3 NiF4 CuBr2

ZrI4 NbI5 MoF6 TcCl6 RuF6 RhF6 PdF4 AgF3

HfI4 TaI5 WF6 ReF7 OsF6 IrF6 PtF6 AuCl6 Exercise (Appraise yourself):

• Two of the group 8 metals (Fe, Ru, and Os) form stable oxides in the +8 oxidation state. Identify these metals; describe the general physical and chemical properties,, and physical state of the oxides; and decide whether they are acidic or basic oxides.

• Ans. The +8 oxidation state corresponds to a stoichiometry of MO4. Because the heavier transition metals tend to be stable in higher oxidation states, we expect Ru and Os to form the most stable tetroxides.

• Because oxides of metals in high oxidation states are generally covalent compounds, RuO4 and OsO4 should be volatile solids or liquids.

• Finally, because oxides of transition metals in high oxidation states are usually acidic, RuO4 and OsO4 should dissolve in strong aqueous base to form oxoanions. Understanding importance of relativistic effect:

0 aB is Bohr radius aB is relativistic Bohr radius

• S orbitals are “core-penetrating” and have therefore a probability to be in the vicinity of the nucleus. this interaction will not only contract and stabilise the inner orbitals (s and p) but it will also affect the outer ones. • This contraction and stabilisation effect is called the “direct” relativistic effect. • d and f orbitals have no “core-penetrating” parts and therefore no spatial probability at the nucleus. • These d and f orbitals are influenced by the so-called “indirect” relativistic effect which is caused by the contraction of the s and p core shell, leading to a more effective shielding of the nuclear Coulomb potential, which destabilises and expands the d- and f- orbitals. • This destabilizing effect of the valence d-orbitals affects the next higher s- and p-orbitals by a small additional indirect stabilisation. On 5d elements Going from 3d to 4d, we expect broader bands, and for 5d, relativistic effects. The changes in the radial distribution function correspond to scalar relativistic effects.

In addition, for heavier elements (usually beyond La), spin-orbit coupling (also referred to as “fully relativistic” effects mix up quantum numbers. Energy levels with and without relativistic effects.. Energy levels with and without relativistic effects.. higher O.S. stabilization in 4d and 5d conventional

Lower relativistic ionization potential

Destabilized d orbitals where d electrons feel less attracted to nucleus • Thermodynamic stability of Hg(IV) over Hg(II)

• Exercise : a) The ground state structure of PtF6 has a regular octahedral structure and a closed-shell ground state. Still why this compound has a high electron affinity (ca 7 eV) ? Contrarily, No PdF6 complex has been reported so far ?

Ans. Tendency to attain higher O.S. ; efficient destabilization of the 5d orbitals - 2- Can form both PtF6 and PtF6

Apparently relativistic destabilization of 4d orbitals are not sufficient to permit their oxidation. • Exercise : b) Au(III) compounds with oxygen and the halogens are well known, and less reactive than Cu(III) or Ag(III) analogs ?

Ans. Tendency to attain higher O.S. ; efficient destabilization of the 5d orbitals i.e. - Au2F10, AuF6 exist but no such for Ag counterpart c) CsAu is stable as separate entity but isolation of Ag- needs electrochemical methods in liq. Ammonia soln as well as alkaide ion isolation needs stabilizing counter ligands?

Ans. Enhanced stability of 6s orbitals render attainment of such unusual O.S. Colour :