Question

Account for the following:
i. Manganese shows maximum number of oxidation states in 3d series.
ii. ${{E}^{\circ }}$ value for $M{{n}^{+3}}/M{{n}^{+2}}$ couple is much more positive than that for $C{{r}^{+3}}/C{{r}^{+2}}$
iii. $T{{i}^{+4}}$ is colourless whereas ${{V}^{+4}}$ is coloured in an aqueous solution.

Answer 1

The electronic configuration of the respective elements is responsible for all the characteristics asked in the question. Manganese has the maximum number of the unpaired electrons in its excited state as it has half-filled configuration and V has electrons present for excitation while Ti does not have them. So, the same thing is responsible for all these.

Complete step by step solution:
-The electrons are responsible for the characteristics of the compounds and the molecules as they are loosely held and can be easily removed, gained or shared. They are filled according to different rules like aufbau’s principle, hund’s rule and pauli’s principle.
There are 4 quantum numbers n, l, m and s. The principal quantum number (n) tells us about the shells of the atom. The azimuthal quantum number (l) tells us about the subshells of the atom. The magnetic quantum number (m) tells us about the orbitals of the atom and the spin quantum number (s) tells us about the orientation of the electrons in the orbital.
-Third quantum number is magnetic quantum number and is denoted by m. It gives us the exact orbitals. Its total value can be given as ${{n}^{2}}$ or as (2l+1) since it lies in the range (-l to +l).
-The configuration of Mn is $3{{d}^{5}}4{{s}^{2}}$. The last electrons enter the 4s subshell and are removed from it when Mn forms cation by losing two electrons. So the unpaired electrons left in it are 5 which are maximum for d-subshell as it contains 5 orbitals only.
-As the unpaired electrons are maximum in Mn ion, the oxidation states are maximum for this element as it can form any number of bonds due to these unpaired electrons. The first oxidation state is +2 for it as the cation is formed by removing 2 electrons of the 4s subshell.
-5 electrons are present in Mn d-subshell which count for the oxidation state. So 5 different oxidation states can exist in Mn which range from +2 to +7. This is why it is a good oxidizing agent.
-As the number of unpaired electrons are maximum in Mn, it has complete half-filled orbitals of the d-subshell. Thus it attains the half-filled configuration which is next to the full-filled configuration in stability. So it needs more energy of activation than Cr as it is extra stable. Cr has the configuration as $3{{d}^{5}}4{{s}^{1}}$ and so its ions will have 3 or 4 electrons only in d-subshell which is not full-filled or half-filled.
-The configuration of Ti is $3{{d}^{2}}4{{s}^{2}}$ and so when 4 electrons are removed from it, there will not be any electrons left in the d-subshell of Ti which can undergo transition to attain colour of the compound. Vanadium has electrons present for d-d transition and so shows colour. Thus we see that we can account for all of the above by the electronic configuration.

Note: Many elements do not follow this law and so their configuration is different from that depicted by the law. Lanthanoid and actinoid series do not follow this law. Also elements like Nb, Cu, Cr, Mo, Tc, Ru, Rh, Pd, Ag, Au, Pt do not follow this law.