Question

M. Laing has suggested a new arrangement of the periodic table (J.Chem.Ed.1989, 66, 746).  Answer the following questions after reading his arguments.

a) Laing's periodic table places carbon, silicon, titanium, and zirconium in the same column.  How is this justified on the basis of electron configuration?

b) The formulas of the common binary chlorides of beryllium and magnesium are BeCl2 and MgCl2.  Predict the formula for the binary chloride of cadmium

c) What (if any) value do you see in aligning F and Cl with Mn, Tc, and Re?

Answer 1

a) Laing's periodic table emphasizes on the importance of subgroup relationships and makes it simple to obtain the maximum amount of chemical information from a glance at the periodic table. This periodic table is based on the electronic configuration of the elements. For example,

6C 1s22s2 2p2 or [He] 2s2 2p2

14Si 1s2 2s2 2p63s2 3p2 or [Ne] 3s2 3p2

22Ti 1s2 2s2 2p6 3s2 3p63d2 4s2 or [Ar] 3d2 4s2

40Zr 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p64d25s2 or [Kr] 4d2 5s2

From the electronic configuration of these 4 elements we can see that they have similar number of valence electrons. Thus the chemical properties of the elements are also expected to be similar which in turn justifies the position/arrangement of the elements in the periodic table.

b) The electronic configuration of Beryllium, magnesium and cadmium are as follows

4Be 1s22s2  or [He] 2s2

12Mg 1s2 2s2 2p63s2 or [Ne] 3s2

48Cd 1s2 2s2 2p6 3s2 3p6 3d10 4s24p64d10 4s2 or [Kr] 4d10 4s2

from these electronic configurations we can see that the valence shell of the elements is completely filled with the valence electrons first entering the d-orbital followed by the s-orbital. All these elements have a tendency to loose the s-orbital electrons and become di-positive ions. Now since Be and Mg form BeCl2 and MgCl2 respectively, Cd is expected to form CdCl2. Hence Cd will form CdCl2.

c) The electronic configuration of F, Cl, Mn, Tc and Re are as follows

9F 1s22s2 2p5 or [He] 2s2 2p5

17Cl 1s2 2s2 2p6 3s2 3p5 or [Ne] 3s2 3p5

25Mn 1s2 2s2 2p6 3s2 3p6 3d5 4s2 or [Ar]3d5 4s2

43Tc 1s2 2s2 2p6 3s2 3p6 3d54s2 4p64d5 5s2 or [Kr] 4d5 5s2

75Re 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p64d10 5s2 5p6 4f14 5d5 6s2 or [Xe] 4f145d5 6s2

Depending on the number of electrons (ie 7 in each case) in the valence shell of the elements, these elements expected to show similar properties. But there is a catch, the valence electron enters the p-orbital in case of F and Cl whereas Mn, Tc and Re have the valence electrons accommodated in the d-orbital of the elements. So, as a result of this F and Cl require 1 electron to complete the octet and be stable. Thus they have a high tendency to form uni-negative ions. whereas for the Mn, Tc and Re since there are 5 electrons in the d-orbitals, it is already half filled and hence stable. so , it does not have a tendency to gain one electron and become uni-negative anion. Rather, they can donate their s-orbital electrons and form a di-positive cation and be stabilised. So, the behaviour of the F and Cl and Mn, Tc and Re are not similar. This could be one of the demerits of the Laing's periodic table.

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