Average values of the Duration of Electronic Transitions

Transcription

1 Average values of the Duration of Electronic Transitions Kid Spoon Forward 18 s Backward 35 s Fever Strip Forward -14 s Backward 22 s Nail Forward 4s Backward 28 s Ironing of color paper Pink paper Forward: Instantaneous Backward: 3.2 min

2 Conclusions 1. All the four material investigated are Thermoluminescent materials. They absorb heat and emit as light. 2. The backward color change is always slower than the forward color change The backward color change involves several phenomena whereas the forward is single and straightforward.

3 Atom, Cation, and Anion Cation Positively charged Smaller than the Parent Atom Anion Atom Negatively charged Bigger than the Parent Atom

4 Periodic Trend in Ionic Radii: The Group Ionic radius increases as we move down the group. The reasons for this trend are: (1) Number of shells increases; (2) Shielding also increases Periodic Trend in Ionic Radii: The Period Ionic radius decreases as we move from Left to Right in a Period. The reasons for this trend is: electrons are added to the same shell; therefore, nucleus attracts them continuously.

5 Shielding of the Valence Electron by the Inner Electrons The plot of atomic number versus atomic radius gives a measure of the shielding effect. The shielding effect is the reduction in the effective nuclear charge on the outer electrons, due to the inner electrons. It is also referred to as the screening effect (or) atomic shielding. The shielding effect also explains why valence-shell electrons are more easily removed from the atom. The Shielding Effect can be defined as: Diminishing of the force or the control of the nucleus on outer electrons by the inner electrons, which act as a curtain between the nucleus and the outer electrons.

6 Shielding Effect and Effective Nuclear Charge Slater s Rule The effective nuclear charge is the net positive charge experienced by valence electrons. It can be given approximated by the equation: Z eff = Z S, where Z is the atomic number and S is the number of shielding electrons. 1. Calculate Z eff for a valence electron in an oxygen atom Calculate Z eff for the 4s electron in a copper atom, Cu Calculate Z eff for a 3d electron in a copper atom, Cu. (1) O 8 Valence Electron (2) Cu 29 4s electron (3) Cu 29 3d electron n Z eff = Z S = 8 2 = 6 Inner electrons are electrons in the 1 st shell. n Z eff = Z S = = 2 Inner electrons are electrons in the 1 st, 2 nd, and 3 rd shells Z eff = Z S = = 17 Inner electrons are electrons in the 1 st and 2 nd shells

7 The graph above represents the change in ionic radius with atomic number. Ionic Radius is the distance between the center of the nucleus and the valence shell. In both graphs, we see that there is decrease in ionic radius with atomic number. The left graph is for Lanthanides, which are f block elements. The right graph is for fourth period elements, which include Calcium (which is a s block element) and Sc through Zn (which are d block elements called Transition Elements). For the lanthanides, there is a dramatic fall in the ionic radius whereas for the fourth period elements, the decrease is not steep. The steep decrease for lanthanides is due to lanthanide contraction, which is reduction in the size of lanthanide ions. In lanthanides, electrons are continuously added to the 4f orbitals, which are in the 4 th shell much inner to the valence shell, which is the 6 th shell. For the fourth period elements, there is only a small fall in the ionic radius, this is because after Calcium, electrons are added to the 3d orbitals, which are in the 3 rd shell just inner to the valence shell, which is the 6 th shell.

8 A is the correct answer as anion is bigger than the cation and more the charge, the smaller is the cation and the bigger is the anion.

9 ISOLECTRONIC SPECIES Isoelectronic species are atoms or cation, and anion which have the same number of electrons. Write the possible isoelectronic species for Helium H Li + Be 2+ Write the possible isoelectronic species for Krypton Se 2 Br Rb + Sr 2+ Note: For the Noble Gas, take two elements prior to it and two elements after. The prior elements are nonmetals add the correct number of electrons and convert them into anions.. The later elements are metals, remove the correct number of electrons and convert them into cations.

10 Ionic radius decrease across periods because effective nuclear charge increases. That is, the net positive charge experienced by an electron in the ion increases as a result of the number of protons in the nucleus increasing.

11 IONIZATION ENERGY Ionization Energy in general is the energy that must be supplied to an atom to form cations. Ionization Energy can be First Ionization Energy or Second Ionization Energy or Third Ionization Energy corresponding successively to the formation of Unipositive Cation from atom; Dipositive cation from Unipositive Cation; and Tripositive cation from Dipositive cation. First Ionization Energy {Atom to Unipositive Cation} The amount of energy required to completely remove an electron from a gaseous atom to form a unipositive cation. X (g) + IE 1 X + + e- The second ionization energy {Unipositive Cation to Dipositive Cation} X + (g) + IE 2 X 2+ (g) + e- The third ionization energy {Dipositive Cation to Tripositive Cation} X 2+ (g) + IE 3 X 3+ (g) + e- More energy required to remove 2 nd electron, and still more energy required to remove 3 rd electron Generally, IE 3 > IE 2 > IE 1

12 Give the Group Name for the elements that occupy the apex of the curve and the bottom of the curve Why do Noble Gases have the highest IE value? Elements occupying the Apex:. Elements occupying the bottom:.. The noble gases possess very high ionization energies because of their full valence shells as indicated in the graph. Note that helium has the highest ionization energy of all the elements. Why do alkali metals have the least IE values? Alkali Metals are the Most Positive Metals and they readily lose the one valence electron they have to attain the Noble Gas Configuration.

13 Why does the value of IE 1 decrease within a group and increase within a period? TRENDS The ionization energy of the elements within a period generally increases from left to right. This is due to valence shell stability. The ionization energy of the elements within a group generally decreases from top to bottom. This is due to electron shielding.

14 Electron Affinity Electron affinity is the energy released when an electron is added to a gaseous atom to form the anion. The apex points are occupied by halogens; they have the highest electron affinity. This is because halogens are the most electronegative elements as they are next to the Noble Gases and are in need of just one electron to become the Noble Gases. Nitrogen is a non-metal and it is expected to have higher Electron Affinity as non-metals are electronegative. Nitrogen although a non-metal has low Electron Affinity; this can be explained as follows: The bottom points are occupied by (i) Noble Gases (ii) Alkaline Earth Metals (iii) Mg, Mn, Zn, Cd, Hf, and Hg, which are stable transition metals (iv) Nitrogen Noble gases have fully filled valence shell; so they do not need electrons and therefore they have very low electron affinity. Alkaline earth metals and the stable transition elements; they are electropositive and do not need electrons; therefore, they have very low electron affinity. p, d, and f orbitals when half-filled are very stable electronic configurations; elements which have this configuration are very stable. Nitrogen has p 3 configuration; therefore it is stable and does not have affinity for the electrons. In reality, nitrogen is an inert gas and is not very reactive.