Chemistry Spl

Hi Manish,
Slater's Rules:
1) Write the electron configuration for the atom using the following design;
(1s)(2s,2p)(3s,3p) (3d) (4s,4p) (4d) (4f) (5s,5p)
2) Any electrons to the right of the electron of interest contributes no shielding. (Approximately correct statement.)
3) All other electrons in the same group as the electron of interest shield to an extent of 0.35 nuclear charge units
4) If the electron of interest is an s or p electron: All electrons with one less value of the principal quantum number shield to an extent of 0.85 units of nuclear charge. All electrons with two less values of the principal quantum number shield to an extent of 1.00 units.
5) If the electron of interest is an d or f electron: All electrons to the left shield to an extent of 1.00 units of nuclear charge.
6) Sum the shielding amounts from steps 2 through 5 and subtract from the nuclear charge value to obtain the effective nuclear charge.
Examples:
Calculate Z* for a valence electron in fluorine.
(1s2)(2s2,2p5)
Rule 2 does not apply; 0.35 6 + 0.85 2 = 3.8
Z* = 9 3.8 = 5.2 for a valence electron.
Calculate Z* for a 6s electron in Platinum.
(1s2)(2s2,2p6)(3s2,3p6) (3d10) (4s2,4p6) (4d10) (4f14) (5s2,5p6) (5d8) (6s2)
Rule 2 does not apply; 0.35 1 + 0.85 16 + 60 1.00 = 73.95
Z* = 78 73.95 = 4.15 for a valence electron.
Shielding:
The first ionization energy for hydrogen is 1310 kJmol1 while the first ionization energy for lithium is 520 kJmol1. The IE for lithium is lower for two reasons;
1) The average distance from the nucleus for a 2s electron is greater than a 1s electron;
2) The 2s1 electron in lithium is repelled by the inner core electrons, so the valence electron is easily removed.
For reason #2 the inner core electrons shield the valence electron from the nucleus so the outer most electron only experiences an effective nuclear charge. In the case of the lithium the bulk of the 1s electron density lies between the nucleus and the 2s1 electron. So the valence electron 'sees' the sum of the charges or approximately +1. In reality the charge the valence electron experiences is greater than 1 because the radial distribution show their is some probability of finding the 2s electron close to the nucleus.
All the best.

Hi Mahmoud,
SnO2 nanoparticles/graphene (SnO2/GP) nanocomposite was synthesized by a facile microwave method. The X-ray diffraction (XRD) pattern of the nanocomposite corresponded to the diffraction peak typical of graphene and the rutile phase of SnO2 with tetragonal structure. The field emission scanning electron microscope (FESEM) images revealed that the graphene sheets were dotted with SnO2 nanoparticles with an average size of 10 nm. The X-ray photoelectron spectroscopy (XPS) analysis indicated that the development of SnO2/GP resulted from the removal of the oxygenous groups on graphene oxide (GO) by Sn2+ ions. The nanocomposite modified glassy carbon electrode (GCE) showed excellent enhancement of electrochemical performance when interacting with mercury(II) ions in potassium chloride supporting electrolyte. The current was increased by more than tenfold, suggesting its potential to be used as a mercury(II) sensor.
All the best.