From RJC Chem Tutorial (2000/2001)
Explain the following observations as fully as you can:
(a) When a beam of alpha particles is fired at a thin sheet of gold foil, the majority pass through the foil and are either not deflected at all or deflected through only a small angle; however a few are deflected through large angles, some even greater than 90 degrees.
(b) The first ionization energy of Nitrogen is higher than that of either the elements immediately preceding or following it in the Periodic Table.
(c) Once the 3s and 3p orbitals are filled, subsequent electrons enter the 4s orbital first, rather than 3d orbitals. However, when transition metals form ions, electrons are lost from 4s orbital first then 3d orbitals
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Answer:
(a) There are large spaces in the gold atoms for the alpha particles to pass through undeflected of only slightly repelled by the nucleus. ( alpha particles are positively charged ) However, when some of the alpha particles come into contact with the nucleus, they will be deflected at various angles depending on the angle of collisions.
(b) The effective nuclear charge on the valence electrons is greater in N than in C, therefore more energy required to ionize N.
*In O , the removal of 1 electron in the paired electrons would lower electron-electron-repulsion, enabling a more stable configuration. Thus, less energy is required to achieve this preferred configuration. Therefore, more energy required to ionize N in comparison.
(c) 4s is at a lower energy level than 3d, therefore they are filled up first. However, 3d is closer to the nucleus than 4s. Once both 3d and 4s are occupied by electrons, 3d electrons repel the 4s electrons even further away from the nucleus and up to a higher energy level, becoming more unstable. Therefore 4s electrons will be removed first during ionization.
