Question

explain the tetrahedral structure of chlorine oxide with a bond angle more than 109°28'

Answer 1

This process continues until the entire 3d shell is filled. Thus at Zn we have the
configuration [~r]4s~3dlO. Thereafter, the next lowest available orbitals are 4p which get
filled in the next six elements. This same sequence of events for the filling of 5s and 4d
orbitals is repeated again in the elements following krypton in the second transition series.
10 2 This series starts with Y and is completed at Cd having the configuration [Kr]4d 5s . After
10 2 6
xenon, [Kr]4d 5s 5p , the next available orbitals are 45 5d, 6s and'6p orbitals. The 4f
orbitals are so slightly penetrating with respect to the xenon core that they have scarcely
gained any stability, while the mbre penetrating 6s and 6p levels have gained a good deal of
stability. Hence, in the next two elements, electrons are added to 6s orbitals giving Cs and
Ba, respectively. However, the 6s electrons do not shield the 4forbitals effectively, so the
latter abruptly feel an increase in effective nuclear charge and thus suffer a steep drop in
energy (Fig. 12.2). At the'same time, with the addition of electrons in the 6s orbital. the 5d
orbitals also drop in energy in the same manner as the 3d ones, This creates a situation in
which 5d and 4f orbitals are of almost the same energy. The next electron in lanthanum thus
enters the 5d orbital, but in the following element cerium, the electronic configuration is
[xe]6s2 5d1 4f. The electrons then continue to be added to the 4f orbital till we reach
2 4 ytterbium which has the configuration [Xe]6s 4f . Now with the 6s and 4f shells full, the
next lowest levels are the 5d's. Hence from luteciurn onwards, the electrons enter the 5d
2 4 10 , orbital. This continues till we reach mercury which has the configuration [Xe]6s 4f 5d . The electronic configurations of transition element