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a. In the sixth period, after filling of 6p orbitals, the next electron (i.e. 57th) enters the 5d-orbital against aufbau principal and there after the filling of seven 4f-orbitals starts with cerium (Z = 71). Explain this anomalous behaviour. b. In the seventh period, after the filling of 7s-orbital, the next two electrons (i.e. 89th and 90th) enter the 6d-orbital against Aufbau principle and there after the filling of seven 5f-orbitals begins with proactinium (Pr, Z = 91) and ends up with lawrencium (Lr, Z = 103). Explain this anomalous behaviour. |
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Answer» Solution :This can be explained on the basis of GREATER stability of the xenon (inert gas) core. After barium `(Ba, Z = 56)`, the addition of the next electron (i.e. `57th)`, should occur in `4f`-orbitals lie inside the core, and will tend to destatabilise the xenon core. `Z = 54, rArr [kr]4d^(10)4f^(0)5s^(2)5p^(6)5d^(0)` Therefore, the `57th` electron prefers to enter `5d`-orbitals which lies outside the xenon core and whose energy is only sightly greater than that of `4f`-orbitals. Thus, stability of the atom due to xenon core compensates more than the SLIGHT instability caused by the addition of one electron to the higher energy `5d`-orbital instead of the lower energy `4f`-orbital. Thus, the outer electronic configuration of La `(Z = 57)` is `5d^(1)6s^(2)` rather than the expected `4f'6s^(2)`. Once `5d`-orbital has one electron, the next electron (i.e. `58th)` instead of entering the outer `5d`-orbital, enters the inner `4f`-orbnital. This is due to greater nuclear charge and thereafter the continuous filling of the `4f`-sub shell occurs till it is complete at lutetium `(Lu, Z = 71, 4f^(14)5d^(1)6s^(2))` b. The anomalous behaviour is due to: i. The SMALLER energy difference between `5f`- and `6d`-orbitals than between `4f`- and `5d`-orbitals. ii. Due to the greater stability of radon `(Rn, Z = 86)`, the next two electrons (i.e. `89th` and `90th)` after filling the `7s`-orbital prefer to enter `6d`-orbitals before filling of `5f`-orbitals begin with proactinium `(Pa, Z = 91)` and continues till it is complete with lawrencium `(Lr, Z = 103)`. |
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