Why atomic radii of zirconium and hafnium are almost same?
Answers
This is a reason of poor screening by 4f electrons. The 6s electrons are drawn towards the nucleus, thus resulting in a smaller atomic radius.
In hydrogenic atoms, the subshell determines the average separation of an electron from the nucleus, and this decreases with increasing charge on the nucleus which in turn leads to a decrease in atomic radius.
In multi-electron species, the decrease in radius brought about by an increase in nuclear charge is partially offset by increasing electrostatic repulsion among electrons. In particular, a "screening effect" operates: i.e., as electrons are added in outer shells, electrons already present shield the outer electrons from nuclear charge, making them experience a lower effective charge on the nucleus. The shielding effect exerted by the inner electrons decreases in the order s > p > d > f. Usually, as a particular subshell is filled in a period, atomic radius decreases. This effect is particularly pronounced in the case of lanthanides, as the 4f subshell which is filled across these elements is not very effective at shielding the outer shell (n=5 and n=6) electrons. Thus the shielding effect is less able to counter the decrease in radius caused by increasing nuclear charge. This leads to "lanthanide contraction". The ionic radius drops from 103 pm for lanthanum(III) to 86.1 pm for lutetium(III).
About 10% of the lanthanide contraction has been attributed to relativistic effects.
The results of the increased attraction of the outer shell electrons across the lanthanide period may be divided into effects on the lanthanide series itself including the decrease in ionic radii, and influences on the following or post-lanthanide elements.
Due to the lanthanide contraction, the atomic size of the post–lanthanide elements (elements following the last element of the lanthanide series) becomes small.
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