Question

An increase in both atomic and ionic radii with atomic number occurs in any group of the periodic table and in accordance with this the ionic radii of $Ti(IV)$ and $Zr(IV)$ ions are $0.68A{}^\circ$ and $0.74A{}^\circ$ respectively; but for $Hf(IV)$ ion, the ionic radius is $0.75A{}^\circ$ which is almost the same as $Zr(IV)$ ion. This is due to ____.

Answer 1

Hint:** The question has given periodic trends of atomic size and ionic radii when we move from top to bottom in a group. Locate the position block of $Ti$, $Zr$ and $Hf$ in the periodic table and also write their valence shell electronic configuration. Think about the effect which reduces the ionic size of $Hf (IV)$ and makes it approximately equal to $Zr (IV)$.

Complete step by step solution:
- The elements given in the question are d-block elements, Titanium ($Ti$), Zirconium ($Zr$) and Hafnium ($Hf$). All three are elements of the same group in the periodic table which lie one below another respectively.
- Titanium has atomic number 22, zirconium has atomic number 40 and hafnium has atomic number 72. Their valence shell electronic configuration is,
${}^{22}Ti=[Ar]3{{d}^{2}}4{{s}^{2}}$
${}^{40}Zr=[Kr]4{{d}^{2}}5{{s}^{2}}$
${}^{72}Hf=[Xe]4{{f}^{14}}5{{d}^{2}}6{{s}^{2}}$
- Now, from the electronic configuration we can see that hafnium has completely filled 4f orbitals.
- In the question, all three elements are present in +4 oxidation state. So, titanium and zirconium ions will have noble gas configuration of argon and krypton respectively. But, hafnium will not attain noble gas configuration due to the presence of completely filled 4f-orbitals.
- Due to the presence of completely filled 4f-orbitals, there will be a greater force of nuclear attraction due to addition of more protons to the nucleus. So, the fourth shell will experience a greater pull from the nucleus which will result in contraction of size.
- This phenomena is known as lanthanide contraction because hafnium is the next element after lanthanide series and it has size nearly equal to zirconium.

Therefore, for $Hf(IV)$ ions, the ionic radius is $0.75A{}^\circ$ which is almost the same as $Zr(IV)$ ion due to lanthanide contraction.

Note: Remember atomic and ionic radii of zirconium is approximately equal to that of hafnium due to lanthanide contraction. Lanthanide contraction is the reduction in atomic or ionic size caused due to greater force of nuclear attraction as the electrons are filled in the same shell.