
Give reason, \[Zr\] and \[Hf\] have almost identical atomic radii?
Answer
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Hint:Atomic radius is defined as the distance between the nucleus and the boundary of the valence shell of electrons. Thus the atomic radius is dependent on the number of electrons in an element.
Complete step by step answer:
Zirconium is an element in the periodic table with atomic number \[40\] and electronic configuration \[\left[ {Kr} \right]4{d^2}5{s^2}\] . It is a transition metal which lies in the group \[4\] and period \[5\] of the periodic table. The atomic radius of the Zirconium atom is \[159pm\] .
Hafnium is an element in the periodic table with atomic number \[72\] and electronic configuration\[\left[ {Xe} \right]4{f^{14}}5{d^2}6{s^2}\] . It is also a transition metal which lies in the group \[4\] and period \[6\] of the periodic table. The atomic radius of a Hafnium atom is \[156pm\] .
As we move down the group in periodic table from period \[5\] to period \[6\], the \[4f\] subshell is added to the total number of electrons. The \[4f\] shell is due to the presence of fourteen lanthanide elements between Zirconium and Hafnium.
The shielding effect of the electron present in various shells varies in the order\[s > p > d > f\] . This indicates that the electrons present in the \[f\] shell exhibit very poor shielding effect. Due to such poor shielding effect the attraction of the nucleus for the valence shell electrons increases as the valence shell electrons feel more attraction. This results in a decrease in ionic radius and the size of the atom.
Thus the \[Hf\] is more attracted towards the nucleus due to the presence of fourteen \[f\] electrons than the nucleus of \[Zr\]. Thus the electrons present in \[6s\] shell of \[Hf\] are more attracted than the \[5s\] electrons of \[Zr\]. This type of decrease in atomic size and atomic radii is known as lanthanide contraction. Hence \[Zr\] and \[Hf\] have almost identical atomic radii.
Note:
\[Zr\] is a pre lanthanide element and \[Hf\] is a post lanthanide element. Due to lanthanide contraction, the atomic radii of post lanthanide elements are similar to pre lanthanide elements.
Complete step by step answer:
Zirconium is an element in the periodic table with atomic number \[40\] and electronic configuration \[\left[ {Kr} \right]4{d^2}5{s^2}\] . It is a transition metal which lies in the group \[4\] and period \[5\] of the periodic table. The atomic radius of the Zirconium atom is \[159pm\] .
Hafnium is an element in the periodic table with atomic number \[72\] and electronic configuration\[\left[ {Xe} \right]4{f^{14}}5{d^2}6{s^2}\] . It is also a transition metal which lies in the group \[4\] and period \[6\] of the periodic table. The atomic radius of a Hafnium atom is \[156pm\] .
As we move down the group in periodic table from period \[5\] to period \[6\], the \[4f\] subshell is added to the total number of electrons. The \[4f\] shell is due to the presence of fourteen lanthanide elements between Zirconium and Hafnium.
The shielding effect of the electron present in various shells varies in the order\[s > p > d > f\] . This indicates that the electrons present in the \[f\] shell exhibit very poor shielding effect. Due to such poor shielding effect the attraction of the nucleus for the valence shell electrons increases as the valence shell electrons feel more attraction. This results in a decrease in ionic radius and the size of the atom.
Thus the \[Hf\] is more attracted towards the nucleus due to the presence of fourteen \[f\] electrons than the nucleus of \[Zr\]. Thus the electrons present in \[6s\] shell of \[Hf\] are more attracted than the \[5s\] electrons of \[Zr\]. This type of decrease in atomic size and atomic radii is known as lanthanide contraction. Hence \[Zr\] and \[Hf\] have almost identical atomic radii.
Note:
\[Zr\] is a pre lanthanide element and \[Hf\] is a post lanthanide element. Due to lanthanide contraction, the atomic radii of post lanthanide elements are similar to pre lanthanide elements.
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