
Zr and Hf have almost equal atomic and ionic radii because of :
(A) diagonal relationship .
(B) lanthanoid contraction .
(C) actinoid contraction .
(D) belonging to the same group .
Answer
573k+ views
Hint: As we know the distance between an atom’s nucleus and outer electron shell is called the atomic radius. This is influenced by the number of electrons present between the outermost shell and the nucleus.
Complete step by step answer:
Firstly, the atomic radius of the metal zirconium Zr, is 159pm . It is a transition metal that lies in the 5th period and that of hafnium (Hf), it is the element of period -6 and its atomic radii is 156pm .
As ,Zr and Hf have almost similar atomic and ionic radii it can be explained on the basis of shielding effect . Due to lanthanide contraction. Moreover, the electrons which are present in inner shells, shields outer electrons from nuclear charge , after that they experience a low effective nuclear charge .
The shielding effect which is exerted by the electrons decreased in the order such as $s > p > d > f$ . In case of lanthanide elements , like $Hf,4f$ subshell which is field and the outer elements are not very much effective at the shielding effect . Therefore , $Zr$ and $Hf$ have almost similar atomic radii . As a result of lanthanide contraction , the atomic line of $Hf$ is similar to $Zr$ (elements of previous period).
As in the lanthanide contraction , when we move from lanthanum to lutetium then the ionic radius gets decreased due to poor shielding effect. Poor shielding effect takes place in 4f electrons . Due to this 6s electrons get attracted towards the nucleus and the size gets decreased .
As , due to lanthanide contraction size of Hafnium is almost getting equal with Zirconium .
Hence the correct option is C.
Note:
As u move down a column (group ) of the periodic table atomic and ionic radius increase because a shell of electrons is added to the atoms . And as you move across a row , atomic size decreases because the number of protons also increases and it exerts a stronger pull on the electrons .
Complete step by step answer:
Firstly, the atomic radius of the metal zirconium Zr, is 159pm . It is a transition metal that lies in the 5th period and that of hafnium (Hf), it is the element of period -6 and its atomic radii is 156pm .
As ,Zr and Hf have almost similar atomic and ionic radii it can be explained on the basis of shielding effect . Due to lanthanide contraction. Moreover, the electrons which are present in inner shells, shields outer electrons from nuclear charge , after that they experience a low effective nuclear charge .
The shielding effect which is exerted by the electrons decreased in the order such as $s > p > d > f$ . In case of lanthanide elements , like $Hf,4f$ subshell which is field and the outer elements are not very much effective at the shielding effect . Therefore , $Zr$ and $Hf$ have almost similar atomic radii . As a result of lanthanide contraction , the atomic line of $Hf$ is similar to $Zr$ (elements of previous period).
As in the lanthanide contraction , when we move from lanthanum to lutetium then the ionic radius gets decreased due to poor shielding effect. Poor shielding effect takes place in 4f electrons . Due to this 6s electrons get attracted towards the nucleus and the size gets decreased .
As , due to lanthanide contraction size of Hafnium is almost getting equal with Zirconium .
Hence the correct option is C.
Note:
As u move down a column (group ) of the periodic table atomic and ionic radius increase because a shell of electrons is added to the atoms . And as you move across a row , atomic size decreases because the number of protons also increases and it exerts a stronger pull on the electrons .
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