Which element is larger in size?
(A) Co
(B) Ni
(C) Cu
(D) Zn
Answer
249.6k+ views
Hint: In the end from left to right across the periodic table, the atomic radius increases in case of d block elements. A greater size variation is observed in transitional elements due to electron-electron repulsion and increase in shielding effect.
Complete step-by-step solution:
In case of transitional elements that belong to the d block, atomic number increases as we move left to right across the period. This results in an increase of electrons and thereby increasing the nuclear charge. But the electrons are added to (n-1) penultimate shells, the electron cloud density of inner shells increases due to increase in screening effect. Therefore, there is a decrease in atomic radius but the variation is small as compared to s-block and p-block elements.
The point to be noted is atomic radius in d block decreases initially then remains constant and finally increases at the end. This can be understood by considering the nuclear attractions and inter electronic repulsions, where the first one supports decrease in atomic radii and the other one supports increase in atomic radii.
From Sc to Zn, as the new electron adds in the same orbital, nuclear charge increases by one factor. As the number of electrons are low in the energy cell and the lower shielding power of d orbital results in inter electronic repulsions less than nuclear charge, thus decreasing atomic radii. In the middle, the two factors operate equally which leads to constant size.
As the number of electrons in inner orbital increases, the outer electrons feel more repulsion and are pushed away at the end of the transition series. This results in an increase in the atomic radii.
Co < Ni < Cu < Zn
Hence, the correct options are (D).
Note: This can also be understood by looking at s orbital which is spherical in shape and shields the outermost electron effectively from the nucleus in comparison to d block. From nickel onwards, the new electron is added to the s orbital and hence the radii increases from nickel to zinc.
Complete step-by-step solution:
In case of transitional elements that belong to the d block, atomic number increases as we move left to right across the period. This results in an increase of electrons and thereby increasing the nuclear charge. But the electrons are added to (n-1) penultimate shells, the electron cloud density of inner shells increases due to increase in screening effect. Therefore, there is a decrease in atomic radius but the variation is small as compared to s-block and p-block elements.
The point to be noted is atomic radius in d block decreases initially then remains constant and finally increases at the end. This can be understood by considering the nuclear attractions and inter electronic repulsions, where the first one supports decrease in atomic radii and the other one supports increase in atomic radii.
From Sc to Zn, as the new electron adds in the same orbital, nuclear charge increases by one factor. As the number of electrons are low in the energy cell and the lower shielding power of d orbital results in inter electronic repulsions less than nuclear charge, thus decreasing atomic radii. In the middle, the two factors operate equally which leads to constant size.
As the number of electrons in inner orbital increases, the outer electrons feel more repulsion and are pushed away at the end of the transition series. This results in an increase in the atomic radii.
Co < Ni < Cu < Zn
Hence, the correct options are (D).
Note: This can also be understood by looking at s orbital which is spherical in shape and shields the outermost electron effectively from the nucleus in comparison to d block. From nickel onwards, the new electron is added to the s orbital and hence the radii increases from nickel to zinc.
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