
Why is the atomic radius of Li larger than that of Be?
Answer
501.3k+ views
Hint: We can define atomic radius as a factor which is used to measure the size of an atom. We can say that atomic radius is the distance that is the center of a nucleus to the barrier of the enclosed shells of electrons. We can measure atomic radius in terms of van Der Waals radius, ionic radius, covalent radius, and metallic radius.
Complete answer:
We know that both Lithium, Li, and beryllium, Be, are both situated in period $2$ of the periodic table, in group $1$ and group $2$ , individually.
For both these components, the furthest electrons are situated on the second energy level during the $2s$ -subshell. But these valence electrons, which at last decide the atomic radius, are nearer to the core for atoms of beryllium.
That happens on the grounds that beryllium has a higher effective nuclear charge, Zeff when compared to lithium.
The effective nuclear charge is essentially a proportion of the net positive charge that influences the valence electrons in a particle.
For lithium's situation, the furthest electron is being screened from the core by two center electrons. The equivalent is valid for the two valence electrons in beryllium, they are screened by two center electrons.
Beryllium has $4$ protons in its nucleus, instead of $3$ protons which are available in the core of a lithium atom.
This implies that the core of a beryllium atom would pull on the furthest electrons with more power, successfully packing the distance among itself and these electrons.
That is the reason lithium has a bigger atomic radius when compared to beryllium. Indeed, atomic radius diminishes as you move from left to directly across a period.
Note:
We have to know when there is an increased electron shell, the principal and azimuthal quantum number increases, so there would increase in the atomic radius. Another factor that results in increase in atomic radius is shielding, with increase with shielding, the number of electrons found in the inner shell would also increase and so, atomic radii increases.
Complete answer:
We know that both Lithium, Li, and beryllium, Be, are both situated in period $2$ of the periodic table, in group $1$ and group $2$ , individually.
For both these components, the furthest electrons are situated on the second energy level during the $2s$ -subshell. But these valence electrons, which at last decide the atomic radius, are nearer to the core for atoms of beryllium.
That happens on the grounds that beryllium has a higher effective nuclear charge, Zeff when compared to lithium.
The effective nuclear charge is essentially a proportion of the net positive charge that influences the valence electrons in a particle.
For lithium's situation, the furthest electron is being screened from the core by two center electrons. The equivalent is valid for the two valence electrons in beryllium, they are screened by two center electrons.
Beryllium has $4$ protons in its nucleus, instead of $3$ protons which are available in the core of a lithium atom.
This implies that the core of a beryllium atom would pull on the furthest electrons with more power, successfully packing the distance among itself and these electrons.
That is the reason lithium has a bigger atomic radius when compared to beryllium. Indeed, atomic radius diminishes as you move from left to directly across a period.
Note:
We have to know when there is an increased electron shell, the principal and azimuthal quantum number increases, so there would increase in the atomic radius. Another factor that results in increase in atomic radius is shielding, with increase with shielding, the number of electrons found in the inner shell would also increase and so, atomic radii increases.
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