Answer
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Hint: The shielding effect is also known as atomic shielding or electron shielding. It describes the attractions between an electron and the nucleus in any atom with more than one electron.
The shielding effect can be defined as the reduction in the effective nuclear charge on the electron cloud.
Complete step by step answer:
The shielding effect can be defined as the reduction in effective nuclear charge on the electron cloud and it is due to the difference in the attraction forces on the electron in an atom. It is a special case of electric field screening.
The wider the electrons shells in the space, the weaker is the electronic interaction between the electron and the nucleus due to screening. In general, we can order the electron shells as $s,\,p,\,d,\,f$
$S(s) > S(p) > S(d) > S(f)$
Where S is the screening strength that a given orbital provides to the rest of the electrons.
The angular momentum quantum number $l$ takes integer value of zero up to $(n - 1)$ where n is principal quantum number. So, if $n = 3$ then $l$ has value 0,1 and 2. These correspond to s, p and d orbitals. If you see the normalized electron probability function plotted against distance from the nucleus you will see that the lower the value of $l$ the greater the penetrating power of electrons hence the greater shielding ability which is in the order $s > p > d > f$.
So, the correct answer is “Option A-True”.
Note:
The size of $s\,and\,p$ orbital are very less as compared to the $d\,and\,f$ because the electrons of $d\,and\,f$ are spread in a large area and hence there is less effective shielding. Structures of orbitals are also responsible for that.
Shielding effect depends on many factors such as size of atoms, charge density, orbitals involved and effective nuclear charge.
The shielding effect can be defined as the reduction in the effective nuclear charge on the electron cloud.
Complete step by step answer:
The shielding effect can be defined as the reduction in effective nuclear charge on the electron cloud and it is due to the difference in the attraction forces on the electron in an atom. It is a special case of electric field screening.
The wider the electrons shells in the space, the weaker is the electronic interaction between the electron and the nucleus due to screening. In general, we can order the electron shells as $s,\,p,\,d,\,f$
$S(s) > S(p) > S(d) > S(f)$
Where S is the screening strength that a given orbital provides to the rest of the electrons.
The angular momentum quantum number $l$ takes integer value of zero up to $(n - 1)$ where n is principal quantum number. So, if $n = 3$ then $l$ has value 0,1 and 2. These correspond to s, p and d orbitals. If you see the normalized electron probability function plotted against distance from the nucleus you will see that the lower the value of $l$ the greater the penetrating power of electrons hence the greater shielding ability which is in the order $s > p > d > f$.
So, the correct answer is “Option A-True”.
Note:
The size of $s\,and\,p$ orbital are very less as compared to the $d\,and\,f$ because the electrons of $d\,and\,f$ are spread in a large area and hence there is less effective shielding. Structures of orbitals are also responsible for that.
Shielding effect depends on many factors such as size of atoms, charge density, orbitals involved and effective nuclear charge.
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