Answer
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Hint: The screening effect is also called the shielding effect. It is known as the attraction between an electron and a nucleus in an atom with more than one electron. The shielding effect is also defined as the reduction in the effective nuclear charge on the electron cloud.
Complete step by step solution:
-The shielding effect is also defined as the reduction in the effective nuclear charge on the electron cloud. It is a special type of electric field screening.
-The wider the electron shells in the space, the weaker the electronic interaction between the electron and the nucleus due to the screening. In general, the screening effect in the shells is:
$s>p>d>f$
-The angular momentum quantum l takes integer values of zero up to (n-1) where n is the principal quantum number.
-So, if n = 3, then l has values 0,1 and 2. These correspond to s,p and d orbitals respectively.
-If you see the normalised electron probability function plotted against the distance from the nucleus you will see that the lower value of l the greater the penetrating power of electrons and hence the greater shielding ability which is in the order of $s>p>d>f$.
Clearly, the answer is (A).
Note: The size of s and p orbitals are very less compared to the d and f orbitals because the electrons of d and f are spread in a larger area and hence there is less shielding. The structures of the orbitals are also a responsible factor of this.
Complete step by step solution:
-The shielding effect is also defined as the reduction in the effective nuclear charge on the electron cloud. It is a special type of electric field screening.
-The wider the electron shells in the space, the weaker the electronic interaction between the electron and the nucleus due to the screening. In general, the screening effect in the shells is:
$s>p>d>f$
-The angular momentum quantum l takes integer values of zero up to (n-1) where n is the principal quantum number.
-So, if n = 3, then l has values 0,1 and 2. These correspond to s,p and d orbitals respectively.
-If you see the normalised electron probability function plotted against the distance from the nucleus you will see that the lower value of l the greater the penetrating power of electrons and hence the greater shielding ability which is in the order of $s>p>d>f$.
Clearly, the answer is (A).
Note: The size of s and p orbitals are very less compared to the d and f orbitals because the electrons of d and f are spread in a larger area and hence there is less shielding. The structures of the orbitals are also a responsible factor of this.
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