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the order of screening effect of electrons of s p d f of a given shell of an atom on its outer shell electrons is ______________

the order of screening effect of electrons of s p d f of a given shell of an atom on its outer shell electrons is ______________

Grade:11

4 Answers

Sunil Kumar FP
askIITians Faculty 183 Points
9 years ago

It is described as a decrease in shielding effort between the nucleus and last orbital due to electrons present between them i.e., greater the screening effect, easier the removal of electron. In multi-electron atoms, the electrons present in the outermost shell do not experience the complete attraction by the nuclear charge because of the inner electrons. In other words, the force of attraction experienced by the valence electrons is less than that experienced by the inner electrons. Thus, the outermost electrons are shielded or screened from the nucleus by the inner electrons. This is known asshielding effectorscreening effect.

-s orbitals have the largest screening effect for a given n value since s electrons are closer to the nucleus.
-p orbital's have the next highest screening effect and then comes d and then f orbital's. Thus in simple terms the screening effect decreases in order :s orbitals > p orbitals> d orbitals> f orbitals.
Balwant
21 Points
7 years ago
Sir;with increase in shielding effect ionisation enthalpy decreases and down the group ionisation enthalpy decreases its means shieldind efect increases but d and have the poor shielding please clear my confusion
Akash dhuware
13 Points
6 years ago
It`s because d orbitals have such a shape (double dumbbell shape) that their electron density is low and therefore they provide less repulsion and hence less screening effect
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the answer to your question below.
 
The screening effect of the orbitals in the decreasing order is
s orbitals > p orbitals> d orbitals> f orbitals
The reason for this behaviour depends on the closeness and the shape of the orbital. (You can find the detailed explanation in Mr. Sunil’s answer)
 
Thanks and regards,
Kushagra

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