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Arrange s, p and d sub-shells of a shell in the decreasing order to effective nuclear charge (${ Z }_{ eff }$) experienced by the electron present in them.
(A) $s>p>d$
(B) $p>s>d$
(C) $d>p>s$
(D) none of the above

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Last updated date: 22nd Jun 2024
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Answer
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Hint: In order to solve this question we need to understand shielding and penetration power of the subshells. Shielding occurs due to the partial neutralization of nuclear charge by the core electrons and electron-electron repulsions. The principal quantum number and the azimuthal quantum number determine how close an electron can approach the nucleus and this is called penetration.

Complete step by step solution:
Due to the negative charge of electrons, they are pulled pretty close to the nucleus due to its positive charge. The electrons are attracted to the nucleus while simultaneously repelled by each other due to the same negative charge. So, there is a fine balance between the attractive and repulsive forces which results in shielding. The principal quantum number and the azimuthal quantum number determine how close an electron can approach the nucleus and this is called penetration.
In theory, every electron in an atom should feel the same amount of pull from the nucleus. However, that is not the case. The electrons closest to the nucleus feel almost the same nuclear charge (called Effective Nuclear Charge) which is closest or equal to the actual nuclear charge. But, as we move away from the core electrons to the outer valence electrons, ${ Z }_{ eff }$ decreases significantly. This is due to the shielding. The electrons closest to the nucleus cause a decrease in the amount of nuclear charge exerted on the outer electrons. This phenomenon occurs due to the partial neutralization of nuclear charge by the core electrons and electron-electron repulsions.
Also, the amount of nuclear charge exerted on an electron depends on its distance from the nucleus. If an electron is a present closer to the nucleus i.e. it is more penetrated, it will experience stronger attraction to the nucleus. Core electrons penetrate more than the outer valence electrons and hence feel more of the nuclear charge.
For a multi-electron system, we define the electron penetration by its relative electron density (probability density) near the nucleus of the atom. Different orbitals have different wave functions which give rise to different radial distributions and probabilities which depend upon the value of the shell (n) and subshell (${ m }_{ l }$). For example, the 2s electron has more electron density near the nucleus than a 2p electron which implies that the electrons in the 2s orbital penetrate the nucleus of the atom more than the 2p electrons.
If the value of n is the same, then the penetrating power of the subshells decreases in the order: $s>p>d$. But if the value of n and l are different then the following trend is observed:
$1s>2s>2p>3s>3p>4s>3d>4p>5s>4d>5p>6s>4f....$

Hence the correct answer is (A) $s>p>d$.

Hint: Electrons which are present in orbitals that have greater penetration power will experience stronger attraction to the nucleus. Due to this, they will be less shielded and hence experience more effective nuclear charge. These electrons also shield outer electrons.