Question
Answers

The shielding constant \[\sigma \] for \[{}_7N\] atom is:
A.3.10
B.3.40
C.3.80
D.4.20

Answer Verified Verified
Hint:
Shielding constant basically aims at determining the shielding or screening a given electron in an atom is experiencing due to all the other electrons present in the same and lower orbitals of the same atom.
Formula used:
\[\sigma = \Sigma {n_i}{S_i}\]

Complete step by step answer:
Now to calculate this shielding constant, there are a few rules that we need to follow known as Slater’s Rules:
Identify the electronic configuration of the given atom and write it down in the format:
(1s)(2s,2p)(3s,3p)(3d)(4s,4p)(4d)(4f)(5s,5p)(5d)(5f)…
Narrow down on which electron you want consider the shielding constant for. After this, eliminate all the other higher value orbitals.
Slater’s Rules are now divided into two parts:
If the shielding is experienced by an electron in s or p orbital
Electrons present in the same group shield 0.35, with the exception of 1s which shields 0.30
Electrons present in (n-1) group shields 0.85
Electrons present in (n-2) and all the other lower groups shield 1.00
 If the shielding is experienced by an electron in d or f orbital
Electrons present in the same group shield 0.35
Electrons present in lower groups shield 1.00
The shielding effect can then be calculate using the formula:
\[\sigma = \Sigma {n_i}{S_i}\]
Where \[{n_i}\]corresponds to the number of electrons in then given shell, while\[{S_i}\] is the shielding of the shell in accordance to Slater’s Rules.
Following these rules, let us consider \[{}_7N\] atom.
Electronic configuration = \[1{s^2}2{s^2}2{p^3}\]
Now let us consider an electron in the 2p orbital.
Hence,
\[\sigma = \Sigma {n_i}{S_i}\]
\[\sigma \]= (0.85)(2) + (0.35)(4)
\[\sigma \]=1.7 + 1.4
\[\sigma \]=3.10

Hence, Option A is the correct option.

Note:
Shielding effect takes place when the lower valence electrons provide a sort of a resistance to the outer electrons from the nuclear force of attraction. Hence when calculating shielding effect, the outer electrons are not considered because they do not contribute to the shielding of the given electron.


Bookmark added to your notes.
View Notes
×