# What is an effective nuclear charge and the periphery of nitrogen atom when an extra electron is added in the formation of an anion?(A)- 1.20(B)- 2.45 (C)- 3.55 (D)- 5.95

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Hint: Effective nuclear charge refers to the charge possessed by the outermost valence electron. This concept also takes into account the shielding effect provided by the electrons surrounding the nucleus. For the calculation of effective nuclear charge, we use Slater’s rules.

-Electron shielding describes the ability of an atom’s inner electron to shield the outermost valence electron from the strong electrostatic attraction of the nucleus. The trend of shielding effect on the periodic table increases across the period and down a group due to the addition of the number of electrons across the period and the addition of shells along with the groups.

-For the calculation of effective nuclear charge, the steps to be followed are as follows-
(i) First, write down the electronic configuration of the atom according to the Aufbau principle,
Electronic configuration of Nitrogen atom in excited state is $1{{s}^{2}}2{{s}^{2}}2{{p}^{4}}$ .
(ii) Identify the electron of interest and ignore all electrons in higher groups.
(iii) Slater’s rule is now broken into two cases:
For shielding experienced by as- or p-electron,
(a) electrons within the same group shields 0.35, except the 1s which shields 0.30.
(b) electrons within the n-1 group shields 0.85.
(c) electrons within the n-2 or lower groups shield 1.00.

For shielding experienced by nd or nf valence electrons,
(a) electrons within the same group shield 0.35.
(b) electrons within the lower groups shield 1.00.
With the help of these values, the screening constant, $\sigma =(0.35\times 5)+(0.85\times 2)$
$\sigma =3.45$
(iv) Finally calculating the effective nuclear charge using the following formula,
${{Z}_{eff}}=Z-\sigma$
where ${{Z}_{eff}}$ is the effective nuclear charge,
Z is the number of protons existing in the nucleus
$\sigma$ is the screening constant.
For ${{N}^{-}}$ anion, ${{Z}_{eff}}=Z-\sigma$
=7-3.45=3.55
So, the correct answer is 3.55
So, the correct answer is “Option C”.

Note: Carefully calculate the screening constant. Screening constant the average amount of density between the nucleus and the electron.