Chemistry, 09.10.2019 20:30 robert7248

What is the estimated effective nuclear charge, z eff, experienced by an electron in a 3p orbital of a chlorine atom?

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Answer from: yungkxng57

The estimated effective nuclear charge experienced by a 3 p electron of chlorine is $\boxed{{\text{6}}{\text{.1}}}$ .

Further Explanation:

Effective Nuclear Charge:

It is the charge experienced by the outermost valence electrons of any many-electron atom. This is done by considering the number of shielding electrons that are present around the nucleus.

Slater’s Rule:

It is used to evaluate the effective nuclear charge experienced by an electron in the many-electron atom. According to this rule, the actual nuclear charge experienced by an electron is less than the actual nuclear charge due to screening by the other electrons present in the atom.

The electronic configuration is the distribution of electrons of an atom in the atomic orbitals.

The effective nuclear charge is calculated as follows:

${Z_{{\text{eff}}}}=Z-s$ …… (1)

Here,

Z is the actual nuclear charge.

S is the screening constant.

Rules for determining the screening constant for an electron are as follows:

1. The electronic configuration of the atom is written and is arranged in groups as follows,

$\left({1s}\right)\left({2s,2p}\right)\left({3s,3p}\right)\left({3d}\right)\left({4s,4p}\right)\left({4d}\right)\left({4f}\right)\left({5s,5p}\right)....$

2. Identify the group in which the considered electron is present. Ignore the electrons present in (n+1) or higher group as there is no contribution of these electrons in the screening of that electron. The screening constant for each group is given as follows:

(a) The screening due to another electron present in the same group is 0.35. Except for the screening due to other electron is 0.30.

(b) If the considered electron is present in (ns, np) group then the screening due to the electrons present in (n-1) group is 0.85 and the screening due to the electrons present in (n-2) or less group is 1.00.

(c) If the considered electron is present in (nd) or (nf) group then the screening due to the electrons present in the same (n-1) or less group is 1.00.

3. The sum of the screening constant from all the electrons is calculated.

The atomic number of chlorine is 17. So its electronic configuration is $\left({1{s^2}}\right)\left({2{s^2}2{p^6}}\right)\left({3{s^2}3{p^5}}\right)$

The formula to calculate the screening constant for a 3p electron is as follows:

$s=\left[\begin{gathered}\left({0.35}\right)\left({{\text{number of electrons in same group}}}\right)\hfill\\+\left({0.85}\right)\left({{\text{number of electrons in }}\left({n-1}\right)}\right)\hfill\\+\left({1.00}\right)\left({{\text{number of electrons in }}\left({n-2}\right){\text{or less group}}}\right)\hfill\\\end{gathered}\right]$ …… (2)

Substitute 6 for the number of electrons in the same group, 8 for the number of electrons in (n-1) group and 2 for the number of electrons in (n-2) or less group in equation (2).

$\begin{aligned}s&=\left[{\left({0.35}\right)\left(6\right)+\left({0.85}\right)\left(8\right)+\left({1.00}\right)\left(2\right)}\right]\\&=\left[{2.1+6.8+2}\right]\\&=10.9\\\end{aligned}$