Question

The effective nuclear charge for the outermost electron of an oxygen atom is.
A: $4.55$
B: $5.35$
C: $6.68$
D: None of these

Answer 1

Hint : Effective nuclear charge is the net positive charge experienced by an electron in an atom. It is reduced by some value as the inner electrons will prevent outer electrons from complete pull of the nucleus, which is called screening.
Formula used: ${Z_{eff}} = Z - \sigma$ where ${Z_{eff}}$ is effective nuclear charge, $Z$ is atomic number and $\sigma$ is screening constant.

Complete step by step solution :
As we know, an effective nuclear charge is a net positive charge experienced by an electron in an atom. Term effective is used because inner electrons of an atom prevent outermost electrons of an atom from experiencing full nuclear charge. The reduction produced in the nuclear charge by inner electrons is called the shielding effect. Core charge is the effective nuclear charge experienced by an electron. We can calculate effective nuclear charge by reducing $\sigma$ which is a screening constant from the atomic number of that element. We can find screening constant as follows:
a: Each of the remaining electrons in the ${n^{th}}$ shell makes a contribution of $0.35$ to the value of the screening constant.
b: Each of the electrons in ${\left( {n - 1} \right)^{th}}$ shell electrons makes a contribution of $0.85$ to value of screening constant.
c: Each of the remaining shell electrons make a contribution of $1.0$ to the value of screening constant. Value of the screening constant can be found by adding the value of the screening constant from each of the electrons.
Applying this to oxygen: electronic configuration of oxygen is
$1{s^2}2{s^2}2{p^4}$
${n^{th}}$ shell is $2$
${\left( {n - 1} \right)^{th}}$ shell is $1$
Number of electrons in ${n^{th}}$ shell is $4 + 2 = 6$ , this means remaining electrons in this shell is $5$
Number of electrons in ${\left( {n - 1} \right)^{th}}$ shell $= 2$, from $1{s^2}$
Value of $\sigma$ is $\left( {5 \times 0.35} \right) + \left( {2 \times 0.85} \right) = 3.45$
As we know, ${Z_{eff}} = Z - \sigma$ and for oxygen $Z$ is $8$
So, ${Z_{eff}} = 8 - 3.45$
${Z_{eff}} = 4.55$
Therefore net positive charge on the outermost electron of oxygen is $4.55$ . So, the answer to this question is option A that is $4.55$.
Additional information: Lanthanide contraction is the greater than expected decrease in ionic radii of the elements in the lanthanide series from lanthanum to lutetium. This is the result of poor shielding of nuclear charge by $4f$ electrons.

Note : There is an exception in shielding effect of different subshells of an atom. d and f subshells have poor shielding effect as compared to s and p subshells. This poor shielding is the cause of lanthanide and actinide contraction.