Screening effect and effective nuclear charge Questions in English

(C) The screening effect (also known as shielding effect) depends on the shape of the orbitals. The order of screening power is $s > p > d > f$.
Since $d$-orbitals have a more diffuse shape compared to $p$-orbitals,they are less effective at shielding the outer electrons from the nuclear charge.
Therefore,the screening effect of $d$-electrons is less than that of $p$-electrons.

(A) The screening effect (or shielding effect) of inner electrons reduces the effective nuclear charge experienced by the valence electrons.
This reduction in effective nuclear charge decreases the electrostatic attraction between the nucleus and the outermost electrons.
As a result,the energy required to remove the outermost electron,known as the ionisation potential,decreases.

(A) The screening effect (or shielding effect) depends on the penetration power of the orbitals.
Electrons in $s$-orbitals are closest to the nucleus and have the highest penetration power,followed by $p$,$d$,and $f$ orbitals.
Therefore,the ability of these orbitals to shield the outer electrons from the nuclear charge follows the order: $s > p > d > f$.

(D) The effective nuclear charge $(Z_{eff})$ is defined as the net positive charge experienced by an electron in a multi-electron atom. It is calculated using the formula: $Z_{eff} = Z - \sigma$,where $Z$ is the nuclear charge (atomic number) and $\sigma$ is the shielding constant (or screening constant). Therefore,it depends on both the nuclear charge and the shielding constant.

(A) $1$. In $Sc$ $(Z=21)$,the electronic configuration is $[Ar] 3d^1 4s^2$. The effective nuclear charge $(Z_{eff})$ is calculated as $Z - \sigma$. Due to different shielding effects of $3d$ and $4s$ electrons,their $Z_{eff}$ values are not the same. Thus,statement $A$ is incorrect.
$2$. The increase in $Z_{eff}$ across a period is generally $0.65$ per step,which is a standard observation in Slater's rules,making statement $B$ correct.
$3$. For $H^-$ $(1s^2)$ and $He$ $(1s^2)$,the screening constant $\sigma$ for one electron due to the other is $0.30$. Thus,statement $C$ is correct.
$4$. $Z_{eff}$ is fundamentally dependent on the shielding constant,which varies with the principal quantum number $(n)$ and orbital type,making statement $D$ correct.

(D) The core charge is defined as the nuclear charge minus the number of inner-shell electrons (core electrons).
For oxygen $(Z = 8)$,the electronic configuration is $1s^2 2s^2 2p^4$.
The number of core electrons (electrons in the $1s$ orbital) is $2$.
The nuclear charge is $+8$.
Therefore,the core charge $= +8 - 2 = +6$.

(B) The effective nuclear charge is given by the formula $Z_{eff} = Z - \sigma$,where $Z$ is the atomic number and $\sigma$ is the screening constant.
For a hydrogen atom $(H)$,there is only one electron,so there are no other electrons to provide shielding.
Therefore,the screening constant $\sigma = 0$.
Thus,$Z_{eff} = Z - 0 = Z$.

(D) The effective nuclear charge $(Z_{eff})$ generally increases across a period from left to right due to the addition of electrons in the same shell while the nuclear charge increases.
For option $A$: $Z_{eff}$ of $Be$ $(1.95)$ < $Z_{eff}$ of $B$ $(2.60)$.
For option $B$: $Z_{eff}$ of $C$ $(3.25)$ < $Z_{eff}$ of $N$ $(3.90)$.
For option $C$: $Z_{eff}$ of $Na$ $(2.20)$ > $Z_{eff}$ of $K$ ($2.20$ is not correct,$Na$ is $2.20$ and $K$ is $2.20$ approximately,but $Na$ is in a higher group period context,actually $Z_{eff}$ decreases down the group).
For option $D$: $Z_{eff}$ of $F$ $(5.20)$ > $Z_{eff}$ of $O$ $(4.55)$.
Therefore,in the pair $(F, O)$,the first element $(F)$ has a higher $Z_{eff}$ than the second element $(O)$.

(C) The effective nuclear charge $Z_{eff}$ is calculated using Slater's rules: $Z_{eff} = Z - S$,where $Z$ is the atomic number and $S$ is the shielding constant.
For Oxygen $(Z=8)$: The electronic configuration is $1s^2 2s^2 2p^4$. The shielding constant $S = (2 \times 0.85) + (5 \times 0.35) = 1.7 + 1.75 = 3.45$.
Thus,$Z_{eff} = 8 - 3.45 = 4.55$.
For Fluorine $(Z=9)$: The electronic configuration is $1s^2 2s^2 2p^5$. The shielding constant $S = (2 \times 0.85) + (6 \times 0.35) = 1.7 + 2.1 = 3.80$.
Thus,$Z_{eff} = 9 - 3.80 = 5.20$.

(A) The energy of an orbital in a multi-electron system depends on the effective nuclear charge $(Z_{eff})$.
As the nuclear charge $(Z)$ increases,the attraction between the nucleus and the electrons increases,which stabilizes the orbital and lowers its energy.
Among the given elements,the atomic numbers are: $H$ $(Z=1)$,$Li$ $(Z=3)$,$Na$ $(Z=11)$,and $K$ $(Z=19)$.
Since $K$ has the highest nuclear charge $(Z=19)$,the $2s$ orbital in $K$ experiences the strongest attraction,resulting in the lowest energy.

(C) In $Lithium$ $(Z=3)$,the $1s^2$ electrons shield the $2s$ electron from the full nuclear charge.
Due to this shielding effect,the $2s$ electron experiences an effective nuclear charge $(Z_{eff})$ which is less than the actual nuclear charge of $ 3$.
The formula is $Z_{eff} = Z - \sigma$,where $\sigma$ is the shielding constant.

(D) The atomic number of Boron is $5$. The electronic configuration is $1s^2 2s^2 2p^1$.
Using Slater's rules,the shielding constant $\sigma$ for the $2p$ electron is $(2 \times 0.85) + (2 \times 0.35) = 1.7 + 0.7 = 2.4$.
Thus,$Z_{eff} = Z - \sigma = 5 - 2.4 = 2.6$. Let this be $x$.
The atomic number of Oxygen is $8$. The electronic configuration is $1s^2 2s^2 2p^4$.
Using Slater's rules,the shielding constant $\sigma$ for a $2p$ electron is $(2 \times 0.85) + (5 \times 0.35) = 1.7 + 1.75 = 3.45$.
Thus,$Z_{eff} = 8 - 3.45 = 4.55$.
Comparing the two values: $4.55 - 2.6 = 1.95$.
Therefore,the $Z_{eff}$ of oxygen is $x + 1.95$.

(C) According to Slater's rules,the screening constant $(\sigma)$ for an electron in an $ns$ or $np$ orbital is calculated based on the electrons in the same shell and inner shells.
For an electron in the $3s$ orbital,the screening contribution from other electrons in the same $n=3$ shell is $0.35$ per electron.
Similarly,for an electron in the $3p$ orbital,the screening contribution from other electrons in the same $n=3$ shell is also $0.35$ per electron.
Since both $3s$ and $3p$ orbitals belong to the same principal quantum number $(n=3)$ and follow the same rules for calculating the screening constant for electrons within the same shell,their screening constant values are identical.

(D) The electronic configuration of $Na$ $(Z=11)$ is $1s^2 2s^2 2p^6 3s^1$.
In this atom,the valence shell is the $n=3$ shell.
The screening effect (or shielding effect) is caused by all the inner electrons present in the shells $n=1$ and $n=2$.
Therefore,the electrons in the $1s$,$2s$,and $2p$ orbitals all contribute to the screening of the valence electron.

(N/A) Effective nuclear charge $(Z_{eff})$ is the net positive charge experienced by an electron in a multi-electron atom. The closer an orbital is to the nucleus,the greater is the $Z_{eff}$ experienced by the electron$(s)$ in it.
$(i)$ The $2s$ orbital is closer to the nucleus than the $3s$ orbital,so $2s$ experiences a greater $Z_{eff}$.
$(ii)$ The $4d$ orbital is closer to the nucleus than the $4f$ orbital,so $4d$ experiences a greater $Z_{eff}$.
$(iii)$ The $3p$ orbital is closer to the nucleus than the $3d$ orbital,so $3p$ experiences a greater $Z_{eff}$.

(B) Nuclear charge is defined as the net positive charge experienced by an electron in a multi-electron atom.
The higher the atomic number,the higher is the nuclear charge. Silicon $(Si)$ has $14$ protons while aluminium $(Al)$ has $13$ protons.
Since the atomic number of $Si$ $(Z=14)$ is greater than that of $Al$ $(Z=13)$,the valence electrons in $Si$ experience a greater pull from the nucleus compared to $Al$.
Therefore,the electrons in the $3p$ orbital of $Si$ experience a more effective nuclear charge than those in $Al$.

(N/A) In multielectron atoms,the outermost electrons are shielded or screened from the nucleus by the inner electrons. This is known as the shielding or screening effect.
Due to the shielding effect,the valence electron is held less tightly by the nucleus because of the intervening core electrons.
For example: The $2s$ electron in lithium $(Li)$ is shielded from the nucleus by the inner core of $1s$ electrons. As a result,the valence electron experiences a net positive charge (effective nuclear charge) which is less than the actual nuclear charge of $+3$.
Group: Shielding effect increases down a group as the number of inner shells increases. If inner orbitals are completely filled,the shielding effect is significant.
Period: In a period,the number of inner shells remains constant while the nuclear charge increases. Consequently,the shielding effect is less dominant compared to the increase in nuclear charge. Regarding orbital penetration,$2s$ electrons are closer to the nucleus than $2p$ electrons,meaning $2s$ electrons experience more effective nuclear charge and less shielding compared to $2p$ electrons.

(C) The effective nuclear charge $(Z_{eff})$ experienced by an electron depends on its distance from the nucleus and the shielding effect of inner electrons.
The $4p$ orbital has the highest principal quantum number $(n=4)$ compared to $2p$ $(n=2)$ and $3p$ $(n=3)$.
Electrons in the $4p$ orbital are farthest from the nucleus and are shielded by all the inner electrons $(1s, 2s, 2p, 3s, 3p, 3d, 4s)$.
Due to the greater distance and increased shielding,the $4p$ electrons experience the lowest effective nuclear charge among the given options.

(A) The effective nuclear charge $(Z_{eff})$ experienced by an electron depends on its distance from the nucleus and the shielding effect. Orbitals closer to the nucleus experience a greater effective nuclear charge.
$(i)$ For $2s$ and $3s$,$2s$ is closer to the nucleus than $3s$,so $2s$ experiences a larger $Z_{eff}$.
$(ii)$ For $4d$ and $4f$,$4d$ is closer to the nucleus than $4f$ due to the poor shielding of $f$-orbitals,so $4d$ experiences a larger $Z_{eff}$.
$(iii)$ For $3d$ and $3p$,$3p$ is closer to the nucleus than $3d$,so $3p$ experiences a larger $Z_{eff}$.
Thus,the orbitals experiencing larger effective nuclear charge are $2s$,$4d$,and $3p$.

(B) The electrons of $Si$ experience a higher effective nuclear charge.
Electronic configuration of $Al$ $(Z=13)$: $[Ne] 3s^{2} 3p^{1}$
Electronic configuration of $Si$ $(Z=14)$: $[Ne] 3s^{2} 3p^{2}$
Both elements have their valence electrons in the same shell $(n=3)$ and the same subshell $(3p)$.
However,the atomic number $(Z)$ of $Si$ $(14)$ is greater than that of $Al$ $(13)$.
As the number of protons increases,the nuclear charge increases,leading to a greater effective nuclear charge $(Z_{eff})$ experienced by the valence electrons in $Si$ compared to $Al$.

(A) The shielding effect of orbitals follows the order: $s > p > d > f$.
Since the effective nuclear charge $(Z_{eff})$ is inversely related to the shielding effect,the electrons in orbitals with lower shielding experience a higher effective nuclear charge.
Therefore,the increasing order of $Z_{eff}$ experienced by electrons in these subshells is $f < d < p < s$.

(N/A) In any given subshell,the energy of orbitals decreases as the effective nuclear charge $(Z_{\text{eff}})$ increases.
For example,the energy of $2s$ orbitals for $He$,$Be$,and $Mg$ follows the order: $E_{2s}(He) > E_{2s}(Be) > E_{2s}(Mg)$.

(B) The shielding effect depends on the penetration power of the orbitals. The order of penetration power is $s > p > d > f$.
Since the $2s$ orbital is more penetrating than the $2p$ orbital,the $2s$ electrons are closer to the nucleus and are less shielded by inner electrons.
Conversely,the $2p$ electrons are less penetrating,further from the nucleus,and experience a greater shielding effect from the inner shell electrons.

(D) Effective nuclear charge $(Z_{eff})$ increases across a period from left to right due to an increase in atomic number while the shielding effect remains relatively constant.
Among the given elements ($Li$,$Be$,$O$,$F$),all belong to the second period.
Fluorine $(F)$ has the highest atomic number $(Z = 9)$,which results in the strongest attraction between the nucleus and the valence electrons,making its effective nuclear charge the maximum.

(A) $1$. The shielding effect is most effective when the orbitals in the inner shells are completely filled,as they provide a dense electron cloud to screen the valence electrons from the nucleus. Thus,statement $(1)$ is correct.
$2$. Inner shells exhibit a significant shielding effect. Thus,statement $(2)$ is incorrect.
$3$. As we move down a group,the shielding effect increases due to the addition of new shells,which leads to a decrease in ionization energy. Thus,statement $(3)$ is incorrect.
$4$. The shielding effect depends on the number of inner electrons,not directly on the nuclear charge. An increase in nuclear charge actually increases the effective nuclear charge $(Z_{eff})$,which is the opposite of the shielding effect. Thus,statement $(4)$ is incorrect.

(A) The screening effect (or shielding effect) is the reduction in the effective nuclear charge on an electron due to the presence of other electrons in the inner shells or the same shell.
For a given shell,the $s$-orbital is closest to the nucleus and has the highest electron density near the nucleus,making it the most effective at shielding outer electrons.
The order of penetration power and screening ability is $s > p > d > f$.
This is because $s$-orbitals are more spherical and closer to the nucleus,while $f$-orbitals are more diffused and further away,providing the least shielding.
Therefore,the correct order of the screening effect is $s > p > d > f$.
Hence,the correct option is $A$.