Question bank — Effective nuclear charge Z_eff — Slater's rules

False. $S$ is defined as repulsion from other electrons only; a charge cannot push on itself. Always use (group population $-1$) for the same group.

True. Since $S=\sum (\text{positive contributions})\ge 0$, we always have $Z_{eff}=Z-S\le Z$. Shielding can only reduce the felt charge, never amplify it.

False. Each electron has its own $S$: a deep $1s$ electron is barely shielded (high $Z_{eff}$), an outer $3s$ electron is heavily shielded (low $Z_{eff}$). $Z_{eff}$ is per-electron, not per-atom.

False. For a $d$ (or $f$) target, every group to its left — including same-$n$ $s,p$ — contributes the full $1.00$. The $0.85$ tier exists only for $s/p$ targets.

False. The proton adds $+1$ to $Z$ but the new same-shell electron adds only $\approx 0.35$ to $S$, so $Z_{eff}$ climbs by $\approx 0.65$ each step. This net rise drives the whole period trend.

False. Larger $Z_{eff}$ = the electron is pulled harder, so it is harder to remove → higher Ionization Energy. Easy ionization comes from low $Z_{eff}$ (e.g. Na's $3s$ at $2.20$).

False. $3s$ and $3p$ share a group, but $3d$ is a separate group placed after it: $(3s,3p)(3d)$. Grouping by $n$ alone is the classic error.

The $1s$ shell is $n-2$ relative to the $n=3$ target, not $n-1$. Shells $\le n-2$ shield at $1.00$, so it's $2\times 1.00$. Only the $n=2$ shell would be the $0.85$ tier.

For a $d$ target only groups to the left (deeper/lower in the Slater order) count. $4s$ sits to the right of $3d$, so it contributes $0$ to the $3d$ electron's shielding.

Being beside you is not being between you and the nucleus. Same-group electrons are at roughly the same radius, so they screen poorly — only $0.35$, the weakest nonzero tier.

Inside the $1s$ group the same-group value is the special $0.30$, not $0.35$. So $S=0.30$ and $Z_{eff}=2-0.30=1.70$.

Filling order tracks total energy, not $Z_{eff}$. Once both are occupied, the $3d$ actually feels a higher $Z_{eff}$ than $4s$; that is exactly why $4s$ leaves first. See Aufbau and 4s vs 3d filling.

Adding an electron raises $S$ (more same-shell repulsion) while $Z$ is fixed, so $Z_{eff}$ drops. That is why anions are larger than their neutral atoms.

That rule is about what shields the $d$ electron, not what the $d$ electron shields. As a shielder of an outer $ns$, a $3d$ electron is treated as an $(n-1)$ neighbour at $0.85$ — poor penetration means poor screening.

The $n-1$ shell is mostly inside the target but its outer edge overlaps the target's own region, so it doesn't sit fully between the electron and nucleus — it screens strongly but incompletely.

A same-shell partner shares the target's radius, so it spends only a fraction of its orbit genuinely between the electron and the nucleus; $0.35$ is the empirical value that best reproduces that partial time-inside across many measured atoms. The smallness is physical; the exact digits are tuned.

They are poorly penetrating and sit further out for a given $n$, so they neither shield nor are shielded like the $s,p$ of the same shell — a separate group with all-left $=1.00$ captures that different behaviour. See Shielding and Penetration.

Many-electron atoms can't be solved exactly, so the numbers ($0.35$, $0.85$, $1.00$) were tuned to reproduce measured energies — they encode real physics but aren't outputs of the Schrödinger equation.

A rising $Z_{eff}$ pulls the same-shell electrons inward more tightly each step, contracting the electron cloud even as electrons are added.

The $3d$ feels $Z_{eff}\approx 8.85$ vs the $4s$'s $\approx 4.35$, so the $4s$ is held far more loosely and is energetically cheapest to lose first.

If same-shell electrons shielded a full $1.00$, each added proton would be exactly cancelled and $Z_{eff}$ would stay flat — the incomplete $0.35$ leaves a net $\approx 0.65$ gain that grows the trend.

There are no other electrons, so $S=0$ and $Z_{eff}=1-0=1$. A one-electron atom feels the full nucleus — the ceiling case of no shielding.

Only the same-group $0.30$ can — there is nothing "to the left" of $1s$, so no $0.85$ or $1.00$ term exists for a $1s$ target.

No electron means no electron to feel a charge, so $Z_{eff}$ is undefined; it is always the charge felt by a chosen electron.

The one with more protons has the higher $Z_{eff}$ if the extra electron went into the same shell (net $+0.65$); Slater predicts the heavier one holds its valence electron tighter.

It rises sharply — deep electrons have almost nothing "between" them and the nucleus, so their $S$ is tiny and $Z_{eff}$ approaches $Z$.

No. Each of the other electrons in the group still contributes its $0.35$; a full shell simply means the count of same-group shielders is maximal, so same-shell shielding is at its largest, not zero.