Exercises — Variable oxidation states — reasons

Rule: oxygen $=-2$, fluorine and chlorine (bonded to metal) $=-1$, and all OS in the species add to the overall charge.

• (a) $2\text{Cr}+7(-2)=-2\Rightarrow 2\text{Cr}=12\Rightarrow \text{Cr}=+6$.
• (b) $\text{Mn}+4(-2)=-1\Rightarrow \text{Mn}=+7$.
• (c) $\text{Ti}+4(-1)=0\Rightarrow \text{Ti}=+4$.
• (d) $\text{V}+5(-1)=0\Rightarrow \text{V}=+5$.

Mn is element 25: $[\text{Ar}]3d^{5}4s^2$. $3d$ holds $5$, $4s$ holds $2$, so 7 valence electrons ($5+2$). This "7" is exactly why Mn can reach $+7$ — you have seven electrons to play with.

Count valence electrons (electrons beyond the Ar core):

All five hit their theoretical ceiling — this is the rising arm of the curve (see figure).

For $+2$: cost $=I{E}_1+I{E}_2=700+1500=2200$. Release $=3600$. Since $2200<3600$, $\Delta H_{\text{form}}<0$ → $+2$ is stable (net release $1400$). For $+3$: cost $=I{E}_1+I{E}_2+I{E}_3=700+1500+2800=5000$. Release $=5100$. Since $5000<5100$, $\Delta H_{\text{form}}<0$ → $+3$ is also stable (net release $100$, only just). Notice the razor-thin margin at $+3$: this is why higher states are fragile — one big jump in $IE$ tips the balance to unrecoverable.

Rising arm (Sc→Mn): each added $3d$ electron is unpaired and sits at roughly the same energy as $4s$. Successive $IE$s rise only gently (see L3.2), so every valence electron can still be paid for by strong M–O/M–F bonds. Max OS = valence count. Turning point (Mn): at $3d^{5}4s^2$ all seven are available and half-filled $d$ has no pairing penalty yet — the peak. Falling arm (Fe→Zn): past $d^5$, new electrons must pair up in already-occupied $d$ orbitals, and $Z_{\text{eff}}$ keeps rising because each added proton is poorly shielded by same-shell $d$ electrons. Both effects pull $3d$ down in energy and inward → the $IE$ needed to strip the top electrons balloons → those high states can no longer be repaid. So the realised ceiling drops, until Zn ($3d^{10}4s^2$) offers only its two $4s$ electrons → $+2$ only.

Look at the jumps (ratios), not the raw values.

• X: $I{E}_3/I{E}_2=7730/1450\approx 5.3$ — a huge cliff after the 2nd electron. That cliff means the 3rd electron comes from a deep filled shell → X is s-block (like Mg). It stops dead at +2.
• Y: $I{E}_2/I{E}_1\approx 2.05$, $I{E}_3/I{E}_2\approx 2.07$, $I{E}_4/I{E}_3\approx 1.35$ — smooth, gentle rises, no cliff. That gradual rise is the $(n-1)d\approx ns$ signature → Y is a transition metal, comfortable in +2, +3, and possibly +4. The diagnostic is the ratio: an s-block metal shows one dramatic jump right after its group's electron count; a transition metal shows a slow ramp.

Why ${\text{Mn}}_2{\text{O}}_7$ works: reaching $+7$ means Mn must lose/share seven electrons. Only a small, hard, highly electronegative partner can grip that much shared electron density and form strong bonds — oxygen qualifies (small, $-2$, forms strong M=O bonds). Why ${\text{MnI}}_7$ fails: iodide ${\text{I}}^{-}$ is huge and its outer electrons are loosely held → easily oxidised. A metal centre hungry for seven electrons will simply rip electrons off ${\text{I}}^{-}$, converting $2{\text{I}}^{-}\to {\text{I}}_2+2e^{-}$ and collapsing to a low OS iodide. So instead of ${\text{MnI}}_7$ you get ${\text{MnI}}_2$ (+ ${\text{I}}_2$). ${\text{MnF}}_7$ vs ${\text{MnCl}}_7$: fluorine is smaller and far harder to oxidise than chlorine, so ${\text{MnF}}_7$-type high-OS fluorides are the more plausible direction (fluorides and oxides are where the highest OS live). Chlorine, larger and more oxidisable, tops out lower — consistent with the general rule O, F → high OS; heavy halides → low OS. See Acidic vs Basic character of Oxides across OS for the follow-on that ${\text{Mn}}_2{\text{O}}_7$ is strongly acidic.

Use $\Delta H_{\text{form}}=\Delta H_{\text{atom}}+\sum I{E}_i+\Delta H_{\text{anion}}+U_{\text{lattice}}$. ${\text{MCl}}_2$: $150+2150+(-700)+(-2500)=-900{\text{kJ mol}}^{-1}$. ${\text{MCl}}_3$: $150+5450+(-1050)+(-5600)=-1050{\text{kJ mol}}^{-1}$. Both are negative (both form), but ${\text{MCl}}_3$ is more negative by $150$ → $+3$ is the favoured state here, because the much larger lattice release ($-5600$ vs $-2500$) of the higher-charged ion outpaces the extra $I{E}_3$ cost. This is the quantitative face of "$\sum I{E}_i$ rises slowly so several $n$ satisfy the inequality."

Sc configuration: $[\text{Ar}]3d^{1}4s^2$.

• Removing two electrons gives ${\text{Sc}}^{2+}$: $[\text{Ar}]3d^1$ — a lone, high-energy $d$ electron with nothing stabilising it.
• Removing the third gives ${\text{Sc}}^{3+}$: $[\text{Ar}]$ — the noble-gas core, extremely stable. The lattice/hydration release for the small, triply-charged ${\text{Sc}}^{3+}$ is large and easily repays $I{E}_3$. So Sc "rolls all the way to $+3$": $+2$ is a shallow, transient stop; $+3$ is a deep energetic well. The Ca analogy fails because Ca's third electron would come from the filled $3p$ core (an enormous $I{E}_3$ cliff, s-block behaviour), whereas Sc's third electron is the easily-lost $3d^1$. Fe contrast: $[\text{Ar}]3d^{6}4s^2$. Losing two → ${\text{Fe}}^{2+}$ ($3d^6$) is stable; losing one more → ${\text{Fe}}^{3+}$ ($3d^5$, half-filled, extra-stable by exchange energy). Here both landing points are genuinely favourable, so Fe truly shows $+2$ and $+3$ (see Stability of +2 and +3 states (Fe, Mn)). The difference: for Sc only $+3$ hits a stable core; for Fe two adjacent states are both stable.

(a) Zn is element 30: $[\text{Ar}]3d^{10}4s^2$. (b) No $+3$: removing a third electron must come out of the filled, low-lying $3d^{10}$ shell. That is a genuine $I{E}_3$ cliff (like an s-block jump) — no lattice/hydration term can repay it. So ${\text{Zn}}^{3+}$ doesn't form. (c) No common $+1$: removing just one $4s$ leaves $3d^{10}4s^1$, an unpaired, un-noble configuration; removing both $4s$ reaches the fully symmetric $3d^{10}$ closed set, which the doubly-charged ion's large lattice/hydration energy repays well. $+2$ is the stable well, so $+1$ is skipped. Generalisation: variable OS survives only while successive electrons come from the near-degenerate $(n-1)d/ns$ pool at gradual cost. The moment the next electron must leave a filled inner shell (Zn's $3d^{10}$, or any s-block core), $IE$ jumps by a cliff, cannot be repaid, and the element locks to a single OS. This is the same principle that fixes Na at $+1$ — it just happens at the end of the $d$-series instead of the start of a period.

Diagnostic that tells an s-block from a d-block metal from IE data ::: a sudden huge jump (cliff) in successive IEs marks the s-block's inner-shell wall; a gradual ramp marks the d-block's $(n-1)d\approx ns$ pool. Why does the observed max OS peak at Mn ::: up to Mn all $(3d+4s)$ electrons are unpaired and cheap to remove; after Mn pairing + rising $Z_{\text{eff}}$ bury the $d$ electrons. Why ${\text{Mn}}_2{\text{O}}_7$ exists but ${\text{MnI}}_7$ does not ::: O is small/hard and stabilises high shared charge; ${\text{I}}^{-}$ is large and easily oxidised, so it reduces the metal to a low OS. Why does Zn show only $+2$ ::: its third electron would leave the filled $3d^{10}$ shell — an IE cliff no release term can repay.