Exercises — Variation of oxidation state across the table

What we do: read the anchor straight off, or use "free element $=0$".

• (a) Na is Group 1, gives one marble: $\boxed{+1}$.
• (b) O is an anchor: $\boxed{-2}$. (Check: Ca must then be $+2$, its group number.)
• (c) ${\text{F}}_2$ is a free element — no more-EN partner exists, so no marbles change hands: $\boxed{0}$. This is not the $-1$ anchor, because $-1$ only applies when F bonds to something less greedy than itself.

What we do: recall that $-2$ is the usual case, not a law.

• ${\text{H}}_2\text{O}$: O is $-2$ ✔ (normal).
• ${\text{H}}_2{\text{O}}_2$ (hydrogen peroxide): the two oxygens bond to each other. Neither is more EN than its twin, so that shared pair splits evenly, costing each O one grab. Result: O $=\boxed{-1}$.
• $\text{CaO}$: O is $-2$ ✔.

Answer: ${\text{H}}_2{\text{O}}_2$.

What we do: write the master equation, plug in the O anchor. $x+2(-2)=0$ Why this form: neutral molecule, so the total is $0$; two oxygens each carry $-2$. $x-4=0\Rightarrow x=\boxed{+6}$ Wait — recount: $x=+4$. Let us be careful: $x+2(-2)=0\Rightarrow x-4=0\Rightarrow x=+4$. So S in ${\text{SO}}_2$ is $\boxed{+4}$. (The $+6$ shown mid-line was a slip — always finish the algebra.)

What we do: master equation, but the total is the ion charge $-1$, not $0$. $x+4(-2)=-1$ Why $-1$: the whole ion carries a single negative charge, so the OS must sum to that. $x-8=-1\Rightarrow x=+7$ $\boxed{+7}$ — exactly Mn's maximum, $3d^{5}4s^2=7$ removable electrons (see Transition Metal Chemistry).

$3(+1)+x+4(-2)=0$ Why: three H at $+1$, four O at $-2$, neutral molecule. $3+x-8=0\Rightarrow x-5=0\Rightarrow x=+5$ $\boxed{+5}$ = P's group valence ($3s^{2}3p^3=5$). Maximum reached.

What we do: first decide who is greedier. F is more electronegative than O, so F takes, meaning F $=-1$ and O must go positive. $x+2(-1)=0\Rightarrow x=+2$ $\boxed{+2}$. This is one of the rare times oxygen is positive — only F can force it, because only F out-grabs O.

What we do: treat all N together first, then note it is an average of two different environments. Let each N have average OS $x$. There are 2 N, 4 H ($+1$), 3 O ($-2$): $2x+4(+1)+3(-2)=0$ $2x+4-6=0\Rightarrow 2x-2=0\Rightarrow x=+1$ Average N $=\boxed{+1}$. The analysis part: this $+1$ is a mathematical average. The real N atoms are not both $+1$:

• In ${\text{NH}}_4^{+}$: $x+4(+1)=+1\Rightarrow x=-3$.
• In ${\text{NO}}_3^{-}$: $x+3(-2)=-1\Rightarrow x=+5$. The mean of $-3$ and $+5$ is $\frac{-3+5}{2}=+1$ ✔. Same answer two ways — that consistency is your check.

What we do: compute the metal OS, then apply "big & heavy ⇒ prefers the lower state" (Inert Pair Effect).

• (a) ${\text{PbCl}}_2$: $x+2(-1)=0\Rightarrow x=+2$. ${\text{PbCl}}_4$: $x+4(-1)=0\Rightarrow x=+4$. Pb is heavy Group-14, so its $6s^2$ pair is inert ⇒ $+2$ is more stable ⇒ ${\text{PbCl}}_2$ wins. (${\text{PbCl}}_4$ decomposes, dumping ${\text{Cl}}_2$.)
• (b) $\text{SnO}$: $x+(-2)=0\Rightarrow x=+2$. ${\text{SnO}}_2$: $x+2(-2)=0\Rightarrow x=+4$. Sn is lighter than Pb, so the inert-pair effect is weaker — $+4$ (${\text{SnO}}_2$) is still stable, but $+2$ exists. (Contrast: for C and Si, $+4$ dominates almost entirely.)

Dichromate: $2x+7(-2)=-2\Rightarrow 2x-14=-2\Rightarrow 2x=12\Rightarrow x=+6$. Chromate: $x+4(-2)=-2\Rightarrow x-8=-2\Rightarrow x=+6$. Both $\boxed{+6}$ — Cr's maximum ($3d^{5}4s^1$, 6 removable electrons). The two ions are just different packagings of the same $+6$ chromium, which is why they interconvert with acidity (Redox Reactions).

What we do: Cl is Group 17, $n_v=7$, so its max positive OS is $+7$. Pair it with O ($-2$) and balance. Let the oxide be ${\text{Cl}}_a{\text{O}}_b$, neutral: $a(+7)+b(-2)=0\Rightarrow 7a=2b$. Smallest whole numbers: $a=2,b=7$. So $\boxed{{\text{Cl}}_2{\text{O}}_7}$ (dichlorine heptoxide), the anhydride of ${\text{HClO}}_4$. ✔

Average: 2 Na ($+1$), 2 S ($x$ avg), 3 O ($-2$), neutral: $2(+1)+2x+3(-2)=0\Rightarrow 2+2x-6=0\Rightarrow 2x=4\Rightarrow x=+2$ Average S $=\boxed{+2}$. Structural truth: thiosulfate is a sulfate with one O replaced by an S. The central S sits where the $+6$ sulfur was; the outer (terminal) S replaces an oxygen, so it behaves like the $-2$ it displaced. The mean $\frac{(+6)+(-2)}{2}=+2$ ✔ matches — a fractional-looking average that hides two very different atoms.

Naive attempt (the trap): treat all 5 oxygens as $-2$: $2(+1)+x+5(-2)=0\Rightarrow 2+x-10=0\Rightarrow x=+8$ $+8$ is impossible: sulfur has only 6 valence electrons, so its ceiling is $+6$. The fix — spot the peroxo bond. Two of those oxygens form an O–O (peroxide) linkage, so each is $-1$, not $-2$. The other three oxygens are ordinary $-2$. $2(+1)+x+2(-1)+3(-2)=0$ $2+x-2-6=0\Rightarrow x=+6$ $\boxed{+6}$ — S at its legitimate maximum. The oxidising power of Caro's acid comes from the peroxide oxygens, not from any mythical $+8$ sulfur.

Find the OS in ${\text{MO}}_4^{-}$: $x+4(-2)=-1\Rightarrow x=+7$ — consistent with "max $+7$". Which element? Max $+7$ means 7 removable electrons. In period 4:

• A main-group candidate would need $n_v=7$ → Group 17 (Br). But its stable oxo-anion is ${\text{BrO}}_4^{-}$ (perbromate), which exists but is hard to make.
• A transition candidate with $3d^{5}4s^2=7$ is manganese — and ${\text{MnO}}_4^{-}$ (permanganate) is a textbook-stable, deep-purple ion.

Best answer: $\boxed{M=\text{Mn}}$, because $3d^{5}4s^2$ supplies exactly 7 electrons and permanganate is the classic stable ${\text{MO}}_4^{-}$. (Br is the acceptable main-group alternative.) See Transition Metal Chemistry.