Oxidation Number Calculator with Steps | StatsUnlock

To calculate an oxidation number, give the atoms with fixed values their numbers (group 1 metals +1, oxygen −2, hydrogen +1, fluorine −1), then choose the unknown so that all the oxidation numbers add up to the charge on the compound or ion. For chromium in potassium dichromate, K₂Cr₂O₇:

Rule of thumb: the sum of the oxidation numbers equals the overall charge: 0 for a neutral compound, −2 for Cr₂O₇²⁻. An oxidation number that goes up in a reaction means oxidation; one that goes down means reduction.

An oxidation number (or oxidation state) is a bookkeeping charge. Imagine every bond in a compound broken so that the more electronegative atom takes all the shared electrons: the charge left on each atom is its oxidation number. IUPAC defines it this way in its 2016 recommendation (Karen, McArdle & Takats, 2016). In water, oxygen takes both bonding pairs, so O is −2 and each H is +1.

An oxidation number calculator applies the standard rules in the order textbooks teach them, then uses one fact to finish the job: the oxidation numbers of all the atoms add up to the overall charge. Whatever is left unknown is solved from that sum. The calculator also tracks oxidation numbers across a reaction to find what is oxidised and reduced, works out the metal in a coordination compound from its ligands, and counts bonds to give each carbon in an organic molecule its own oxidation number.

A worked one-liner: in KMnO₄, K is +1 and each O is −2, so +1 + x + 4(−2) = 0 and Mn is +7.

Who uses it: oxidation numbers are taught in NCERT Class 11 (“Redox Reactions”) and asked every year in JEE and NEET; GCSE, A level, AP Chemistry and IB courses use them to balance redox equations and name compounds such as iron(III) chloride. Analysts use them to work out redox titrations, and electrochemists to count the electrons in a cell reaction.

What this free oxidation number calculator solves

One tool covers every oxidation number calculation in a typical course. It works as an oxidation number calculator with steps for compounds and ions, a change in oxidation number calculator for redox reactions, an oxidation number calculator for coordination compounds, and an oxidation number calculator for organic compounds that gives each carbon separately. Press Try it to load an example into the calculator, then press the button to see the full working.

Free, with no sign-up. Everything runs in your browser: nothing you type is sent to a server or stored.

En español: ¿cómo se calcula el número de oxidación? Asigna los valores fijos (O −2, H +1, metales del grupo 1 +1) y despeja la incógnita para que la suma sea igual a la carga: en K₂Cr₂O₇, 2(+1) + 2x + 7(−2) = 0, así que el Cr es +6.

How accurate is this oxidation number calculator?

The rules follow the IUPAC definition and the priority order used in textbooks: fixed values first, then the sum. Results are exact fractions, never rounded decimals, so Fe₃O₄ gives +8/3 rather than 2.67. Every answer is checked before it gets the Verified badge: the atoms × oxidation numbers must add up to the charge, and in a balanced redox equation the electrons lost must equal the electrons gained. The calculator also warns when a value is above an element’s group maximum (a sign of peroxide bonds, as in CrO₅) or when it had to assume a value. The R and Python scripts in Sections 10 and 11 return the same +6 for chromium in K₂Cr₂O₇.

Studies of how students learn redox find that oxidation numbers are easy to calculate but often confused with ionic charges and with the gain or loss of real electrons (Garnett & Treagust, 1992; Österlund, Berg & Ekborg, 2010). That is why the results label each value with the rule it came from and show where it sits on the number line.

4.1 Positive, negative and zero

A positive oxidation number means the atom has, on paper, given electrons to more electronegative partners; a negative one means it has taken them. Zero means neither, as in every free element (O₂, I₂, S₈, Fe). The size tells you how many: Mn at +7 in KMnO₄ has given away all seven of its valence electrons, which is why permanganate is such a strong oxidising agent.

4.2 The highest and lowest possible values

For main-group elements the highest oxidation number is the number of valence electrons (the group number, or group − 10 for groups 13 to 18), and the lowest is that number minus 8. Sulfur ranges from −2 (H₂S) to +6 (H₂SO₄); nitrogen from −3 (NH₃) to +5 (HNO₃). Transition metals reach their group number at most: Cr +6, Mn +7. A species at its maximum can only be reduced; at its minimum it can only be oxidised.

4.3 Fractional oxidation numbers are averages

An oxidation number such as +8/3 for iron in Fe₃O₄ does not mean any atom carries a fraction of an electron. Magnetite contains one Fe²⁺ and two Fe³⁺: (2 + 3 + 3) ÷ 3 = 8/3. The same happens in S₄O₆²⁻ (average +5/2) and in KO₂ (O at −1/2). The algebraic method always gives the average; the structure gives each atom.

4.4 A value above the maximum means a peroxide

If the sum puts an element above its group number, the rule “O is −2” has failed somewhere. CrO₅ gives Cr +10, which is impossible: the molecule has two O–O peroxide groups, so four oxygens are −1 and chromium is really +6. Peroxodisulfate (S₂O₈²⁻) gives +7 for S for the same reason; the true value is +6. The calculator flags these cases.

4.5 Oxidation number, charge, valency and formal charge

These four are easy to mix up (Parkin, 2006). The charge is real and belongs to a whole ion. The oxidation number is a bookkeeping charge per atom with every bond made ionic. Valency is the number of bonds an atom forms, with no sign. Formal charge splits each bond equally instead of giving it to the more electronegative atom. In carbon monoxide, C has oxidation number +2 but formal charge −1, and the molecule has no charge at all.

4.6 Oxidation and reduction: OIL RIG

Oxidation is loss of electrons, so the oxidation number goes up; reduction is gain, so it goes down. The oxidising agent contains the element that is reduced, and the reducing agent contains the element that is oxidised. In a balanced equation the electrons lost equal the electrons gained: in MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O, one Mn gains 5 and five Fe lose 1 each.

4.7 Coordination compounds and d-electron counts

In a complex the metal’s oxidation number is the charge of the complex ion minus the ligand charges. Neutral ligands (NH₃, H₂O, CO, en) add nothing; anionic ones (Cl⁻, CN⁻, C₂O₄²⁻) add their charge. Once you have it, the d-electron count is group number minus oxidation number: Fe(II) in K₄[Fe(CN)₆] is d⁶, and Co(III) in [Co(NH₃)₆]³⁺ is also d⁶.

4.8 Organic carbon: the oxidation ladder

Each bond from carbon to a more electronegative atom (O, N, halogen) counts +1, each bond to H counts −1, and C–C bonds count 0. That gives the ladder used in organic chemistry: CH₄ (−4), CH₃OH (−2), HCHO (0), HCOOH (+2), CO₂ (+4). Turning ethanol into acetic acid moves the CH₂OH carbon from −1 to +3, an oxidation (Menzek, 2002).

4.9 A field note on where these problems go wrong

In marking oxidation number work, the same slips appear again and again: forgetting to multiply by the number of atoms (writing 2 + x − 2 = 0 for K₂Cr₂O₇), setting the sum to 0 for an ion, giving oxygen −2 in a peroxide, and treating the oxidation number as a real ionic charge (Cr⁶⁺ does not exist in dichromate). Concept-inventory research finds these errors persist into university (Brandriet & Bretz, 2014). The calculator shows the multiplication in every step and checks the sum against the charge.

▶ Solve a problem above to fill all five cards with your own answer.

The rules for calculating oxidation numbers are applied in order: a rule higher on the list wins when two disagree. The final rule, that the oxidation numbers add up to the charge, turns the problem into one equation with one unknown.

This flowchart shows how to calculate the oxidation number of an element in a compound, on the same problem used in the Excel, R, Python and by-hand sections: calculate the oxidation number of Cr in K₂Cr₂O₇.

Check any box yourself: the Excel sheet in Section 9, the R script in Section 10 and the Python script in Section 11 all return +6 for this problem.

1. Choose what you want to find. Pick a compound or ion, a change in a reaction, the metal in a complex, or each carbon in an organic molecule.
2. Type the formula or name. Write K2Cr2O7 or potassium dichromate; the preview shows the atoms and the charge as you type.
3. Add the charge for an ion. Write it in the formula as Cr2O7^2- or in the charge box as 2-.
4. Pick the element to find. Choose the underlined atom from the list, or leave it on every element.
5. For a reaction, type the equation. Use -> for the arrow and choose acidic or basic solution for ionic redox equations.
6. For a complex, use square brackets. Write K4[Fe(CN)6] or [Co(NH3)6]Cl3 with ligands in round brackets.
7. For an organic compound, write it condensed. Write CH3CH2OH, with = and # for double and triple bonds.
8. Press the button. Nothing is calculated before you press it, and changing any input clears the old answer.
9. Read the rule for each element and the check. The Verified badge means the oxidation numbers add up to the charge.
10. Copy, share or batch. Copy the answer, share a link, download a report, or solve a whole worksheet on the Batch tab.

Excel turns the sum rule into one small table: one row per element with its number of atoms and its known oxidation number, a column for atoms × oxidation number, and one cell that solves for the unknown. The sheet below uses the headline problem, so it must return +6 for chromium in K₂Cr₂O₇, the same as the oxidation number calculator above.

9.1 The functions you need

9.2 Step by step

9.3 Awkward cases

• Peroxides and hydrides. Type the exception in column C (O = −1 in H₂O₂, H = −1 in NaH) before solving.
• Ammonium salts. Solve each ion on its own row block: NH₄⁺ with charge +1 and NO₃⁻ with charge −1.
• Two unknowns. Excel needs one unknown per equation. Split the compound into its ions (Fe₂(SO₄)₃ into Fe³⁺ and SO₄²⁻) or use the calculator.

9.4 Why your Excel answer might differ

9.5 Charting in Excel

1. Select the elements and their contributions (A2:A4, D2:D4).
2. Insert → Charts → Clustered Column. Positive bars (electrons given away) and negative bars (electrons taken) balance at zero, as in Chart 2 above.

9.6 Excel errors you will probably hit

This base R script reads a formula (with brackets and an ion charge), applies the fixed rules in priority order, solves the remaining element from the sum rule, prints each element with the rule it used, checks nine textbook compounds and saves one 300 dpi chart. For the headline problem it prints Cr = +6, matching the oxidation number calculator on this page.

10.1 The complete script

R, base only, saves oxidation_number.png

# ============================================================
# Oxidation number calculator in R (base R only):
# fixed rules first, then solve the unknown from "sum = charge"
# StatsUnlock - https://statsunlock.com/oxidation-number-calculator/
# ============================================================

# ---- 1. Your compound or ion ----------------------------------------
formula <- "K2Cr2O7"     # write ions as "Cr2O7^2-" or "MnO4^-"
find    <- "Cr"          # the element you want (the underlined atom)

group1 <- c("Li", "Na", "K", "Rb", "Cs")
group2 <- c("Be", "Mg", "Ca", "Sr", "Ba")
fixed  <- c(Al = 3, Zn = 2, Ag = 1)
metals <- c(group1, group2, "Al", "B")   # partners that make H a hydride (-1)

parse_formula <- function(f) {
  charge <- 0
  m <- regmatches(f, regexec("\\^([0-9]*)([+-])$", f))[[1]]
  if (length(m)) {
    charge <- ifelse(m[2] == "", 1, as.numeric(m[2])) * ifelse(m[3] == "+", 1, -1)
    f <- sub("\\^[0-9]*[+-]$", "", f)
  }
  # expand brackets such as Ca(NO3)2 -> CaN2O6 counts
  while (grepl("\\(", f)) {
    m2 <- regmatches(f, regexec("\\(([^()]*)\\)([0-9]*)", f))[[1]]   # innermost bracket and its multiplier
    k <- ifelse(m2[3] == "", 1, as.numeric(m2[3]))
    parts <- regmatches(m2[2], gregexpr("[A-Z][a-z]?[0-9]*", m2[2]))[[1]]
    rep <- paste0(sapply(parts, function(p) {
      el <- sub("[0-9]+$", "", p); n <- sub("^[A-Z][a-z]?", "", p)
      paste0(el, ifelse(n == "", 1, as.numeric(n)) * k)
    }), collapse = "")
    f <- sub(m2[1], rep, f, fixed = TRUE)
  }
  parts <- regmatches(f, gregexpr("[A-Z][a-z]?[0-9]*", f))[[1]]
  counts <- setNames(numeric(0), character(0))
  for (p in parts) {
    el <- sub("[0-9]+$", "", p); n <- sub("^[A-Z][a-z]?", "", p)
    n <- ifelse(n == "", 1, as.numeric(n))
    if (el %in% names(counts)) counts[el] <- counts[el] + n else counts[el] <- n
  }
  list(counts = counts, charge = charge)
}

oxidation_numbers <- function(f) {
  p <- parse_formula(f); counts <- p$counts; z <- p$charge
  els <- names(counts)
  ox <- setNames(rep(NA_real_, length(els)), els)
  rule <- setNames(rep("", length(els)), els)
  if (length(els) == 1) {
    ox[1] <- z / counts[1]; rule[1] <- "element or simple ion"
    return(list(counts = counts, z = z, ox = ox, rule = rule))
  }
  n_unknown <- function() sum(is.na(ox))
  for (e in els) {
    if (e == "F") { ox[e] <- -1; rule[e] <- "F is always -1" }
    else if (e %in% group1) { ox[e] <- 1; rule[e] <- "group 1 is +1" }
    else if (e %in% group2) { ox[e] <- 2; rule[e] <- "group 2 is +2" }
    else if (e %in% names(fixed)) { ox[e] <- fixed[[e]]; rule[e] <- "fixed value" }
  }
  if ("H" %in% els && is.na(ox["H"]) && n_unknown() > 1) {
    hydride <- all(setdiff(els, "H") %in% metals)
    ox["H"] <- ifelse(hydride, -1, 1); rule["H"] <- ifelse(hydride, "H is -1 in a metal hydride", "H is +1")
  }
  if ("O" %in% els && is.na(ox["O"]) && n_unknown() > 1) { ox["O"] <- -2; rule["O"] <- "O is -2" }
  for (e in c("Cl", "Br", "I")) {
    if (e %in% els && is.na(ox[e]) && n_unknown() > 1 && !("O" %in% els) && !("F" %in% els)) { ox[e] <- -1; rule[e] <- "halogen is -1" }
  }
  left <- els[is.na(ox)]
  if (length(left) != 1) stop("More than one unknown: ", paste(left, collapse = ", "))
  known <- sum(counts[!is.na(ox)] * ox[!is.na(ox)])
  ox[left] <- (z - known) / counts[left]; rule[left] <- "solved from sum = charge"
  list(counts = counts, z = z, ox = ox, rule = rule)
}

frac_txt <- function(v) {                    # 2.666667 -> "8/3" without extra packages
  for (d in 1:12) if (abs(v * d - round(v * d)) < 1e-9) return(if (d == 1) as.character(round(v)) else paste0(round(v * d), "/", d))
  format(v, digits = 6)
}
sgn <- function(v) ifelse(v > 0, paste0("+", frac_txt(v)), frac_txt(v))

# ---- 2. Solve and show the working ------------------------------------
r <- oxidation_numbers(formula)
cat("Formula ", formula, ", charge ", r$z, "\n", sep = "")
for (e in names(r$counts)) {
  cat(sprintf("  %-2s  atoms %d  ON %5s  contribution %5s  (%s)\n", e, r$counts[[e]],
              sgn(r$ox[[e]]), sgn(r$counts[[e]] * r$ox[[e]]), r$rule[[e]]))
}
total <- sum(r$counts * r$ox)
cat("  Sum of contributions =", total, ifelse(abs(total - r$z) < 1e-9, "= charge", "!= charge"), "\n")
cat("Oxidation number of", find, "in", formula, "=", sgn(r$ox[[find]]), "\n")

# ---- 3. Checks ----------------------------------------------------------
stopifnot(abs(total - r$z) < 1e-9)
tests <- list(c("K2Cr2O7", "Cr", 6), c("KMnO4", "Mn", 7), c("H2SO4", "S", 6),
              c("Cr2O7^2-", "Cr", 6), c("Na2S2O3", "S", 2), c("Na5P3O10", "P", 5),
              c("H2O2", "O", -1), c("NaH", "H", -1), c("OF2", "O", 2))
for (t in tests) stopifnot(abs(oxidation_numbers(t[1])$ox[[t[2]]] - as.numeric(t[3])) < 1e-9)
stopifnot(abs(oxidation_numbers("Fe3O4")$ox[["Fe"]] - 8 / 3) < 1e-9)
cat("Verdict: Verified (all test compounds agree)\n")

# ---- 4. Figure: each element's contribution to the charge ---------------
vals <- r$counts * r$ox
png("oxidation_number.png", width = 2100, height = 1350, res = 300)
par(mar = c(4, 4.5, 3, 1))
b <- barplot(vals, names.arg = names(vals), col = ifelse(vals > 0, "#ea580c", "#2563eb"),
             ylab = "Atoms x oxidation number", main = paste0(formula, ": contributions add up to ", r$z),
             ylim = range(c(0, vals)) * 1.25)
abline(h = 0, col = "#334155")
text(b, vals, labels = paste0(r$counts, " x (", sapply(r$ox, sgn), ")"), pos = ifelse(vals >= 0, 3, 1), cex = 0.8)
invisible(dev.off())
cat("Saved oxidation_number.png\n")

Output: Cr atoms 2 ON +6 contribution +12 (solved from sum = charge), Oxidation number of Cr in K2Cr2O7 = +6, Verdict: Verified. With formula <- "Fe3O4" it prints +8/3.

10.2 R traps

• Escape the caret in regular expressions: "\\^", or the charge is not read.
• Named vectors keep the element order of the formula; counts[el] on a missing name gives NA, so test with %in% first.
• The script stops with a message when two elements have no rule (for example Fe₂(SO₄)₃); split the salt into its ions.

10.3 What each part does

10.4 The figure

The script saves oxidation_number.png at 2100 × 1350 pixels and 300 dpi: one bar per element showing atoms × oxidation number, orange above zero and blue below.

This script does the same job in plain Python with exact fractions: it parses the formula, applies the fixed rules, solves the unknown from the sum rule, prints the working, checks ten compounds (including Fe₃O₄ at +8/3) and saves a 300 dpi chart. It is the same method the oxidation number calculator runs in your browser.

Install once: pip install matplotlib

11.1 The complete script

Python, saves oxidation_number.png

# ============================================================
# Oxidation number calculator in Python:
# fixed rules first, then solve the unknown from "sum = charge"
# StatsUnlock - https://statsunlock.com/oxidation-number-calculator/
# pip install matplotlib
# ============================================================
import re
from fractions import Fraction
import matplotlib
matplotlib.use("Agg")
import matplotlib.pyplot as plt

# ---- 1. Your compound or ion ----------------------------------------
formula = "K2Cr2O7"     # write ions as "Cr2O7^2-" or "MnO4^-"
find = "Cr"             # the element you want (the underlined atom)

GROUP1 = {"Li", "Na", "K", "Rb", "Cs"}
GROUP2 = {"Be", "Mg", "Ca", "Sr", "Ba"}
FIXED = {"Al": 3, "Zn": 2, "Ag": 1}
METALS = GROUP1 | GROUP2 | {"Al", "B"}          # partners that make H a hydride (-1)

def parse(f):
    """Return (atom counts, charge) for a formula such as Ca(NO3)2 or SO4^2-."""
    charge = 0
    m = re.search(r"\^(\d*)([+-])$", f)
    if m:
        charge = (int(m.group(1)) if m.group(1) else 1) * (1 if m.group(2) == "+" else -1)
        f = f[:m.start()]
    def group(s, i):
        out, order = {}, []
        while i < len(s) and s[i] != ")":
            if s[i] == "(":
                inner, i, sub_order = group(s, i + 1)
                i += 1                                  # skip ")"
                n = re.match(r"\d*", s[i:]).group(); i += len(n)
                for k, v in inner.items():
                    out[k] = out.get(k, 0) + v * (int(n) if n else 1)
                order += [k for k in sub_order if k not in order]
            else:
                el = re.match(r"[A-Z][a-z]?", s[i:]).group(); i += len(el)
                n = re.match(r"\d*", s[i:]).group(); i += len(n)
                out[el] = out.get(el, 0) + (int(n) if n else 1)
                if el not in order: order.append(el)
        return out, i, order
    counts, _, order = group(f, 0)
    return {k: counts[k] for k in order}, charge

def oxidation_numbers(f):
    counts, z = parse(f)
    ox, rule = {}, {}
    if len(counts) == 1:                             # element or monatomic/homonuclear ion
        el = next(iter(counts))
        ox[el] = Fraction(z, counts[el]); rule[el] = "element or simple ion"
        return counts, z, ox, rule
    def unknown():
        return [e for e in counts if e not in ox]
    for e in counts:
        if e == "F": ox[e], rule[e] = Fraction(-1), "F is always -1"
        elif e in GROUP1: ox[e], rule[e] = Fraction(1), "group 1 is +1"
        elif e in GROUP2: ox[e], rule[e] = Fraction(2), "group 2 is +2"
        elif e in FIXED: ox[e], rule[e] = Fraction(FIXED[e]), "fixed value"
    if "H" in counts and "H" not in ox and len(unknown()) > 1:
        hydride = all(e in METALS for e in counts if e != "H")
        ox["H"], rule["H"] = (Fraction(-1), "H is -1 in a metal hydride") if hydride else (Fraction(1), "H is +1")
    if "O" in counts and "O" not in ox and len(unknown()) > 1:
        ox["O"], rule["O"] = Fraction(-2), "O is -2"
    for e in ("Cl", "Br", "I"):
        if e in counts and e not in ox and len(unknown()) > 1 and "O" not in counts and "F" not in counts:
            ox[e], rule[e] = Fraction(-1), "halogen is -1"
    left = unknown()
    if len(left) != 1:
        raise ValueError("More than one unknown: " + ", ".join(left) + ". Use the online calculator.")
    x = left[0]
    known = sum(counts[e] * ox[e] for e in ox)
    ox[x] = (z - known) / counts[x]; rule[x] = "solved from sum = charge"
    return counts, z, ox, rule

def sign(v):
    return ("+" if v > 0 else "") + str(v)

# ---- 2. Solve and show the working ------------------------------------
counts, z, ox, rule = oxidation_numbers(formula)
print(f"Formula {formula}, charge {z}")
for e, n in counts.items():
    print(f"  {e:2s}  atoms {n}  ON {sign(ox[e]):>5s}  contribution {sign(n * ox[e]):>5s}  ({rule[e]})")
total = sum(counts[e] * ox[e] for e in counts)
print("  Sum of contributions =", total, "= charge" if total == z else "!= charge")
print(f"Oxidation number of {find} in {formula} = {sign(ox[find])}")

# ---- 3. Checks ----------------------------------------------------------
assert total == z, "oxidation numbers must add up to the charge"
for f, el, expected in [("K2Cr2O7", "Cr", 6), ("KMnO4", "Mn", 7), ("H2SO4", "S", 6),
                        ("Cr2O7^2-", "Cr", 6), ("Na2S2O3", "S", 2), ("Na5P3O10", "P", 5),
                        ("H2O2", "O", -1), ("NaH", "H", -1), ("OF2", "O", 2)]:
    assert oxidation_numbers(f)[2][el] == expected, f
assert oxidation_numbers("Fe3O4")[2]["Fe"] == Fraction(8, 3)
print("Verdict: Verified (all test compounds agree)")

# ---- 4. Figure: each element's contribution to the charge ---------------
els = list(counts)
vals = [float(counts[e] * ox[e]) for e in els]
fig, ax = plt.subplots(figsize=(7, 4.5))
bars = ax.bar(els, vals, color=["#ea580c" if v > 0 else "#2563eb" for v in vals])
ax.axhline(0, color="#334155", linewidth=1)
for b, e in zip(bars, els):
    ax.annotate(f"{counts[e]} x ({sign(ox[e])})", (b.get_x() + b.get_width() / 2, b.get_height()),
                ha="center", va="bottom" if b.get_height() >= 0 else "top", fontsize=10)
ax.set_ylabel("Atoms x oxidation number")
ax.set_title(f"{formula}: contributions add up to {z}")
fig.tight_layout()
fig.savefig("oxidation_number.png", dpi=300)
print("Saved oxidation_number.png")

Output: K atoms 2 ON +1 contribution +2 (group 1 is +1), Oxidation number of Cr in K2Cr2O7 = +6, Verdict: Verified.

11.2 Python traps

• Use Fraction, not floats: 8/3 stays exact and the sum check is an exact equality.
• Dictionary keys are case sensitive: "Co" is cobalt, "CO" is carbon monoxide.
• The regular expression \^(\d*)([+-])$ reads the charge only at the end of the formula.

11.3 Packages

pip install pymatgen gives Composition("K2Cr2O7").oxi_state_guesses(), which guesses oxidation states from a table of common values, and chempy parses formulas and balances reaction equations.

11.4 What each part does

11.5 The figure

The script saves oxidation_number.png at 300 dpi with each element’s contribution to the charge.

12.1 Rules for calculating oxidation numbers, with exceptions

12.2 Common oxidation numbers of the elements

The most common oxidation numbers of elements met in courses, with their range (lowest to highest) and their Pauling electronegativity.

12.3 Oxidation numbers in common compounds and polyatomic ions

Every row was computed by the same engine as the oxidation number calculator.

12.4 Calculating oxidation numbers: practice worksheet with answers

Twenty-five questions in three groups: compounds, polyatomic ions, and changes in oxidation number. Every answer was checked by the calculator engine. Cover the last column and try them first.

Eight examples run through the calculator, from potassium dichromate to the carbons of ethanol. Each figure is drawn from the computed numbers: bars to the right of zero are positive oxidation numbers, bars to the left negative.

This is the written method examiners expect, on the same problem as Sections 7, 9, 10 and 11: calculate the oxidation number of Cr in K₂Cr₂O₇.

14.1 How to calculate the oxidation number of an element in a compound, step by step

1. Write the overall charge. K₂Cr₂O₇ is neutral, so the sum is 0.
2. Let the unknown be x. Cr = x.
3. Give the fixed values. K = +1 (group 1), O = −2.
4. Multiply by the number of atoms. 2 × (+1) = +2; 2 × x = 2x; 7 × (−2) = −14.
5. Add and set equal to the charge. +2 + 2x − 14 = 0.
6. Solve. 2x = 12, so x = +6.
7. Check. 2(+1) + 2(+6) + 7(−2) = 2 + 12 − 14 = 0, and +6 is chromium’s group number, its highest possible value.

14.2 How to calculate the oxidation number of an ion

Set the sum equal to the ion’s charge instead of 0. Dichromate: 2x + 7(−2) = −2, so 2x = 12 and Cr = +6, the same as in K₂Cr₂O₇. Permanganate: x + 4(−2) = −1, so Mn = +7. For the ammonium ion, x + 4(+1) = +1 gives N = −3.

14.3 How to calculate the oxidation number from a Lewis structure

Draw the structure, give both electrons of each bond to the more electronegative atom (split bonds between identical atoms equally), count the electrons each atom ends up with, and subtract from its valence electrons. In H₂O₂ (H–O–O–H), each O keeps its 4 lone-pair electrons, takes both O–H electrons and one of the O–O pair: 7 electrons, so 6 − 7 = −1. This is the only method that gives the two different S atoms in thiosulfate (+5 and −1) instead of their average, +2.

14.4 How to calculate the oxidation number of coordination compounds

Find the charge of the complex ion from the counter-ions, then subtract the ligand charges. K₄[Fe(CN)₆]: four K⁺ make the complex −4; six CN⁻ give −6; so Fe + (−6) = −4 and Fe = +2. [Co(NH₃)₆]Cl₃: three Cl⁻ make the complex +3; NH₃ is neutral; so Co = +3.

14.5 How to calculate the oxidation number of carbon in organic compounds

For each carbon, count +1 for every bond to O, N, S or a halogen (a double bond counts 2), −1 for every bond to H, and 0 for bonds to other carbons. In ethanol, CH₃–CH₂OH, the CH₃ carbon is 0 − 3 = −3 and the CH₂OH carbon is 1 − 2 = −1. Their average, −2, is what the formula method gives for C₂H₆O: 2x + 6(+1) + (−2) = 0.

14.6 How to calculate the change in oxidation number

Find the oxidation number of each element before and after, and subtract. In MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺, Mn changes by +2 − (+7) = −5 (reduced, gains 5 electrons) and Fe by +3 − (+2) = +1 (oxidised, loses 1). To balance the electrons, five Fe²⁺ are needed for every MnO₄⁻.

14.7 On a scientific calculator

On a Casio fx-991EX or fx-82MS, type the sum rule directly: ( 0 − ( 2 × 1 + 7 × (−) 2 ) ) ÷ 2 = gives 6. For Fe₃O₄, ( 0 − 4 × (−) 2 ) ÷ 3 = shows 8/3; press S⇔D to see 2.666666667.

Use the negative key (−), not the subtract key, for −2. Key names differ between models; check your manual.

14.8 Do it in the lab: a permanganate titration of iron(II)

Example data. The 1 : 5 ratio comes straight from the change in oxidation number: MnO₄⁻ + 5 Fe²⁺ + 8 H⁺ → Mn²⁺ + 5 Fe³⁺ + 4 H₂O. The iron(II) concentration is 0.0938 mol dm⁻³, or 5.24 g dm⁻³ of iron.

14.9 How to calculate oxidation number: practice problems with answers

Ten practice problems, easy to hard. Open each one to see the answer.

14.10 Mistakes that lose marks

Research on redox learning finds students often treat oxidation numbers as real charges and confuse them with the ion charge on the whole species (Garnett & Treagust, 1992; Rahayu et al., 2011).

• Forgetting to multiply by the number of atoms: writing 1 + x − 2 = 0 for K₂Cr₂O₇.
• Setting the sum to 0 for an ion: Cr₂O₇²⁻ then gives +7 instead of +6.
• Giving O −2 in a peroxide or H +1 in a metal hydride.
• Writing the sign after the number (6+) for an oxidation number; that is the style for ionic charges.
• Saying the oxidising agent is oxidised: it is the substance that is reduced.
• Reporting 2.67 for Fe₃O₄ without saying it is an average of +2 and +3.
• Using the formula method for carbon in an organic compound when the question asks about one particular carbon.

Four real examples

Decision tree

1. Is it one compound or ion? Yes: compound mode. No: go on.
2. Is there an arrow, with reactants and products? Yes: change-in-oxidation-number mode.
3. Is there a metal with ligands in square brackets? Yes: coordination mode.
4. Do you need each carbon in an organic molecule? Yes: organic mode with a condensed formula.

Limitations. An oxidation number calculator works from the formula, so when atoms of one element sit in different environments it gives their average. The IUPAC definition is based on the bonded structure (Karen, McArdle & Takats, 2016), and for species such as metal nitrosyls, polysulfides or metal clusters the formula alone cannot settle every atom’s value. Electronegativities are Pauling values (Pauling, 1932; Allred, 1961).

What the answer is

An oxidation number is the charge an atom would have if every bond were ionic. Give the fixed values, then solve the unknown so that the total equals the charge. For K₂Cr₂O₇, 2(+1) + 2x + 7(−2) = 0 gives Cr = +6.

What it means chemically

The oxidation number shows how far an atom has been oxidised. An element at its maximum, such as Cr(VI) or Mn(VII), can only gain electrons, which makes dichromate and permanganate strong oxidising agents.

Watch out for

Multiply by the number of atoms, use the ion’s charge for ions, check for peroxides and hydrides, and remember that a fraction is an average.

Next steps

Practise with the worksheet in Section 12.4 and the ten problems in Section 14.9, then use the change in oxidation number to balance redox equations and to work out titration ratios.

Solving for x is algebra, and comparing titration results is statistics; these tools cover both.

Peer-reviewed sources for the methods and data used in this oxidation number calculator: the IUPAC definition of oxidation state, electronegativity scales, atomic weights, and research on how students learn redox chemistry.

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