Complex Ions, Oxoanions & Mixed

CHEExam Practice & Reasoning

Transition metals readily form complexes and high-oxidation oxoanions because of small size, high charge density and available $d$ -orbitals. This subtopic also gathers fast assignment tasks like finding oxidation states in oxoanions and comparing the oxidising power of $MnO_4^-$ versus $Cr_2O_7^{2-}$ .

Oxidation-state balance in an oxoanion

x + n(-2) = (\text{ion charge})

x

is the oxidation state of the metal,

n

the number of O atoms (each

-2

); e.g. for

MnO_4^-

x + 4(-2) = -1 \Rightarrow x = +7

Comparison of standard reduction potentials

E^\circ(MnO_4^-/Mn^{2+}) = +1.51\ \text{V} \;>\; E^\circ(Cr_2O_7^{2-}/Cr^{3+}) = +1.33\ \text{V}

the larger (more positive)

E^\circ

marks the stronger oxidising agent.

Transition metals form many complexes due to small cationic size, high effective nuclear charge, high charge density and availability of vacant $d$ -orbitals to accept lone pairs from ligands.
Oxidation state of Mn: $MnO_2$ is $+4$ , $K_2MnO_4$ is $+6$ , $KMnO_4$ is $+7$ , $MnCl_2$ is $+2$ (O counts $-2$ , K counts $+1$ , Cl counts $-1$ ).
Oxidation state of Cr: $Cr_2O_3$ is $+3$ , $CrO_3$ is $+6$ , $K_2Cr_2O_7$ is $+6$ , $K_2CrO_4$ is $+6$ .
$MnO_4^-$ ( $Mn$ is $+7$ ) is intensely purple from a charge-transfer (ligand $\rightarrow$ metal) transition, whereas $Mn^{2+}$ ( $d^5$ ) is almost colourless because its $d\text{-}d$ transitions are spin-forbidden and therefore very weak.
$MnO_4^-$ is a stronger oxidising agent than $Cr_2O_7^{2-}$ : $E^\circ(MnO_4^-/Mn^{2+}) = +1.51$ V exceeds $E^\circ(Cr_2O_7^{2-}/Cr^{3+}) = +1.33$ V, partly because reduction of $MnO_4^-$ gives the extra-stable half-filled $d^5$ $Mn^{2+}$ .
Actinoids show a wider range of oxidation states than lanthanoids because the $5f$ , $6d$ and $7s$ energy levels lie close together, so more electrons take part in bonding (lanthanoids are dominated by the stable $+3$ state).
High oxidation states ( $+6$ , $+7$ ) survive only in oxoanions such as $MnO_4^-$ , $Cr_2O_7^{2-}$ , $CrO_4^{2-}$ where small, highly electronegative oxygen stabilises the charge through M=O bonding.
Complex formation is favoured by ligands such as $CN^-$ , $NH_3$ , $H_2O$ , $Cl^-$ that have lone pairs to donate; the metal supplies empty $(n-1)d$ , $ns$ and $np$ orbitals for coordinate bonds.
To assign oxidation state, work systematically: O is $-2$ , H is $+1$ , group-1 metal is $+1$ , halide is $-1$ , then solve for the transition metal so the sum equals the ion charge.
In $MnO_4^-$ and $Cr_2O_7^{2-}$ the metal is at its $d^0$ centre, so the intense colour cannot be a $d\text{-}d$ transition — it must be charge transfer, and that is why the bands are so strong.
$MnO_4^-$ and $Cr_2O_7^{2-}$ are strong oxidisers only in acidic medium; in alkaline medium dichromate converts to chromate ( $CrO_4^{2-}$ ) and permanganate is reduced only to $MnO_2$ or $MnO_4^{2-}$ , so their full oxidising strength needs plentiful $H^+$ .

Where the marks go

Forgetting that O is $-2$ and miscounting atoms: in $Cr_2O_7^{2-}$ there are two Cr, so $2x + 7(-2) = -2$ gives $x = +6$ (not $+12$ ) — divide by the number of metal atoms.
Stating that $MnO_4^-$ is coloured by $d\text{-}d$ transitions; with $Mn$ at $d^0$ there are no $d$ electrons, so the colour is charge-transfer (ligand-to-metal).
Comparing oxidising power the wrong way round — a more positive $E^\circ$ (not a less positive one) means the stronger oxidiser; $+1.51$ V beats $+1.33$ V.
Confusing manganate $MnO_4^{2-}$ ( $Mn$ is $+6$ , green) with permanganate $MnO_4^-$ ( $Mn$ is $+7$ , purple) — different charge, colour and oxidation state.
Claiming all transition metals reach their group oxidation state as simple ions; states like $+7$ exist only in covalent oxoanions, never as a bare $Mn^{7+}$ cation.

How the board asks it

Numericaloxidation-state assignment in oxoanions
Calculate the oxidation state of manganese in $KMnO_4$ and of chromium in $K_2Cr_2O_7$ , showing how the charges balance (taking $O$ as $-2$ and $K$ as $+1$ ).
Give reasonsstandard reduction potentials of $MnO_4^-$ and $Cr_2O_7^{2-}$
Account for the fact that $MnO_4^-$ is a stronger oxidising agent than $Cr_2O_7^{2-}$ , given $E^\circ(MnO_4^-/Mn^{2+}) = +1.51$ V and $E^\circ(Cr_2O_7^{2-}/Cr^{3+}) = +1.33$ V.
Give reasonscharge-transfer colour at a $d^0$ centre
Give reasons: $MnO_4^-$ is intensely coloured even though manganese in it has a $d^0$ configuration and can have no $d\text{-}d$ transitions.
Distinguishmanganate $MnO_4^{2-}$ vs permanganate $MnO_4^-$
Distinguish between the manganate ion $MnO_4^{2-}$ and the permanganate ion $MnO_4^-$ on the basis of the oxidation state of manganese and their colours.
Give reasonsoxoanion stabilisation of high oxidation states
Explain why the high oxidation states $+6$ and $+7$ of transition metals are found only in oxoanions such as $CrO_4^{2-}$ and $MnO_4^-$ , and not as free $Cr^{6+}$ or $Mn^{7+}$ cations.
Give reasonscomplex formation by transition metal ions
Account for the strong tendency of transition metal ions to form complexes with ligands such as $CN^-$ and $NH_3$ .

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.