Q.Write the IUPAC name of $\text{K}_4[\text{Fe}(\text{CN})_6]$.

Co$^{3+}$ + 6 $NO_2^-$ gives a $-3$ complex ion, balanced by 3 $Na^+$: $Na_3[Co(NO_2)_6]$.

'Hexanitrocobaltate(III)' tells us the complex ion contains Co in the +3 oxidation state, coordinated to 6 nitrito ($NO_2^-$) ligands: $[Co(NO_2)_6]$.

Charge on the complex ion $= (+3) + 6\times(-1) = -3$, i.e. $[Co(NO_2)_6]^{3-}$.

…

$K_3[Al(C_2O_4)_3]$'s complex ion is anionic, so IUPAC naming requires: cation named first (potassium), ligand named with multiplying prefix and 'o' ending (trioxalato), then the metal name with the '-ate' suffix (aluminate) and its oxidation state in Roman numerals, (III).

Determine the oxidation state of Al: each oxalato ligand ($C_2O_4^{2-}$) carries a $-2$ charge; with 3 of them, total ligand charge $=-6$. The complex ion charge is $-3$ (balanced by 3 $K^+$ outside). So $x+(-6)=-3 \Rightarrow x=+3$, i.e. Al(III).

Apply IUPAC naming rules for coordination compounds: (i) name the cation (potassium) first, then the anion (the complex); (ii) ligands are named in alphabetical order with a multiplying prefix — oxalate as a ligand is named 'oxalato', and with three of them the prefix 'tri-' is used (giving 'trioxalato'); (iii) since the complex ion is an anion, the metal's name takes the suffix '-ate': aluminium becomes 'aluminate'; (iv) the oxidation state of the metal is given in Roman numerals in parentheses immediately after the metal name: (III). Assembling: Potassium trioxal …

Pt(IV) + 4 neutral $NH_3$ + 2 $Br^-$ ligands gives a $+2$ complex cation, balanced by 2 outer $Br^-$ ions.

'Tetraamminedibromoplatinum(IV) bromide': the coordination sphere contains 4 ammine ($NH_3$, neutral) ligands and 2 bromo ($Br^-$) ligands around $Pt^{4+}$, so the complex ion is $[PtBr_2(NH_3)_4]$.

Charge on complex ion $=$ (oxidation state of Pt) $+$ (charge from ligands) $= (+4) + (2\times-1) = +2$.

…

Iron(III) with 6 cyanide ligands gives a −3 complex ion, needing three sodium ions to balance charge.

Sodium hexacyanoferrate(III) has iron in the +3 oxidation state (Roman numeral III) coordinated to 6 cyanide ($CN^-$) ligands. Charge on the complex ion: $(+3) + 6(-1) = -3$, i.e. …