Q.Write the IUPAC names of the following coordination compounds:

Concept: Werner Coordination Theory – ligands are named before the metal, with oxidation state in Roman numerals; anionic ligands end in -o, neutral ligands retain their name, and the counter ion is named last.

Step 1 – Identify the ligands and their names

1. $NH_3$ = ammine, $Cl$ = chlorido, $NO_2$ (nitro, N-bonded) = nitrito-$\kappa N$ (commonly "nitro").
2. $C_2O_4^{2-}$ = oxalato.
3. $Cl$ = chlorido, $en$ = ethane-1,2-diamine (commonly called ethylenediamine).
4. $NH_3$ = ammine, $CO_3^{2-}$ = carbonato.
5. $SCN^-$ (S-bonded) = thiocyanato-$\kappa S$ (commonly "thiocyanato"). Step 2 – Determine oxidation state of the metal

(a) Pt: $x + 2(0) + (-1) + (-1) = 0 \Rightarrow x = +2$.

(b) Cr: $3(+1) + x + 3(-2) = 0 \Rightarrow x = +3$.

(c) Co: $x + 2(-1) + 2(0) + (-1) = 0 \Rightarrow x = +3$.

(d) Co: $x + 5(0) + (-2) + (-1) = 0 \Rightarrow x = +3$.

(e) Co: $(+1) + x + 4(-1) = 0 \Rightarrow x = +3$ (Hg is taken as Hg(I), per the NCERT textbook's own worked answer for this example).

Step 3 – Write names in alphabetical order of ligands (ignoring prefixes), then metal + oxidation state, then counter ion

(a) ammine ligands first, then chlorido, then nitrito-N → diamminechloridonitrito-N-platinum(II) …

The key to naming coordination compounds is to remember that the cation is named first (whether it’s the complex or the counter-ion), then the anion. Within the complex, ligands are named in alphabetical order (ignoring numerical prefixes like di-, tri-), followed by the metal name (with its oxidation state in parentheses). Anionic ligands end in -o, neutral ligands keep their name (except water = aqua, ammonia = ammine, carbon monoxide = carbonyl), and cationic ligands end in -ium.

Let’s work through each one.

(a) $[Pt(NH_3)_2Cl(NO_2)]$

1. Identify the complex ion. This is a neutral complex — no counter-ions outside the square brackets. The central metal is platinum (Pt).

2. Determine the oxidation state of Pt.

   • $NH_3$ is neutral (0).
   • $Cl^-$ is anionic (–1).
   • $NO_2^-$ is anionic (–1). Let oxidation state of Pt be $x$. $x + 2(0) + (-1) + (-1) = 0 \implies x = +2$. So platinum is in the +2 state.

3. Name the ligands alphabetically.

   • $Cl^-$ is chlorido (anionic ligand, ends in -o).
   • $NH_3$ is ammine (neutral, special name).
   • $NO_2^-$ can bind through N or O. Here, it’s written as $NO_2$ (not $ONO$), so it’s nitrito-κN (the κN indicates nitrogen is the donor atom). If it were O-bound, it would be nitrito-κO. Alphabetical order: ammine (a) comes before chlorido (c), which comes before nitrito-κN (n). So: ammine, chlorido, nitrito-κN.

4. Add numerical prefixes. Two ammines → diammine. One chlorido → no prefix. One nitrito → no prefix.

5. Name the metal. For a neutral complex, the metal is named as the element itself (no suffix). So: platinum.

6. Add oxidation state. In parentheses: (II).

Final name:

diamminechloridonitrito-κN-platinum(II)

(b) $K_3[Cr(C_2O_4)_3]$

1. Identify the complex ion. The complex is inside the brackets: $[Cr(C_2O_4)_3]^{3-}$. The counter-ions are three $K^+$ ions.

2. Determine the oxidation state of Cr.

   • $C_2O_4^{2-}$ (oxalate) is a bidentate ligand with charge –2 each. Three oxalates give total –6.
   • Let Cr oxidation state be $x$. $x + 3(-2) = -3 \implies x - 6 = -3 \implies x = +3$. So chromium is in the +3 state.

3. Name the ligands.

   • $C_2O_4^{2-}$ is oxalato (anionic ligand, ends in -o).
   • Three oxalato ligands → trioxalato.

4. Name the metal in the anion. For an anionic complex, the metal name ends in -ate. Chromium becomes chromate. So: chromate(III).

5. Name the cation first. Potassium is named as potassium.

Final name:

potassium trioxalatochromate(III)

(c) $[CoCl_2(en)_2]Cl$

1. Identify the complex ion. The complex cation is $[CoCl_2(en)_2]^+$, and the counter-anion is $Cl^-$.

2. Determine the oxidation state of Co.

   • $en$ (ethane-1,2-diamine) is neutral (0).
   • Each $Cl^-$ inside the complex is –1. Two chlorides give –2.
   • The overall complex charge is +1 (since one $Cl^-$ outside). Let Co oxidation state be $x$. $x + 2(0) + 2(-1) = +1 \implies x - 2 = +1 \implies x = +3$. So cobalt is in the +3 state.

3. Name the ligands alphabetically.

   • $Cl^-$ is chlorido.
   • $en$ is ethane-1,2-diamine (neutral, use full IUPAC name; common name “ethylenediamine” is also accepted but IUPAC prefers the systematic name). Alphabetical order: chlorido (c) comes before ethane-1,2-diamine (e). So: chlorido, ethane-1,2-diamine.

4. Add numerical prefixes. Two chlorido → dichlorido. Two en → bis(ethane-1,2-diamine) (use bis- for bidentate ligands to avoid ambiguity; di- would be fine but bis- is clearer).

5. Name the metal in the cation. Cobalt is named as cobalt.

6. Add oxidation state. (III).

7. Name the counter-anion. The outside $Cl^-$ is chloride.

Final name:

dichloridobis(ethane-1,2-diamine)cobalt(III) chloride

(d) $[Co(NH_3)_5(CO_3)]Cl$

1. Identify the complex ion. The complex cation is $[Co(NH_3)_5(CO_3)]^+$, and the counter-anion is $Cl^-$.

2. Determine the oxidation state of Co.

   • $NH_3$ is neutral (0). Five ammines → 0.
   • $CO_3^{2-}$ (carbonate) is –2.
   • Overall complex charge is +1 (one $Cl^-$ outside). Let Co oxidation state be $x$. $x + 5(0) + (-2) = +1 \implies x - 2 = +1 \implies x = +3$. So cobalt is in the +3 state.

3. Name the ligands alphabetically.

   • $NH_3$ is ammine.
   • $CO_3^{2-}$ is carbonato (anionic ligand). …

Werner Coordination Theory — IUPAC Naming Method

Method: IUPAC Nomenclature of Coordination Compounds

This method follows the additive nomenclature rules from IUPAC, which are directly based on Werner's concept of primary and secondary valencies.

Steps for Naming

1. Identify the complex ion — cation or anion (or neutral).
2. Name the ligands first (alphabetically, ignoring prefixes like di-, tri-):
   • Anionic ligands end in -o, in the modern -ido style (e.g., Cl⁻ → chlorido; NO₂⁻ → nitrito-N when N-bonded, nitrito-O when O-bonded).
   • Neutral ligands keep their name (e.g., NH₃ → ammine, en → ethane-1,2-diamine).
3. Name the central metal:
   • If the complex is anionic, add -ate to the metal name (e.g., cobalt → cobaltate).
   • If cationic or neutral, use the metal name as is.
4. Indicate oxidation state of the metal in Roman numerals in parentheses.
5. Write counter ions (outside the coordination sphere) as separate words.

Solutions

(a) $[Pt(NH_3)_2Cl(NO_2)]$

• Ligands: 2 ammine (NH₃), 1 chlorido (Cl⁻), 1 nitrito-N (NO₂⁻, N-bonded) — alphabetically: ammine, chlorido, nitrito-N.
• Metal: Platinum (neutral complex, no -ate).
• Oxidation state: Let Pt = $x$; NH₃ neutral, Cl⁻ = −1, NO₂⁻ = −1 → $x - 2 = 0$ → $x = +2$.
• Name: diamminechloridonitrito-N-platinum(II)

Note: The ligand is written $NO_2$ (nitrogen first) in the formula, which specifically signals N-bonding — so it is nitrito-N (older name "nitro"), not the O-bonded nitrito-O. This locant is not optional or interchangeable; it must match how the formula writes the ligand.

(b) $K_3[Cr(C_2O_4)_3]$

• Counter ion: 3 K⁺ (potassium).
• Complex anion: $[Cr(C_2O_4)_3]^{3-}$.
• Ligand: Oxalate (C₂O₄²⁻) → oxalato (3 of them → trioxalato).
• Metal: Chromium → chromate (anionic complex).
• Oxidation state: Cr = $x$; each oxalate = −2 → $x - 6 = -3$ → $x = +3$.
• Name: potassium trioxalatochromate(III)

(c) $[CoCl_2(en)_2]Cl$

• Counter ion: 1 Cl⁻ outside → chloride.
• Complex cation: $[CoCl_2(en)_2]^+$.
• Ligands: 2 chlorido (Cl⁻), 2 ethane-1,2-diamine (en, neutral) — alphabetically: chlorido, ethane-1,2-diamine.
• Metal: Cobalt (cationic complex).
• Oxidation state: Co = $x$; Cl⁻ = −1 each, en neutral → $x - 2 = +1$ → $x = +3$.
• Name: dichloridobis(ethane-1,2-diamine)cobalt(III) chloride

Exam tip: "en" is written in full as ethane-1,2-diamine — the printed NCERT name (ethylenediamine is the common synonym; never write the bare abbreviation "en" in a name). Use bis because the ligand's own name already contains "di".

(d) $[Co(NH_3)_5(CO_3)]Cl$

• Counter ion: 1 Cl⁻ → chloride.
• Complex cation: $[Co(NH_3)_5(CO_3)]^+$.
• Ligands: 5 ammine (NH₃), 1 carbonate (CO₃²⁻) — alphabetically: ammine, carbonate.
• Metal: Cobalt.
• Oxidation state: Co = $x$; NH₃ neutral, CO₃²⁻ = −2 → $x - 2 = +1$ → $x = +3$.
• Name: pentaamminecarbonatocobalt(III) chloride

(e) $Hg[Co(SCN)_4]$ …

(a) $[Pt(NH_3)_2Cl(NO_2)]$

Common Mistakes

• Wrong order of ligands: Writing chlorido before ammine or nitrito before chlorido. Why it's wrong: Ligands are named alphabetically (by ligand name, not by symbol or charge). Correct alphabetical order: ammine → chlorido → nitrito.
• Dropping the multiplying prefix: There are TWO $NH_3$ ligands here, so the name needs di-ammine, not just "ammine" — an easy point to lose.
• Confusing nitrito-N vs nitrito-O: The ligand $NO_2^-$ can bind through N (nitrito-N, older name "nitro") or through O (nitrito-O). The formula here writes the ligand as $NO_2$ (nitrogen first), which signals N-bonding. How to avoid: $NO_2$ (N first) → N-bonded → nitrito-N; $ONO$ (O first) → O-bonded → nitrito-O. Always include the locant, don't drop it.
• Missing oxidation state: For platinum, you must calculate and show the oxidation state in Roman numerals.

Correct IUPAC Name

Diamminechloridonitrito-N-platinum(II)

(Alphabetical: ammine, chlorido, nitrito-N; two NH₃ → "di-"; Pt oxidation state = +2)

(b) $K_3[Cr(C_2O_4)_3]$

Common Mistakes

• Naming the anion incorrectly: The complex ion is $[Cr(C_2O_4)_3]^{3-}$. Students often forget to change the ending of the metal to -ate for anionic complexes. How to avoid: If the complex is negative, the metal name ends in -ate (e.g., chromate, ferrate, cobaltate).
• Wrong ligand name: $C_2O_4^{2-}$ is oxalato, not “oxalate” (ligand names end in -o).
• Missing potassium counterion: The cation $K^+$ is named first as potassium.

Correct IUPAC Name

Potassium trioxalatochromate(III)

(Oxidation state of Cr = +3; three oxalato ligands → trioxalato)

(c) $[CoCl_2(en)_2]Cl$

Common Mistakes

• Not recognizing ethylenediamine (en) as a bidentate ligand: It’s written as (en)_2, meaning two bidentate ligands. Students sometimes forget to use bis, tris, tetrakis for chelating ligands. How to avoid: For bidentate or polydentate ligands, use bis (2), tris (3), tetrakis (4) instead of di, tri, tetra.
• Forgetting the counterion: The $Cl^-$ outside the bracket is a separate ion — name it last as chloride.
• Wrong oxidation state: Co is +3 here (two $Cl^-$ inside + one outside = total -3, so Co must be +3).

Correct IUPAC Name

Dichloridobis(ethane-1,2-diamine)cobalt(III) chloride

(Alphabetical: chlorido before ethane-1,2-diamine; use bis for en, since its name already contains "di")

(d) $[Co(NH_3)_5(CO_3)]Cl$

Common Mistakes

• Naming carbonate incorrectly: $CO_3^{2-}$ can be carbonato (if monodentate) or carbonato-κO (if bidentate). Here, it’s likely monodentate (only one O bonded), so just carbonato. How to avoid: If unsure, use carbonato — but check if the problem specifies bonding mode.
• Wrong ligand order: Alphabetical order: ammine before carbonato.
• Missing oxidation state: Co is +3 (five neutral $NH_3$, one $CO_3^{2-}$, one $Cl^-$ outside → net +1, so Co = +3).

Correct IUPAC Name

Pentaamminecarbonatocobalt(III) chloride

(Alphabetical: ammine, carbonato; penta for five)

(e) $Hg[Co(SCN)_4]$

Common Mistakes

• Assuming Hg must be +2 here: It's tempting to default to $Hg^{2+}$, but with one Hg and one complex ion per formula unit, the NCERT textbook's own worked answer for this exact example balances it as $Hg^{+}$ (mercury(I)) with a $-1$ complex anion, not $Hg^{2+}$ with a $-2$ anion. How to avoid: Solve the charge balance from the actual 1:1 stoichiometry rather than assuming a "usual" oxidation state — here that means Co ends up as +3, not +2. …