Q.Identify the optically active compounds from the following:

We need the structure whose Column II name is wrong. The method is to name each complex ourselves: name the cation before the anion, list ligands alphabetically (ignoring multiplying prefixes), use the correct oxidation state in Roman numerals, and use bis/tris/tetrakis when a ligand name itself contains di/tri or is complex.

1. Structure A: $\left[\mathrm{CoBr}_2(\mathrm{en})_2\right]\mathrm{Cl}$, name P

• Complex ion $[\mathrm{CoBr}_2(\mathrm{en})_2]^+$ with $\mathrm{Cl}^-$ as counter-ion.
• Oxidation state: Co $+ 2(-1) + 2(0) = +1$ overall, so Co is +3.
• Ligands alphabetically: bromido before ethane-1,2-diamine.
• Because "ethane-1,2-diamine" already contains the prefix "di", two of them must be written as bis(ethane-1,2-diamine). The correct name is Dibromidobis(ethane-1,2-diamine)cobalt(III) chloride.
• The given name P uses di-(ethane-1,2-diamine) instead of bis(...) — this is the error. Structure A is named incorrectly.

2. Structure B: $\mathrm{Na}\left[\mathrm{PtBrCl}(\mathrm{NO}_2)(\mathrm{NH}_3)\right]$, name Q

• Anionic complex $[\mathrm{PtBrCl}(\mathrm{NO}_2)(\mathrm{NH}_3)]^-$ with $\mathrm{Na}^+$.
• Oxidation state: Pt $+ (-1) + (-1) + (-1) + 0 = -1$, so Pt is +2.
• Ligands alphabetically: ammine, bromido, chlorido, nitrito-N. The N-bonded $\mathrm{NO}_2^-$ is correctly written as nitrito-N.
• Name Q lists the ligands in the right order with the correct oxidation state, so it is acceptable.

3. Structure C: $\left[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_2(\mathrm{NH}_3)_4\mathrm{Cl}\right]\mathrm{SO}_4$, name R

• Complex ion $[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_2(\mathrm{NH}_3)_4\mathrm{Cl}]^{2+}$ with $\mathrm{SO}_4^{2-}$.
• Oxidation state: Cr $+ 4(0) + 2(0) + (-1) = +2$ overall, so Cr is +3. …

The question tests your ability to match a coordination compound's formula with its correct IUPAC name. Four areas commonly trip students up: the oxidation state of the central metal, the alphabetical order of ligand names, the correct use of ligand abbreviations (like "en" for ethylenediamine), and the proper placement of anionic ligands in the name. Let's examine each option carefully.

1. Option (A): $\mathrm{K}_3[\mathrm{Al}(\mathrm{C}_2\mathrm{O}_4)_3]$ – Potassium trioxalatoaluminate (III)

   • The complex ion is $[\mathrm{Al}(\mathrm{C}_2\mathrm{O}_4)_3]^{3-}$. Oxalate ($\mathrm{C}_2\mathrm{O}_4^{2-}$) is a bidentate ligand, so three of them give a coordination number of 6. Aluminium here is in the +3 oxidation state (since each oxalate is -2, total ligand charge = -6, and the complex ion charge is -3, so Al must be +3). The name "trioxalatoaluminate (III)" is correct, and the counterion is potassium. This combination is correct.

2. Option (B): $[\mathrm{Pt}(\mathrm{NH}_3)_2\mathrm{Cl}(\mathrm{NO}_2)]$ – Diamminnechloridonitrito – N – platinum (II)

   • The ligands are: two ammines ($\mathrm{NH}_3$), one chloride ($\mathrm{Cl}^-$), and one nitrito ($\mathrm{NO}_2^-$). The name lists them alphabetically: "ammine" before "chlorido" before "nitrito". The "–N" indicates that the nitrito ligand is bonded through nitrogen (nitrito-N), which is correct. The oxidation state: $\mathrm{NH}_3$ is neutral, $\mathrm{Cl}^-$ is -1, $\mathrm{NO}_2^-$ is -1, so total ligand charge = -2. The complex is neutral, so Pt must be +2. The name "diamminnechloridonitrito-N-platinum(II)" is correct. (Note: "diamminne" is a minor spelling variant; the standard is "diammine", but this is not the error here.) This combination is correct.

3. Option (C): $[\mathrm{CoCl}_2(\mathrm{en})_2]\mathrm{Cl}$ – Dichloridodiethylenediamine cobalt (II) chloride …

Concept: IUPAC nomenclature of coordination compounds.

[CoCl2(NO2)(NH3)3] - no counter-ion, so the complex is NEUTRAL.

Oxidation state of Co: x + 2(-1) + (-1) + 3(0) = 0 -> x = +3, i.e. cobalt(III).

Because the complex is neutral (not anionic), the metal keeps its ordinary name 'cobalt', NOT 'cobaltate'.

Ligands, named alphabetically (ammine before chlorido before nitrito):

3 NH3 = triammine

2 Cl- = dichlorido (modern IUPAC uses 'chlorido', not 'chloro')

NO2- bound through N = nitrito-N (nitro) …

The question asks for the IUPAC name of cis-platin, a well-known anticancer drug. Its formula is $[Pt(NH_3)_2Cl_2]$. The name must reflect the ligands, their order, the metal, and its oxidation state.

Concept first: In coordination chemistry, the IUPAC name is built systematically. For a neutral complex, the ligands are named alphabetically (ignoring prefixes like di-, tri-), followed by the metal name with its oxidation state in Roman numerals in parentheses. The oxidation state is crucial — it determines the metal's charge and often the compound's properties.

Let's work through it step by step.

1. Identify the complex and its charge.

   cis-Platin is a neutral molecule: $[Pt(NH_3)_2Cl_2]$. The square brackets indicate the coordination sphere. Since it's neutral, the sum of the charges on the metal and ligands must be zero.

2. Determine the oxidation state of platinum.

   • Ammonia ($NH_3$) is a neutral ligand — charge 0.
   • Chloride ($Cl^-$) is an anionic ligand — each has charge $-1$. Let the oxidation state of Pt be $x$. Then:

So platinum is in the +2 oxidation state.

3. Name the ligands in alphabetical order.

   • The ligands are: ammine ($NH_3$) and chloride ($Cl^-$).
   • Alphabetically, "ammine" comes before "chlorido" (IUPAC uses "ammine" for $NH_3$ and "chlorido" for $Cl^-$ as a ligand).
   • The prefixes "di-" indicate two of each: diammine and dichlorido.
   • So the ligand part is: diamminedichlorido.

4. Name the metal with its oxidation state.

   • Since the complex is neutral, the metal is named as the element itself: platinum.
   • Add the oxidation state in Roman numerals: (II).

5. Assemble the full name.

   Combine: diamminedichloridoplatinum(II).

   Note: IUPAC rules place the metal name immediately after the ligands, without spaces or hyphens (except the oxidation state in parentheses). …

1. Work out the oxidation state of cobalt

The compound is $[\mathrm{Co(NH_3)_5(CO_3)}]\mathrm{Cl}$. The chloride is outside the coordination sphere, so the complex ion carries $+1$.

Let the oxidation number of Co be $x$. Ammonia is neutral; carbonate is $\mathrm{CO_3^{2-}}$:

So cobalt is Co(III) — written as a Roman numeral in parentheses.

2. Apply the IUPAC naming rules in order

1. Cation first, anion second — the complex here is the cation, so it is named first and "chloride" comes last.
2. Ligands are named before the metal, in alphabetical order (the multiplying prefix penta- is ignored when alphabetising). Compare the ligand names: ammine vs carbonato ⇒ ammine comes first.
   • $\mathrm{NH_3}$ as a ligand = ammine (two m's — amine with one m is an organic $\mathrm{-NH_2}$ compound).
   • $\mathrm{CO_3^{2-}}$ as a ligand = carbonato (anionic ligands end in -o).
3. Prefix for number: five ammines ⇒ pentaammine (both a's are kept: pentaammine).
4. Metal name: the complex ion is a cation, so the metal keeps its normal English name, cobalt. The suffix -ate (cobaltate) is used only when the complex ion is an anion.
5. Oxidation state of the metal in Roman numerals: (III).

Assembling: penta + ammine + carbonato + cobalt(III) + chloride …

[!TLDR]

The anion [Fe(CN)$_5$NO]$^{3-}$ has five cyanido ligands and one nitrosyl (NO) ligand, so the compound is potassium pentacyanonitrosylferrate, matching option (A).

Concept

When naming coordination compounds (CBSE/NCERT Class 12 Coordination Compounds), ligands are named alphabetically with their standard names: CN$^-$ = cyanido (cyano), and the NO group bound through nitrogen is nitrosyl. For an anionic complex the metal takes its Latin root plus -ate (iron $\rightarrow$ ferrate).

Solution

The complex ion is $[\mathrm{Fe(CN)_5NO}]^{3-}$ with three K$^+$ counter-ions.

• Five CN$^-$ ligands $\rightarrow$ "pentacyano(nitrosyl)".
• The nitrogen-bonded NO group $\rightarrow$ nitrosyl (not "nitro" = NO$_2$, and not "nitrito" = ONO).
• The complex carries a $-3$ charge, so the metal centre is named as the anion ferrate. …

• (A) $Na_3[Co(ONO)_6]$: $ONO$ bonded through O is the nitrito ligand; Co is $+3$ → sodium hexanitritocobaltate(III). Correct.
• (B) $K_4[Mn(CN)_6]$: Mn $+2$, cyanido ligands → potassium hexacyanomanganate(II). Correct.
• (C) $Cs[FeCl_4]$: Fe $+3$, chlorido ligands → caesium tetrachloroferrate(III). Correct. …

The key to naming this compound lies in recognising that it contains two separate coordination spheres — one positively charged and one negatively charged. The formula $[\text{Pt}(\text{NH}_3)_4][\text{PtCl}_4]$ tells you there is a cationic complex $[\text{Pt}(\text{NH}_3)_4]^{2+}$ and an anionic complex $[\text{PtCl}_4]^{2-}$.

Why are both charges $2+$ and $2-$? Because the overall compound must be neutral. Ammonia ($\text{NH}_3$) is a neutral ligand, so the charge on the cation comes entirely from the metal. For the anion, each chloride ligand carries a $-1$ charge, so four chlorides give $-4$; to get a net $-2$, the platinum must be in the $+2$ oxidation state. So both platinum centres are in the +2 oxidation state.

Now, the IUPAC rules for naming such a salt are straightforward:

• Name the cation first (as a separate coordination entity), then the anion.
• For the cation: ligands are named in alphabetical order (ignoring prefixes like tetra-), then the metal name is used as is (with its oxidation state in Roman numerals in parentheses).
• For the anion: the same ligand order, but the metal name ends in -ate (platinum → platinate), followed by the oxidation state.

Let’s apply this step by step.

1. Identify the cation: $[\text{Pt}(\text{NH}_3)_4]^{2+}$.

   Ligand: tetraammine (four $\text{NH}_3$ groups). Metal: platinum. Oxidation state: $+2$.

   So the cation is tetraammineplatinum(II).

2. Identify the anion: $[\text{PtCl}_4]^{2-}$.

   Ligand: tetrachlorido (four $\text{Cl}^-$ ligands). Metal: platinum → platinate. Oxidation state: $+2$.

   So the anion is tetrachloridoplatinate(II).

3. Combine them: cation first, then anion.

   The full name is tetraammineplatinum(II) tetrachloridoplatinate(II). …

The concept here is the spectrochemical series — an experimentally determined ordering of ligands by their ability to split the d-orbital energies of a transition metal ion (the crystal field splitting parameter $\Delta$). A "strong" ligand produces a large $\Delta$, while a "weak" ligand produces a small $\Delta$.

But why do some ligands cause a larger splitting? Two factors matter:

1. σ-donation: The ligand donates a lone pair to the metal. Better σ-donors raise the energy of the d-orbitals more, increasing $\Delta$.
2. π-interactions: If the ligand has empty π* orbitals (like CO, CN⁻), it can accept electron density from the metal via back-donation. This stabilises the metal's t₂g orbitals, further increasing $\Delta$. If the ligand has filled π orbitals (like halides), it donates into the t₂g, reducing $\Delta$.

The strongest ligands are those that are good σ-donors and strong π-acceptors. Let's evaluate each option.

1. CN⁻ (cyanide) — A strong σ-donor and a good π-acceptor (via its empty π* orbitals). It sits very high in the spectrochemical series, but not at the absolute top.

2. CO (carbon monoxide) — A moderate σ-donor (the lone pair on carbon is not as basic as that on nitrogen in NH₃), but an excellent π-acceptor. The empty π* orbitals on CO are low-lying and overlap very effectively with filled metal d-orbitals. This strong back-donation dramatically increases $\Delta$, placing CO at the very top of the spectrochemical series.

3. NH₃ (ammonia) — A good σ-donor but not a π-acceptor (it has no empty low-energy orbitals). It is a moderate-field ligand, well below CO and CN⁻. …

The key to solving this is understanding how coordination compounds are named. The name gives you a precise recipe: first, the ligands inside the coordination sphere (the complex ion) are listed, then the central metal with its oxidation state, and finally the counterions outside the sphere.

"Penta aqua" means five water ($\text{H}_2\text{O}$) molecules as ligands. "Nitrato" means one nitrate ($\text{NO}_3^-$) ion acting as a ligand. So inside the square brackets, we have $\text{Cr}$ with five $\text{H}_2\text{O}$ and one $\text{NO}_3^-$. The oxidation state of chromium is given as (III), so $\text{Cr}^{3+}$.

Now, the charge of the complex ion inside the brackets must balance. The five water ligands are neutral. The one nitrate ligand carries a $-1$ charge. So the total charge from ligands is $-1$. To get a $\text{Cr}^{3+}$ ion, the complex ion's net charge must be $+2$ (because $+3 - 1 = +2$). That means the complex ion is $[\text{Cr(H}_2\text{O})_5\text{NO}_3]^{2+}$.

The name ends with "nitrate" — this tells us the counterion outside the brackets is nitrate ($\text{NO}_3^-$). To balance the $+2$ charge of the complex ion, we need two nitrate ions. So the full formula is $[\text{Cr(H}_2\text{O})_5\text{NO}_3](\text{NO}_3)_2$.

Let's check each option:

1. Option (A): $[\text{Cr(H}_2\text{O})_6](\text{NO}_3)_3$ — This has six water ligands, not five, and no nitrate ligand inside. This would be called "hexaaquachromium(III) nitrate," not penta aquanitrato.

2. Option (B): $[\text{Cr(H}_2\text{O})_5\text{NO}_3](\text{NO}_3)_2$ — Exactly matches our reasoning: five waters, one nitrate inside, two nitrates outside. This is correct.

3. Option (C): $[\text{Cr(H}_2\text{O})_6](\text{NO}_2)_2$ — This has nitrite ($\text{NO}_2^-$) as counterion, not nitrate. Also has six waters and no nitrate ligand. Completely wrong. …

Step 1 — Split the compound into ions.

The square brackets enclose the coordination sphere: $[\mathrm{Co(NH_3)_4Cl(NO_2)}]^{n+}$. The chloride written outside the brackets is an ionisable counter ion, $\mathrm{Cl^-}$.

Step 2 — Oxidation state of cobalt.

The overall compound is neutral, so the complex ion must be $+1$. Inside the sphere: $\mathrm{NH_3}$ is neutral, $\mathrm{Cl^-}$ is $-1$, $\mathrm{NO_2^-}$ is $-1$.

So it is cobalt(III) — which already eliminates (B) cobalt(II) and (C) cobalt(I).

Step 3 — Name the ligands. …