Q.Write formula for co-ordination compound Potassium trioxalatochromate (III).

Showing the 12 most recent of 56 on this concept.

1. CBSE 2026Set 56/3/11 markMCQ
   Q.The oxidation number of Co in $[Co(en)_3]_2(SO_4)_3$ is : (A) $+3$ (B) $+2$ (C) $+4$ (D) $+6$

   ›Reveal solutionSolution

   The complex cation $[Co(en)_3]^{n+}$ must balance three sulfate anions ($SO_4^{2-}$); since ethylenediamine is neutral, cobalt carries +3.

   Why oxidation numbers matter in coordination compounds

   Oxidation state tells us the formal charge on the central metal after we've assigned all bonding electrons to the more electronegative atom. In coordination chemistry, we treat ligands as intact units: neutral ligands contribute zero, anionic ligands contribute their charge. The sum of oxidation states in the entire complex must equal its net charge.

   The formula $[Co(en)_3]_2(SO_4)_3$ shows us a salt: two complex cations paired with three sulfate anions. Our job is to figure out what charge the cobalt must carry to make the arithmetic work.

   Step-by-step determination

   1. Identify the ionic components

   The compound dissociates into:

   • Two $[Co(en)_3]^{n+}$ cations (where $n$ is unknown)
   • Three $SO_4^{2-}$ anions

   2. Apply charge neutrality

   The entire salt is neutral, so total positive charge equals total negative charge:

   $2 \times (\text{charge on one cation}) = 3 \times 2$

   $2n = 6$

   $n = +3$

   Each complex cation carries a +3 charge: $[Co(en)_3]^{3+}$.

   3. Determine cobalt's oxidation state within the cation

   Now look inside $[Co(en)_3]^{3+}$. Ethylenediamine ($en = H_2NCH_2CH_2NH_2$) is a neutral bidentate ligand—it donates two lone pairs but carries no charge. Three $en$ ligands contribute:

   $3 \times 0 = 0$

   The oxidation state of cobalt plus the ligand contributions must equal the cation charge:

   $x_{Co} + 0 = +3$

   $x_{Co} = +3$ …

   🔒 Start your 14-day free trial to see the full solution →
2. CBSE 2026Set 56/2/11 markMCQ
   Q.The correct IUPAC name of the complex $[Pt(NH_3)_2Cl_2]$ is : (A) diamminedichloridoplatinum (IV) (B) diamminedichloridoplatinum (II) (C) dichloridodiammineplatinum (IV) (D) dichloridodiammineplatinum (II)

   ›Reveal solutionSolution

   The complex $[Pt(NH_3)_2Cl_2]$ is neutral, so the oxidation state of Pt must be +2. Ligands are named alphabetically (ammine before chlorido), and the metal is named without a suffix. The correct IUPAC name is diamminedichloridoplatinum(II) — option (B).

   The key to naming coordination compounds is to follow the IUPAC rules in order: identify the oxidation state of the metal, list ligands alphabetically (ignoring prefixes like di-, tri-), and then name the metal with its oxidation state in parentheses.

   Let’s break this down step by step.

   1. Determine the oxidation state of platinum.

      The complex $[Pt(NH_3)_2Cl_2]$ is neutral — no overall charge.

      • $NH_3$ is a neutral ligand (charge 0).
      • $Cl$ is a negatively charged ligand (chlorido, charge –1). Let the oxidation state of Pt be $x$. Then: $x + 2(0) + 2(-1) = 0 \implies x - 2 = 0 \implies x = +2$. So platinum is in the +2 oxidation state.

   2. Name the ligands in alphabetical order.

      IUPAC rules: ligands are named alphabetically by their name (not by prefix).

      • $NH_3$ is called ammine (note the double 'm').
      • $Cl$ is called chlorido (the anionic ligand name for chloride). Alphabetically, "ammine" comes before "chlorido". So the ligand order is: diammine then dichlorido.

   3. Name the metal.

      Since the complex is anionic? No — it’s neutral. For neutral complexes, the metal is called by its usual name (platinum), followed by the oxidation state in Roman numerals in parentheses: platinum(II).

   4. Assemble the full name.

      Ligands first (with prefixes di- for two identical ligands), then metal + oxidation state:

      diamminedichloridoplatinum(II). …

   🔒 Start your 14-day free trial to see the full solution →
3. CBSE 2026Set 56/2/11 markMCQ
   Q.Which of the following is heteroleptic complex ? (A) $[Co(NH_3)_6]^{3+}$ (B) $[Cr(NH_3)_6]^{3+}$ (C) $[Ni(H_2O)_6]^{2+}$ (D) $[Co(NH_3)_4Cl_2]^+$

   ›Reveal solutionSolution

   A heteroleptic complex contains more than one type of ligand. Among the given options, only $[Co(NH_3)_4Cl_2]^+$ has two different ligands ($NH_3$ and $Cl^-$), making (D) the answer.

   The distinction between homoleptic and heteroleptic complexes is fundamental to coordination chemistry and comes down to ligand diversity.

   A homoleptic complex (from Greek homo = same, leptos = taking) contains only one kind of ligand attached to the central metal ion. Think of it as a "uniform" coordination sphere where every ligand is identical.

   A heteroleptic complex (from Greek hetero = different) contains two or more different types of ligands. The coordination sphere is "mixed."

   This classification matters because heteroleptic complexes exhibit richer isomerism (geometrical, optical) and more varied chemical behavior than their homoleptic counterparts.

   Now let's examine each option systematically:

   1. Option (A): $[Co(NH_3)_6]^{3+}$

      The cobalt(III) ion is surrounded by six ammonia molecules. Every ligand is $NH_3$—no variation whatsoever. This is a textbook homoleptic complex.

   2. Option (B): $[Cr(NH_3)_6]^{3+}$

      Chromium(III) coordinated to six identical ammonia ligands. Again, uniform ligand environment. Homoleptic.

   3. Option (C): $[Ni(H_2O)_6]^{2+}$

      Nickel(II) surrounded by six water molecules. All ligands are the same. Homoleptic.

   4. Option (D): $[Co(NH_3)_4Cl_2]^+$ …

   🔒 Start your 14-day free trial to see the full solution →
4. CBSE 2026Set ANNUAL1 markMCQ
   Q.The oxidation number of nickel in [Ni(CO)4] will be:

   (a) 1

   (b) 0

   (c) 2

   (d) 3

   ›Reveal solutionSolution

   CO is a neutral ligand, so the oxidation number of Ni in [Ni(CO)₄] is 0.

   In a coordination compound, the oxidation number of the central metal is found by assigning charges to the ligands and balancing against the overall charge of the complex. Carbonyl (CO) is a neutral ligand — it donates a lone pair from carbon without carrying any charge itself. Since [Ni(CO)₄] is a ne …

   🔒 Start your 14-day free trial to see the full solution →
5. CBSE 2026Set ANNUAL1 mark
   Q.Write the formula of the coordination compound tetraamine aquachlorido cobalt (III) chloride.

   ›Reveal solutionSolution

   Build the octahedral coordination sphere from the ligands named, find the complex ion's net charge from the metal's oxidation state, then add counter-ions to balance that charge.

   Naming breakdown: 'tetraammine' $\rightarrow$ 4 $NH_3$ ligands (neutral); 'aqua' $\rightarrow$ 1 $H_2O$ ligand (neutral); 'chlorido' $\rightarrow$ 1 $Cl^-$ ligand (anionic, $-1$); 'cobalt(III)' $\rightarrow$ central metal $Co^{3+}$. Coordination number $=4+1+1=6$ (octahedral), consistent with typical cobalt(III) ammine complexes.

   …

   🔒 Start your 14-day free trial to see the full solution →
6. CBSE 2026Set ANNUAL1 mark
   Q.Write answer in one word/sentence: Write chemical formula of Iron (III) hexacyanidoferrate (II).

   ›Reveal solutionSolution

   Iron(III) hexacyanidoferrate(II) is Fe4[Fe(CN)6]3.

   The complex anion hexacyanidoferrate(II) is [Fe(CN)6]4- (Fe in +2, six CN- ligands). The counter-cation is iron(III), Fe3+.

   …

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2026Set ANNUAL1 mark
   Q.A complex has the composition Co(NH₃)₄BrCl₂. Conductance measurement shows that there are two ions per formula unit and on treatment with silver nitrate it forms a yellow precipitate. Write the IUPAC name of the complex compound.

   ›Reveal solutionSolution

   A yellow AgBr precipitate shows free Br⁻; two ions per formula unit fix the structure as [Co(NH₃)₄Cl₂]Br → tetraamminedichloridocobalt(III) bromide.

   Deducing the structure:

   • The composition is $Co(NH_3)_4BrCl_2$.
   • Conductance shows two ions per formula unit, so the complex ionises into one cation and one anion.
   • With $AgNO_3$ it gives a yellow precipitate, which is $AgBr$ (silver chloride is white). So it is bromide ($Br^-$) that is the free, ionisable counter-ion outside the coordination sphere, while both chloride ions are coordinated (non-ionisable) inside.

   Hence the formula is $[Co(NH_3)_4Cl_2]Br$, giving the two ions $[Co(NH_3)_4Cl_2]^+$ and $Br^-$.

   …

   🔒 Start your 14-day free trial to see the full solution →
8. CBSE 2025Set 56/6/11 markMCQ
   Q.Which of the following complex ion is not optically active ? (A) $[Co(ox)_3]^{3-}$ (B) $cis\text{-}[Co(en)_2Cl_2]^+$ (C) $trans\text{-}[Co(en)_2Cl_2]^+$ (D) $[Co(en)_3]^{3+}$

   ›Reveal solutionSolution

   Optical activity in coordination complexes requires the absence of a plane of symmetry. Among the given options, $trans\text{-}[Co(en)_2Cl_2]^+$ has a centre of symmetry and a plane of symmetry, making it optically inactive. The correct answer is (C).

   Why Optical Activity Matters in Coordination Chemistry

   Optical activity is a property of chiral molecules — those that are non-superimposable on their mirror image. In coordination compounds, chirality arises from the spatial arrangement of ligands around the central metal ion. A complex is optically active if it lacks an improper axis of rotation (specifically, a plane of symmetry or a centre of symmetry). The classic test: if a complex and its mirror image cannot be superimposed, they are enantiomers, and the complex is optically active.

   For octahedral complexes, chirality often appears when:

   • Bidentate ligands (like oxalate, ox$^{2-}$, or ethylenediamine, en) create a helical twist.
   • The arrangement of different ligands breaks symmetry.

   Let’s examine each option systematically.

   ---

   1. $[Co(ox)_3]^{3-}$ — The Tris(oxalato) Complex

   Oxalate ($\text{ox}^{2-}$) is a bidentate ligand that forms a five-membered chelate ring. Three oxalate ions around Co(III) give an octahedral geometry. The complex has a propeller-like shape: each oxalate spans one edge of the octahedron, and the three rings are arranged in a helical fashion.

   Think of it like a three-bladed fan. The complex exists as a pair of enantiomers — left-handed and right-handed helices. There is no plane of symmetry because the chelate rings lock the structure into a chiral twist. Therefore, $[Co(ox)_3]^{3-}$ is optically active.

   Tip

   Any octahedral complex with three identical bidentate ligands (like $[M(AA)_3]$) is always chiral — it’s a classic example of helical chirality. The same applies to $[Co(en)_3]^{3+}$ in option (D).

   ---

   2. $cis\text{-}[Co(en)_2Cl_2]^+$ — The Cis Isomer

   Here, two ethylenediamine (en) ligands and two chloride ligands surround Co(III). The “cis” prefix means the two chlorides are adjacent (90° apart). In this geometry, the two en ligands are not equivalent in space — they create a non-superimposable mirror image.

   Draw the structure: the two en rings lie in roughly perpendicular planes. The cis arrangement of Cl atoms breaks any plane of symmetry. The complex is chiral, and indeed, $cis\text{-}[Co(en)_2Cl_2]^+$ has been resolved into enantiomers. So it is optically active.

   Watch out

   A common mistake is to think that any complex with two identical bidentate ligands is automatically chiral. That’s only true for the cis isomer — the trans isomer is different, as we’ll see next.

   ---

   3. $trans\text{-}[Co(en)_2Cl_2]^+$ — The Trans Isomer …

   🔒 Start your 14-day free trial to see the full solution →
9. CBSE 2025Set ANNUAL1 mark
   Q.Write formula for co-ordination compound Potassium trioxalatochromate (III).

   ›Reveal solutionSolution

   Three bidentate oxalate ligands (each -2) plus Cr3+ gives a -3 complex ion balanced by 3 K+.

   'Trioxalato' means three oxalate ($C_2O_4^{2-}$) ligands (bidentate, each carrying charge $-2$); 'chromate(III)' means the central metal is chromium in the $+3$ oxidation state.

   …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2025Set D1 markMCQ
    Q.The IUPAC name of complex compound [Co(NH3)6]Cl3 is

    (a) Hexa-ammine cobalt (III) chloride

    (b) Hexa-ammine cobalt (II) chloride

    (c) Hexa-ammine trichloridocobalt (III)

    (d) None of these

    ›Reveal solutionSolution

    [Co(NH3)6]Cl3 = hexaamminecobalt(III) chloride.

    Rules of IUPAC nomenclature:

    • Name the cation first, then the anion.
    • Within the complex, ligands are named alphabetically before the metal.
    • NH3 as a ligand is 'ammine' (six of them -> hexaammine).
    • Oxidation state of Co: three Cl- give -3; overall neutral, so Co = +3, written as (III).
    • The chloride outside is the counter-anion. …

    🔒 Start your 14-day free trial to see the full solution →
11. CBSE 2025Set ANNUAL1 mark
    Q.Write IUPAC name of the K3[Cr(C2O4)3] complex.

    ›Reveal solutionSolution

    K3[Cr(C2O4)3] names as potassium tris(oxalato)chromate(III), found by first working out Cr's oxidation state from the ionic charges.

    Step 1 - find oxidation state of Cr:

    Oxalate (C2O4)^2- is a bidentate ligand with charge -2; there are 3 of them: total ligand charge = 3 x (-2) = -6.

    3 K+ balance the complex anion's charge, so the complex ion [Cr(C2O4)3]^3- carries charge -3.

    Let oxidation state of Cr = x: x + (-6) = -3, so x = +3.

    Step 2 - construct the IUPAC name:

    • Cation (potassium) named first, unchanged. …

    🔒 Start your 14-day free trial to see the full solution →
12. CBSE 2025Set ANNUAL1 markMCQ
    Q.The oxidation state of Fe in [Fe(CN)₆]⁻³:

    (a) +3

    (b) +2

    (c) +4

    (d) -3

    ›Reveal solutionSolution

    Since each CN⁻ ligand carries a −1 charge and the complex ion has an overall charge of −3, the oxidation state of Fe works out to +3.

    …

    🔒 Start your 14-day free trial to see the full solution →