The metal atom present in complex MABXL, where A, B, X and L are all different unidentate ligands and M is the central metal atom with sp³ hybridisation. The total number of geometrical isomers possible for this complex is:
A0
B2
C3
D4
✅ Correct: A
For a tetrahedral complex MA₁A₂A₃A₄ (all different monodentate ligands), geometrical isomerism does NOT exist. Tetrahedral complexes with four different ligands show only optical isomerism (chirality), not geometrical isomerism. Answer: 0 geometrical isomers.
Q2
In which of the following coordination complexes will the crystal field stabilisation energy (CFSE) in an octahedral field be equal to zero?
AK₃[Fe(SCN)₆]
B[Fe(en)₃]Cl₃
C[Fe(NH₃)₆]Br₂
DK₄[Fe(CN)₆]
✅ Correct: A
Fe³⁺ in K₃[Fe(SCN)₆]: Fe is +3, d⁵ configuration. SCN⁻ is a weak field ligand → high spin. d⁵ high spin: t₂g³ eg² (each orbital has 1 electron). CFSE = 3(−0.4Δₒ) + 2(+0.6Δₒ) = −1.2Δₒ + 1.2Δₒ = 0. Answer: K₃[Fe(SCN)₆].
Q3
The d-electronic configuration of an octahedral Co(II) complex having a magnetic moment of approximately 3.95 Bohr Magneton (BM) is:
At₂g⁴ eg³
Bt₂g⁶ eg¹
Ct₂g⁵ eg²
Dt₂g³ eg⁰
✅ Correct: C
μ = √(n(n+2)) BM. For 3.95 BM: n = 3 unpaired electrons. Co²⁺ is d⁷. High spin d⁷: t₂g⁵ eg² → 3 unpaired electrons (5+2=7, with 2 unpaired in eg and 1 in t₂g). √(3×5) = √15 ≈ 3.87 BM ≈ 3.95 BM. Configuration: t₂g⁵ eg².
Q4
A transition metal M among Mn, Cr, Co and Fe has the highest standard electrode potential for the M³⁺/M²⁺ couple. It forms a complex of the type [M(CN)₆]⁴⁻. The number of electrons present in the eg orbitals of this complex is:
A0
B1
C2
D3
✅ Correct: B
Highest E° for M³⁺/M²⁺: Co³⁺/Co²⁺ (+1.84 V) — highest among the given metals. Co²⁺ is d⁷. In [Co(CN)₆]⁴⁻: CN⁻ is strong field ligand → low spin d⁷: t₂g⁶ eg¹. Electrons in eg = 1.
Q5
The spin-only magnetic moment value (in BM) for [MnBr₄]²⁻ complex, where Mn is in +2 oxidation state and Br⁻ is a weak field ligand, is:
A√15
B√35
C√24
D√8
✅ Correct: B
Mn²⁺: d⁵. Br⁻ is weak field ligand → high spin → all 5 d electrons are unpaired. n = 5. μ = √(5(5+2)) = √(5×7) = √35 ≈ 5.92 BM.
Q6
What is the IUPAC name of the coordination compound [Co(NH₃)₄Cl₂]Cl?
ATetraamminedichloridocobalt(III) chloride
BTetraamminedichlorocobalt(III) chloride
CDichloridotetraamminecobalt(III) chloride
DCobalt tetraammine dichloride chloride
✅ Correct: A
IUPAC naming: ligands in alphabetical order — chlorido before ammine (C before A — wait: a comes before c alphabetically, so ammine listed first). Actually: ammine (a) comes before chlorido (c) alphabetically. Name: tetraamminedichloridocobalt(III) chloride. Ligands listed alphabetically: ammine (4), chlorido (2). Co is +3.
Q7
The number of geometrical isomers for [Pt(NH₃)(Br)(Cl)(py)] square planar complex is:
A2
B3
C4
D1
✅ Correct: B
Square planar complex with 4 different ligands (NH₃, Br⁻, Cl⁻, py) has 3 geometrical isomers. For [MAbcd] square planar: 3 geometrical isomers exist (a trans to b, a trans to c, or a trans to d).
Q8
Werner's theory of coordination compounds proposed that metals show:
AOnly primary valency
BOnly secondary valency
CBoth primary and secondary valency
DNeither primary nor secondary valency
✅ Correct: C
Werner proposed that transition metals have two types of valency: (1) Primary valency (ionisable, = oxidation state, satisfied by counter ions), (2) Secondary valency (non-ionisable, = coordination number, satisfied by ligands inside coordination sphere). This theory explained many complex compounds.
Q9
Which of the following complexes exhibits optical isomerism?
A[Co(NH₃)₃Cl₃]
B[Co(en)₃]³⁺
C[PtCl₄]²⁻
D[Ni(CN)₄]²⁻
✅ Correct: B
[Co(en)₃]³⁺: tris(ethylenediamine)cobalt(III) — three bidentate en ligands in octahedral arrangement. This complex is chiral (non-superimposable mirror images) → shows optical isomerism (Δ and Λ forms). Square planar complexes like [PtCl₄]²⁻ and [Ni(CN)₄]²⁻ don't show optical isomerism.
Q10
Effective Atomic Number (EAN) of nickel in [Ni(CO)₄] complex is:
A28
B36
C34
D32
✅ Correct: B
EAN = atomic number of metal + electrons donated by ligands − electrons lost as ions. Ni(0) in [Ni(CO)₄]: Ni has 28 electrons. CO is neutral, each donates 2e⁻. 4 CO ligands donate 8 electrons. EAN = 28 + 8 = 36 (equals Kr electron configuration).
Q11
Which of the following ligands is a bidentate ligand?
ANH₃
BCl⁻
Cen (ethylenediamine)
DEDTA
✅ Correct: C
Bidentate ligands have two donor atoms: en (ethylenediamine, H₂N-CH₂-CH₂-NH₂) coordinates through both N atoms. NH₃ and Cl⁻ are monodentate. EDTA is hexadentate (6 donor atoms: 4 O and 2 N).
Q12
In the complex [Fe(CO)₅], the oxidation state of iron is:
A+2
B+5
C0
D+1
✅ Correct: C
CO is a neutral ligand. Complex [Fe(CO)₅] has no charge and CO is neutral → oxidation state of Fe = 0. This is an organometallic complex where Fe is in zero oxidation state. Hybridisation: dsp³ (trigonal bipyramidal).
Q13
The coordination number of platinum in [PtCl₂(NH₃)₂] is:
A2
B4
C6
D3
✅ Correct: B
Coordination number = number of ligand donor atoms directly bonded to central metal. [PtCl₂(NH₃)₂]: 2 Cl⁻ + 2 NH₃ = 4 ligands. Coordination number = 4. Geometry: square planar (Pt²⁺, d⁸ configuration, dsp² hybridisation).
Q14
The complex [CoCl₂(en)₂]⁺ can exhibit which type(s) of isomerism?
AOnly geometrical isomerism
BOnly optical isomerism
CBoth geometrical and optical isomerism
DNo isomerism
✅ Correct: C
[CoCl₂(en)₂]⁺ is an octahedral complex. It shows: (1) Geometrical isomerism: cis (Cl cis to each other) and trans (Cl trans) forms. (2) Optical isomerism: the cis form is chiral — exists as Δ and Λ optical isomers. Trans form is achiral.
Q15
Which of the following is correctly matched for complex [Fe(H₂O)₅(NO)]²⁺ (brown ring complex)?
AFe is +2, NO is neutral
BFe is +1, NO is +1
CFe is +3, NO⁺ is the ligand
DFe is +3, NO is −1
✅ Correct: C
Brown ring complex: It is considered as [Fe(H₂O)₅(NO⁺)]²⁺ where Fe is in +1 oxidation state and NO is +1... Standard convention: Fe is +3 and NO is a ligand acting as NO⁺ (nitrosonium ion). Charge balance: +3 + (+1) − 2 (from charge of +2) — Actually: 3+ + NO charge = 2+. If NO is NO⁺: 3+ + 1+ − 2 = 2+. Correct.
Q16
The stability of coordination compounds is measured by:
AFormation constant (Kf)
BDissociation constant (Kd)
CStability constant and formation constant are same thing
DBoth A and B are correct — Kf = 1/Kd
✅ Correct: D
Stability constant (Kf, formation constant) measures the tendency of complex to form. Kd (dissociation constant) measures tendency to dissociate. They are reciprocals of each other: Kf = 1/Kd. Higher Kf = more stable complex. Both descriptions are valid.
Q17
In crystal field theory (CFT), the splitting of d-orbitals in octahedral field gives:
AThree lower energy t₂g orbitals and two higher energy eg orbitals
BTwo lower energy eg orbitals and three higher energy t₂g orbitals
CAll five d-orbitals at same energy
DTwo groups of equal energy
✅ Correct: A
In an octahedral crystal field: d-orbitals split into three lower energy t₂g (dxy, dxz, dyz) at −0.4Δₒ and two higher energy eg (dx²-y², dz²) at +0.6Δₒ. The energy separation is Δₒ (crystal field splitting parameter). This splitting determines colour and magnetic properties.
Q18
Which of the following has zero crystal field stabilisation energy (CFSE)?
Ad³ high spin
Bd⁵ high spin
Cd⁶ low spin
Dd⁸ octahedral
✅ Correct: B
d⁵ high spin in octahedral: t₂g³ eg². CFSE = 3(−0.4Δₒ) + 2(+0.6Δₒ) = −1.2 + 1.2 = 0. This is why Mn²⁺ (d⁵) and Fe³⁺ (d⁵) high spin complexes have zero CFSE and are relatively less stable.
Q19
The linkage isomers differ in:
AThe ligands present in the coordination sphere
BThe way an ambidentate ligand coordinates to the metal
CThe number of ligands
DThe geometry of the complex
✅ Correct: B
Linkage isomers arise when an ambidentate ligand (one that can coordinate through different atoms) coordinates through different donor atoms. Example: NO₂⁻ can bind through N (nitro, M-NO₂) or O (nitrito, M-ONO). [Co(NH₃)₅(NO₂)]²⁺ — nitro vs nitrito isomers.
Q20
The chelate effect refers to:
AThe tendency of polydentate ligands to form more stable complexes than monodentate ligands
BThe tendency of chelates to react with each other
CThe ring closure reaction in coordination chemistry
DThe crystal field stabilisation from chelate rings
✅ Correct: A
Chelate effect: complexes with polydentate (chelating) ligands are more stable than those with equivalent monodentate ligands. This is mainly due to entropy — forming one large chelate ring releases more free ligands (water molecules), increasing entropy significantly. [Co(en)₃]³⁺ is much more stable than [Co(NH₃)₆]³⁺.
R
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Expert · 5 Years Experience
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