Online Test — Coordination Compounds
20 Questions • 15 min • Chapter MCQ
15:00
Question 1 of 20
A coordination compound differs from a double salt because it:
Dissociates completely in water
Retains its complex ion in solution
Has no metal
Is always ionic
Explanation: The complex ion of a coordination compound stays intact in solution; a double salt dissociates fully.
Question 2 of 20
The primary valence in Werner’s theory equals the:
Coordination number
Oxidation state
Atomic mass
Number of ligands
Explanation: Primary (ionisable) valence corresponds to the oxidation state.
Question 3 of 20
The coordination number of Co in [Co(en)3]3+ is:
3
4
5
6
Explanation: en is bidentate; three en supply six donor atoms, so CN = 6.
Question 4 of 20
The IUPAC name of K3[Fe(CN)6] is:
Potassium hexacyanidoferrate(III)
Potassium hexacyanidoferrate(II)
Potassium ferricyanide(II)
Tripotassium ironhexacyanide
Explanation: Fe is +3 ($x-6=-3$); anionic complex → ferrate(III).
Question 5 of 20
Which is an ambidentate ligand?
en
NO2−
NH3
C2O42−
Explanation: NO2− binds through N (nitrito-N) or O (nitrito-O).
Question 6 of 20
EDTA4− acts as a ligand of denticity:
2
4
6
3
Explanation: EDTA4− has six donor atoms (2 N + 4 O), hexadentate.
Question 7 of 20
The pair [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2·H2O are:
Linkage isomers
Ionisation isomers
Solvate (hydrate) isomers
Optical isomers
Explanation: Water moves between inside and outside the coordination sphere — solvate isomerism.
Question 8 of 20
Square planar geometry corresponds to the hybridisation:
$sp^3$
$dsp^2$
$d^2sp^3$
$sp^3d^2$
Explanation: Square planar four-coordinate complexes use $dsp^2$.
Question 9 of 20
An inner-orbital octahedral complex has hybridisation:
$sp^3$
$dsp^2$
$d^2sp^3$
$sp^3d^2$
Explanation: Inner-orbital (low-spin) octahedral complexes use inner $(n-1)d$: $d^2sp^3$.
Question 10 of 20
[NiCl4]2− is:
Square planar, diamagnetic
Tetrahedral, paramagnetic
Octahedral, paramagnetic
Linear, diamagnetic
Explanation: Weak-field Cl− → $sp^3$ tetrahedral, $d^8$ with 2 unpaired e−.
Question 11 of 20
The spin-only moment of [CoF6]3− ($d^6$, high-spin) is about:
0 BM
1.73 BM
2.83 BM
4.90 BM
Explanation: 4 unpaired electrons: $\mu=\sqrt{4(4+2)}=4.90$ BM.
Question 12 of 20
Which complex can show optical isomerism?
[Co(NH3)6]3+
trans-[Co(en)2Cl2]+
cis-[Co(en)2Cl2]+
[PtCl4]2−
Explanation: cis-bis(en) is chiral (no symmetry plane) — d/l enantiomers.
Question 13 of 20
In an octahedral field the lower-energy set of d orbitals is:
$e_g$
$t_{2g}$
$t_2$
$a_{1g}$
Explanation: $t_{2g}$ points between ligands and is stabilised by $0.4\,\Delta_o$.
Question 14 of 20
The value of $\Delta_t$ relative to $\Delta_o$ is:
$\frac{4}{9}\Delta_o$
$\frac{9}{4}\Delta_o$
$\Delta_o$
$2\Delta_o$
Explanation: $\Delta_t=\frac{4}{9}\Delta_o$ for the same metal and ligands.
Question 15 of 20
A low-spin octahedral complex forms when:
$\Delta_o > P$
$\Delta_o < P$
$\Delta_o = 0$
the ligand is weak field
Explanation: Pairing (low-spin) needs the splitting to exceed the pairing energy.
Question 16 of 20
Which ion is colourless?
$\text{Cu}^{2+}$
$\text{Sc}^{3+}$
$\text{Fe}^{3+}$
$\text{Ni}^{2+}$
Explanation: Sc3+ is $d^0$; no d–d transition is possible.
Question 17 of 20
The metal present in vitamin B12 is:
Fe
Mg
Co
Zn
Explanation: Vitamin B12 (cyanocobalamin) contains cobalt.
Question 18 of 20
The CFSE of a low-spin octahedral $d^6$ ion (ignoring pairing terms) is:
$0$
$-0.4\,\Delta_o$
$-1.2\,\Delta_o$
$-2.4\,\Delta_o$
Explanation: $t_{2g}^6 e_g^0$: CFSE $=6\times(-0.4)\,\Delta_o=-2.4\,\Delta_o$.
Question 19 of 20
The oxidation state of platinum in [Pt(NH3)2Cl2] is:
+2
+1
+4
0
Explanation: $x+2(0)+2(-1)=0\Rightarrow x=+2$.
Question 20 of 20
The number of moles of AgCl precipitated by 1 mol [Co(NH3)4Cl2]Cl with excess AgNO3 is:
3
2
1
0
Explanation: Only one Cl is outside the sphere (ionisable), so 1 mol AgCl.