Chemistry
Coordination Chemistry
325 Questions
Coordination chemistry focuses on coordination compounds, their structures, and their magnetic properties. The questions cover primary and secondary valencies, ligand types, and geometries of complexes. This topic is heavily tested in chemistry competitive exams and requires a good grasp of molecular structures.
primary and secondary valencycomplex geometriesligand field theorymagnetic properties of complexesisomerism in complexes
Coordination Chemistry Questions
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Optical
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Ionisation
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Geometrical
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Linkage
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Hydrated
D
Correct answer
Explanation
This option is correct because complex pentamminenitrito chromium (III) chloride exhibits the linkage isomerism.
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WF6
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ZrF84-
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ReH92-
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IF8-
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IF7
C
Correct answer
Explanation
The capped square antiprismatic molecular geometry describes the shape of compounds where nine atoms, groups of atoms, or ligands are arranged around a central atom, defining the vertices of a gyroelongated square pyramid. Example is ReH92-.
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magnesium
-
iron
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copper
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nitrogen
B
Correct answer
Explanation
The heme group is a prosthetic group containing a porphyrin ring with a central iron atom (Fe2+). This iron is crucial for oxygen binding in hemoglobin and myoglobin, and for electron transfer in cytochromes.
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linear
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trigonal planar
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tetrahedral
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square planar
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octahedral
D
Correct answer
Explanation
In square planar complexes, the geometry is prevalent for transition metal complexes with d8 configuration. Thus, the central atom (Pt) in carboplatin has dsp2 hybridisation with square planar geometry.
E
Correct answer
Explanation
In a weak field ligand system, the electron filling of splitting of d-orbital corresponds to Hund’s rule, followed by pairing. In strong field ligand system, the electron filling of splitting of d-orbital corresponds to pairing at initial stage.
For Fe3+ electronic configuration is d5.
For d5 in weak field ligand system,
CFSE = 3 (- 4) + 2 (6) = - 12 + 12 = 0 Dq (All electrons is unpaired, 3 in t2g and 2 in eg)
For d5 in strong field ligand system,
CFSE = 5 (- 4) + 0 = - 20 Dq + 2P
(two paired electrons and one electron is unpaired in t2g and 0 electron in eg orbital)
Hence, the CFSE for octahedral complexes of Fe3+ in a weak field and strong field ligand systems are 0 and -20 Dq+2P respectively.
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3D3/2
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3F3/2
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4D3/2
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4F3/2
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4F9/2
D
Correct answer
Explanation
V2+ has d3 electronic configuration and it is empowered in a weak field ligand.
The term symbol is calculated by the formula-
2s+1 LJ
For d3,
n = 3
s = n/2 = 3/2
2s + 1 = 4
L = 3 = F
J = (L + s) to (L - s) = 2, 1, 0
term symbols = 2s + 1 LJ = 4F9/2, 4F7/2, 4F5/2, 4F3/2
The ground state term for V2+ is 4F3/2.
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Only A
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Only B
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Only C
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Only A and B
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Only B and C
E
Correct answer
Explanation
Ground state term for Mn4+ is 6S5/2.
Ground state term for Fe2+ is 5D4.
Ground state term for Co3+ is 5D4.
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6S5/2 and 5D4
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6S3/2 and 5D3
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5S5/2 and 5D4
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6S3/2 and 4D3/2
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6S5/2 and 5D3/2
A
Correct answer
Explanation
Fe3+ and Fe2+ have d5 and d6 electronic configurations and both are in high spin ligand system.
The term symbol is calculated by the formula-
2s+1 LJ
For d5,
n = 5
s = n/2 = 5/2
2s + 1 = 6
L = 0 = S
J = (L + s) to (L - s) = 5/2
Possible term symbol for Fe3+ = 2s + 1 LJ = 6S5/2
Similarly, possible term symbols for Fe2+ = 5D4, 5D3, 5D2, 5D1, 5D0
The ground state term for Fe3+ is 6S5/2.
The ground state term for Fe2+ is 5D4.
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(η-C5H5)2Ti
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(η-C5H5)2V
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(η-C5H5)2Cr
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(η-C5H5)2Fe
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(η-C5H5)2Co
A
Correct answer
Explanation
(η-C5H5)2Ti has zero number of unpaired electrons and possesses a diamagnetic nature.
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S2-, SCN-, NO3-
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NO2-, CN-, PPh3
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Cl-, N3-, NO2-
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NO3-, NO2-, I-
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PPh3, S2-, SCN-
A
Correct answer
Explanation
On the basis of ligand field theory, the spectrochemical series is an empirically-derived list of ligands ordered by the size of the splitting CFSE that they produce. It can be seen that the low-field ligands are all pi-donors, the high field ligands are pi-acceptors, and ligands such as H2O and NH3, which are neither of the two, are in the middle.
The spectrochemical series is give as:
I− < Br− < S2− < SCN− < Cl− < NO3− < N3− < F− < OH− < C2O42− < H2O < NCS− < CH3CN < py (pyridine) < NH3 < en (ethylenediamine) < bipy (2,2'-bipyridine) < phen (1,10-phenanthroline) < NO2− < PPh3 < CN− < CO
Hence, S2-, SCN- and NO3- are all weak field ligands.
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[FeF6]3-and [Mn(H2O)6]3+
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[Cr(NH3)6]3+ and [Mn(CN)6]3-
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[CoF6]3-and [Cr(NH3)6]3+
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[Co(NH3)6]3+and [Mn(CN)6]3-
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[Mn(CN)6]3- and [Mn(H2O)6]3+
A
Correct answer
Explanation
[FeF6]3- and [Mn(H2O)6]3+ both have sp3d2 hybridisation and hence, are outer orbital or high spin complexes.
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Only A
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Only B
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Only C
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Only A and B
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Only B and C
E
Correct answer
Explanation
A and B both have a paramagnetic nature with 2 unpaired electrons respectively.
Hence, these are identical in the number of unpaired electrons.
Match the coordination complexes in Column-I with corresponding coordination numbers and select the correct answer.
| |
|
| Column-I |
|
|
Column-II |
| P. [Mn(NH3)6]2+ |
1. 7 |
| Q. [(Ph4As)2Mn(NO2)4] |
2. 8 |
| R. [Mn(EDTA)(NH3)]2- |
3. 5 |
| S. [Mn(S2C6H3Me)2]- |
4. 4 |
|
5. 6 |
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P - 3, Q - 5, R - 1, S - 2
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P - 4, Q - 2, R - 3, S - 5
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P - 5, Q - 2, R - 1, S - 4
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P - 3, Q - 2, R - 4, S - 1
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P - 2, Q - 5, R - 1, S - 3
C
Correct answer
Explanation
The coordination complexes [Mn(NH3)6]2+, [(Ph4As)2Mn(NO2)4], [Mn(EDTA)(NH3)]2- and [Mn(S2C6H3Me)2]- have the coordination numbers of 6, 8, 7 and 4 respectively.
Hence, the representation of the codes P - 5, Q - 2, R - 1, S - 4 is the correct answer.
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Cobalt (II) bromide
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Cobalt (II) chloride
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Cobalt (II) carbonate
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Cobalt (II) sulfate
C
Correct answer
Explanation
Cobalt (II) carbonate is the inorganic compound with the formula CoCO3. This reddish paramagnetic solid is an intermediate in the hydrometallurgical purification of cobalt from its ores, as an inorganic pigment and as a precursor to catalysts.
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Aromaticity
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Polarity
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Hapticity
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Stacking
C
Correct answer
Explanation
The term hapticity is used to describe how a group of contiguous atoms of a ligand are coordinated to a central atom.