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

Multiple choice
  1. Hydrogen

  2. Nitrogen

  3. Chlorine

  4. Cobalt

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In coordination compounds, donor atoms are ligand atoms that directly coordinate to the central metal ion. In [Co(NH3)6]Cl3, ammonia (NH3) ligands coordinate through nitrogen atoms. Chlorine is outside the coordination sphere as a counterion.

Multiple choice
  1. [Co (NH3)3 Cl3]

  2. [Co (en)2 Cl2]+

  3. [Co (NH3)6] Cl3

  4. [Co (edta)]

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Optical activity requires non-superimposable mirror images (chirality). [Co(NH3)6]Cl3 has six identical ammonia ligands in octahedral geometry with perfect symmetry - it has multiple planes and axes of symmetry, making it achiral. The other compounds have asymmetric arrangements.

Multiple choice
  1. 1

  2. 2

  3. 4

  4. 6

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The oxalate ion (C2O4)2- is a bidentate ligand, coordinating through two oxygen atoms. In [Cr(C2O4)3]3-, three oxalate ligands each donate two electron pairs, giving 3×2=6 donor atoms bonded to chromium, so CN=6.

Multiple choice
  1. Square planar

  2. Tetrahedral

  3. Trigonal planar

  4. Trigonal bipyramidal

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Fe(CO)5 is iron pentacarbonyl, a neutral coordination complex. With five CO ligands around a central metal ion, the only stable geometry that minimizes repulsion while satisfying the 18-electron rule is trigonal bipyramidal. Square planar would require only 4 ligands, tetrahedral geometry with 5 ligands is unstable, and trigonal planar only accommodates 3 ligands.

Multiple choice
  1. [Co(NH3)5ONO]2+ and [Co(NH3)5NO2]2+

  2. [Co(NH3)4Cl2] NO2 and [Co(NH3)4 (Cl) NO2] Cl

  3. [Co(py)2 (H2O)2Cl2]Cl and [Co(py)2 (H2O)Cl3] H2O

  4. [Co(NH3)4 Cl2]NO2 and [Co(NH3)4(Cl)NO2]Cl

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Linkage isomerism occurs when ambidentate ligands can bind through different donor atoms. The nitrite ion (NO2-) is a classic ambidentate ligand that can bind through nitrogen (nitro: -NO2) or through oxygen (nitrito: -ONO). The two complexes [Co(NH3)5ONO]2+ and [Co(NH3)5NO2]2+ differ only in which atom binds to cobalt, making them linkage isomers.

Multiple choice
  1. A and R are both correct and R is the correct explanation of A.

  2. A and R are both correct and R is not the correct explanation of A.

  3. A is correct and R is incorrect.

  4. A is incorrect and R is correct.

  5. A and R are both incorrect.

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This option is correct because in case of [Mn(H2O)6]+2, the d-d transitions are spin forbidden because each of the d-orbitals is singly occupied (half filled) and the spin multiplicity changes when the electron goes to an excited state from ground state.

Multiple choice
  1. 3F2

  2. 5D0

  3. 2D5/2

  4. 1S0

  5. 4F3/2

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

S = ½ x number of unpaired electrons = ½ x 2 = 1 2S + 1 = 2 x 1 + 1 = 3 L = 3 for d2 ion configuration for which the ground term is F J = L - S = 3 - 1 = 2 The ground state term is (2S + 1)LJ = 3F2 Thus, this option is correct.

Multiple choice
  1. A and R are both correct and R is the correct explanation of A.

  2. A and R are both correct and R is not the correct explanation of A.

  3. A is correct and R is incorrect.

  4. A is incorrect and R is correct.

  5. A and R are both incorrect.

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Complex [Co(NH3)3F]+2 absorbs in the far ultra-violet region but in complex [Co(NH3)3Br]+2, absorption occurs at a longer wavelength. It is so because the position of the charge transfer band depends on the nature of the metal and the ligand.

Multiple choice
  1. Only 1

  2. Only 2

  3. Only 3

  4. 1 and 2

  5. 2 and 3

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This option is correct because the intensity of spin forbidden transition is about one-hundredth of that for a spin allowed transition and for d2 configuration, the ground state terms 3F and 3P are considered.

Multiple choice
  1. 1 - d, 2 - c, 3 - a, 4 - b

  2. 1 - c, 2 - b, 3 - d, 4 - a

  3. 1 - b, 2 - a, 3 - c, 4 - d

  4. 1 - b, 2 - a, 3 - d, 4 - c

  5. 1 - a, 2 - b, 3 - c, 4 - d

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This option is correct because [Ti(H2O)6]+3, KMnO4, Ni(CO)4 and [Cu(en)2]+ are examples of d-d transition, ligand to metal charge transfer, metal to ligand charge transfer and interligand charge transfer, respectively.

Multiple choice
  1. A and R are both correct and R is the correct explanation of A.

  2. A and R are both correct and R is not the correct explanation of A.

  3. A is correct and R is incorrect.

  4. A is incorrect and R is correct.

  5. A and R both are incorrect.

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The mixing of d and p orbitals does not occur in octahedral complexes which have centre of symmetry, like [Ni(NH3)6]+2 and [Co(NH3)6]+3, because in these cases, the metal ligand bonds vibrate so that the ligands spend most of their time out of their centrosymmetric equilibrium position. Thus, this option is correct.

Multiple choice
  1. Only A

  2. Only B

  3. Only C

  4. A and B

  5. B and C

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Substitution in octahedral complexes, which destroys the center of symmetry of the ligand, gives higher absorption.