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

Which of the following complexes has square planar geometry?

  1. MnO4-

  2. FeCl4-

  3. CoCl42-

  4. PdCl42-

  5. ZnCl42-
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

With a second row d8 metal ion such as Pd2+ (which already generates a strong field), even a weak field ligand Cl- leads the formation of a square planar complex. Example: PdCl42-

Multiple choice

Which of the following is a pair of linkage isomers?

  1. [Co(NH3)6][Cr(C2O4)3] and [Co(C2O4)3][Cr(NH3)6]

  2. [PtBr(NH3)3]NO2 and [Pt(NO2)(NH3)3]Br

  3. [CrCl(H2O)5]Cl2.H2O and [Cr(H2O)6]Cl3

  4. [Co(ONO)(NH3)5]Cl and [Co(NO2)(NH3)5]Cl

  5. cis-[PtCl2(NH3)2] and trans-[PtCl2(NH3)2]
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

It is a pair of linkage isomers.

Multiple choice

Wilkinson's catalyst is the common name for

  1. trans-chlorocarbonylbis(triphenylphosphine) iridium(I).

  2. chlorotris(triphenylphosphine)rhodium(I)

  3. N,N'-bis-1,2-cyclohexanediaminomanganese chloride

  4. dicyanidobis(ethylenediamine)cadmium(II)

  5. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Wilkinson's catalyst is the common name for chlorotris(triphenylphosphine)rhodium(I).

Multiple choice

Which of the following is a pair of ionisation iosmers?

  1. [Pd(NCS)2(PPh3)2] and [Pd(SCN)2(PPh3)2]

  2. [Cu(NH3)4][PtCl4] and [Pt(NH3)4][CuCl4]

  3. [Co(NH3)5Br]SO4 and [Co(NH3)5(SO4)]Br

  4. [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6]

  5. [Co(NH3)5(NO2)]2+ and [Co(NH3)5(ONO)]2+
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

It is a pair of ionisation isomers.

Multiple choice

Which of the following is a set of tetrahedral complexes?

  1. [PtCl4]2-, [AuBr4]-, [Co(CN)4]2-

  2. [PtCl4]2-, [CoCl4]2-, [NiCl4]2-

  3. [CoCl4]2-, [MnO4]-, [NiCl4]2-

  4. [AuBr4]-, [Co(CN)4]2-, [MnO4]-

  5. [Co(CN)4]2-, [CoCl4]2-, [AuBr4]-
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

[CoCl4]2-, [MnO4]-, [NiCl4]2-are tetrahedral complexes.

Multiple choice

Which of the following d-electronic configurations will give maximum CFSE in an octahedral complex?

  1. Low spin-d4

  2. High spin-d4

  3. Low spin-d5

  4. High spin-d5

  5. High spin-d7
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In octahedral complexes, CFSE is given by the formula:  CFSE = value of t2g (n) + value of eg (n) Dq, Now, CFSE for low-spin d4 = -4 (4) + 6 (0) = -16 + 0 = -16 Dq CFSE for high-spin d4 = -4 (3) + 6 (1) = -12 + 6 = -6 Dq CFSE for low-spin d5 = -4 (5) + 6 (0) = -20 + 0 = -20 Dq CFSE for high-spin d5 = -4 (3) + 6 (2) = -12 + 12 = 0 Dq CFSE for low-spin d7 = -4 (5) + 6 (2) = -20 + 12 = -8 Dq Hence, low-spin d5 will give maximum CFSE in an octahedral complex.

Multiple choice

Which metal ion represents zero value of CFSE in octahedral complexes if it associated with a weak field ligand?

  1. Fe(III)

  2. Fe(II)

  3. Cr(II)

  4. Mn(I)

  5. Mn(III)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Correct answer. In weak field ligand, the filling of electrons are in accordance with Hund’s rule. The CFSE is given by the formula:  CFSE = value of t2g (n) + value of eg (n) Dq, Fe(III) corresponds to the d5 configuration. Now, CFSE for Fe(III) (d5) = -4 (3) + 6 (2) = -12 + 12 = 0 Dq CFSE for Fe(II) (d6) = -4 (4) + 6 (2) = -16 + 12 = -4 Dq CFSE for Cr(II) (d4) = -4 (3) + 6 (1) = -12 + 6 = -6 Dq CFSE for Mn(I) (d6) = -4 (4) + 6 (2) = -16 + 12 = -4 Dq CFSE for Mn(III) (d4) = -4 (3) + 6 (1) = -12 + 6 = -6 Dq Hence, the metal ion with zero CFSE is Fe(III) when it associated with a weak field ligand.

Multiple choice

Match the following salt names with their molecular formula ||| |---|---| |(i) Fischer salt|(a) [Pt(NH3)4][PtCl4]| |(ii) Zeise’s salt|(b) Fe2[Fe(CN)6]| |(iii) Magnus’ green salt|(c) K3[Co(NO2)6]| |(iv) Turnbull’s blue|(d) K3[Fe(CN)6]| |(v) Gmelin’s salt|(e) K[PtCl3(C2H4)]|

  1. (i) - c, (ii) - e, (iii) - a, (iv) - b, (v) - d

  2. (i) - c, (ii) - e, (iii) - a, (iv) - d, (v) - b

  3. (i) - d, (ii) - e, (iii) - a, (iv) - c, (v) - b

  4. (i) - d, (ii) - a, (iii) - e, (iv) - c, (v) - d

  5. (i) - c, (ii) - a, (iii) - e, (iv) - b, (v) - d
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In this option, all salt names are correctly matched with their molecular formulae.

Multiple choice

A compound is formed by two elements M and N. Atoms of the element N (as anions) make ccp and those of the element M (as cations) occupy all the octahedral voids. What is the formula of the compound?

  1. MN3

  2. M3N2

  3. MN2

  4. M2N

  5. MN
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

The ccp lattice is formed by the element N. The number of octahedral voids generated would be equal to the number of atoms of Y present in it. Since all the octahedral voids are occupied by the atoms of M, so their number would also be equal to that of the element N. Thus, the atoms of elements M and N are present in equal numbers or 1 : 1 ratio. Therefore, the formula of the compound is MN.

Multiple choice

Which of the following sets represents paramagnetic metal ions only?

  1. Sc3+, Ti4+, Cu2+, Zn2+

  2. Sc3+, Fe3+, Ni2+, Cu2+

  3. Mn3+, Fe3+, Ni2+, Cu2+

  4. Ti4+, Ni2+, Cu2+, Zn2+

  5. Mn3+, Fe3+, Ni2+, Zn2+
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Mn3+: 3d4, 4 unpaired electrons; paramagnetic Fe3+: 3d5, Five unpaired electrons; paramagnetic Ni2+: 3d8, 2 unpaired electrons; paramagnetic Cu2+: 3d9, One unpaired electrons; paramagnetic

Multiple choice

Which of the following substances show anti-ferromagnetism?

  1. Fe3O4

  2. MgFe2O4

  3. MnO

  4. CrO2

  5. NaCl
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Substances like MnO showing anti-ferromagnetism have domain structure similar to ferromagnetic substance, but their domains are oppositely oriented and cancel out each other's magnetic moment.

Multiple choice

Which of the following complexes is NOT correctly named?

  1. [Co(NH3)6]Cl3: Hexaamminecobalt(III) chloride

  2. [Cu(H2O)4]SO4: Tetraaquacopper(II) sulphate

  3. [Cr(en)3]Cl3:Triethylenediamminechromium(III) chloride

  4. K4[Fe(CN)6]: Potassium hexacyanoferrate(II)

  5. Li[AlH4]: Lithium tetrahydridoaluminate(III)
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Correct IUPAC name of [Cr(en)3]Cl3 is tris(ethylenediammine)chromium(III) chloride.