Q.1: Which of the following is a chelating ligand?
(a) NH₃
(b) Cl⁻
(c) en (ethylenediamine)
(d) CO
Answer: (c) en (ethylenediamine)
Explanation: Ethylenediamine (en) is a bidentate ligand, meaning it can form two bonds to a metal ion, making it a chelating ligand.
Q.2: The oxidation state of cobalt in [Co(NH₃)₆]³⁺ is:
(a) +1
(b) +2
(c) +3
(d) +4
Answer: (c) +3
Explanation: The complex [Co(NH₃)₆]³⁺ has an overall charge of +3, and NH₃ is a neutral ligand, so the oxidation state of cobalt is +3.
Q.3: The IUPAC name of [Pt(NH₃)₂Cl₂] is:
(a) Diamminedichloroplatinum(II)
(b) Dichlorodiammineplatinum(II)
(c) Diamminedichloroplatinum(IV)
(d) Dichlorodiammineplatinum(IV)
Answer: (a) Diamminedichloroplatinum(II)
Explanation: According to IUPAC nomenclature, the name of the complex [Pt(NH₃)₂Cl₂] is Diamminedichloroplatinum(II).
Q.4: The geometry of [Ni(CO)₄] is:
(a) Tetrahedral
(b) Square planar
(c) Octahedral
(d) Linear
Answer: (a) Tetrahedral
Explanation: [Ni(CO)₄] has a tetrahedral geometry as it involves dsp³ hybridization of the central metal atom.
Q.5: Which of the following ligands can cause the most splitting in the d-orbitals of the central metal ion in an octahedral complex?
(a) H₂O
(b) NH₃
(c) CN⁻
(d) Cl⁻
Answer: (c) CN⁻
Explanation: CN⁻ is a strong field ligand and causes the maximum splitting of d-orbitals in an octahedral field.
Q.6: The number of unpaired electrons in [CoF₆]³⁻ is:
(a) 0
(b) 1
(c) 3
(d) 4
Answer: (d) 4
Explanation: [CoF₆]³⁻ is a high-spin complex with Co³⁺ in the 3d⁶ configuration, resulting in four unpaired electrons.
Q.7: The type of isomerism exhibited by [Co(NH₃)₄Cl₂]⁺ is:
(a) Geometrical isomerism
(b) Optical isomerism
(c) Linkage isomerism
(d) Ionization isomerism
Answer: (a) Geometrical isomerism
Explanation: [Co(NH₃)₄Cl₂]⁺ exhibits geometrical isomerism as the chloride ligands can be arranged in cis- or trans- positions.
Q.8: The color of [Ti(H₂O)₆]³⁺ is:
(a) Blue
(b) Green
(c) Violet
(d) Yellow
Answer: (c) Violet
Explanation: The complex [Ti(H₂O)₆]³⁺ is violet in color due to d-d transitions.
Q.9: The effective atomic number (EAN) of [Fe(CN)₆]³⁻ is:
(a) 35
(b) 36
(c) 37
(d) 38
Answer: (b) 36
Explanation: The EAN of [Fe(CN)₆]³⁻ is 36, calculated as 26 (atomic number of Fe) + 6×2 (12 from 6 CN⁻ ligands) – 3 (charge) = 35.
Q.10: The hybridization of the central metal ion in [Ni(CN)₄]²⁻ is:
(a) sp³
(b) dsp²
(c) d²sp³
(d) sp²
Answer: (b) dsp²
Explanation: The complex [Ni(CN)₄]²⁻ has a square planar geometry, indicating dsp² hybridization.
Q.11: The coordination number of Pt in [Pt(en)₂Cl₂]²⁺ is:
(a) 2
(b) 4
(c) 6
(d) 8
Answer: (b) 4
Explanation: In [Pt(en)₂Cl₂]²⁺, Pt is coordinated to two ethylenediamine (en) ligands and two chloride ions, making the coordination number 4.
Q.12: The complex [Co(NH₃)₅Cl]²⁺ can exhibit:
(a) Optical isomerism
(b) Linkage isomerism
(c) Geometrical isomerism
(d) Ionization isomerism
Answer: (d) Ionization isomerism
Explanation: [Co(NH₃)₅Cl]²⁺ can exhibit ionization isomerism as the counter ion can change, resulting in different compounds.
Q.13: Which of the following complexes is diamagnetic?
(a) [CoF₆]³⁻
(b) [Fe(CN)₆]⁴⁻
(c) [MnCl₆]⁴⁻
(d) [Cr(H₂O)₆]³⁺
Answer: (b) [Fe(CN)₆]⁴⁻
Explanation: [Fe(CN)₆]⁴⁻ is diamagnetic because Fe²⁺ in a low-spin state has no unpaired electrons.
Q.14: The geometry of [Fe(CN)₆]⁴⁻ is:
(a) Tetrahedral
(b) Square planar
(c) Octahedral
(d) Linear
Answer: (c) Octahedral
Explanation: [Fe(CN)₆]⁴⁻ has an octahedral geometry due to the arrangement of six cyanide ligands around the central iron ion.
Answer: (a) Geometrical isomerism
Explanation: [Cr(NH₃)₄Cl₂]Cl can show cis-trans isomerism due to the different possible arrangements of the chloride ligands.
Q.16: The oxidation state of the central metal ion in [MnO₄]⁻ is:
(a) +2
(b) +3
(c) +5
(d) +7
Answer: (d) +7
Explanation: The oxidation state of Mn in [MnO₄]⁻ is +7 as oxygen has a -2 oxidation state, giving a total of -8, and the overall charge is -1.
Q.17: The ligand which causes the maximum crystal field splitting is:
(a) H₂O
(b) NH₃
(c) CN⁻
(d) F⁻
Answer: (c) CN⁻
Explanation: CN⁻ is a strong field ligand and causes maximum splitting of the d-orbitals in a crystal field.
Q.18: The shape of [Fe(H₂O)₆]³⁺ is:
(a) Tetrahedral
(b) Square planar
(c) Octahedral
(d) Trigonal bipyramidal
Answer: (c) Octahedral
Explanation: [Fe(H₂O)₆]³⁺ has an octahedral shape due to the arrangement of six water molecules around the central iron ion.
Q.19: Which of the following does not show geometrical isomerism?
(a) [Pt(NH₃)₂Cl₂]
(b) [Co(NH₃)₄Cl₂]⁺
(c) [Ni(NH₃)₄]²⁺
(d) [Pt(en)₂]²⁺
Answer: (c) [Ni(NH₃)₄]²⁺
Explanation: [Ni(NH₃)₄]²⁺ does not show geometrical isomerism as all the ligands are the same and it has a symmetrical tetrahedral shape.
Q.20: The type of hybridization in [Co(CN)₆]³⁻ is:
(a) sp³
(b) dsp²
(c) d²sp³
(d) sp²
Answer: (c) d²sp³
Explanation: [Co(CN)₆]³⁻ has an octahedral geometry which corresponds to d²sp³ hybridization.
Q.21: Which of the following is an ambidentate ligand?
(a) NO₃⁻
(b) SCN⁻
(c) OH⁻
(d) CN⁻
Answer: (b) SCN⁻
Explanation: SCN⁻ is an ambidentate ligand as it can coordinate through either sulfur or nitrogen atoms.
Q.22: The complex [Ni(CO)₄] is:
(a) Paramagnetic
(b) Diamagnetic
(c) Ferromagnetic
(d) Antiferromagnetic
Answer: (b) Diamagnetic
Explanation: [Ni(CO)₄] is diamagnetic as it has no unpaired electrons in its electronic configuration.
Q.23: The IUPAC name of [Cr(H₂O)₄Cl₂]Cl is:
(a) Tetraamminedichlorochromium(III) chloride
(b) Tetraaquadichlorochromium(III) chloride
(c) Tetraamminedichlorochromium(IV) chloride
(d) Tetraaquadichlorochromium(IV) chloride
Answer: (b) Tetraaquadichlorochromium(III) chloride
Explanation: According to IUPAC nomenclature, the name of the complex [Cr(H₂O)₄Cl₂]Cl is Tetraaquadichlorochromium(III) chloride.
Q.24: The coordination number of Ni in [Ni(NH₃)₆]²⁺ is:
(a) 4
(b) 5
(c) 6
(d) 8
Answer: (c) 6
Explanation: In [Ni(NH₃)₆]²⁺, Ni is coordinated to six ammonia ligands, making the coordination number 6.
Q.25: The type of hybridization in [PtCl₄]²⁻ is:
(a) sp³
(b) dsp²
(c) d²sp³
(d) sp²
Answer: (b) dsp²
Explanation: [PtCl₄]²⁻ has a square planar geometry which corresponds to dsp² hybridization.
Q.26: Which of the following is a weak field ligand?
(a) CN⁻
(b) NH₃
(c) CO
(d) Cl⁻
Answer: (d) Cl⁻
Explanation: Cl⁻ is a weak field ligand and causes less splitting of the d-orbitals in a crystal field.
Q.27: The geometry of [Pt(NH₃)₄]²⁺ is:
(a) Tetrahedral
(b) Square planar
(c) Octahedral
(d) Trigonal bipyramidal
Answer: (b) Square planar
Explanation: [Pt(NH₃)₄]²⁺ has a square planar geometry due to dsp² hybridization.
Q.28: The complex [Fe(CN)₆]³⁻ is:
(a) Paramagnetic
(b) Diamagnetic
(c) Ferromagnetic
(d) Antiferromagnetic
Answer: (a) Paramagnetic
Explanation: [Fe(CN)₆]³⁻ is paramagnetic as it has unpaired electrons in its electronic configuration.
Q.29: The type of isomerism exhibited by [Co(NH₃)₄Cl₂]⁺ is:
(a) Geometrical isomerism
(b) Optical isomerism
(c) Linkage isomerism
(d) Ionization isomerism
Answer: (a) Geometrical isomerism
Explanation: [Co(NH₃)₄Cl₂]⁺ exhibits geometrical isomerism as the chloride ligands can be arranged in cis- or trans- positions.
Q.30: The hybridization of the central metal ion in [Ni(CO)₄] is:
(a) sp³
(b) dsp²
(c) d²sp³
(d) sp²
Answer: (a) sp³
Explanation: The complex [Ni(CO)₄] has a tetrahedral geometry, indicating sp³ hybridization.
Q.31: Which of the following ligands can cause the most splitting in the d-orbitals of the central metal ion in an octahedral complex?
(a) H₂O
(b) NH₃
(c) CN⁻
(d) Cl⁻
Answer: (c) CN⁻
Explanation: CN⁻ is a strong field ligand and causes the maximum splitting of d-orbitals in an octahedral field.
Q.32: The color of [Ni(H₂O)₆]²⁺ is:
(a) Blue
(b) Green
(c) Yellow
(d) Red
Answer: (b) Green
Explanation: The complex [Ni(H₂O)₆]²⁺ is green in color due to d-d transitions.
Q.33: The IUPAC name of [Pt(NH₃)₄]²⁺ is:
(a) Tetraamminoplatinum(II)
(b) Tetraammineplatinum(II)
(c) Tetraamminoplatinum(IV)
(d) Tetraammineplatinum(IV)
Answer: (b) Tetraammineplatinum(II)
Explanation: According to IUPAC nomenclature, the name of the complex [Pt(NH₃)₄]²⁺ is Tetraammineplatinum(II).
Q.34: The coordination number of Cu in [Cu(NH₃)₄]²⁺ is:
(a) 2
(b) 4
(c) 6
(d) 8
Answer: (b) 4
Explanation: In [Cu(NH₃)₄]²⁺, Cu is coordinated to four ammonia ligands, making the coordination number 4.
Q.35: The complex [Ag(CN)₂]⁻ is:
(a) Paramagnetic
(b) Diamagnetic
(c) Ferromagnetic
(d) Antiferromagnetic
Answer: (b) Diamagnetic
Explanation: [Ag(CN)₂]⁻ is diamagnetic as it has no unpaired electrons in its electronic configuration.
Q.36: The IUPAC name of [CoCl₂(en)₂]⁺ is:
(a) Dichlorobis(ethylenediamine)cobalt(III) ion
(b) Dichlorobis(ethylenediamine)cobalt(II) ion
(c) Dichlorobis(ethylenediamine)cobalt(IV) ion
(d) Dichlorobis(ethylenediamine)cobalt(I) ion
Answer: (a) Dichlorobis(ethylenediamine)cobalt(III) ion
Explanation: According to IUPAC nomenclature, the name of the complex [CoCl₂(en)₂]⁺ is Dichlorobis(ethylenediamine)cobalt(III) ion.
Q.37: The geometry of [Cr(NH₃)₆]³⁺ is:
(a) Tetrahedral
(b) Square planar
(c) Octahedral
(d) Trigonal bipyramidal
Answer: (c) Octahedral
Explanation: [Cr(NH₃)₆]³⁺ has an octahedral geometry due to the arrangement of six ammonia molecules around the central chromium ion.
Q.38: The type of isomerism exhibited by [Pt(NH₃)₂Cl₂] is:
(a) Geometrical isomerism
(b) Optical isomerism
(c) Linkage isomerism
(d) Ionization isomerism
Answer: (a) Geometrical isomerism
Explanation: [Pt(NH₃)₂Cl₂] exhibits geometrical isomerism as the chloride ligands can be arranged in cis- or trans- positions.
Q.39: The hybridization of the central metal ion in [Zn(NH₃)₄]²⁺ is:
(a) sp³
(b) dsp²
(c) d²sp³
(d) sp²
Answer: (a) sp³
Explanation: The complex [Zn(NH₃)₄]²⁺ has a tetrahedral geometry, indicating sp³ hybridization.
Q.40: Which of the following complexes is paramagnetic?
(a) [Ni(CO)₄]
(b) [Fe(CN)₆]⁴⁻
(c) [CoF₆]³⁻
(d) [MnCl₆]⁴⁻
Answer: (c) [CoF₆]³⁻
Explanation: [CoF₆]³⁻ is paramagnetic as it has unpaired electrons in its electronic configuration.
Q.41: The effective atomic number (EAN) of [Ni(CO)₄] is:
(a) 34
(b) 35
(c) 36
(d) 37
Answer: (c) 36
Explanation: The EAN of [Ni(CO)₄] is 36, calculated as 28 (atomic number of Ni) + 4×2 (8 from 4 CO ligands) = 36.
Q.42: The ligand which causes the maximum crystal field splitting is:
(a) H₂O
(b) NH₃
(c) CN⁻
(d) F⁻
Answer: (c) CN⁻
Explanation: CN⁻ is a strong field ligand and causes maximum splitting of the d-orbitals in a crystal field.
Q.43: The type of hybridization in [Fe(CO)₅] is:
(a) sp³
(b) dsp²
(c) d²sp³
(d) sp³d
Answer: (d) sp³d
Explanation: [Fe(CO)₅] has a trigonal bipyramidal geometry which corresponds to sp³d hybridization.
Q.44: The color of [Co(H₂O)₆]²⁺ is:
(a) Blue
(b) Green
(c) Yellow
(d) Red
Answer: (a) Blue
Explanation: The complex [Co(H₂O)₆]²⁺ is blue in color due to d-d transitions.
Q.45: The IUPAC name of [Zn(NH₃)₄]²⁺ is:
(a) Tetraamminozinc(II) ion
(b) Tetraamminezinc(II) ion
(c) Tetraamminozinc(IV) ion
(d) Tetraamminezinc(IV) ion
Answer: (b) Tetraamminezinc(II) ion
Explanation: According to IUPAC nomenclature, the name of the complex [Zn(NH₃)₄]²⁺ is Tetraamminezinc(II) ion.
Q.46: The geometry of [CuCl₄]²⁻ is:
(a) Tetrahedral
(b) Square planar
(c) Octahedral
(d) Trigonal bipyramidal
Answer: (a) Tetrahedral
Explanation: [CuCl₄]²⁻ has a tetrahedral geometry due to the arrangement of four chloride ligands around the central copper ion.
Q.47: The type of isomerism exhibited by [Fe(NH₃)₂(CN)₄]⁻ is:
(a) Geometrical isomerism
(b) Optical isomerism
(c) Linkage isomerism
(d) Ionization isomerism
Answer: (a) Geometrical isomerism
Explanation: [Fe(NH₃)₂(CN)₄]⁻ exhibits geometrical isomerism due to different possible arrangements of the ligands around the central metal ion.
Q.48: The hybridization of the central metal ion in [Fe(CN)₆]⁴⁻ is:
(a) sp³
(b) dsp²
(c) d²sp³
(d) sp²
Answer: (c) d²sp³
Explanation: The complex [Fe(CN)₆]⁴⁻ has an octahedral geometry, indicating d²sp³ hybridization.
Q.49: Which of the following complexes is diamagnetic?
(a) [Fe(CN)₆]³⁻
(b) [Co(NH₃)₆]³⁺
(c) [Ni(CO)₄]
(d) [MnCl₆]⁴⁻
Answer: (c) [Ni(CO)₄]
Explanation: [Ni(CO)₄] is diamagnetic as it has no unpaired electrons in its electronic configuration.
Q.50: The IUPAC name of [Co(en)₃]³⁺ is:
(a) Tris(ethylenediamine)cobalt(III) ion
(b) Tris(ethylenediamine)cobalt(II) ion
(c) Tris(ethylenediamine)cobalt(IV) ion
(d) Tris(ethylenediamine)cobalt(I) ion
Answer: (a) Tris(ethylenediamine)cobalt(III) ion
Explanation: According to IUPAC nomenclature, the name of the complex [Co(en)₃]³⁺ is Tris(ethylenediamine)cobalt(III) ion.
