Here are 20 multiple-choice questions (MCQs) from the chapter "Coordination Compounds" (Class 12 NCERT):
1. Which of the following ligands is a bidentate ligand?
a) CO
b) Cl⁻
c) Ethylenediamine (en)
d) NH₃
2. The oxidation state of the central metal in [Fe(CN)₆]³⁻ is:
a) +3
b) +2
c) +6
d) +4
3. The coordination number of cobalt in [Co(NH₃)₆]³⁺ is:
a) 3
b) 4
c) 5
d) 6
4. Which of the following is not a chelating ligand?
a) Oxalate
b) Ethylenediamine
c) Acetate
d) Glycine
5. The geometrical shape of [PtCl₂(NH₃)₂] is:
a) Tetrahedral
b) Square planar
c) Octahedral
d) Trigonal bipyramidal
6. The IUPAC name of [Co(NH₃)₅Cl]Cl₂ is:
a) Pentamminechloridocobalt(III) chloride
b) Chloropentamminecobalt(III) chloride
c) Hexaamminecobalt(III) chloride
d) Pentaamminecobalt(II) chloride
7. What is the effective atomic number (EAN) of cobalt in [Co(NH₃)₆]³⁺?
a) 36
b) 33
c) 34
d) 35
8. In [Fe(CN)₆]⁴⁻, the hybridization of Fe is:
a) sp³d
b) dsp²
c) d²sp³
d) sp²
9. Which of the following statements is correct for a high-spin octahedral complex?
a) It has a large crystal field splitting energy.
b) It involves pairing of electrons.
c) It shows maximum unpaired electrons.
d) It is low-spin complex.
10. The splitting of d-orbitals in tetrahedral complexes leads to:
a) Three orbitals with lower energy
b) Two orbitals with lower energy
c) Equal energy of all orbitals
d) Four orbitals with higher energy
11. Which of the following ions is colorless?
a) [Fe(H₂O)₆]³⁺
b) [Cu(H₂O)₆]²⁺
c) [Sc(H₂O)₆]³⁺
d) [CoCl₄]²⁻
12. In crystal field theory, Δ₀ stands for:
a) Ligand field stabilization energy
b) Bond dissociation energy
c) Crystal field splitting energy
d) Ionization energy
13. Which of the following complex ions will have maximum paramagnetism?
a) [Fe(CN)₆]³⁻
b) [Co(NH₃)₆]³⁺
c) [Mn(CN)₆]³⁻
d) [Cu(NH₃)₄]²⁺
14. The hybridization and geometry of [Ni(CO)₄] are:
a) sp³ and tetrahedral
b) dsp² and square planar
c) sp² and trigonal planar
d) d²sp³ and octahedral
15. The metal ion present in chlorophyll is:
a) Fe
b) Co
c) Mg
d) Zn
16. In a coordination complex, the donor atom is:
a) Always a metal
b) Always non-metal
c) The ligand atom donating the electron pair
d) None of the above
17. Which of the following is not a common ligand in coordination chemistry?
a) CN⁻
b) CO
c) H₂O
d) Na⁺
18. The crystal field stabilization energy (CFSE) for [Co(H₂O)₆]²⁺ (a weak field ligand complex) is:
a) 0 Δ₀
b) 2.4 Δ₀
c) 1.6 Δ₀
d) 1.2 Δ₀
19. The term “ambidentate ligand” refers to a ligand that:
a) Can donate two pairs of electrons
b) Can attach through two different atoms
c) Can form two complexes simultaneously
d) Can form bonds with two metal atoms
20. Which of the following statements is correct for a square planar complex?
a) It always has a coordination number of 4.
b) It is formed by sp³ hybridization.
c) It is most commonly found in 3d-series metals.
d) It is always diamagnetic.
These questions cover various aspects of coordination compounds such as ligands, coordination numbers, geometries, and crystal field theory.
Here are the answers to the MCQs on "Coordination Compounds":
1. c) Ethylenediamine (en)
2. a) +3
3. d) 6
4. c) Acetate
5. b) Square planar
6. a) Pentamminechloridocobalt(III) chloride
7. a) 36
8. c) d²sp³
9. c) It shows maximum unpaired electrons
10. b) Two orbitals with lower energy
11. c) [Sc(H₂O)₆]³⁺
12. c) Crystal field splitting energy
13. c) [Mn(CN)₆]³⁻
14. a) sp³ and tetrahedral
15. c) Mg
16. c) The ligand atom donating the electron pair
17. d) Na⁺
18. a) 0 Δ₀
19. b) Can attach through two different atoms
20. a) It always has a coordination number of 4
These answers align with the NCERT syllabus for Class 12, Chemistry, Chapter "Coordination Compounds."
Here are the explanations for the answers to the MCQs on "Coordination Compounds":
1. c) Ethylenediamine (en)
Ethylenediamine (en) is a bidentate ligand because it has two donor atoms (nitrogen atoms), both of which can donate electron pairs to form coordinate bonds with a central metal atom/ion.
2. a) +3
In the complex ion [Fe(CN)₆]³⁻, cyanide (CN⁻) is a monodentate ligand with a charge of -1. Since the overall charge of the complex is -3, and there are 6 cyanide ions, the oxidation state of Fe must be +3 to balance the charge.
3. d) 6
In [Co(NH₃)₆]³⁺, there are six ammonia (NH₃) ligands bound to the central cobalt ion. Therefore, the coordination number of cobalt is 6.
4. c) Acetate
Acetate (CH₃COO⁻) is a monodentate ligand because it binds to the central metal atom through only one oxygen atom, while other ligands like oxalate and ethylenediamine are bidentate, forming chelates by using two donor atoms.
5. b) Square planar
The complex [PtCl₂(NH₃)₂] has a square planar geometry, which is typical for d⁸ complexes (such as platinum) in the presence of strong field ligands. The platinum ion undergoes dsp² hybridization to form this square planar structure.
6. a)Pentamminechloridocobalt(III) chloride
In the IUPAC name for [Co(NH₃)₅Cl]Cl₂, the prefix "pentammine" refers to five ammonia (NH₃) ligands, "chloro" refers to the chlorido ligand in the coordination sphere, and the oxidation state of cobalt is +3, indicated as cobalt(III).
7. a) 36
The effective atomic number (EAN) is calculated by adding the number of electrons in the metal ion to the number of electrons donated by the ligands. For [Co(NH₃)₆]³⁺, cobalt has an atomic number of 27, loses 3 electrons, leaving 24 electrons, and each ammonia ligand donates 2 electrons (6 × 2 = 12). Therefore, EAN = 24 + 12 = 36.
8. c) d²sp³
In [Fe(CN)₆]⁴⁻, the cyanide ion is a strong field ligand, causing pairing of electrons in the lower-energy d orbitals of Fe²⁺. This results in d²sp³ hybridization, giving an octahedral geometry.
9. c) It shows maximum unpaired electrons
In a high-spin octahedral complex, the crystal field splitting energy (Δ₀) is small, so electrons occupy the higher-energy eg orbitals rather than pairing in the lower t₂g orbitals. This results in the maximum number of unpaired electrons, characteristic of a high-spin complex.
10. b) Two orbitals with lower energy
In a tetrahedral crystal field, the d-orbitals split into two sets: two orbitals (eg) with lower energy and three orbitals (t₂g) with higher energy, the reverse of the splitting in octahedral complexes.
11. c) [Sc(H₂O)₆]³⁺
Scandium (Sc³⁺) has an empty d-orbital configuration (3d⁰), meaning it cannot absorb visible light, resulting in a colorless complex. The other complexes have partially filled d-orbitals, leading to color due to d-d transitions.
12. c) Crystal field splitting energy
Δ₀ represents the crystal field splitting energy, which is the energy difference between the t₂g and eg sets of d-orbitals in an octahedral field.
13. c) [Mn(CN)₆]³⁻
Manganese in [Mn(CN)₆]³⁻ has a +3 oxidation state and a d⁴ configuration. Since cyanide is a strong field ligand, it causes electron pairing, but the complex will still have unpaired electrons, making it paramagnetic. The number of unpaired electrons is higher compared to other options.
14. a) sp³ and tetrahedral
In [Ni(CO)₄], nickel is in the zero oxidation state with an sp³ hybridization, leading to a tetrahedral geometry. CO is a strong field ligand that results in this specific geometry.
15. c) Mg
The central metal ion in chlorophyll is magnesium (Mg²⁺), which is essential for photosynthesis in plants.
16. c) The ligand atom donating the electron pair
In coordination compounds, the donor atom is the atom in the ligand that donates an electron pair to form a coordinate bond with the metal ion.
17. d) Na⁺
Na⁺ is not a ligand, as it is a cation that cannot donate electron pairs. Ligands such as CN⁻, CO, and H₂O donate electron pairs to form coordinate bonds with metal ions.
18. a) 0 Δ₀
In a weak field complex like [Co(H₂O)₆]²⁺, the crystal field stabilization energy (CFSE) is very small, and with no electron pairing, the CFSE can be effectively considered as 0 Δ₀.
19. b) Can attach through two different atoms
An ambidentate ligand can bind to a metal atom through two different atoms. For example, SCN⁻ can bind through sulfur (S) or nitrogen (N).
20. a) It always has a coordination number of 4
Square planar complexes generally have a coordination number of 4. This geometry is common in d⁸ metal ions, such as platinum(II), palladium(II), and nickel(II).
No comments:
Post a Comment