Coordination Compounds - NEET/JEE/KEAM
1) Each correct answers have 4 Marks
2) Each wrong answers have -1 Mark
3) Total Time - 15 Minutes
Questions and Answers
- 1.
The name of complex ion, [Fe(CN)6]3- is,
- A.
Hexacyanoiron (III) ion
- B.
Hexacyanitoferrate (III) ion
- C.
Tricyanoferrate (III) ion
- D.
Hexacyanidoferrate (III) ion
Correct Answer
D. Hexacyanidoferrate (III) ionExplanation
The complex ion [Fe(CN)6]3- consists of a central iron (III) ion (Fe3+) surrounded by six cyanide (CN-) ligands. The name "Hexacyanidoferrate (III) ion" accurately describes the composition and charge of the complex. "Hexa" refers to the six cyanide ligands, "cyanido" indicates the presence of the CN- ligands, "ferrate" indicates the central iron ion, and "(III)" represents the charge of the ion. This name follows the IUPAC naming rules for coordination compounds.Rate this question:
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- 2.
An excess of AgNO​​​3​​ is added to 100 ml of a 0.01 M solution of dichlorotetraaquachromium (III) chloride. The number of moles of AgCl precipitated would be:
- A.
0.002
- B.
0.003
- C.
0.01
- D.
0.001
Correct Answer
D. 0.001Explanation
When excess AgNO3 is added to dichlorotetraaquachromium (III) chloride, a precipitation reaction occurs, forming AgCl. The balanced equation for this reaction is: CrCl3 + 4AgNO3 → Cr(NO3)3 + 3AgClFrom the balanced equation, we can see that for every 1 mole of CrCl3, 3 moles of AgCl are formed. Since the initial solution contains 0.01 moles of CrCl3 (0.01 M), the number of moles of AgCl precipitated would be 3 times that, which is 0.03 moles. However, since there is an excess of AgNO3, all the CrCl3 will react, and the limiting reagent is CrCl3. Therefore, the number of moles of AgCl precipitated would be 0.01 moles (3 times 0.01), which is equal to 0.001.Rate this question:
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- 3.
The complex, [Pt(Py) (NH​​​​​3​)BrCl] will have how many geometrical isomers?
- A.
3
- B.
4
- C.
0
- D.
2
Correct Answer
A. 3Explanation
The complex [Pt(Py)(NH​​​​​3​)BrCl] contains two different ligands, Py and NH​​​​​3​, which can occupy different positions around the central Pt atom. Since there are two different ligands, there can be two different arrangements of these ligands, resulting in two geometrical isomers. Additionally, the Br and Cl ligands can also interchange positions, resulting in another isomer. Therefore, the complex [Pt(Py)(NH​​​​​3​)BrCl] can have a total of three geometrical isomers.Rate this question:
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- 4.
Oxidation number of Ni in [Ni (C​​​​​​2​​​​O​​​​4)3]4-
- A.
3
- B.
4
- C.
2
- D.
6
Correct Answer
C. 2Explanation
The oxidation number of Ni in [Ni(C2O4)3]4- is 2. This is because the overall charge of the complex ion is 4-, and since there are three C2O4 ligands, each with a charge of 2-, the total charge contributed by the ligands is 6-. Therefore, the Ni ion must have a charge of +2 to balance out the overall charge of the complex ion.Rate this question:
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- 5.
The total number of possible isomers for the complex compound [Cu​​​​​II​​​ (NH​​​​3)4] [Pt​​​II Cl​​​​​4]
- A.
3
- B.
6
- C.
5
- D.
4
Correct Answer
D. 4Explanation
The complex compound [Cu​​​​​II​​​ (NH​​​​3)4] [Pt​​​II​​ Cl​​​​​4] consists of two coordination complexes, one with copper and the other with platinum. The copper complex has four ammonia ligands, while the platinum complex has four chloride ligands. The number of possible isomers for each complex can be determined using the formula 2^n, where n is the number of ligands. For the copper complex, there are 2^4 = 16 possible isomers. However, since the ligands are all the same (ammonia), these isomers are not distinguishable. Therefore, there is only one isomer for the copper complex. For the platinum complex, there are 2^4 = 16 possible isomers, but again, since the ligands are all the same (chloride), there is only one isomer. Therefore, the total number of possible isomers for the complex compound is 1 x 1 = 1.Rate this question:
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- 6.
The anion of acetylacetone (acac) forms Co(acac)3 chelate with Co​​​​​3+. The rings of the chelate are,
- A.
Three membered
- B.
Five membered
- C.
Four membered
- D.
Six membered
Correct Answer
D. Six memberedExplanation
The anion of acetylacetone (acac) forms a six-membered chelate with Co3+. This is because the acetylacetone molecule has two carbonyl groups that can act as donor atoms to bind with the metal ion. The resulting chelate ring is formed by coordinating the metal ion with both carbonyl groups and additional ligands. In this case, the chelate formed with Co3+ has a six-membered ring structure.Rate this question:
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- 7.
Red precipitate is obtained when ethanol solution of dimethylglyoxime is added to ammoniacal Ni(II). Which of the following statements is not true?
- A.
Red complex has a sugar planar geometry
- B.
Complex has symmetrical H-bonding
- C.
Red complex has a tetrahedral geometry
- D.
Dimethylglyoxime functions as bidentate ligand
Correct Answer
C. Red complex has a tetrahedral geometryExplanation
The red complex formed when ethanol solution of dimethylglyoxime is added to ammoniacal Ni(II) does not have a tetrahedral geometry. This is because the coordination of dimethylglyoxime as a bidentate ligand results in a planar geometry for the complex. The statement "Red complex has a sugar planar geometry" is true and refers to the planar arrangement of the ligand around the central Ni(II) ion. The statement "Complex has symmetrical H-bonding" is also true, indicating the presence of hydrogen bonding within the complex. Therefore, the only statement that is not true is "Red complex has a tetrahedral geometry".Rate this question:
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- 8.
CN​​​​​-​​​​ is a strong field ligand. This is due to the fact that,
- A.
It carries -ve charge
- B.
It is a puseudohalide
- C.
It can accept electrons from metal species
- D.
It forms high spin complexes with metal species
Correct Answer
B. It is a puseudohalideExplanation
CN- is a strong field ligand because it is a pseudohalide. Pseudohalides are ligands that have similar properties to halides, such as their ability to form coordination complexes with metal species. In the case of CN-, it can form strong bonds with metal ions by donating its lone pair of electrons. This results in a high degree of electron pairing and a low spin state in the metal complex, making it a strong field ligand.Rate this question:
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- 9.
The number of unpaired electrons in the complex ion [CoF​6]3- is (Atomic no: Co= 27)
- A.
0
- B.
2
- C.
3
- D.
4
Correct Answer
D. 4Explanation
The complex ion [CoF6]3- contains a cobalt ion (Co3+) surrounded by six fluoride ions (F-). Cobalt has an atomic number of 27, which means it has 27 electrons. The cobalt ion has a +3 charge, indicating it has lost 3 electrons. Each fluoride ion contributes 1 electron, so the total number of electrons contributed by the fluoride ions is 6. Since there are 3 fewer electrons than the total number of electrons in the neutral cobalt atom, there are 3 unpaired electrons in the complex ion.Rate this question:
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- 10.
The number of unpaired electrons in the complex [Cr(NH​​​​​3)6] Br​​​​​3​​​​ is (Atomic number, Cr= 24)
- A.
4
- B.
1
- C.
2
- D.
3
Correct Answer
D. 3Explanation
The complex [Cr(NH3)6]Br3 contains a central chromium atom surrounded by six ammonia ligands and three bromide ions. The coordination number of chromium in this complex is 6, which means that it can form six bonds with the surrounding ligands. Each ammonia ligand donates a pair of electrons to form a coordinate bond with the chromium atom. Since there are six ammonia ligands, there are a total of 12 electrons donated to the chromium atom. The chromium atom itself has 24 electrons, so subtracting the 12 donated electrons gives us 12 unpaired electrons. Therefore, the correct answer is 3.Rate this question:
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- 11.
Which of the following is common donor atom in ligands?
- A.
A) Arsenic
- B.
B) Nitrogen
- C.
C) Oxygen
- D.
D) Both 'b' and 'c'
Correct Answer
D. D) Both 'b' and 'c'Explanation
Both nitrogen and oxygen are common donor atoms in ligands. This means that they can donate a pair of electrons to form a coordinate bond with a metal ion. Nitrogen is commonly found in ligands such as ammonia (NH3) and amines, while oxygen is commonly found in ligands such as water (H2O) and alcohols. Therefore, option d) is the correct answer.Rate this question:
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- 12.
The complex ion [Co (NH​​​​​3)6]3+ is formed by sp​​​​​3​​​​d​2 hybridisation. Hence the ion should possess
- A.
Octahedral geometry
- B.
Tetrahedral geometry
- C.
Square planner geometry
- D.
Tetragonal geometry
Correct Answer
A. Octahedral geometryExplanation
The complex ion [Co(NH3)6]3+ is formed by sp3d2 hybridization. This means that the central cobalt atom is surrounded by six ligands, which are ammonia molecules. In sp3d2 hybridization, the central atom forms six sigma bonds and has no lone pairs of electrons. This arrangement of bonds and no lone pairs corresponds to an octahedral geometry, where the ligands are arranged around the central atom in a symmetrical manner. Therefore, the correct answer is octahedral geometry.Rate this question:
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- 13.
An eg: of a sigma bonded organometallic compound is:
- A.
Grignard's reagent
- B.
Ferrocene
- C.
Cobaltocene
- D.
Ruthenocene
Correct Answer
A. Grignard's reagentExplanation
Grignard's reagent is an example of a sigma bonded organometallic compound. It is formed by the reaction of an alkyl or aryl halide with magnesium metal. The resulting compound contains a carbon-metal bond, with the carbon atom bonded to the metal atom through a sigma bond. Grignard's reagent is widely used in organic synthesis for the formation of new carbon-carbon bonds.Rate this question:
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- 14.
Which of the following has longest C-O bond length? (Free C-O bond length in CO is 1.128A)
- A.
Ni (CO)4
- B.
[Co (CO)4]-
- C.
[Fe (CO)4]2-
- D.
[Mn (CO)6]+
Correct Answer
C. [Fe (CO)4]2-Explanation
The [Fe (CO)4]2- ion has the longest C-O bond length. This is because the Fe(II) ion has a larger ionic radius compared to Ni(0), Co(0), and Mn(II) ions. The larger ionic radius of Fe(II) allows for a greater separation between the Fe atom and the oxygen atoms in the CO ligands, resulting in longer C-O bond lengths.Rate this question:
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- 15.
Which of the following carbonyls will have the strongest C-O bond?
- A.
Mn (CO)6+
- B.
Cr (CO)6
- C.
V (CO)6-
- D.
Fe (CO)5
Correct Answer
A. Mn (CO)6+Explanation
The strength of a C-O bond is determined by the electronegativity of the central metal atom. In this case, Mn (manganese) has the highest electronegativity among the given options, making it capable of forming a stronger bond with oxygen. Therefore, Mn (CO)6+ will have the strongest C-O bond.Rate this question:
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