Coordination Compounds
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When 1 mol of CrCl3.6H2O is treated with excess of AgNO3, 3 mol of AgCl is obtained. The formula of the complex is
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[CrCl3(H2O)3].3H2O
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[CrCl2(H2O)4]Cl.2H2O
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[CrCl(H2O)5]Cl2.H2O
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[Cr(H2O)6]Cl3
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Coordination complexes in chemistry consist of atoms, which are metallic in nature and an array of boundless ions that are known as complexing agents. Find out about this and more through the quiz below. All the best.
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- 2.
The compounds [Co(SO4)(NH3)5]Br and [Co(SO4)(NH3)5]Cl represent:
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Linkage isomerism
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Ionisation isomerism
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Coordination isomerism
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No isomerism
Correct Answer
A. No isomerismExplanation
The compounds [Co(SO4)(NH3)5]Br and [Co(SO4)(NH3)5]Cl do not represent any form of isomerism. Isomerism refers to the existence of different compounds with the same molecular formula but different structural arrangements or bonding patterns. In this case, both compounds have the same molecular formula and the same coordination sphere around the central cobalt ion. The only difference is the counterion, which is Br- in one compound and Cl- in the other. Since the coordination sphere and the overall structure of the compounds remain unchanged, there is no isomerism present.Rate this question:
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- 3.
2,4- Dinitrophenyl hydrazine is an example for:
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Tridentate ligand
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Monodentate ligand
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Polydentate ligand
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Didentate ligand
Correct Answer
A. Monodentate ligandExplanation
Monodentate ligands are ligands that can form only one bond with a metal ion. In the case of 2,4-Dinitrophenyl hydrazine, it can form a single bond with a metal ion through one of its functional groups. Therefore, it is classified as a monodentate ligand.Rate this question:
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- 4.
Which of the following complex ion shows geometrical isomerism?
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[Cr(H2O)4Cl2]+
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[Pt(NH3)3Cl]
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[Co(NH3)6]3+
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[Co(CN)5(NC)]3-
Correct Answer
A. [Cr(H2O)4Cl2]+Explanation
The complex ion [Cr(H2O)4Cl2]+ shows geometrical isomerism. This is because the two chloride ligands can be either cis (on the same side) or trans (on opposite sides) to each other. This results in two possible isomers, cis-[Cr(H2O)4Cl2]+ and trans-[Cr(H2O)4Cl2]+, which have different spatial arrangements.Rate this question:
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- 5.
The oxidation state of Fe in the brown ring complex [Fe(H2O)5NO]SO4 is:
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+ 1
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+ 2
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+ 3
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+ 4
Correct Answer
A. + 1 -
- 6.
Which of the following can exhibit geometrical isomerism?
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[MnBr4]2-
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[Pt(NH3)3Cl]+
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[PtCl2P(C2H5)3]2
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[Fe(H2O)5NO]2+
Correct Answer
A. [PtCl2P(C2H5)3]2Explanation
The compound [PtCl2P(C2H5)3]2 can exhibit geometrical isomerism because it contains a square planar coordination geometry around the platinum atom. In this compound, the two ethyl groups (C2H5) can be arranged either cis or trans to each other. This results in two different isomers with distinct spatial arrangements. The other compounds listed do not have the necessary structural features to exhibit geometrical isomerism.Rate this question:
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- 7.
According to IUPAC nomenclature, sodium nitroprusside is named as:
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Sodium nitroferricyanide
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Sodium nitroferrocyanide
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Sodium pentacyano nitrosylferrate (II)
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Sodium pentacyano nitrosylferrate (III)
Correct Answer
A. Sodium pentacyano nitrosylferrate (III)Explanation
According to IUPAC nomenclature, the compound sodium nitroprusside is named as Sodium pentacyano nitrosylferrate (III). This name indicates the composition and oxidation state of the compound. "Sodium" refers to the presence of sodium ions, "pentacyano" indicates the presence of five cyanide ligands, "nitrosyl" indicates the presence of a nitrosyl ligand, and "ferrate (III)" indicates the oxidation state of iron in the compound.Rate this question:
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- 8.
Hybridisation and shape of K3[Co(CO3)3] is:
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D2sp3and Octahedral
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Sp3d2 and Octahedral
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Dsp2 and square planar
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Sp3 and tetrahedral
Correct Answer
A. Sp3d2 and OctahedralExplanation
The hybridization and shape of K3[Co(CO3)3] is sp3d2 and Octahedral. In this compound, the central cobalt atom is bonded to six ligands, three of which are carbonate ions (CO3) and three are potassium ions (K). The cobalt atom undergoes hybridization to form six hybrid orbitals, consisting of one s orbital, three p orbitals, and two d orbitals. These hybrid orbitals then overlap with the ligand orbitals to form six sigma bonds, resulting in an octahedral shape.Rate this question:
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- 9.
Total number of possible isomers for the complex compound [CuII(NH3)4][PtIICl4] are:
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3
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4
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5
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6
Correct Answer
A. 4Explanation
The complex compound [CuII(NH3)4][PtIICl4] consists of two parts: the copper complex [CuII(NH3)4] and the platinum complex [PtIICl4]. The copper complex contains a central copper ion surrounded by four ammonia ligands, while the platinum complex contains a central platinum ion surrounded by four chloride ligands. The number of possible isomers for each complex can be determined using the coordination number and geometries of the ligands. For the copper complex, the coordination number is 4 and the ligands are all the same, resulting in only one possible isomer. For the platinum complex, the coordination number is also 4, but the ligands are different, allowing for different possible arrangements. Therefore, the total number of possible isomers for the complex compound is 1 * number of isomers for the platinum complex, which is 4.Rate this question:
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- 10.
Square planar complexes of the type MABXL (where A, B, X and L are unidentates) show:
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Two cis and one trans isomer
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Two trans and one cis isomer
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Two cis and two trans isomer
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One cis and one trans isomer
Correct Answer
A. Two cis and one trans isomerExplanation
Square planar complexes have a square planar geometry, meaning that there are four ligands arranged around the central metal atom in a flat plane. In this case, the complex is MABXL, where A, B, X, and L are unidentate ligands. The two cis isomers refer to the arrangement of the ligands where two of them are located on the same side of the square plane, while the other two are on the opposite side. The trans isomer refers to the arrangement where two ligands are located opposite to each other across the square plane. Therefore, the correct answer is two cis and one trans isomer.Rate this question:
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- 11.
In spectrochemical series, chlorine is above water i.e., Cl > H2O, this is due to:
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Good π-acceptor properties of Cl
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Strong σ-donor and good π-acceptor properties of Cl
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Good π-donor properties of Cl
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Larger size of Cl than H2O
Correct Answer
A. Good π-donor properties of ClExplanation
The correct answer is "Good π-donor properties of Cl." In the spectrochemical series, the position of a ligand is determined by its ability to donate electron density to the metal center. Chlorine has good π-donor properties, meaning it can donate electron density to the metal through its p-orbitals. This makes it a strong ligand and places it above water in the spectrochemical series.Rate this question:
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- 12.
Which of the following shall form an octahedral complex?
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D4 (low spin)
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D8 (high spin)
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D6 (low spin)
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All of these
Correct Answer
A. D8 (high spin)Explanation
An octahedral complex refers to a coordination compound with six ligands surrounding the central metal ion, forming an octahedral shape. In the given options, the d8 (high spin) configuration is the only one that can form an octahedral complex. This is because d8 refers to a metal ion with eight d-electrons, which allows for the formation of six coordination bonds with ligands. The high spin configuration indicates that the electrons will occupy the available d-orbitals in a way that maximizes their spin, resulting in a high spin complex. Therefore, d8 (high spin) is the correct answer.Rate this question:
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- 13.
Which of the following is diamagnetic in nature?
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Co3+, octahedral complex with weak field ligand
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Co3+, octahedral complex with strong field ligand
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Co2+ in tetrahedral complex
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Co2+ in square planar complex
Correct Answer
A. Co3+, octahedral complex with strong field ligandExplanation
Diamagnetic substances are those that do not have any unpaired electrons and are not attracted to a magnetic field. In octahedral complexes, strong field ligands cause pairing of electrons in the d-orbitals, resulting in all electrons being paired and making the complex diamagnetic. Therefore, Co3+ in an octahedral complex with a strong field ligand is diamagnetic.Rate this question:
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- 14.
The value of n in the carbonyl: (CO)n – Co – Co – (CO)n
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4
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5
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6
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7
Correct Answer
A. 4Explanation
The given structure represents a carbonyl compound, with CO groups attached to a central metal atom Co. The number of CO groups on each side of the metal atom is represented by the value of n. In this case, the structure shows that there are 4 CO groups on each side of the metal atom, resulting in a value of n equal to 4.Rate this question:
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- 15.
If the length of CO bond in carbon monoxide is 1.128 Ǻ, then what is the value of CO bond length in Fe(CO)5?
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1.15 Ǻ
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1.128 Ǻ
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1.72 Ǻ
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1.118 Ǻ
Correct Answer
A. 1.15 ǺExplanation
The correct answer is 1.15 Ǻ. In Fe(CO)5, the CO bond length will be slightly longer than the CO bond length in carbon monoxide alone. This is because the presence of the iron atom in Fe(CO)5 can slightly weaken the CO bond, causing it to stretch and increase in length. Therefore, the CO bond length in Fe(CO)5 is slightly longer than 1.128 Ǻ, but not as long as 1.72 Ǻ or 1.118 Ǻ. The closest option is 1.15 Ǻ, which is the correct answer.Rate this question:
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Current Version
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Mar 21, 2023Quiz Edited by
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Dec 19, 2013Quiz Created by
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