Important Questions Class 12 Chemistry Chapter 9

Important Questions Class 12 Chemistry Chapter 9

Important Questions for CBSE Class 12 Chemistry Chapter 9 – Coordination Compounds

Class 12 Chapter 9 Coordination Compounds discusses the various transition metal compounds. Some of the topics covered here include naming coordination compounds in the IUPAC system and determining the corresponding chemical formulas when given an IUPAC name. Class 12 Chemistry Chapter 9 also covers important terms like ligands, central atoms, coordination entities and coordination numbers.

The Important Questions for Class 12 Chemistry Chapter 9 Coordination Compounds presented here assist students in learning about “coordination compounds” and comprehending the concepts associated with them.

Students can use the Important Questions for Class 12 Chemistry Chapter 9 to understand the key topics and questions which can be asked in board exams to get good grades. Extramarks subject matter experts have developed these solutions in accordance with the most recent CBSE Syllabus 2022-23.

CBSE Class 12 Chemistry Chapter-9 Important Questions

Important Questions for Class 12 Chemistry Chapter 9 – Coordination Compounds

Chemistry is frequently regarded as a difficult subject, and students frequently struggle to achieve high marks in this subject due to the numerous equations and formulas. That is why it is critical to regularly practise all of the important questions and equations to gain a thorough understanding of the subject.

Students can access the entire set of important questions from the link given.

Very Short Answer Questions – 1 mark

Q1. What exactly is a coordination compound?

A. Coordination compounds are those that contain complicated ions and have a core metal atom or ion that is linked by coordinate bonds to a number of ions or neutral molecules.

Examples:K4[Fe(CN)6],[Cu(NH3)4]SO4,Ni(CO)4

Q2. Mention the IUPAC name of Tollens’ reagent.

A. Tollens’ reagent has the chemical formula Ag(NH) OH 3 2 and the IUPAC name Ammoniacal Silver Nitrate.

Q3. What is the IUPAC name for [PtCl2(NH2CH3)(NH3)2]Cl?

A. The IUPAC name of the given coordination compound is diamine chlorido methylamine platinum (II) chloride.

Q4. What is a chelate ligand? Give an example.

A. Chelating ligands are ligands that have a large number of donor sites and can form stable five- to six-membered rings with metals. Chelating ligands include ethylenediamine, acetylacetone and ethylenediaminetetraacetic acid (EDTA).

Q5. Name one homogeneous catalyst used in alkene hydrogenation.

A. The most thoroughly studied homogeneous hydrogenation catalyst used in the hydrogenation of alkenes is the four-coordinated rhodium complex Rh[(C6H5)3P]3Cl. G. is named after its discoverer. Wilkinson’s catalyst is the name given to this catalyst.

Q6. How many geometrical isomers can the tetrahedral complex Ni(CO)4 have?

Since the relative positions of the unidentate ligands linked to the central metal atom are identical, no geometrical isomers will form for the specified coordination complex.

Q7. Name the compound used to calculate the volumetric hardness of water.

A. The compound ethylenediaminetetraacetate is used to estimate the hardness of water using volumetric techniques (EDTA).

Short Answer Type Questions – 2 Mark

Q1. Give main postulates of  Werner’s Theory.

A. The main postulates of Werner’s Theory are as follows:

(i) The central metal ion represents two types of valencies, primary and secondary.

(ii) The primary valency of a compound is its oxidation state, and the secondary valency of a complex is its coordination number.

(iii) Because the secondary valency of each species is fixed, the coordination number is also fixed.

(iv) The metal atom fulfils both the primary and secondary valencies of a metal atom. A negative ion fulfils the main valency. Secondary valencies are filled by negative ions and neutral molecules.

Q2. With the help of an example, write two differences between a double salt and a coordination compound.

A. The distinction between a double salt and a coordination compound is as follows:

A double salt is a compound that consists of two distinct salt components. In water, they completely dissociate into their ions. Potash Alum K2 SO4.Al2 (SO4)3.24H2 O is an example.

A complex salt is a chemical composed of a core metal atom and ligands that form coordination bonds with it. Coordination compounds do not completely dissociate into their ions in water. Potassium ferrocyanide, K Fe(CN) 4 6.

Q3. The [Mn(H2 O)6]2+ ion contains five unpaired electrons, whereas the [Mn(CN)6]4- ion contains only one. Use the Crystal Field Theory to explain.

A. Mn2+ has the electronic configuration 3d5 4s0. In the case of hexaaquamanganese (II) ion, the ligand is water, which is a weak field ligand that is unable to couple up the electrons of the 3d subshell, and the configuration is t2g3 and eg2. The CN in the case of the hexacyano ion is a two strong field ligand that causes electron pairing in the d orbitals, leaving only one unpaired electron in t2g5 eg0.

Q4. Mention the applications of the following complex compounds:

         a. Silver, gold, or other noble metal electroplating

A. Complex ions of specific cations are formed for two reasons: To maintain the stability of the cation. In their pure state, some metal cations, such as gold are not stable. They become significantly more stable when complexed with a ligand.

        b. In photography

A. A key component of photographic film is light sensitive silver bromide. To make silver salts, sodium bromide is added to a silver nitrate solution. On the sensitive surface of ordinary film is a gelatin coating containing silver bromide crystals.

Q5. How does EDTA aid in the treatment of lead poisoning?

A. To remove lead from our bodies, we use a ligand that can form metal complexes with lead. The chemical EDTA is used to treat lead poisoning. The abbreviation for ethylene diamine tetraacetic acid (EDTA) is ethylene diamine tetraacetic acid. It contains two nitrogen atoms and four oxygen atoms, all of which can donate electrons to metals. In the body, lead replaces calcium in the Ca-EDTA complex. Urine excretes lead-EDTA, a more soluble complex.

Short Answer Type Questions – 3 mark 

Q1. Give an account for the following:

(i) [NiCl4]2 is paramagnetic, whereas Ni(CO)4 is diamagnetic, even though both 4 are tetrahedral.

A. Despite the fact that they are both tetrahedral, [NiCl4]2 is paramagnetic and Ni(CO)4 is diamagnetic. It has the structure 3d8 4s2, which indicates that it is in the zero oxidation state. CO, on the other hand, is an extremely potent field ligand. As a result, it is diamagnetic and induces the pairing of unpaired 3d electrons; however, because Cl is a weak ligand and is unable to pair up the unpaired electrons, [NiCl4]2 is paramagnetic.

(ii) [Ni(NH3)6]2+ is weakly paramagnetic, whereas Fe(H2O)6]3 is strongly paramagnetic.

A. In the [Ni(NH3)6]2+ complex ion, the ligand NH3 is neither a strong nor a weak field ligand. In fact, it is a weak strong field ligand. The ligand NH3 acts as a weak field ligand in the [Ni(NH3)6]2+ ion because the crystal field stabilisation energy is less than the pairing energy. As a result, under the influence of an octahedral crystal field, the electronic configuration is t2 g6eg2. According to the aforementioned electrical structure, the complex has two unpaired electrons. As a result, the [Ni(NH3)6]2+ complex has a weak paramagnetic character.

The oxidation state of Fe in the complex [Fe(H2O)6]3+ is +3, with the structure 3 d5. Water (H2O) is a weak ligand that prevents 3 d electrons from coupling. The sp3d2 hybridisation produces an octahedral complex with 5 unpaired electrons in the outer orbits. As a result, it becomes extremely paramagnetic.

(iii) [Co(NH3)6] 3+ belongs to the inner orbital complex, whereas [Ni(NH3)6]2+ belongs to the outer orbital complex.

A. Since the oxidation state of cobalt is +3 in [Co(NH3)6]3+, its d-orbital will have six electrons, resulting in a d 6 system. Because NH3 is a strong field ligand, the pairing will begin in three dimensions. Two subshells of 3 d, 4 s, and 4p empty orbitals are involved in the hybridisation. As a result, the hybridisation is d2sp3. As a result, [Co(NH3)6]3+ is an octahedral complex with an inner orbit.

The oxidation state of nickel in [Ni(NH3)6]2+ is +2, and its outer shell electronic configuration is 3 d84 s0 due to the presence of two unpaired electrons in the 3 d subshell. The hybridisation involves empty 4 s, 4 p, and 4 d orbitals, which is why it is an outer orbital octahedral complex.

Q2. A metal complex with the formula Cr(NH3)4 Cl2 Br has been isolated in two forms, ‘A’ and ‘B.’ When form ‘A’ reacts with AgNO3 solution, it produces a white precipitate that is readily soluble in dilute aqueous ammonia, whereas form ‘B’ produces a pale yellow precipitate that is soluble in concentrated ammonia solution. Write the ‘A’ and ‘B’ formulas. Mention the isomerism that occurs between ‘A’ and ‘B.’

A. Compound A’sformula is [Cr(NH3)4ClBr]Cl. It reacts with AgNO3 as follows:

[Cr(NH3)4ClBr]Cl+AgNO3→AgCl+[Cr(NH3)4ClBr]+NO−3

Silver chloride is the white precipitate (AgCl). This white silver chloride dissolves in ammonia solution as follows:

AgCl+2NH4OH→[Ag(NH3)2Cl]+2H2O

[Cr(NH3)4Cl2]Br is compound B. It reacts with AgNO3 as follows:

[Cr(NH3)4Cl2]Br+AgNO3→AgBr+[Cr(NH3)4Cl2]+NO−3

Silver bromide (AgBr) precipitate is formed, which is soluble in concentrated ammonia solution:

AgBr+2NH4OH→[Ag(NH3)4Br]+2H2O

As a result, compound A is 3-4 Cr(NH)ClBr. Cland compound B has the formula 3 4 2 Cr(NH)ClBr. They are both ionisation isomers of one another.