Do you know why blood appears red and emerald appears deep blue, what makes medicines effective against certain diseases, and how vitamin B12 helps in metabolism? The answers to these numerous questions lie in coordination compounds. These compounds are made up of a central metal atom or ion bonded to surrounding molecules or ions, known as ligands. These ligands define their structure and reactivity in fascinating ways.
NCERT Solutions for Class 12 Chemistry Chapter 5 Coordination compounds
In this article, students can find the NCERT Class 12 Chemistry Solutions. Our subject experts design them to provide a systematic and structured approach to these important concepts, helping students develop a clear understanding of the critical concepts. These NCERT solutions are a valuable resource to enhance candidates performance in board exams as well as in competitive exams like NEET, JEE, etc. The coordination compounds ncert solutions is important from the exam point of view as it carries good weightage in CBSE board exams. Below, we have provided selected HOTS questions as well. It will enhance conceptual clarity and problem-solving ability.
NCERT Solutions for Class 12 Chemistry Chapter 5: Download PDF
Students can download the class 12 chemistry chapter 5 coordination compounds solutions pdf for free by clicking on the provided link. These solutions are designed to help you understand the fundamental concepts and solve textbook questions with ease.
NCERT Solutions for Class 12 Chemistry Chapter 5 Coordination Compounds (Intext Questions From 5.1 to 5.10 )
These solutions will help in better concept clarity and exam preparation. These class 12 chemistry chapter 5 coordination compounds question answer provide clear and accurate solutions, helping students understand the fundamental concepts of coordination chemistry. These solutions of NCERT simplify complex topics like nomenclature, bonding, and isomerism, making it easier to grasp and apply in exams.
In both compounds, the oxidation state of Nickel is +2. So it has d8 configuration.
Now, on the basis of ligand pairing of electrons will occur. Since CN - is a strong ligand so pairing will occur, but in case of Cl - pairing will not be there as it is a weak ligand. So, the configuration of both the compounds looks like :-
Thus $[Ni(CN)_{4}]^{2-}$ is a square planer and diamagnetic and $[NiCl_{4}]^{2-}$ has tetrahedral geometry and is paramagnetic.
The difference in the magnetic behaviour is due to the nature of ligands present. In case of $[NiCl_{4}]^{2-}$ the oxidation state of nickel is +2 and also Cl - is a weak ligand. Thus its configuration becomes:-
So it is paramagnetic and tetrahedral in nature.
In the case of $[Ni(CO)_{4}]$ , the oxidation state of nickel is 0. So its configuration is 3d 8 4s2 . We also know that CO is a strong ligand, thus the configuration of nickel becomes:-
Hence the given compound is diamagnetic but tetrahedral in nature.
The oxidation state of Pt in the given compound is +2. Also, it is given that the compound has square planar geometry, i.e., it has dsp2 hybridisation (d8).
CN - is a strong ligand so the pairing of electron will occur.
So there are no unpaired electrons in the given compound.
Consider Hexaqua manganese (II) :- In this compound, the oxidation state of Mn is +2, and its electronic configuration is d5.
H2O is a weak ligand and the crystal field is octahedral, so the arrangement of electrons will be t2g3 eg2.
So the total number of unpaired electrons is 5.
Now consider hexacyanoion:- In this compound, the oxidation state of Mn is +2. It is surrounded by the strong ligands CN-, so pairing will be there.
Its arrangement will be t2g5 eg0 .
Thus, the number of unpaired electrons will be 1.
NCERT Solutions for Class 12 Coordination Compounds (Exercise Questions )
These class 12 chemistry chapter 5 coordination compounds solutions offer detailed step by step answers to all the exercise questions from the chapter on coordination compounds. They help students strengthen their understanding of key concepts, practise problem-solving, and prepare thoroughly for board and competitive exams.
Werner his theory of coordination compounds and gave some postulates. The main postulates are:
1. In coordination compounds, metals show two types of linkages or valences, namely primary valency and secondary valency.
2. The primary valences are generally ionisable and are satisfied or balanced by negative ions.
3. The secondary valences are non-ionisable. These are satisfied by either neutral molecules or by negative ions. The secondary valence is equal to the coordination number (No. of atoms surrounding the metal) and is constant for a metal.
4. According to different coordination numbers, the ions/groups bound by the secondary linkages to the metal have characteristic spatial arrangements.
The major difference between both the compounds is that the first compound is a salt and the other one is a coordination compound. In case of double salt compounds (Mohr's salt), the compound breaks into its constituent ions when dissolved in water, therefore it gives a positive test for the presence of Fe+2. But in case of coordination compounds, they maintain their identity in both solid and dissolved state. Thus the individual property of each constituent is lost. And therefore it doesn't give a positive test for Cu+2.
(i) Coordination entity:- It is an electrically charged species carrying either a positive charge or a negative charge. In a coordination entity, the central atom or ion is surrounded by some number of neutral molecules or negative ions ( called ligands) accordingly.
(ii) Ligand:- Ligands are the neutral molecules or negatively charged ions that surround the metal atom in a coordination entity according to the holding capacity of central metal ion are known as ligands. NH3 and H2O are two neutral ligands.
(iii) Coordination number:- The total number of metals that surrounds the central metal ion is known as coordination number.
For e.g,(a) In case of $\left [ PtCl_6 \right ]^{2-}$ six chlorine atoms are attached to Pt, thus the coordination number of the given compound is 6.
(b) In case of $\left [ Ni(NH_3)_4 \right ]^{2-}$ the central metal ion Ni is surrounded by 4 atoms of ligand, so its coordination number is 4.
(iv) Coordination polyhedron:- It is defined as the spatial arrangement of the ligands which are directly attached to the central metal ion/atom.
E.g. square planar, tetrahedral.
(v) Homoleptic:- Homoleptic compounds are defined as the compounds in which the donor/ligand attached to central metal atom/ion is of one kind.
E.g. $\left [ PtCl_6 \right ]^{2-}$ and $\left [ Ni(NH_3)_4 \right ]^{2-}$
(vi) Heteroleptic:- These are the coordination compounds in which central atoms are attached with more than one type of ligand.
E.g. $\left [ Co(NH_3)_5Cl \right ]^{2+}$ and $\left [ Co(NH_3)_4 Cl_2\right ]^{+1}$
The geometrical isomers of the compound are given below:-
We know that the given compound has tetrahedral geometry, so it can be optically active only when it has unsymmetric chelating agents. Hence the given compound doesn't have any optically active isomer.
(i) a green precipitate with aqueous potassium fluoride and (ii) a bright green solution with aqueous potassium chloride. Explain these experimental results.
Answer :
We know that strong ligands can replace weak ligands from its solution.
(i) In this case F - ions can replace H 2 O from aqueous copper sulphate solution.
When KCN is passed through an aqueous solution of copper sulphate, CN - being a strong ligand, will replace water and form $K_2\left [ Cu(CN)_4 \right ]$ .
It is known that in stable coordination compounds, the individual identity of each constituent is lost i.e., Cu +2 is not available freely.
Thus no precipitate of copper sulphide is obtained in the given conditions.
In this splitting dx2 y2 and dz2 experience a rise in energy and make the eg level, while dxy , dyz and dzx experience a fall in energy and generate the t2g level.
The arrangement of ligands in the increasing order of their crystal-field splitting energy (CFSE) values is known asthespectrochemical series. $\begin{aligned} \mathrm{I}^{-}<\mathrm{Br}^{-}<\mathrm{SCN}^{-}<\mathrm{Cl}^{-}<\mathrm{S}^{2-}<\mathrm{F}^{-} & <\mathrm{OH}^{-}<\mathrm{C}_2 \mathrm{O}_4{ }^{2-}<\mathrm{H}_2 \mathrm{O}<\mathrm{NCS}^{-} \\ & <\text {edta }^{4-}<\mathrm{NH}_3<\text { en }<\mathrm{CN}^{-}<\mathrm{CO}\end{aligned}$
The ligands on the right side of the series, strong field ligands are present, whereas on the left-hand side, weak field ligands are present.
The strong field ligands are capable of splitting d orbitals to a higher extent as compared to weak field ligands.
It is known that the degenerated d-orbitals split into two levels - eg and t2g . The splitting of the degenerate levels due to the presence of ligands is called the crystal-field splitting and the energy difference between the two levels (eg and t2g ) is called the crystal-field splitting energy (CFSE).
The CFSE is denoted by Δo .
After splitting of orbitals, the filling of the electrons starts. After 1 electron has been filled in each of the three t2g orbitals, the fourth electron can enter the eg orbital ( t2g3 eg1 like electronic configuration) or the pairing of the electrons can take place in the t2g orbitals ( t2g4 eg0 like electronic configuration).
If the CFSE value or Δ o value of a ligand is less than the pairing energy (P), then the electrons enter into the eg orbital. And, if the Δ o value of a ligand is more than the pairing energy (P), then the electrons will enter the t2g orbital.
In $[Cr(NH_{3})_{6}]^{3+}$ the oxidation state of the compound is +3. Its electronic configuration is d3 . Also, NH3 is a weak field ligand so the pairing of electrons will not occur.
So this compound is paramagnetic in nature.
In $[Ni(CN)_{4}]^{2-}$ the oxidation state of the Ni is +2. Its electronic configuration is d8. Also, CN - is a strong field ligand so the pairing of electrons will occur.
Hence, the above compound is diamagnetic in nature.
In the case of $[Ni(CN)_{4}]^{2-}$ , we have CN - as a strong field ligand. So, the pairing of electrons will occur. The electronic configuration of Ni+2 is d6. Since all the electrons will be paired thus d-d electronic transition is not possible on this case. Whereas in case of $[Ni(H_{2}O)_{6}]^{2+}$ we have a weak field ligand (H2O). The pairing of electrons will not occur. Thus electrons from a lower state of energy can transit to a higher state of energy and thus will give some colour.
In both compounds, the oxidation state of Fe is +2. Also, in $[Fe(H_{2}O)_{6}]^{2+}$ we have weak field ligand whereas in $[Fe(CN)_{6}]^{4-}$ we have strong field ligand. So there is a difference in CFSE value in both compounds. As a result, the colour shown by both compounds is different.
The metal-carbon bond in metal carbonyls has both σ and π character. Basically the M–C σ bond is generated due to the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal. Whereas the M–C π bond is formed due to the donation of a pair of electrons from a filled d orbital of metal into the vacant/empty antibonding π* orbital of carbon monoxide. As a result, this metal-to-ligand bonding leads to a synergic effect which strengthens the bond between CO and the metal.
Central metal ion : Cr.
Coordination number $=6$.
We know,
The oxidation state is:
$
\begin{aligned}
& x+2(0)+2(-1)=+1 \\
& x-2=-1 \\
& x=+3
\end{aligned}
$ The d orbital occupation: $\mathrm{t}_{2 \mathrm{~g}}{ }^3$.
The oxidation state of Cr in the given complex is +3. The coordination number of Cr is 6. The d orbital occupation for the central metal ion Cr3+ is t2g3.
Coordination number $=4$.
We know,
The oxidation state is:
$
\begin{aligned}
& x-4=-2 \\
& x=+2
\end{aligned}
$ The $d$ orbital occupation: $e_g{ }^4 t_{2 g}{ }^3$.
The oxidation state of Co in the given coordination compound is +2. Also, the coordination number of Co is 4. The d orbital occupation for the central metal ion Co 2+ is eg4 t2g3 .
Central metal ion: Mn .
Coordination number $=6$.
We know,
The oxidation state is :
$
\begin{aligned}
& x+0=2 \\
& x=+2
\end{aligned}
$ The d orbital occupation: $\mathrm{t}_{2 \mathrm{~g}}{ }^3 \mathrm{eg}^2$.
In the given complex compound the oxidation state of Mn is +2. Also, the coordination number of Mn is 6. The d orbital occupation for the central metal ion Mn+2 is t2g3 eg2.
In the ground state, Ti has 23 electrons with electronic configuration 3d3 4s2.
The oxidation state of Ti in the given compound is +3.
Hence, it will now have the configuration 3d2. Since it has 2 unpaired electrons and has the ability to undergo d-d transition, the given complex gives violet colour.
When a ligand is attached to the metal ion in such a manner that it forms a ring-like structure, then the metal-ligand bond is found to be more stable, i.e., complexes containing chelate rings are more stable than complexes without rings. The formation of such rings is known as the chelate effect.
Coordination compounds play a great role in biological systems. The pigment which is responsible for photosynthesis, chlorophyll, is a coordination compound of magnesium. Haemoglobin (which acts as oxygen carrier) the red pigment of blood is a coordination compound of iron. Vitamin B 12, and cyanocobalamine, the anti-pernicious anaemia factor, are few coordination compounds of cobalt which have biological importance.
The role of coordination compounds in the medicine industry is very significant such as the use of chelate therapy in medicinal chemistry. The excess of copper and iron is removed by the chelating ligands D–penicillamine and desferrioxamine B via the formation of coordination compounds. Nowadays, some coordination compounds of platinum (such as cis–platin and related compounds) effectively inhibit the growth of tumours.
(iii) In analytical chemistry, the familiar colour reactions given by metal ions with a number of ligands, generally chelating ligands. The formation of coordination entities gives the basis for their detection and estimation by classical and instrumental methods of analysis.
In the metal extraction process of metals, like silver and gold, make use of complex formation. For example, gold combines with cyanide in the presence of oxygen and water to form the coordination entity [Au(CN)2] in aqueous solution, which can be further separated by addition of zinc.
We know that due to the chelation effect stability of the chelating compound is more than the simple compound. Thus it is easy to notice that $[Fe(C_{2}O_{4})_{3}]^{3-}$ is most stable among all given compounds.
Class 12 Chemistry NCERT Chapter 5: Higher Order Thinking Skills (HOTS) Questions
These coordination compounds class 12 question answer are designed to challenge students analytical and application skills beyond basic textbook knowledge. NCERT Solutions for Class 12 encourage critical thinking and help students tackle complex problems confidently in exams.
Question: Given below are two statements : Statement (I) : In octahedral complexes, when $\Delta_{\mathrm{o}}<\mathrm{P}$ high spin complexes are formed. When $\Delta_{\mathrm{o}}>\mathrm{P}$ low spin complexes are formed. Statement (II) : In tetrahedral complexes because of $\Delta_t<\mathrm{P}$, low spin complexes are rarely formed.
In the light of the above statements, choose the most appropriate answer from the options given below :
1) Statement I is correct but Statement II is incorrect.
2) Both Statement I and Statement II are incorrect
3) Statement I is incorrect but Statement II is correct
4) Both Statement I and Statement II are correct
Answer:
In octahedral complex $(\mathrm{CN}=6)$
If $\Delta_0<$ P.E. , then high spin complexes are formed
If $\Delta_0>$ P.E. , then low spin complexes are formed
But in tetrahedral complex $(\mathrm{CN}=4)$
$\Delta_{\mathrm{t}}<$ P.E. , then mainly high spin complexes are formed and rarely low spin complexes are formed.
Hence, the correct answer is option (4).
Question: The number of paramagnetic complex among $\left[\mathrm{FeF}_6\right]^3,\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-},\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-},\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$, $\left[\mathrm{MnCl}_6\right]^{3-}$ and $\left[\mathrm{CoF}_6\right]^{3-}$, which involved $\mathrm{d}^2 \mathrm{sp}^3$ hybridization is ____________
Answer:
The octahedral complexes with strong field ligands generally form inner orbital complexes.
Only $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ and $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-} \quad$ are paramagnetic and $\mathrm{d}^2 \mathrm{sp}^3$ hybridisation of metal.
Hence, the answer is 2.
Question: Given below are two statements : Statement I: A homoleptic octahedral complex, formed using monodentate ligands, will not show stereoisomerism.
Statement II : cis- and trans- platin are heteroleptic complexes of Pd.
In light of the above statements, choose the correct answer from the options given below.
(1) Both statement I and Statement II are false.
(2) Statement I is false but Statement II is true.
(3) Both statement I and Statement II are true.
(4) Statement I is true but Statement II is false.
Answer:
Homolytic cleavage: Bond breaks evenly, each atom gets one electron, forming free radicals.
Heterolytic cleavage: Bond breaks unevenly, one atom gets both electrons, forming ions.
[Ma6] type complex will not show stereoisomerism, where $a$ is monodentate ligand
Cis and trans-platin are heterolytic complexes of $\mathrm{Pt}($ Platinum $)$. Formula is $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$
Hence, the correct answer is option (4).
CBSE Class 12th Syllabus: Subjects & Chapters
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Approach to Solve Questions of Chapter 5 Coordination Compounds
Sometimes, problems related to class 12 chemistry chapter 5 coordination compounds solutions seem difficult, but once we understand the basic formulas, it becomes very easy to solve all the questions. We can follow the steps given below to solve the questions based on this chapter.
1. Understand the key concepts
This is the basic yet crucial step. Some of the important concepts are listed below-
Ligands and coordination number
Nomenclature rules Bonding theories
Isomerism
Stability and applications of coordination compounds
2. Follow a stepwise strategy to solve the questions
A. Nomenclature questions
Identify the ligands, metal and their oxidation states and follow IUPAC rules.
B. Structure and formula
Determine oxidation number and coordination number and predict the geometry based on the hybridization. Use VBT or CFT to justify geometry and magnetic properties.
C. Bonding and Hybridization
Use Valence Bond Theory to determine inner or outer orbital complex and hybridization then apply Crystal Field Theory for splitting of d-orbitals in octahedral/tetrahedral fields and magnetic behavior (low spin vs high spin)
D. Magnetic Moment and Color
Apply formula to get the magnetic moment of the compound $\mu=\sqrt{n(n+2)} \mathrm{BM}$ where n = number of unpaired electrons.
Relate unpaired electrons to color and magnetism.
3. Refer to Solved Examples
Go through the coordination compounds ncert solutions as they illustrate standard methods. Use these methods to solve the questions.
4. Practice NCERT Exercises
Attempt in-text questions after each concept and solve all the exercises. You can also prepare tables and charts to revise the concepts.
Topics of NCERT Syllabus Class 12 Chemistry Coordination Compounds
This section outlines all the important topics covered under the class 12 chemistry chapter 5 coordination compounds solutions. The topics given in this chapter are given below:
5.7 Importance and Applications of Coordination Compounds
What Extra Should Students Study Beyond NCERT for JEE/NEET?
Along with the NCERT solutions for class 12 chemistry chapter 5 Coordination Compounds students preparing for JEE/NEET should explore advanced reference books and additional practice materials. The table below will help you classify the topics based on board exams and competitive exams.
What Students Learn from NCERT Solutions for Class 12 Chemistry Chapter 5 Coordination Compounds
These class 12 chemistry chapter 5 coordination compounds question answer help students understand the core principles of coordination chemistry in a structured way. Given below some points on key learnings of this chapter:
Here students will learn about how to name coordination entities according to IUPAC rules.
Basic theories of coordination and types of linkages are explained well in these solutions.
These coordination compounds class 12 question answer help students to identify ligands, their denticity, and determine coordination numbers.
These solutions explain different types of isomerism such as structural and stereoisomerism.
Valence Bond Theory and Crystal Field Theory are explained well in these solutions.
Stability constants and factors affecting the stability of complexes are explained in detail using these solutions through a series of solved examples.
NCERT Solutions Class 12 Chemistry Chapterwise
Apart from the class 12 chemistry chapter 5 coordination compounds solutions , students can also explore the links provided below for solutions to other chapters.
The NCERT books and syllabus are designed in a way that helps build strong fundamentals in students. Reading the NCERT books and syllabus prepares them to perform well in board exams as well as in competeive exams.
Q: What is NCERT solutions for class 12 chemistry chapter 5 Coordination Compounds?
A:
Class 12 Chemistry Chapter 5 NCERT solutions provide step by step answers to all the textbook questions from the Coordination Compounds chapter. They help students understand concepts such as nomenclature, isomerism, bonding theories, and the applications of coordination compounds.
Q: How to get NCERT solutions for Class 12 chemistry Chapter 5 coordination compounds?
A:
Students can access Class 12 Chemistry Chapter 5 NCERT solutions from trusted educational websites, learning platforms, or reference books that provide solved NCERT questions.
Q: What is the crystal field splitting energy?
A:
The crystal field splitting energy is the energy difference between the d orbitals of a metal ion in a coordination compound, caused by the presence of ligands. It arises due to the interaction between the metal ion's d electrons and the ligand's electric field.
Q: What is the difference between a unidentate and a multidentate ligand?
A:
A unidentate ligand is a ligand that binds to a metal ion through a single donor atom, while a multidentate ligand is a ligand that binds to a metal ion through multiple donor atoms. For example, ammonia is a unidentate ligand, while ethylenediamine is a bidentate ligand.
Q: What is the magnetic moment of a coordination compound?
A:
The magnetic moment of a coordination compound is a measure of its magnetic properties, which depends on the number and arrangement of unpaired electrons in the compound. It is expressed in units of Bohr magneton (μB).
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Yes, you can switch from Science in Karnataka State Board to Commerce in CBSE for 12th. You will need a Transfer Certificate from your current school and meet the CBSE school’s admission requirements. Since you haven’t studied Commerce subjects like Accountancy, Economics, and Business Studies, you may need to catch up before or during 12th. Not all CBSE schools accept direct admission to 12th from another board, so some may ask you to join Class 11 first. Make sure to check the school’s rules and plan your subject preparation.
For the 12th CBSE Hindi Medium board exam, important questions usually come from core chapters like “Madhushala”, “Jhansi ki Rani”, and “Bharat ki Khoj”.
Questions often include essay writing, letter writing, and comprehension passages. Grammar topics like Tenses, Voice Change, and Direct-Indirect Speech are frequently asked.
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Previous years’ question papers help identify commonly asked questions.
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