NCERT solutions for class 12 chemistry chapter 9 Coordination Compounds 2021

NCERT solutions for class 12 chemistry chapter 9 Coordination Compounds

Edited By Apoorva Singh | Updated on Aug 16, 2022 12:54 PM IST | #CBSE Class 12th

NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds - Coordination compounds is a conceptual chapter related to chemical reactions and atoms so you must pay special attention to the NCERT solutions for Class 12 Chemistry chapter 9 Coordination Compounds which are provided here for free. Coordination compounds are an important part of inorganic chemistry and coordination compounds. As per the NCERT, coordination compounds are the foundation of modern Inorganic Chemistry. In this NCERT solution, there are 11 in-text questions and 32 questions in the exercise and you will find all the NCERT Solutions for Class 12 Chemistry Chapter 9 when you scroll down.

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NCERT solutions for class 12 chemistry chapter 9 Coordination Compounds

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Find NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds Below:

Solutions to In-Text Questions Ex 9.1 to 9.10

Question 9.1(i) Write the formulas for the following coordination compounds:

(i) Tetraamminediaquacobalt(III) chloride

Answer :

The chemical formula for the coordination compound Tetraamminediaquacobalt(III) chloride is

$[CO(H_2O)_2(NH_3)_4]Cl_3$

Question 9.1(ii) Write the formulas for the following coordination compounds:

(ii) Potassium tetracyanidonickelate(II)

Answer :

The formula for the coordination compound Potassium tetracyanidonickelate II is :

$K_2[Ni(CN)_4]$

Question 9.1(iii) Write the formulas for the following coordination compounds:

(iii) Tris(ethane–1,2–diamine) chromium(III) chloride

Answer :

The formula for the coordination compound Tris(ethane–1,2–diamine) chromium(III) chloride is :

$[Cr(en)_3]Cl_3$

Question 9.1(iv) Write the formulas for the following coordination compounds:

(iv) Amminebromidochloridonitrito-N-platinate(II)

Answer :

The formula for the coordination compound Amminebromidochloridonitrito-N-platinate(II) :

$[Pt(NH_3)BrCl(NO_2)]^{-}$

Question 9.1(v) Write the formulas for the following coordination compounds:

(v) Dichloridobis(ethane–1,2–diamine)platinum(IV) nitrate

Answer :

The chemical formula for the coordination compounds: Dichloridobis(ethane–1,2–diamine)platinum(IV) nitrate

$[PtCl_2(en)_2](NO_3)_2$

Question 9.1(vi) Write the formulas for the following coordination compounds:

(vi) Iron(III) hexacyanidoferrate(II)

Answer :

The formula for the coordination compound Iron(III) hexacyanidoferrate(II):

$Fe_4[Fe(CN)_6]_3$

Question 9.2(i) Write the IUPAC names of the following coordination compounds:

$(i)\left [ Co\left ( NH_{3} \right )_{6} \right ]Cl_{3}$

Answer :

The IUPAC name of the coordination compound is :

Hexaamminecobalt(III) chloride

Question 9.2(ii) Write the IUPAC names of the following coordination compounds:

$(ii) [Co(NH_{3})_{5}Cl]Cl_{2}$

Answer :

The IUPAC name of the coordination compound is

Pentaamminechloridocobalt(III) chloride

Question 9.2(iii) Write the IUPAC names of the following coordination compounds:

$(iii) K_{3}[Fe(CN)_{6}]$

Answer :

The IUPAC name of the coordination compound is

Potassium hexacyanoferrate(III).

Question 9.2(iv) Write the IUPAC names of the following coordination compounds:

$(iv)K_{3}[Fe(C_{2}O_{4})_{3}]$

Answer :

The IUPAC name of the following coordination compound is

Potassium trioxalatoferrate(III)

Question 9.2(v) Write the IUPAC names of the following coordination compounds:

$(v)K_{2}[PdCl_{4}]$

Answer :

The IUPAC names of the coordination compound is

Potassium tetrachloridopalladate(II).

Question 9.2(vi) Write the IUPAC names of the following coordination compounds:

$(vi)[Pt(NH_{3})_{2}Cl(NH_{2}CH_{3})]Cl$

Answer :

The IUPAC name of the following coordination compound is :

Diamminechlorido(methylamine)platinum(II) chloride.

Question 9.3(i) Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers:

$(i) K[Cr (H_{2}O)_{2}(C_{2}O_{4})_{2}$

Answer :

Both geometrical (cis-, trans-) and optical isomers exists for the given compound.

1649396267533

Question 9.3(ii) Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers:

$(ii) [Co(en)_{3}]Cl_{3}$

Answer :

Two optical isomers can exist, Dextro and Laevo.

1649396307716

Question 9.3(iii) Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers:

$(iii) [Co(NH_{3})_{5}(NO_{2})](NO_{3})_{2}$

Answer :

$[Co(NH_{3})_{5}(NO_{2})](NO_{3})_{2}$

There are 10 possible isomers.There are geometrical, ionisation and linkage isomers possible.

A pair of optical isomer

1649396358804

it also shows ionization isomerism

$[Co(NH_{3})_{5}(NO_{2})](NO_{3})_{2}$ AND $[Co(NH_{3})_{5}(NO_{3})](NO_{3})(NO_2)$

It can also show linkage isomerism

$[Co(NH_{3})_{5}(NO_{2})](NO_{3})_2$ AND $[Co(NH_{3})_{5}(ONO)](NO_{3})_2$

Question 9.3(iv) Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers:

$(iv)\; [Pt(NH_{3})(H_{2}O)Cl_{2}]$

Answer :

Geometrical (cis-, trans-) isomers can exists for

1649396402281

Question 9.4(i) Give evidence that following is ionisation isomer.

$(i) [Co(NH_{3})_{5}Cl]SO_{4}$

Answer :

The ionisation isomers dissolve in water to yield different ions and thus react differently to various reagents:

[Co(NH3)5Br]SO4 + Ba2+ → BaSO4 (s)

[Co(NH3)5SO4]Br + Ba2+ → No reaction

[Co(NH3)5Br]SO4 + Ag+ → No reaction

[Co(NH3)5SO4]Br + Ag+ → AgBr (s)

Question 9.4(ii) Give evidence that following is ionisation isomer.

$(ii) [Co(NH_{3})_{5}(SO_{4})]Cl$

Answer :

The ionisation isomers dissolve in water to yield different ions and thus react differently to various reagents:

[Co(NH3)5Br]SO4 + Ba2+ → BaSO4 (s)

[Co(NH3)5SO4]Br + Ba2+ → No reaction

[Co(NH3)5Br]SO4 + Ag+ → No reaction

[Co(NH3)5SO4]Br + Ag+ → AgBr (s)

Question 9.5 Explain on the basis of valence bond theory that ion with square planar structure is diamagnetic and the $[NiCl_{4}]^{2-}$ ion with tetrahedral geometry is paramagnetic.

Answer :

In both the compounds, the oxidation state of Nickel is +2. So it has d ⁸ 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 :-

1649396442693

1649396465299

1649396484196

Thus is a square planer and diamagnetic and has tetrahedral geometry and is paramagnetic.

Question 9.6 is paramagnetic while is diamagnetic though both are tetrahedral. Why?

Answer :

The difference in the magnetic behaviour is due to the nature of ligands present. In case of the oxidation state of nickel is +2 and also Cl ^- is a weak ligand. Thus its configuration becomes:-

1649396517195

So it is paramagnetic and tetrahedral in nature.

In the case of _, the oxidation state of nickel is 0. So its configuration is 3d ⁸ 4s ² . We also know that CO is a strong ligand, thus the configuration of nickel becomes:-

1649396537524

Hence the given compound is diamagnetic but tetrahedral in nature.

Question 9.7 is strongly paramagnetic whereas is weakly paramagnetic. Explain.

Answer :

In both the compounds Fe has +3 oxidation state i.e., d ⁵ configuration.

1649396605801

In the case of strong ligand (CN ^- ), the pairing of electron will occur. So number of electrons left unpaired will be 1.

In case of weak ligand (H ₂ O), pairing of electron will not there. Thus number of electrons unpaired will be 5.

We know that paramagnetic strength is directly proportional to the number of unpaired electrons.

Hence paramagnetism will be more in case of

Question 9.8 Explain is an inner orbital complex whereas is an outer orbital complex.

Answer :

Firstly consider _:-

The oxidation state of cobalt is +3. So the electronic configuration of it will be d ⁶ .

Since (NH ₃ ) is a strong ligand so the pairing of electron will be there.

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So, it has d ² sp ³ hybridisation and an inner orbital complex.

Now consider,

The oxidation state of nickel is +3. So its electronic configuration will be d ⁸ .

Also (NH ₃ ) is a strong ligand so pairing of electrons will be seen.

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Thus is an outer orbital complex.

Question 9.9 Predict the number of unpaired electrons in the square planar ion.

Answer :

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 dsp ² hybridisation (d ⁸ ).

CN ^- is a strong ligand so the pairing of electron will occur.

So there are be no unpaired electrons in the given compound.

Question 9.10 The hexaqua manganese(II) ion contains five unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory.

Answer :

Consider Hexaqua manganese (II) :- In this compound, the oxidation state of Mn is +2 and its electronic configuration is d ⁵ .

H ₂ O is a weak ligand and crystal field is octahedral so the arrangement of electrons will be t ₂ ³ g eg ² .

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 t _2g ⁵ eg ⁰ .

Thus the number of unpaired electrons will be 1.

NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds- Exercise Questions

Question 9.1 Explain the bonding in coordination compounds in terms of Werner’s postulates.

Answer :

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.

Question 9.2 solution mixed with solution in 1:1 molar ratio gives the test of $Fe^{2+}$ ion but solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of $Cu^{2+}$ ion. Explain why?

Answer :

1594906643433

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 .

Question 9.3 Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.

Answer :

(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.

$E.g\ \left [ PtCl_4 \right ]^{2-}\ and\ \left [ Ag(CN)_2 \right ]^{-}$

(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. NH ₃ and H ₂ O 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 six chlorine atoms are attached to Pt, thus the coordination number of the given compound is 6.

(b) In case of 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. and

(vi) Heteroleptic:- These are the coordination compounds in which central atoms are attached with more than one type of ligand.

E.g. and

Question 9.4 What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.

Answer :

Unidentate ligands:- The ligand which has only one donor site are known as unidentate ligands. E.g. Cl ^- , NH ₃

Bidentate(or didentate):- The ligands which have two donor sites are known as bidentate ligands. E.g. Ethane - 1,2 diamine, Oxalate ion

Ambidentate ligands:- The ligands which can attach themselves to the central metal through two different atoms. E.g NO ₂ , SCN

Question 9.5 (i) Specify the oxidation numbers of the metals in the following coordination entities:

$(i)[Co(H_{2}O)(CN)(en)_{2}]^{2+}$

Answer :

Let us assume that coordination number of cobalt is x.

$x\ +\ 0\ +\(-1)\ + 2(0)\ =\ +2$

Thus $x = 3$

Hence coordination number of cobalt is +3.

Question 9.5(ii) Specify the oxidation numbers of the metals in the following coordination entities:

$(ii) [CoBr_{2}(en)_{2}]^{+}$

Answer :

Let us assume that the coordination number of cobalt is x.

Then according to the question :

$x\ + 2(-1)\ +\ 2(0)\ =\ 1$

Hence the coordination number of cobalt is 3.

Question 9.5(iii) Specify the oxidation numbers of the metals in the following coordination entities:

$(iii)[PtCl_{4}]^{2-}$

Answer :

Let us assume the oxidation state of Pt to be x.

Then the equation becomes :-

$\\x\ +\ 4(-1)\ =\ -2\\x=+2$

Hence the oxidation state of Pt metal is 2.

Question 9.5(iv) Specify the oxidation numbers of the metals in the following coordination entities:

$(iv)K_{3}[Fe(CN)_{6}]$

Answer :

Let the oxidation state of Fe be x.

Then the equation will be :-

$\\x\ +\ 6(-1)\ =\ -3\\x=+3$

Thus oxidation state of Fe in this coordination compound is 3.

Question 9.5(v) Specify the oxidation numbers of the metals in the following coordination entities:

$(v) [Cr(NH_{3})_{3}Cl_{3}]$

Answer :

Let us assume the oxidation state of Cr to be x.

Then according to the question we get,

$x\ +\ 3(0)\ +\ 3(-1)\ =\ 0$

$x=+3$

So the oxidation state of Cr will be 3.

Question 9.6(i) Using IUPAC norms write the formulas for the following:

(i) Tetrahydroxidozincate(II)

Answer :

Using the IUPAC rules :-

$\left [ Zn(OH)_4 \right ]^{2-}$

Question 9.6(ii) Using IUPAC norms write the formulas for the following:

(ii) Potassium tetrachloridopalladate(II)

Answer :

The required compound is :-

$K_2\left [ PdCl_4 \right ]$

Question 9.6(iii) Using IUPAC norms write the formulas for the following:

(iii) Diamminedichloridoplatinum(II)

Answer :

The required chemical formula of compound is :-

Question 9.6(iv) Using IUPAC norms write the formulas for the following:

(iv) Potassium tetracyanidonickelate(II)

Answer :

The required compound is :-

Question 9.6(v) Using IUPAC norms write the formulas for the following:

(v) Pentaamminenitrito-O-cobalt(III)

Answer :

The compound is :-

Question 9.6(vi) Using IUPAC norms write the formulas for the following:

(vi) Hexaamminecobalt(III) sulphate

Answer :

The required compound is :-

Question 9.6(vii) Using IUPAC norms write the formulas for the following:

(vii) Potassium tri(oxalato)chromate(III)

Answer :

The required compound is :-

Question 9.6(viii) Using IUPAC norms write the formulas for the following:

(viii) Hexaammineplatinum(IV)

Answer :

The required compound is :-

Question 9.6(ix) Using IUPAC norms write the formulas for the following:

(ix) Tetrabromidocuprate(II)

Answer :

The required compound is

Question 9.6(x) Using IUPAC norms write the formulas for the following:

(x) Pentaamminenitrito-N-cobalt(III)

Answer :

The required compound is :-

Question 9.7(i) Using IUPAC norms write the systematic names of the following:

$(i)[Co(NH_{3})_{6}]Cl_{3}$

Answer :

Using IUPAC norms the name of the given compound is :-

Hexaamminecobalt(III) chloride

Question 9.7(ii) Using IUPAC norms write the systematic names of the following:

$(ii) \; [Pt(NH_{3})_{2}Cl(NH_{2}CH_{3})]Cl$

Answer :

According to the IUPAC norms :-

Diamminechlorido(methylamine) platinum(II) chloride

Question 9.7(iii) Using IUPAC norms write the systematic names of the following:

$(iii)\; [Ti(H_{2}O)_{6}]^{3+}$

Answer :

Using the nomenclature rules, the required name of the compound is:- Hexaquatitanium(III) ion

Question 9.7(iv) Using IUPAC norms write the systematic names of the following:

$(iv)\; [Co(NH_{3})_{4}Cl(NO_{2})]Cl$

Answer :

Using IUPAC norms, the name of the given compound is :-

Tetraamminichloridonitrito-N-Cobalt(III) chloride

Question 9.7(v) Using IUPAC norms write the systematic names of the following:

$(v)\; [Mn(H_{2}O)_{6}]^{2+}$

Answer :

The IUPAC name of the givne compound is :-

Hexaquamanganese(II) ion

Question : 9.7(vi) Using IUPAC norms write the systematic names of the following:

$(vi)\; [NiCl_{4}]^{2-}$

Answer :

According to the IUPAC norms, the systematic name of the given compound is:- Tetrachloridonickelate(II) ion

Question 9.7(vii) Using IUPAC norms write the systematic names of the following:

$(vii)[Ni(NH_{3})_{6}]Cl_{2}$

Answer :

The IUPAC name of the given compound is :- Hexaamminenickel(II) chloride

Question 9.7(viii) Using IUPAC norms write the systematic names of the following:

$(viii)[Co(en)_{3}]^{3+}$

Answer :

The name of the given compound is :- Tris(ethane-1, 2-diammine) cobalt(III) ion

Question 9.7(ix) Using IUPAC norms write the systematic names of the following:

$(ix)[Ni(CO)_{4}]$

Answer :

The IUPAC name of the given compound is :- Tetracarbonylnickel(0)

Question 9.8 List various types of isomerism possible for coordination compounds, giving an example of each.

Answer :

Two main types of isomerism are known in case of coordination compounds which can be further divided into subgroups:-

(a) Stereoisomerism :-

(i) Geometrical isomerism E.g

. 1594906673186

(ii) Optical isomerism

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(b) Structural isomerism :-

(i) Linkage isomerism E.g and

(ii) Ionisation isomerism E.g and

(iii) Coordination isomerism E.g and

(iv) Solvate isomerism E.g and

Question : 9.9(i) How many geometrical isomers are possible in the following coordination entities?

$(i)\; [Cr(C_{2}O_{4})_{3}]^{3-}$
Answer:

No geometrical isomers are possible since the given compound is a bidentate compound.

Question 9.9(ii) How many geometrical isomers are possible in the following coordination entities?

$(ii) [Co(NH_{3})_{3}Cl_{3}]$

Answer :

The facial (fac) and meridional (mer) isomers are possible for the given compound.

1649396758911

Question 9.10(i) Draw the structures of optical isomers of:

$(i); [Cr(C_{2}O_{4})_{3}]^{3-}$

Answer :

The optical isomers of the given compound are:-

1649396803910

Question 9.10(ii) Draw the structures of optical isomers of:

$(ii)[PtCl_{2}(en)_{2}]^{2+}$

Answer :

The optical isomers of the given compound are given below :-

1594906726312

Question 9.10(iii) Draw the structures of optical isomers of:

$(iii)\; [Cr(NH_{3})_{2}Cl_{2}(en)]^{+}$

Answer :

The optical isomers of the given compound are:-

1594906750068

Question 9.11(i) Draw all the isomers (geometrical and optical) of:

$(i) [CoCl_{2}(en)_{2}]^{+}$

Answer :

The configurational isomers are:-

1594906764565

1594906775457

Question 9.11(ii) Draw all the isomers (geometrical and optical) of:

$(ii) [Co(NH_{3})Cl(en)_{2}]^{2+}$

Answer :

The isomers of the given compound are :-

1594906788215

1594906797729

Question 9.11(iii) Draw all the isomers (geometrical and optical) of:

$(iii) [Co(NH_{3})_{2}Cl_{2}(en)]^{+}$

Answer :

The possible isomers of the given compound are as follows :-

1649396850191

1649396870776

Question 9.12 Write all the geometrical isomers of and how many of these will exhibit optical isomers?

Answer :

The geometrical isomers of the compound are given below:-

1649396910677

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.

Question 9.13 Aqueous copper sulphate solution (blue in colour) gives:

(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.

1649396955245

(ii) In this compound, Cl ^- being stronger ligand will replace H ₂ O and give a bright green solution 1649396973455

Question 9.14 What is the coordination entity formed when excess of aqueous $KCN$ is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when is passed through this solution?

Answer :

When KCN is passed through an aqueous solution of copper sulphate, CN ^- being a strong ligand, will replace water and form .

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.

Question 9.15 Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:

$(i)[Fe(CN)_{6}]^{4-}$

Answer :

In the given compound oxidation state of Fe is +2. The electronic configuration of this compound is 3d ⁶ .

Also CN ^- is a strong field ligand so it will cause the pairing of electrons.

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Therefore the complex is diamagnetic and its geometry is octahedral.

Question 9.15 Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:

$(ii) [FeF_{6}]^{3-}$

Answer :

(ii) In the given complex the oxidation state of Fe is +3. Its electronic configuration is d ⁵ .

Also, the F ^- ions are weak field ligands, therefore, the pairing of electrons will not occur.

1649397108858

Thus its geometry is octahedral and it is paramagnetic in nature.

Question 9.15 Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:

$(iii)[Co(C_{2}O_{4})_{3}]^{3-}$

Answer:

(iii) In the given compound, the oxidation state of Co is +3. Its electronic configuration thus becomes d ⁶ .

Also, oxalate is a weak field ligand therefore pairing of electrons will not occur.

1649397143755

Hence the complex is octahedral and paramgnetic in nature.

Question 9.15 Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:

$(iv)[CoF_{6}]^{3-}$

Answer :

(iv) The oxidation state of Co in the given comound is +3. The electronic configuration of the compound is d ⁶ .

Also, F ^- is a weak field ligand so no pairing of electrons will occur.

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1649397191892

Hence the geometry of compound is octahedral and it is paramagnetic in nature.

Question 9.16 Draw figure to show the splitting of d orbitals in an octahedral crystal field.

Answer :

The splitting of d orbital is shown below:-

1649397251154

In this splitting d _{x ² y ²} and d _{z ²} experience a rise in energy and make the eg level, while d _xy , d _yz and d _zx experience a fall in energy and generate the t _2g level.

Question 9.17 What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.

Answer :

The arrangement of ligands in the increasing order of their crystal-field splitting energy (CFSE) values is known as spectrochemical series .

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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.

Question 9.18 What is crystal field splitting energy? How does the magnitude of decide the actual configuration of d orbitals in a coordination entity?

Answer :

It is known that the degenerated d-orbitals split into two levels - e _g and t _2g . 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 (e _g and t _2g ) is called the crystal-field splitting energy (CFSE).

The CFSE is denoted by Δ _o .

After splitting of orbitals, the filling of the electrons starts. After 1 electron has been filled in each of the three t _2g orbitals, the fourth electron can enter the eg orbital ( t _2g ³ e _g ¹ like electronic configuration) or the pairing of the electrons can take place in the t _2g orbitals ( t _2g ⁴ e _g ⁰ 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 e _g orbital. And, if the Δ _o value of a ligand is more than the pairing energy (P), then the electrons will enter the t _2g orbital.

Question 9.19 is paramagnetic while is diamagnetic. Explain why?

Answer :

In the oxidation state of the compound is +3. Its electronic configuration is d ³ . Also, NH ₃ is a weak field ligand so the pairing of electrons will not occur.

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So this compound is paramagnetic in nature.

In the oxidation state of the Ni is +2. Its electronic configuration is d ⁸ . Also, CN ^- is a strong field ligand so the pairing of electrons will occur.

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Hence the above compound is diamagnetic in nature.

Question 9.20 A solution of is green but a solution of is colourless. Explain .

Answer :

In the case of , we have CN ^- as a strong field ligand. So, the pairing of electrons will occur. The electronic configuration of Ni ⁺² is d ⁶ . Since all the electrons will be paired thus d-d electronic transition is not possible on this case. Whereas in case of we have a weak field ligand (H ₂ O). 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.

Question 9.21 $[Fe(CN)_{6}]^{4-}$ and $[Fe(H_{2}O)_{6}]^{2+}$ are of different colours in dilute solutions. Why?

Answer :

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 the compounds. As a result, the colour shown by both compounds is different.

Question 9.22 Discuss the nature of bonding in metal carbonyls.

Answer :

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.

1649397395194

Question 9.23 Give the oxidation state, d orbital occupation and coordination number of the central metal ion in the following complexes:

$(i) K_{3}[Co(C_{2}O_{4})_{3}]$

Answer :

The oxidation state of Co is +3 and its oxidation number is 6. The d orbital occupation of the given central metal ion is t _2g ⁶ e _g ⁰ .

Question 9.23 Give the oxidation state, d orbital occupation and coordination number of the central metal ion in the following complexes:

$(ii)cis-[CrCl _{2}(en)_{2}]Cl$

Answer :

(ii) The oxidation state of Cr in the given complex is +3. The coordination number of Cr is 6. The d orbital occupation for central metal ion Cr ³⁺ is t _2g ³ .

Question 9.23 Give the oxidation state, d orbital occupation and coordination number of the central metal ion in the following complexes:

$(iii)(NH_{4})_{2}[CoF_{4}]$

Answer :

(iii) 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 e _g ⁴ t _2g ³ .

Question 9.23 Give the oxidation state, d orbital occupation and coordination number of the central metal ion in the following complexes:

$(iv) [Mn(H_{2}O)_{6}]SO_{4}$

Answer :

(iv) 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 t _2g ³ e _g ² .

Question 9.24 Write down the IUPAC name for the following complex and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex:

$(i) K[Cr(H_{2}O)_{2}(C_{2}O_{4})_{2}].3H_{2}O$

Answer :

The IUPAC name of the given compound is Potassium diaquadioxalatochromate (III) trihydrate.

The oxidation state of Cr in this compound is +3.

The electronic configuration is 3d ³ .

The coordination number of this compound is 6.

The magnetic moment of a compound is given by :

$=\ \sqrt{n(n+2)}$ Here n is the number of the unpaired electrons.

or $=\ \sqrt{3(3+2)}\ =\ \sqrt{15}BM$

Stereochemistry:-

1649397447790

$(ii) [Co(NH_{3})_{5}Cl_{-}]Cl_{2}$

Answer :

The IUPAC name of the given compound is Pentaamminechloridocobalt(III) chloride.

The oxidation state of Co in this compound is +3.

The electronic configuration is d ⁶ .

The coordination number of this compound is 6.

The magnetic moment of a compound is given by :

$=\ \sqrt{n(n+2)}$ Here n is the number of the unpaired electrons.

or $=\ \sqrt{0(0+2)}\ =\ 0BM$

Stereochemistry:-

1649397486620

$(iii) [CrCl_{3}(py)_{3}]$

Answer :

The IUPAC name of the given compound is Trichloridotripyridinechromium (III).

The oxidation state of Cr in this compound is +3.

The electronic configuration is d ³ .

The coordination number of this compound is 6.

The magnetic moment of a compound is given by :

$=\ \sqrt{n(n+2)}$ Here n is the number of the unpaired electrons.

or $=\ \sqrt{3(3+2)}\ =\ \sqrt{15}BM$

Stereochemistry:-

1649397530911

$(iv) Cs[FeCl_{4}]$

Answer :

The IUPAC name of the given compound is Caesium tetrachloroferrate (III).

The oxidation state of Cs in this compound is +3.

The electronic configuration is d ⁵ .

The coordination number of this compound is 4.

The magnetic moment of a compound is given by :

$=\ \sqrt{n(n+2)}$ Here n is the number of the unpaired electrons.

or $=\ \sqrt{5(5+2)}\ =\ \sqrt{35}BM=5.92$

Stereochemistry:- Optically inactive.

$(v) K_{4}[Mn(CN)_{6}]$

Answer :

The IUPAC name of the given compound is Potassium hexacyanomanganate(II).

The oxidation state of Mn in this compound is +2.

The electronic configuration is d ⁵ .

The coordination number of this compound is 6.

The magnetic moment of a compound is given by :

$=\ \sqrt{n(n+2)}$ Here n is the number of the unpaired electrons.

or $\\=\ \sqrt{1(1+2)}\ =\ \sqrt{3}BM\\=1.72BM$

Stereochemistry:- Optically inactive.

Question 9.25 Explain the violet colour of the complex on the basis of crystal field theory.

Answer :

In ground state,Ti has 23 electrons with electronic configuration 3d ³ 4s ² .

The oxidation state of Ti in the given compound is +3.

Hence it will now have the configuration 3d ² . Since it has 2 unpaired electrons and has the ability to undergo d-d transition, the given complex gives violet colour.

Question 9.26 What is meant by the chelate effect? Give an example .

Answer :

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.

1649397576241

Question 9.27 Discuss briefly giving an example in each case the role of coordination compounds in:

(i) biological systems

Answer :

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 , cyanocobalamine, the anti-pernicious anaemia factor, are few coordination compounds of cobalt which have biological importance.

Question 9.27 Discuss briefly giving an example in each case the role of coordination compounds in:

(ii) medicinal chemistry

Answer :

The role of coordination compounds in the medicine industry is very huge such as the use of chelate therapy in medicinal chemistry. The excess of copper and iron are 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.

Question 9.27 Discuss briefly giving an example in each case the role of coordination compounds in:

(iii) analytical chemistry

Answer :

(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.

Question 9.27 Discuss briefly giving an example in each case the role of coordination compounds in:

(iv) extraction/metallurgy of metals

Answer :

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.

Question 9.28 How many ions are produced from the complex $Co(NH_{3})_{6}Cl_{2}$ in solution?

(i) 6 (ii) 4 (iii) 3 (iv) 2

Answer :

Total of 3 ions will be produced.

2 of Cl ^- and one ion of [Co(NH ₃ ) ₆ ] ⁺ will be produced.

Question 9.29 Amongst the following ions which one has the highest magnetic moment value?

$(i)[Cr(H_{2}O)_{6}]^{3+}$ $(ii)[Fe(H_{2}O)_{6}]^{2+}$ $(iii)[Zn(H_{2}O)_{6}]^{2+}$

Answer :

(i) No. of unpaired electrons in $[Cr(H_{2}O)_{6}]^{3+}$ is 3.

The magnetic moment is given by :

$=\ \sqrt{n(n+2)}$

or $=\ \sqrt{3(3+2)}\ =\ \sqrt{15}BM$

(ii) Similarly in $[Fe(H_{2}O)_{6}]^{2+}$ the number of unpaired electrons is 4.

So the magnetic moment :

$=\ \sqrt{4(4+2)}\ =\ \sqrt{24}BM$

(iii) In the case of $[Zn(H_{2}O)_{6}]^{2+}$ the number of unpaired electrons is 0. So its magnetic moment is also zero.

Thus $[Fe(H_{2}O)_{6}]^{2+}$ has the highest dipole moment among all.

Question 9.31 Amongst the following, the most stable complex is

$(i)[Fe(H_{2}O_{6})]^{3+}$ $(ii)[Fe(NH_{3})_{6}]^{3+}$ $(iii)[Fe(C_{2}O_{4})_{3}]^{3-}$ $(iv)[FeCl_{6}]^{3-}$

Answer :

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 is most stable among all given compounds.

Question 9.32 What will be the correct order for the wavelengths of absorption in the visible region for the following:

$[Ni(NO_{2})_{6}]^{4-}, [Ni(NH_{3})_{6}]^{2+}, [Ni(H_{2}O)_{6}]^{2+} ?$

Answer :

The order of the wavelength of absorption will be decided from the order of their CFSE values.

The CFSE values increase in the order :- H ₂ O < NH ₃ < NO ₂ ^-

Hence, the order of wavelengths of absorption in the visible region is : [Ni(H ₂ O) ₆ ] ²⁺ > [Ni(NH ₃ ) ₆ ] ²⁺ > [Ni(NO ₂ ) ₆ ] ^4-

NCERT Solutions Class 12 Chemistry - Chapterwise

Chapter 1	NCERT solutions for class 12 chapter 1 The Solid State
Chapter 2	NCERT solutions for class 12 chemistry chapter 2 Solutions
Chapter 3	NCERT Solutions for class 12 chemistry chapter 3 Electrochemistry
Chapter 4	NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics
Chapter 5	NCERT Solutions for class 12 chemistry chapter 5 Surface chemistry
Chapter 6	NCERT solutions for class 12 chemistry General Principles and Processes of isolation of elements
Chapter 7	NCERT solutions for class 12 chemistry chapter 7 The P-block elements
Chapter 8	NCERT Solutions for class 12 chemistry chapter 8 The d and f block elements
Chapter 9	NCERT solutions for class 12 chemistry chapter 9 Coordination compounds
Chapter 10	NCERT Solutions for class 12 chemistry chapter 10 Haloalkanes and Haloarenes
Chapter 11	NCERT solutions for class 12 chemistry Alcohols, Phenols, and Ethers
Chapter 12	NCERT Solutions for class 12 chemistry chapter 12 Aldehydes, Ketones and Carboxylic Acids
Chapter 13	NCERT solutions for class 12 chemistry chapter 13 Amines
Chapter 14	NCERT solutions for class 12 chemistry chapter 14 Biomolecules
Chapter 15	NCERT Solutions for class 12 chemistry chapter 15 Polymers
Chapter 16	NCERT solutions for class 12 chemistry chapter 16 Chemistry in Everyday life

Topics and Sub-topics of NCERT solutions for Class 12 Chemistry Chapter 9 Coordination Compounds-

9.1 Werner's Theory of Coordination Compounds

9.2 Definitions of Some Important Terms Pertaining to Coordination Compounds

9.3 Nomenclature of Coordination Compounds

9.4 Isomerism in Coordination Compounds

9.5 Bonding in Coordination Compounds

9.6 Bonding in Metal Carbonyls

9.7 Importance and Applications of Coordination Compounds

NCERT Solutions for Class 12 Subject-wise

Benefits of NCERT solutions for class 12 chemistry chapter 9 Coordination Compounds

The NCERT Class 12 Chemistry solutions chapter 9 Coordination Compounds are prepared and explained well by subject experts to help you in the preparation of CBSE Board exams.
NCERT solutions for other classes and subjects can aid you in various competitive exams like JEE, NEET, BITS, and, KVPY etc.
The seven sub-topics of NCERT syllabus Chemistry chapter 9 Coordination Compounds covers important concepts like Werner’s Theory of coordination compounds, definitions of some important terms relating to coordination compounds, the rules of nomenclature of coordination compounds etc.
By referring to the Class 12 Chemistry Chapter 9 NCERT book solutions, students can understand all the important concepts and practice questions well enough before their examination.
The comprehensive answers given in the CBSE NCERT solutions for Class 12 Chemistry chapter 9 Coordination Compounds will be helpful in understanding the chapter easily.
The revision will be quite easier because the detailed solutions will help you to remember the concepts and get very good marks in your class.
Homework problems won't bother you anymore, all you need to do is check the detailed NCERT Solutions for Class 12 Chemistry and you are good to go.

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Frequently Asked Questions (FAQs)

1. What is the weightage of NCERT Solutions for Class 12 Chemistry Chapter 9 in CBSE board exam?

Weightage of Coordination Compounds in CBSE Board exam is 6-7 marks. Follow NCERT textbook for a good score in the board exam.

2. What is the weightage of NCERT Class 12 Chemistry chapter 9 in JEE Mains?

Weightage of Coordination Compounds in JEE mains is 10 marks. To practice questions students can refer to NCERT exemplar and JEE Main previous year papers.

3. What is the weightage of NCERT Class 12 Chemistry chapter 9 in NEET?

Coordination compounds is a very important chapter for the NEET exam and it holds a weightage of 9% of the total marks.

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Questions related to CBSE Class 12th

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quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

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Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

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Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
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Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

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Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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kumardurgesh1802 10 July,2024

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Central Board of Secondary Education Class 12th Examination

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT solutions for class 12 chemistry chapter 9 Coordination Compounds

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Topics and Sub-topics of NCERT solutions for Class 12 Chemistry Chapter 9 Coordination Compounds-

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Benefits of NCERT solutions for class 12 chemistry chapter 9 Coordination Compounds

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