NCERT solutions for class 12 chemistry chapter 9 Coordination Compounds

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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.

1649396659090

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.

1649396695953

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

1594906685503

(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.

1649397060679

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 .

1649397299196

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.

1649397357300

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

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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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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

Have a question related to CBSE Class 12th ?

I am interested in graphic designing and coding. how do I make a career .which stream should I go.what engineering subject to take

Hello aspirant,

The purpose of graphic design extends beyond the brand's look. Nevertheless, by conveying what the brand stands for, it significantly aids in the development of a sense of understanding between a company and its audience. The future in the field of graphic designing is very promising.

There are various courses available for graphic designing. To know more information about these courses and much more details, you can visit our website by clicking on the link given below.

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Thank you

Hope this information helps you.

Read Complete Answer

Sajal Trivedi 29 January,2024

i couldn't clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Read Complete Answer

Upasana Barman 24 August,2022

i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

Read Complete Answer

Mayankjha.student2007 23 August,2022

if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

Read Complete Answer

Nuthalapati Vyshnavi 21 August,2022

I got an RT for 2 subjects, so do I have to give exams for all the subjects in the coming year or just the 2. pls help me

Hi,

You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.

Read Complete Answer

Muskan Dhalwal 17 August,2022

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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms^-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

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

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×10⁷J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms⁻² :

Option 1)

2.45×10⁻³ kg

Option 2)

6.45×10⁻³ kg

Option 3)

9.89×10⁻³ kg

Option 4)

12.89×10⁻³ kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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) $20,000 \, \, J - 50,000 \, \, J$

A particle is projected at 60⁰ to the horizontal with a kinetic energy $K$ . The kinetic energy at the highest point

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

In the reaction,

$2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}$

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 .

How many moles of magnesium phosphate, $Mg_{3}(PO_{4})_{2}$ will contain 0.25 mole of oxygen atoms?

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

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

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.

With increase of temperature, which of these changes?

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

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol^-1) is

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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t²) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m² , the number of rotations made by the pulley before its direction of motion if reversed, is

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

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The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

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A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.

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Welding Engineer

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QA Manager

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Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

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An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party.

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Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems.