NCERT Exemplar Class 12 Chemistry Solutions Chapter 9 Coordination Compounds

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NCERT Exemplar Class 12 Chemistry Solutions Chapter 9 Coordination Compounds

Q: 1. What is class 12 Chemistry Chapter 9

- The class 12 coordination compounds solutions will help students to address their doubts and conceptual mistakes. Working on these [problems and questions will make them well prepared for their upcoming board exams.

Q: 2. Who can benefit from these NCERT exemplar Class 12 Chemistry chapter 9 solutions?

- Students who are preparing for the board exams and students preparing for entrance exams, both can equally benefit from this.

Q: 3. How to download these NCERT exemplar Class 12 Chemistry solutions Chapter 9?

- One can download the solutions in PDF format from the solutions page itself by a single link on the given link.

Q: 4. What one can learn from this class 12 chemistry NCERT exemplar solutions chapter 9?

- Students will learn about the coordination compounds and their valency. they will also learn about its types and the chemical reactions with various other compounds.

Edited By Sumit Saini | Updated on Sep 16, 2022 05:46 PM IST | #CBSE Class 12th

NCERT exemplar Class 12 Chemistry solutions chapter 9 is an important chapter which holds great importance in an exam point of view. Class 12 Chemistry NCERT exemplar solutions chapter 9 are carefully prepared by professionals of the field. This NCERT material helps the students and provides them with accurate and precise information. NCERT exemplar Class 12 Chemistry chapter 9 solutions were prepared with the objective of keeping it simple and understandable to the students. Students find it to be reliable and easy to learn and understand the subject concepts. NCERT exemplar Class 12 Chemistry solutions chapter 9 pdf download can be easily accessed online.

This Story also Contains

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: MCQ (Type 1)
NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: MCQ (Type 2)
NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Short Answer Type
NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Matching Type
NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Assertion and Reason Type
NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Long Answer Type
Major Subtopics in NCERT Exemplar Class 12 Chemistry Solutions Chapter 9 Coordination Compounds
NCERT Exemplar Class 12 Chemistry Solutions Chapter 9 Coordination Compounds - Learning Outcome
NCERT Exemplar Class 12 Chemistry Solutions
Important Topics in NCERT Exemplar Class 12 Chemistry Solutions Chapter 9 Coordination Compounds

Also, check - NCERT Solutions for Class 12, other subjects

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: MCQ (Type 1)

Question:1

Which of the following complexes formed by $\text {Cu}^{2+}$ ions is most stable?
(i) $\text {Cu}^{2+}+\text {4NH}_{3}\rightleftharpoons \left [ \text {Cu}\left ( \text {NH}_{3} \right )_{4} \right ]^{2+}\; \; \; \; \; \; \; \text {logK}= 11.6$
(ii) $\text {Cu}^{2+}+\text {4CN}^{-}\rightleftharpoons \left [ \text {Cu}\left ( \text {CN} \right )_{4} \right ]^{2-}\; \; \; \; \; \; \; \text {logK}= 27.3$
(iii) $\text {Cu}^{2+}+\text {2en}\rightleftharpoons \left [ \text {Cu}\left ( \text {en} \right )_{2} \right ]^{2+}\; \; \; \; \; \; \; \text {logK}= 15.4$
(iv) $\text {Cu}^{2+}+\text {4H}_{2}\text {O}\rightleftharpoons \left [ \text {Cu}\left ( \text {H}_{2}\text {O} \right )_{4} \right ]^{2+}\; \; \; \; \; \; \; \text {logK}= 8.9$
Answer:

Option (ii) is the correct answer.
Explanation:
The stability of a compound will be more when the value of log K increases.
For reaction,
$\text {Cu}^{2+}+\text {4CN}^{-}\rightleftharpoons \left [ \text {Cu}\left ( \text {CN} \right )_{4} \right ]^{2-}$
$\text {K}=\frac{\left [ \text {Cu}\left ( \text {CN}_{4} \right ) ^{2-}\right ]}{\left [ \text {Cu}^{2+} \right ]\left [ \text {CN}^{-} \right ]^{4}}\;\text {and} \; \text {log K}=27.3$
Log K has the highest value for this reaction amongst all the four reactions. The value of K will also be higher so, the stability of this complex will be highest.

Question:2

The colour of the coordination compounds depends on the crystal field splitting. What will be the correct order of absorption of wavelength of light in the visible region, for the complexes, $\left [ \text {Co}\left ( \text {NH}_{3} \right )_{6} \right ]^{3+}$ , $\left [ \text {Co}\left ( \text {CN} \right )_{6} \right ]^{3-}$ , $\left [ \text {Co}\left ( \text {H}_{2}\text {O} \right )_{6} \right ]^{3+}$
(i) $\left [ \text {Co}\left ( \text {CN} \right )_{6} \right ]^{3-}> \left [ \text {Co}\left ( \text {NH}_{3} \right )_{6} \right ]^{3+}> \left [ \text {Co}\left ( \text {H}_{2}\text {O} \right )_{6} \right ]^{3+}$
(ii) $\left [ \text {Co}\left ( \text {NH}_{3} \right )_{6} \right ]^{3+}> \left [ \text {Co}\left ( \text {H}_{2}\text {O} \right )_{6} \right ]^{3+}> \left [ \text {Co}\left ( \text {CN} \right )_{6} \right ]^{3-}$
(iii) $\left [ \text {Co}\left ( \text {H}_{2}\text {O} \right )_{6} \right ]^{3+}> \left [ \text {Co}\left ( \text {NH}_{3} \right )_{6} \right ]^{3+}> \left [ \text {Co}\left ( \text {CN} \right )_{6} \right ]^{3-}$
(iv) $\left [ \text {Co}\left ( \text {CN} \right )_{6} \right ]^{3-}> \left [ \text {Co}\left ( \text {NH}_{3} \right )_{6} \right ]^{3+}> \left [ \text {Co}\left ( \text {H}_{2}\text {O} \right )_{6} \right ]^{3+}$
Answer:

Option (iii) is the correct answer.
Explanation :
$\Delta _{0}$ values follow the order : $\left [ \text {Co}\left ( \text {H}_{2}\text {O} \right )_{6} \right ]^{3+}< \left [ \text {Co}\left ( \text {NH}_{3} \right )_{6} \right ]^{3+}< \left [ \text {Co}\left ( \text {CN} \right )_{6} \right ]^{3-}$ and therefore,absorption wavelength follows the order:
$\left [ \text {Co}\left ( \text {H}_{2}\text {O} \right )_{6} \right ]^{3+}> \left [ \text {Co}\left ( \text {NH}_{3} \right )_{6} \right ]^{3+}> \left [ \text {Co}\left ( \text {CN} \right )_{6} \right ]^{3-}$

Question:3

When 0.1 mol $\text {CoCl}_{3}\left ( \text {NH}_{3} \right )_{5}$ is treated with an excess of $\text {AgNO}_{3}$ , 0.2 mol of $\text {AgCl}$ are obtained. The conductivity of the solution will correspond to
(i) 1:3 electrolyte
(ii) 1:2 electrolyte
(iii) 1:1 electrolyte
(iv) 3:1 electrolyte
Answer:

Option (ii) is the correct answer.
Explanation:
One mole of chloride ion gets precipitated by one mole of $\text {AgNO}_{3}$ . $\text {AgCl}$ , is obtained when 0.1 mole of $\text {CoCl}_{3}\left ( \text {NH}_{3} \right )_{5}$ is treated with $\text {AgNO}_{3}$ in excess. So this leaves two free chloride ions in the solution of electrolyte for every 1 mole reaction.
Hence the molecular formula should be $\left [ \text {Co}\left ( \text {NH}_{3} \right )_{5}\text {Cl} \right ]\text {Cl}_{2}$ and the solution of electrolyte should contain $\left [ \text {Co}\left ( \text {NH}_{3} \right )_{5}\text {Cl} \right ]^{2+}$ and two Cl as their constituent ions. Therefore, it is an 1:2 electrolyte.
$\left [ \text {Co}\left ( \text {NH}_{3} \right )_{5}\text {Cl} \right ]\text {Cl}_{2} \rightarrow \left [ \text {Co}\left ( \text {NH}_{3} \right )_{5}\text {Cl} \right ]^{2+} \text {(aq)}+\text {2Cl}^{-} \text {(aq)}$

Question:4

When 1 mol $\text {CrCl}_{3}.\text {6H}_{2}\text {O}$ is treated with an excess of $\text {AgNO}_{3}$ , 3 mol of $\text {AgCl}$ are obtained. The formula of the complex is:
(i) $\left [ \text {CrCl}_{3}\left ( \text {H}_{2}\text {O}\right )_{3} \right ].\text {3H}_{2}\text {O}$
(ii) $\left [ \text {CrCl}_{2}\left ( \text {H}_{2}\text {O}\right )_{4} \right ]\text {Cl}.\text {2H}_{2}\text {O}$
(iii) $\left [ \text {CrCl}\left ( \text {H}_{2}\text {O}\right )_{5} \right ]\text {Cl}_{2}.\text {H}_{2}\text {O}$
(iv) $\left [ \text {Cr}\left ( \text {H}_{2}\text {O}\right )_{6} \right ]\text {Cl}_{3}$
Answer:

Option (iv) is the correct answer.
Explanation: 3 mol of $\text {AgCl}$ indicates that $\text {3Cl}^{-}$ ions are given in the solution. Therefore, the formula should be $\left [ \text {Cr}\left ( \text {H}_{2}\text {O}\right )_{6} \right ]\text {Cl}_{3}$ .

Question:5

The correct $\text {IUPAC}$ name of $\left [ \text {Pt}\left ( \text {NH}_{3} \right )_{2}\text {Cl}_{2} \right ]$ is
(i) Diamminedichloridoplatinum (II)
(ii) Diamminedichloridoplatinum (IV)
(iii) Diamminedichloridoplatinum (0)
(iv) Dichloridodiammineplatinum (IV)
Answer:

Option (i) is the correct answer.
Explanation: diamminedichloridoplatinum (II) is $\left [ \text {Pt}\left ( \text {NH}_{3} \right )_{2}\text {Cl}_{2} \right ]$ .

Question:6

The stabilisation of coordination compounds due to chelation is called the chelate effect. Which of the following is the most stable complex species?
(i) $\left [ \text {Fe}\left ( \text {CO} \right ) _{5}\right ]$
(ii) $\left [ \text {Fe}\left ( \text {CN} \right ) _{6}\right ]^{3-}$
(iii) $\left [ \text {Fe}\left ( \text {C}_{2}\text {O}_{4} \right ) _{3}\right ]^{3-}$
(iv) $\left [ \text {Fe}\left ( \text {H}_{2}\text {O} \right ) _{6}\right ]^{3+}$
Answer:

Option (iii) is the correct answer.
Explanation: Chelation is the formation of cycle of linkages between the metal ion and ligands is responsible for stabilizing the coordination compounds. A ligand which chelates any metal ion is known as chelating ligand.
Here, oxalate ion is the chelating ligand and $\left [ \text {Fe}\left ( \text {C}_{2}\text {O}_{4} \right ) _{3}\right ]^{3-}$ is the coordination compound. Hence, the oxalate ions stabilize the coordination compound by chelating $\text {Fe}^{3+}$ ions.

Question:7

Indicate the complex ion which shows geometrical isomerism.
(i) $\left [ \text {Cr}\left ( \text {H}_{2}\text {O} \right )_{4}\text {Cl}_{2} \right ]^{+}$
(ii) $\left [ \text {Pt}\left ( \text {NH}_{3}\right )_{3}\text {Cl}\right ]$
(iii) $\left [ \text {Co}\left ( \text {NH}_{3}\right )_{6}\right ]^{3+}$
(iv) $\left [ \text {Co}\left ( \text {CN}\right )_{5}\left ( \text {NC} \right )\right ]^{3-}$
Answer:

Option (i) is the correct answer.
Explanation :
(a) $\left [ \text {Cr}\left ( \text {H}_{2}\text {O} \right )_{4}\text {Cl}_{2} \right ]^{+}$

Question:8

The CFSE for octahedral $\left [ \text {CoCl}_{6} \right ]^{4-}$ is 18,000 cm^–1. The CFSE for tetrahedral $\left [ \text {CoCl}_{4} \right ]^{2-}$ will be
(i) 18,000 $\text {cm}^{-1}$
(ii) 16,000 $\text {cm}^{-1}$
(iii) 8,000 $\text {cm}^{-1}$
(iv) 20,000 $\text {cm}^{-1}$
Answer:

Option (iii) is the correct answer.
Explanation :
(c) CFSE for tetrahedral complex is $\Delta _{t}=\left ( \frac{4}{9} \right )\Delta _{0}$
$\Delta _{t}=\frac{4}{9} \times 18000=8,000 \; \text {cm}^{-1}$

Question:9

Due to the presence of ambidentate ligands coordination compounds show isomerism. Palladium complexes of the type $\left [ \text {Pd}\left ( \text {C}_{6}\text {H}_{5} \right )_{2}\left ( \text {SCN} \right )_{2} \right ]$ and $\left [ \text {Pd}\left ( \text {C}_{6}\text {H}_{5} \right )_{2}\left ( \text {NCS} \right )_{2} \right ]$ are
(i) linkage isomers
(ii) coordination isomers
(iii) ionisation isomers
(iv) geometrical isomers
Answer:

Option (i) is the correct answer.
Explanation: Ambident ligands are those ligands which have two different bonding sites. For example: $\text {NCS,NO}_{2}$ etc.
In this case, $\text {NCS}$ contains two binding sites at N and S. Therefore, $\text {NCS}$ (thiocyanate) will be bonding to the metal in these two ways; $\text {M}\leftarrow \text {NCS}$ or $\text {M}\rightarrow \text {SNC}$
Hence, the coordination compounds containing $\text {NCS}$ as ligands can show linkages isomerism; $\left [ \text {Pd}\left ( \text {C}_{6}\text {H}_{5} \right )_{2}\left ( \text {SCN} \right )_{2} \right ]$ and $\left [ \text {Pd}\left ( \text {C}_{6}\text {H}_{5} \right )_{2}\left ( \text {NCS} \right )_{2} \right ]$ are linkage isomers.

Question:10

The compounds $\left [ \text {Co}\left ( \text {SO}_{4}\right ) \left ( \text {NH}_{3} \right ) _{5}\right ]\text {Br}$ and $\left [ \text {Co}\left ( \text {SO}_{4}\right ) \left ( \text {NH}_{3} \right ) _{5}\right ]\text {Cl}$ represent
(i) linkage isomerism
(ii) ionisation isomerism
(iii) coordination isomerism
(iv) no isomerism
Answer:

Option (iv) is the correct answer.
Explanation: $\left [ \text {Co}\left ( \text {SO}_{4}\right ) \left ( \text {NH}_{3} \right ) _{5}\right ]\text {Br}$ and $\left [ \text {Co}\left ( \text {SO}_{4}\right ) \left ( \text {NH}_{3} \right ) _{5}\right ]\text {Cl}$ are different compounds and show no isomerism.

Question:11

A chelating agent has two or more than two donor atoms to bind to a single metal ion. Which of the following is not a chelating agent?
(i) thiosulphate
(ii) oxalate
(iii) glycinato
(iv) ethane-1,2-diamine
Answer:

Option (i) is the correct answer.
Explanation: Thiosulphate or $\text {S}_{2}\text {O}_{3}^{2-}$ is a monodentate ligand and not a chelating agent.

Question:12

Which of the following species is not expected to be a ligand?
(i) $\text {NO}$
(ii) $\text {NH}_{4}^{+}$
(iii) $\text {NH}_{2}\text {CH}_{2}\text {CH}_{2}\text {NH}_{2}$
(iv) $\text {CO}$
Answer:

Option (ii) is the correct answer.
Explanation: To form M-L bonds, the ligand must donate a pair of electrons or any loosely held electron pair to the metal.
e. g
Among all $\text {NH}_{4}^{+}$ does not have any pair of electrons.
Hence $\text {NH}_{4}^{+}$ is not a ligand.

Question:13

What kind of isomerism exists between $\left [ \text {Cr}\left ( \text {H}_{2}\text {O} \right )_{6} \right ]\text {Cl}_{3}\; \text {(violet)}$ and $\left [ \text {Cr}\left ( \text {H}_{2}\text {O} \right )_{5} \text {Cl}\right ]\text {Cl}_{2}.\text {H}_{2}\text {O}\; \text {(greyish-green)}$ ?
(i) linkage isomerism
(ii) solvate isomerism
(iii) ionisation isomerism
(iv) coordination isomerism
Answer:

Option (ii) is the correct answer.
Explanation: The compound has different water molecules in number inside and outside of the coordination sphere.

Question:14

IUPAC name of $\left [ \text {Pt}\left ( \text {NH}_{3} \right )_{2}\text {Cl}\left ( \text {NO}_{2} \right ) \right ]$ is :
(i). Platinum diaminechloronitrite
(ii). Chloronitrito-N-ammineplatinum (II)
(iii). Diamminechloridonitrito-N-platinum (II)
(iv). Diamminechloronitrito-N-platinate (II)
Answer:

Option (iii) is the correct answer.
Explanation: diamminechloridonitrito-N-platinum (II) is $\left [ \text {Pt}\left ( \text {NH}_{3} \right )_{2}\text {Cl}\left ( \text {NO}_{2} \right ) \right ]$

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: MCQ (Type 2)

Question:15

The atomic number of $\text {Mn, Fe and Co}$ are 25, 26 and 27 respectively. Which of the following inner orbital octahedral complexions are diamagnetic?
(i) $\left [ Co(NH_{3})_{6} \right ]^{3+}$
(ii) $\left [ Mn(CN)_{6} \right ]^{3-}$
(iii) $\left [ Fe(CN)_{6} \right ]^{4-}$
(iv) $\left [ Fe(CN)_{6} \right ]^{3-}$
Answer:

Option (i) and (iii) are the correct answers.
Explanation :
(a,c)
(a) Molecular orbital electronic configuration of $Co^{3+}$ in $\left [ Co(NH_{3})_{6} \right ]^{3+}$ is

Number of unpaired electron = 0
Magnetic property = Diamagnetic
(b) Molecular orbital electronic configuration of $Mn^{3+}$ in $\left [ Mn(CN)_{6} \right ]^{3-}$

Number of unpaired electron = 2
Magnetic property = Paramagnetic
(c) Molecular orbital electronic configuration of $Fe^{2+}$ in $\left [ Fe(CN)_{6} \right ]^{4-}$

Number of unpaired electron = 0
Magnetic property = Diamagnetic
(d) Molecular orbital electronic configuration of $Fe^{3+}$ in $\left [ Fe(CN)_{6} \right ]^{3-}$

Number of unpaired electron = 1
Magnetic property = Paramagnetic

Question:16

The atomic number of $\text {Mn, Fe, Co and Ni}$ are 25, 26 27 and 28 respectively. Which of the following outer orbital octahedral complexes have the same number of unpaired electrons?
(i) $\left [ MnCl_{6} \right ]^{3-}$
(ii) $\left [ FeF_{6} \right ]^{3-}$
(iii) $\left [ CoF_{6} \right ]^{3-}$
(iv) $\left [ Ni(NH_{3})_{6} \right ]^{2+}$
Answer:

Option (i) and (iii) are the correct answer.
Explanation :
(a,c)
$[MnCl_{6}]^{3-}: Mn^{3+}(3d^{4})$
$[CoF_{6}]^{3-}: Co^{3+}(3d^{6})$

Question:17

Which of the following options are correct for $[Fe(CN)_{6}]^{3-}$ complex?
(i) $d^{2}sp^{3}$ hybridisation
(ii) $sp^{3}d^{2}$ hybridisation
(iii) paramagnetic
(iv) diamagnetic
Answer:

Option (i) and (iii) are the correct answers.
Explanation :
(a,c) $[Fe(CN)_{6}]^{3-}$

Magnetic nature -paramagnetic.

Question:18

An aqueous pink solution of cobalt(II) chloride changes to deep blue on the addition of an excess of $HCl$ . This is because____________.
(i) $[Co(H_{2}O)_{6}]^{2+}$ is transformed into $[CoCl_{6}]^{4-}$
(ii) $[Co(H_{2}O)_{6}]^{2+}$ is transformed into $[CoCl_{4}]^{2-}$
(iii) tetrahedral complexes have smaller crystal field splitting than octahedral complexes.
(iv) tetrahedral complexes have larger crystal field splitting than octahedral complex.
Answer:

Option (ii) and (iii) are the correct answers.
Explanation: The transition of electrons from t_2g to e_g energy level gives an aqueous pink solution of cobalt (II) chloride in $[Co(H_{2}O)_{6}]^{2+}$ complex and when excess $HCl$ is added to the solution:
(i) $[Co(H_{2}O)_{6}]^{2+}$ is transformed into $[CoCl_{4}]^{2-}$
(ii) Tetrahedral complexes have smaller crystal field spliting than octahedral complexes because $\Delta _{r}=\frac{4}{9}\Delta _{0}$
Hence, options (b) and (c) are correct.

Question:19

Which of the following complexes is homoleptic?
(i) $\left [ Co\left ( NH_{3} \right )_{6} \right ]^{3+}$
(ii) $\left [ Co\left ( NH_{3} \right )_{4}Cl_{2} \right ]^{+}$
(iii) $\left [ Ni\left ( CN \right )_{4} \right ]^{2-}$
(iv) $\left [ Ni\left ( NH_{3} \right )_{4}Cl_{2} \right ]$
Answer:

Option (i) and (iii) are the correct answers.
Explanation: Both $\text {Co and Ni}$ are attached the same kind of ligands in the complexes $\left [ Co\left ( NH_{3} \right )_{6} \right ]^{3+}$ and $\left [ Ni\left ( CN \right )_{4} \right ]^{2-}$ . Hence they are homoleptic.

Question:20

Which of the following complexes are heteroleptic?
(i) $\left [ Cr(NH_{3})_{6} \right ]^{3+}$
(ii) $\left [ Fe(NH_{3})_{4}Cl_{2} \right ]^{+}$
(iii) $\left [ Mn(CN)_{6}\right ]^{4-}$
(iv) $\left [ Co(NH_{3})_{4}Cl_{2} \right ]$
Answer:

Option (ii) and (iv) are the correct answers.
Explanation: Metal is bonded to more than one kind of ligands in complexes $\left [ Fe(NH_{3})_{4}Cl_{2} \right ]^{+}$ and $\left [ Co(NH_{3})_{4}Cl_{2} \right ]$ therefore they are heteroleptic.

Question:21

Identify the optically active compounds from the following: -
(i) $[Co(en)_{3}]^{3+}$
(ii) $trans-[Co(en)_{2}Cl_{2}]^{+}$
(iii) $cis-[Co(en)_{2}Cl_{2}]^{+}$
(iv) $[Cr(NH_{3})_{5}Cl]$

Answer:

Option (i) and (iii) are the correct answers.
(a,c) $[Co(en)_{3}]^{3+}$ and $cis-[Co(en)_{2}Cl_{2}]^{+}$ are optically active compounds because their mirror images are non superimposable isomer.

Non-superimposable isomers of $[Co(en)_{3}]^{3+}$

Non-superimosable isomers of $[Co(en)_{2}Cl_{2}]^{+}$
Hence, (a) and (c) are correct choices.

Question:22

Identify the correct statements for the behaviour of ethane-1, 2-diamine as a ligand.
(i) It is a neutral ligand.
(ii) It is a didentate ligand.
(iii) It is a chelating ligand.
(iv) It is a unidentate ligand.
Answer:

Option (i), (ii) and (iii) are the correct answers.
Explanation :
(a,b,c) Ethane 1, 2-diamine is a neutral and didentate ligand. It is a chelanting agent.

Question:23

Which of the following complexes show linkage isomerism?
(i) $[Co(NH_{3})_{5}(NO_{2})]^{2+}$
(ii) $[Co(H_{2}O)_{5}CO]^{3+}$
(iii) $[Cr(NH_{3})_{5}SCN]^{2+}$
(iv) $[Fe(en)_{2}Cl_{2}]^{+}$

Answer:

Option (i) and (iii) are the correct answers.
Explanation: $NO_{2}$ and $SCN$ show linkage isomerism because they are ambidentate ligands.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Short Answer Type

Question:24

Arrange the following complexes in the increasing order of conductivity of their solution: $[Co(NH_{3})_{3}Cl_{3}], [Co(NH_{3})_{4}Cl_{2}] Cl, [Co(NH_{3})_{6}]Cl_{3} , [Cr(NH_{3})_{5}Cl]Cl_{2}$
Answer:

The correct order of increase in conductivity is given below:
$[Co(NH_{3})_{3}Cl_{3}] < [Co(NH_{3})_{4}Cl_{2}] Cl < [Cr(NH_{3})_{5}Cl]Cl_{2} < [Co(NH_{3})_{6}]Cl_{3}$

Question:25

A coordination compound $CrCl_{3}.4H_{2}O$ precipitates silver chloride when treated with silver nitrate. The molar conductance of its solution corresponds to a total of two ions. Write the structural formula of the compound and name it.
Answer:

There should be one free chlorine atom left outside the coordination sphere if silver chloride is formed in the reaction. $\left [ Cr\left ( H_{2}O \right )_{4}Cl_{2} \right ]Cl$ has to be the formula and it is called tetraaquadichlorido chromium(III) chloride.

Question:26

A complex of the type $[M(AA)_2X_2]^{n+}$ is known to be optically active. What does this indicate about the structure of the complex? Give one example of such complex.
Answer:

If the complex is optically active then its structure has to be cis-octahedral. $[Co(en)_{2}Cl_{2}]^{+}$ is one of the prominent examples of an optically stable compound.

Question:27

The magnetic moment of $\left [ MnCl_{4} \right ]^{2-}$ is $5.92\; BM$ . Explain giving reason.
Answer:

A magnetic moment of 5.92 BM indicates that there are at least 5 unpaired electrons. The formula for Magnetic Moment $=\sqrt{n(n+2)}$
As four ligands are attached to $Mn^{2+}$ , therefore the geometry tetrahedral with 5 unpaired electrons gives a magnetic moment of 5.92 BM.

Question:28

Based on crystal field theory explain why $\text {Co(III)}$ forms a paramagnetic octahedral complex with weak field ligands whereas it forms a diamagnetic octahedral complex with strong field ligands.
Answer:

$t^{4}_{2g} e^{2} _{g}$ will be the electronic configuration where it indicates that it has 4 unpaired electrons and is paramagnetic, with a weak ligand $\Delta _{0}<p$ . The configuration will be $t^{6}_{2g} e^{0}_{g}$ with strong field ligand and there will not be any unpaired electrons. Hence it is diamagnetic.

Question:29

Why are low spin tetrahedral complexes not formed?
Answer:

The crystal field splitting energy is very low for tetrahedral complexes. The crystal field splitting energy is lower than pairing energy, therefore pairing of electrons is not favorable. Therefore, the complexes may not form low spin complexes

Question:30

Give the electronic configuration of the following complexes based on the basis of Crystal Field Splitting theory.
$[CoF_{6}]^{3-}, [Fe(CN)_{6}]^{4-} and \; [Cu(NH_{3})_{6}]^{2+}.$
Answer:

Question:31

Explain why $[Fe(H_{2}O)_{6}]^{3+}$ has a magnetic moment value of 5.92 BM whereas $[Fe(CN)_{6}]^{3-}$ has a value of only 1.74 BM.
Answer:

[Fe(CN)₆]^3- involves d²sp³ hybridisation with one unpaired electron and [Fe(H₂O)₆]³⁺ involves sp³d² hybridisation with five unpaired electrons. This difference is due to the presence of strong ligand CN– and weak ligand H₂O in these complexes

Question:32

Arrange following complex ions in increasing order of crystal field splitting energy ( $\Delta _{o}$ ) :
$[Cr(Cl)_{6}]^{3-}, [Cr(CN)_{6}]^{3-}, [Cr(NH_{3})_{6}]^{3+}.$

Answer:

The correct increasing order of crystal field is:
$[Cr(Cl)_{6}]^{3-}< [Cr(NH_{3})_{6}]^{3+} < [Cr(CN)_{6}]^{3-}$
According to the spectrochemical series, this is also the correct increasing of order of the field strength.

Question:33

Why do compounds having similar geometry have a different magnetic moment?
Answer:

The difference lies in the number of paired and unpaired electrons. Strong field ligand can easily cause the pairing of electrons while the weak field ligands are not able to form pairs. The magnetic moment of a compound depends on the number of unpaired or paired electrons. Hence it is different for compounds having similar geometry.

Question:34

$\text {CuSO}_{4}.\text {5H}_{2}\text {O}$ is blue while $\text {CuSO}_{4}$ is colourless. Why?
Answer:

In $\text {CuSO}_{4}.\text {5H}_{2}\text {O}$ , the extra 5 water act as ligands. They apparently excite the electrons to the higher d orbital which shows the blue color. In $\text {CuSO}_{4}$ , there are no water molecule that can act as ligands, so there is no crystal field slitting happening, hence there is no color.

Question:35

Name the type of isomerism when ambidentate ligands are attached to a central metal ion. Give two examples of ambidentate ligands.
Answer:

The ligands which have two binding sites are known as Ambidendate ligands. Some of the examples are: Nitrite-N, Nitrito-O and Isothiocyanato, Thiocyanato.
Linkage Isomerism is a type of isomerism when ambidentate ligands get attached to central metal ions. This happens because it only differs in the atom that they are linked the central metal ion.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Matching Type

Question:36

In the following question match the items given in Columns I and II.
Match the complex ions given in Column I with the colours given in Column II and assign the correct code :

Column I (Complex ion)	Column II (Colour)
(A) $\left [ Co\left ( NH_{3} \right ) _{6}\right ]^{3+}$	1. Violet
(B) $\left [ Ti\left ( H_{2}O\right ) _{6}\right ]^{3+}$	2. Green
(C) $\left [ Ni\left ( H_{2}O\right ) _{6}\right ]^{2+}$	3. Pale blue
(D) $\left [ Ni\left ( H_{2}O\right ) _{4}(en)\right ]^{2+}(aq)$	4.Yellowish Orange
	5.Blue

(i) A (1) B (2) C (4) D (5)
(ii) A (4) B (3) C (2) D (1)
(iii) A (3) B (2) C (4) D (1)
(iv) A (4) B (1) C (2) D (3)

Answer:

Option (ii) is the correct answer.

Question:37

In the following question match the items given in Columns I and II.
Match the coordination compounds given in Column I with the central metal atoms given in Column II and assign the correct code :

Column I (Coordination Compound)	Column II ( Central metal atom)
A. Chlorophyll	1. rhodium
B. Blood pigment	2. Cobalt
C.Wilkinson catalyst	3.Calcium
D Vitamin B₁₂	4.Iron
	5. magnesium

(i) A (5) B (4) C (1) D (2)
(ii) A (3) B (4) C (5) D (1)
(iii) A (4) B (3) C (2) D (1)
(iv) A (3) B (4) C (1) D (2)

Answer:

Option (i) is the correct answer.

Question:38

In the following question match the items given in Columns I and II.
Match the complex ions given in Column I with the hybridisation and number of unpaired electrons given in Column II and assign the correct code

Column I (Complex ion)	Column II (Hybridisation, number of unpaired electrons)
A. $\left [ Cr\left ( H_{2}O \right )_{6} \right ]^{3+}$	1. $dsp^{2},1$
B. $\left [ Co\left ( CN \right )_{4} \right ]^{2-}$	2. $sp^{3}d^{2},5$
C. $\left [ Ni\left ( NH_{3} \right )_{6} \right ]^{2+}$	3. $d^{2}sp^{3},3$
D. $\left [ MnF_{6} \right ]^{4-}$	4. $sp^{3},4$
	5. $sp^{3}d^{2},2$

(i) A (3) B (1) C (5) D (2)
(ii) A (4) B (3) C (2) D (1)
(iii) A (3) B (2) C (4) D (1)
(iv) A (4) B (1) C (2) D (3)

Answer:

Option (i) is the correct answer.
Explanation : (i) Strong field ligand forms inner orbital complex with hybridisation $d^{2}sp^{3}$
(ii) Weak field ligand forms outer orbital complex with hybridisation $d^{2}sp^{3}$ .
According to VBT, hybridisation and number of unpaired electrons of coordination compounds can be calculated as
(a) $\left [ Cr\left ( H_{2}O \right )_{6} \right ]^{3+}$
MOEC (Molecular orbital electronic configuration) of $Cr^{3+}$ in $\left [ Cr\left ( H_{2}O \right )_{6} \right ]^{3+}$ is

Hybridisation = $d^{2}sp^{3}$
n (number of unpaired electrons) = 3
(b) $\left [ Co\left ( CN \right )_{4} \right ]^{2-}$ is
MOEC of $Co^{2+}$ in $\left [ Co\left ( CN \right )_{4} \right ]^{2-}$ is

Hybridisation = $dsp^{2}$
n (number of unpaired electrons) = 1
(c) $\left [ Ni\left ( NH_{3} \right )_{6} \right ]^{2+}$
MOEC of $Ni^{2+}$ in $\left [ Ni\left ( NH_{3} \right )_{6} \right ]^{2+}$ is

Hybridisation = $sp^{3}d^{2}$
n (number of unpaired electrons) = 2
(d) $\left [ MnF_{6} \right ]^{4-}$
MOEC of $Mn^{2+}$ in $\left [ MnF_{6} \right ]^{4-}$ is

Hybridisation = $sp^{3}d^{2}$
n (number of unpaired electrons) = 5
Hence , correct choice can be represented by (a).

Question:39

In the following question match the items given in Columns I and II.
Match the complex species given in Column I with the possible isomerism given in Column II and assign the correct code :

Column I (Complex species)	Column II (Isomerism)
A. $[Co (NH_{3})_{4}Cl_{2}]^{+}$	1. Optical
B. $cis-[Co (en)_{2}Cl_{2}]^{+}$	2. Ionisation
C. $[Co (NH_{3})_{5}(NO_{2})]Cl_{2}$	3. Coordination
D. $[Co (NH_{3})_{6}]\left [ Cr(CN)_{6} \right ]$	4. Geometrical
	5. Linkage

(i) A (1) B (2) C (4) D (5)
(ii) A (4) B (3) C (2) D (1)
(iii) A (4) B (1) C (5) D (3)
(iv) A (4) B (1) C (2) D (3)

Answer:

Option (iv) is the correct answer.
Explanation : The types of ligands, arrangement of ligands and the geometry of coordination decide the isomerism in coordination compounds.
(a) $[Co (NH_{3})_{4}Cl_{2}]^{+}$ shows geometrical isomerism due to presence of two types of ligand whoes $[Co (NH_{3})_{4}Cl_{2}]^{+}$ arrangement around central metal ion

(b) $cis-[Co (en)_{2}Cl_{2}]^{+}$ shows optical isomer due to its non - superimposable mirror image relationship.

(c) $[Co (NH_{3})_{5}(NO_{2})]Cl_{2}$ shows ionization isomer due to its interchanging ligand from outside the ionization sphere.
(d) $[Co (NH_{3})_{6}]\left [ Cr(CN)_{6} \right ]$ shows coordination isomerdue to interchanging of ligand in between two metal ions from one coordination sphere to another coordination sphere.
Hence, correct choice is (d)

Question:40

In the following question match the items given in Columns I and II.
Match the compounds given in Column I with the oxidation state of cobalt present in it (given in Column II) and assign the correct code.

Column I (Compound)	Column II (Oxidation state of Co)
A. $\left [ Co\left ( NCS \right )\left ( NH_{3} \right )_5 \right ]\left ( SO_{3} \right )$	1. +4
B. $\left [ Co\left ( NH_{3} \right )_{4}Cl_{2} \right ] SO_{4}$	2. 0
C. $Na_{4}\left [ Co(S_{2}O_{3}) _{3}\right ]$	3. +1
D. $[Co_{2}(CO)_{8}]$	4. +2
	5. +3

(i) A (1) B (2) C (4) D (5)
(ii) A (4) B (3) C (2) D (1)
(iii) A (5) B (1) C (4) D (2)
(iv) A (4) B (1) C (2) D (3)

Answer:

Option (iii) is the correct answer.
Explanation: The oxidation state of the whole molecule is considered to calculate the oxidation state of CMI (central metal ion).
(a) $\left [ Co\left ( NCS \right )\left ( NH_{3} \right )_5 \right ]\left ( SO_{3} \right )$
Let oxidation state of $Co$ be x.
$x-1+5\times 0=+2$
$x=+2+1=+3$
(b) $\left [ Co\left ( NH_{3} \right )_{4}Cl_{2} \right ] SO_{4}$
Let oxidation state of $Co$ = x.
$\Rightarrow x+4\times0+2\times(-1)=+2$
$x-2=+2$
$x=4$
(c) $Na_{4}\left [ Co(S_{2}O_{3}) _{3}\right ]$
Let oxidation state of $Co$ = x.
$x+3\times(-2)=-4$
$x-6=-4$
$x=-4+6=+2$
(d) $[Co_{2}(CO)_{8}]$
Let oxidation state of $Co$ = x.
$x-8\times 0=0$
$x=0$
Hence, correct choice is (d)

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Assertion and Reason Type

Question:41

In the following question a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Toxic metal ions are removed by the chelating ligands.
Reason: Chelate complexes tend to be more stable.
(i) Assertion and reason both are true, the reason is the correct explanation of assertion.
(ii) Assertion and reason both are true, but the reason is not the correct explanation of assertion.
(iii) An assertion is true, the reason is false.
(iv) The assertion is false, the reason is true.
Answer:

Option (i) is correct answer.
Explanation: Ligands chelate the metal ions by forming stable complex when solution of chelating ligands is added to a solution containing toxic metals.

Question:42

In the following question a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: $[Cr(H_{2}O)_{6}]Cl_{2}$ and $[Fe(H_{2}O)_{6}]Cl_{2}$ are reducing in nature.
Reason: Unpaired electrons are present in their d-orbitals.
(i) Assertion and reason both are true, the reason is the correct explanation of assertion.
(ii) Assertion and reason both are true, but the reason is not the correct explanation of assertion.
(iii) An assertion is true, the reason is false.
(iv) The assertion is false, the reason is true.
Answer:

Option (ii) is correct answer.
Explanation: In both the complexes, $\text {Cr and Fe}$ Coexist as $\text {Cr}^{2+}$ and $\text {Fe}^{2+}$ respectively, these both complexes become stable when oxidation of metal ion happens to $\text {Cr}^{3+}$ and $\text {Fe}^{3+}$ .

Question:43

In the following question a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Linkage isomerism arises in coordination compounds containing ambidentate ligand.
Reason: Ambidentate ligand has two different donor atoms.
(i) Assertion and reason both are true, the reason is the correct explanation of assertion.
(ii) Assertion and reason both are true, but the reason is not the correct explanation of assertion.
(iii) An assertion is true, the reason is false.
(iv) The assertion is false, the reason is true.
Answer:

Option (i) is correct answer.
Explanation: Two different donor atoms in ambidentate ligand gives rise to linkage isomerism.

Question:44

In the following question a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Complexes of MX₆ and MX₅L type (X and L are unidentate) do not show geometrical isomerism.
Reason: Geometrical isomerism is not shown by complexes of coordination number 6.
(i) Assertion and reason both are true, the reason is the correct explanation of assertion.
(ii) Assertion and reason both are true, but the reason is not the correct explanation of assertion.
(iii) An assertion is true, the reason is false.
(iv) The assertion is false, the reason is true.
Answer:

Option (iii) is correct answer.
Explanation: A different arrangement of ligands is not possible for complexes of MX₆ and MX₅L type. The complexes MA₄B₂, M(AA)₂B₂ and MA₃B₃ with coordination number 6 show geometrical isomerism.

Question:45

In the following question a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: $[Fe(CN)_{6}]^{3-}$ ion shows magnetic moment corresponding to two unpaired electrons.
Reason: Because it has $d^{2}sp^{3}$ type hybridisation.
(i) Assertion and reason both are true, the reason is the correct explanation of assertion.
(ii) Assertion and reason both are true, but the reason is not the correct explanation of assertion.
(iii) An assertion is true, the reason is false.
(iv) The assertion is false, the reason is true.
Answer:

Option (iv) is correct answer.
Explanation: Corresponding to one unpaired ion, $[Fe(CN)_{6}]^{3-}$ shows magnetic moment.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Long Answer Type

Question:46

Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following :(i) $[ CoF_{6}]^{3-}, [Co(H_{2}O)_{6}]^{2+}, [Co(CN)_{6}]^{3-}$
(ii) $[ FeF_{6}]^{3-}, [Fe(H_{2}O)_{6}]^{2+}, [Fe(CN)_{6}]^{4-}$
Answer:

(i)
$[ CoF_{6}]^{3-}$ :
$\text {Co}^{3+}=\text {3d}^{6}$
Number of unpaired electrons =4
Magnetic moment
$=\sqrt{n(n+2)}=\sqrt{4(4+2)}=4.9 B.M$
$[Co(H_{2}O)_{6}]^{2+}$

$\text {Co}^{2+}=\text {3d}^{7}$
Number of unpaired electrons =3
Magnetic moment = $\sqrt{3(3+2)}=3.87 B.M$
$[Co(CN)_{6}]^{3-}$

$\text {Co}^{3+}=\text {3d}^{6}$
Number of unpaired electrons =0
Diamagnetic .
(ii)
$[ FeF_{6}]^{3-}$

$Fe^{3+}=3d^{5}$
Number of unpaired electrons =5
Magnetic moment = $\sqrt{5(5+2)}=5.92\; B.M$
$[Fe(H_{2}O)_{6}]^{2+}$

$Fe^{2+}=3d^{6}$
Number of unpaired electrons =4
Magnetic moment = $\sqrt{4(4+2)}=4.9 \; B.M$
$[Fe(CN)_{6}]^{4-} :$

$Fe^{2+}=3d^{6}$
Diamagnetic.

Question:47

Using valence bond theory, explain the following in relation to the complexes gives below:
$\left [ Mn(CN)_{6} \right ]^{3-},\left [ Co(NH_{3})_{6} \right ]^{3+},\left [ Cr(H_{2}O) _{6}\right ]^{3+},\left [ FeCl_{6} \right ]^{4-}$
(i) Type of hybridisation.
(ii) Inner or outer orbital complex.
(iii) Magnetic behaviour.
(iv) Spin only magnetic moment value.

Answer:

$\left [ Mn(CN)_{6} \right ]^{3-} :$
$Mn^{3+}=3d^{4}$
(i) Hybridisation - $d^{2}sp^{3}$
(ii) Inner orbital complex
(iii) Paramagnetic
(iv) Magnetic moment $\sqrt{2(2+2)}=2.87 \; B.M$
$\left [ Co(NH_{3})_{6} \right ]^{3+}:$
$Co^{3+}=3d^{6}$

(i) Hybridisation - $d^{2}sp^{3}$
(ii) Inner orbital complex
(iii) Diamagnetic
(iv) Magnetic moment=0
$\left [ Cr(H_{2}O) _{6}\right ]^{3+}$
$Cr^{3+}=3d^{3}$

(i) Hybridisation - $d^{2}sp^{3}$
(ii) Inner orbital complex
(iii) Paramagnetic
(iv) Magnetic moment= $\sqrt{3(3+2)}=3.87 \; B.M$
$\left [ FeCl_{6} \right ]^{4-}$
$Fe^{2+}=3d^{6}$

(i) Hybridisation - $sp^{3}d^{2}$
(ii) Outer orbital complex
(iii) Paramagnetic
(iv) Magnetic moment= $\sqrt{4(4+2)}=4.9 \; B.M$

Question:48

$\text {CoSO}_{4}\text {Cl}.\text {5NH}_{3}$ exists in two isomeric forms ‘A’ and ‘B’. Isomer ‘A’ reacts with $\text {AgNO}_{3}$ to give white precipitate, but does not react with $\text {BaCl}_{2}$ . Isomer ‘B’ gives white precipitate with $\text {BaCl}_{2}$ but does not react with $\text {AgNO}_{3}$ . Answer the following questions.

(i) Identify ‘A’ and ‘B’ and write their structural formulas.
(ii) Name the type of isomerism involved.
(iii) Give the IUPAC name of ‘A’ and ‘B’.
Answer:

$\text {CoSO}_{4}\text {Cl}.\text {5NH}_{3}$ :
(i). The first isomer A reacts with $\text {AgNO}_{3}$ and not with $\text {BaCl}_{2}$ , that means it has a $\text {Cl}^{-}$ ion outside the coordination sphere. Therefore, A is $\left [ Co\left ( NH_{3} \right )_{5}SO_{4} \right ]Cl$
B reacts with $\text {BaCl}_{2}$ and does not react with $\text {AgNO}_{3}$ , that means it has $SO^{-}_{4}$ outside the coordination sphere. Therefore B is $\left [ Co\left ( NH_{3} \right )_{5}Cl\right ]SO_{4}$
(ii). Ionisation Isomerism
(iii). A is Pentaamminesulphatocobalt (III) chloride while B is Pentaamminechlorocobalt (III) sulphate.

Question:49

What is the relationship between observed colour of the complex and the wavelength of light absorbed by the complex?
Answer:

Some part of the white light is absorbed when it falls on a complex compound. The crystal field splitting is inversely proportional to the wavelength absorbed by the complex i.e. higher the crystal field splitting, lower is the wavelength absorbed by the complex. The observed color by the complex is the color generated from the wavelength left over.

Question:50

Why are different colours observed in octahedral and tetrahedral complexes for the same metal and same ligands?
Answer:

In octahedral complex, the lower wavelength of light is absorbed more than tetrahedral complex for a same metal and ligand. Formation of the d-orbital splitting is inverted in tetrahedral coordination entity and it is smaller than the octahedral field splitting. It can be shown as below for a same metal ligand.
$\Delta _{t}=\left ( \frac{4}{9} \right )\Delta _{o}.$

As per NCERT exemplar class 12 Chemistry solutions chapter 9 Coordination Compounds, the multiple-choice question helps students to have a varied option while they solve the problems. These questions are taken from the NCERT Exemplar Class 12 Chemistry Solutions Chapter 9; therefore, the students can ace it if they are thorough with the text.

Major Subtopics in NCERT Exemplar Class 12 Chemistry Solutions Chapter 9 Coordination Compounds

Coordination Compounds
Werner’s Theory of Coordination Compounds
Definitions of Some Important Terms Pertaining to Coordination Compounds
Nomenclature of Coordination Compounds
Isomerism in Coordination Compounds
Bonding in Coordination Compounds
Bonding in Metal Carbonyls
Stability of Coordination Compounds
Importance and Applications of Coordination Compounds

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9 Coordination Compounds - Learning Outcome

In Class 12 Chemistry NCERT exemplar solutions chapter 9 the students will learn about the different topics where they will get to know that there are two types of valency. They will also get to know about isomerism, such as optical isomerism and geometrical isomerism exist in coordination compounds. Every topic in this NCERT exemplar solutions for Class 12 Chemistry chapter is fairly detailed and accurately explained.

This helps Students learn about the qualitative analysis in labs for gravimetric analysis and volumetric analysis and calorimetric analysis. This is important to learn as its used agriculture for regulating the concentration of certain metal ions in the soil. It is used while making antidote in medicine for heavy metal poisoning. Overall, NCERT exemplar Class 12 Chemistry solutions chapter 9 apart from being informative, it is also fun and interesting for all those chemistry lovers who love to find different reactions and compounds.

NCERT Exemplar Class 12 Chemistry Solutions

Chapter 1 Solid State

Chapter 2 Solutions

Chapter 3 Electrochemistry

Chapter 4 Chemical Kinetics

Chapter 5 Surface Chemistry

Chapter 6 General Principles and Processes of Isolation of Elements

Chapter 7 The p-Block Elements

Chapter 8 The d & f Block Elements

Chapter 10 Haloalkanes and Haloarenes

Chapter 11 Alcohols, Phenols, and Ethers

Chapter 12 Aldehydes, Ketones, and Carboxylic Acids

Chapter 13 Amines

Chapter 14 Biomolecules

Chapter 15 Polymers

Chapter 16 Chemistry in Everyday Life

Important Topics in NCERT Exemplar Class 12 Chemistry Solutions Chapter 9 Coordination Compounds

· These are a few important aspects of the coordination compounds:

EDTA; 1, 10 -phenanthroline, dimercaprol, D- penicillamine. These topics are provided in detailed in the NCERT exemplar class 12 chemistry chapter 9 solutions.

NCERT exemplar Class 12 Chemistry solutions chapter 9 talks about the addition of molecular compounds which retain their identity in solid state as well as in dissolved state. In these compounds. The central metal atom or ion is linked by ions or molecules with coordinate bonds.

· it provides a detailed insight about the valency and bonding between various compounds. The structures are detailed and fairly good which help the students have a better understanding about the chapter.

NCERT Exemplar Class 12 Solutions

NCERT Exemplar Class 12 Mathematics Solutions

NCERT Exemplar Class 12 Physics Solutions

NCERT Exemplar Class 12 Biology Solutions

Check NCERT Solutions for Class 12 Chemistry

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

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Frequently Asked Question (FAQs)

1. 1. What is class 12 Chemistry Chapter 9

- The class 12 coordination compounds solutions will help students to address their doubts and conceptual mistakes. Working on these [problems and questions will make them well prepared for their upcoming board exams.

2. 2. Who can benefit from these NCERT exemplar Class 12 Chemistry chapter 9 solutions?

- Students who are preparing for the board exams and students preparing for entrance exams, both can equally benefit from this.

3. 3. How to download these NCERT exemplar Class 12 Chemistry solutions Chapter 9?

- One can download the solutions in PDF format from the solutions page itself by a single link on the given link.

4. 4. What one can learn from this class 12 chemistry NCERT exemplar solutions chapter 9?

- Students will learn about the coordination compounds and their valency. they will also learn about its types and the chemical reactions with various other compounds.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

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 will have my boards in Feb 2023 (Class 12) and I still haven't completed 1 chapter in any subject (Physics, chemistry and maths) , My 11th concepts are not that great but its okish, I wanted to know that If I do start studying from 15th of August Can I get above 85?

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

Read Complete Answer

khushi.thakurbbk 07 September,2022

I want to take a admission in class 12 privately.. so can I???

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

Read Complete Answer

Parvati Solanki 02 June,2022

View All

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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Also Read: Career as Nurse

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Resource Links for Online MBA

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A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements.

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The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.

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A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy.

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News and Notifications

October 30, 2023 - 09:26 AM IST :

CBSE Class 12 Time Table Released: Check Details Here

August 01, 2023 - 04:11 PM IST :

CBSE Class 12th supplementary exam result 2023 released. Check now.

July 27, 2023 - 09:43 AM IST :

CBSE Class 12 Compartment result 2023 out soon.

CBSE Board Exams 2024: Class 12 physical education syllabus has 10 units

November 29, 2023 - 01:05 PM IST

Board Exams 2024 Date Sheet Live: CBSE 10th, 12th dates expected soon; UP, ICSE, MP timetable updates

November 28, 2023 - 11:53 AM IST