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NCERT Exemplar Class 12 Chemistry Solutions Chapter 9 Coordination Compounds

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9 Coordination Compounds

Edited By Sumit Saini | Updated on Sep 16, 2022 05:46 PM IST | #CBSE Class 12th
Upcoming Event
CBSE Class 12th  Exam Date : 15 Feb' 2025 - 15 Feb' 2025

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
  1. NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: MCQ (Type 1)
  2. NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: MCQ (Type 2)
  3. NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Short Answer Type
  4. NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Matching Type
  5. NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Assertion and Reason Type
  6. NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Long Answer Type
  7. Major Subtopics in NCERT Exemplar Class 12 Chemistry Solutions Chapter 9 Coordination Compounds
  8. NCERT Exemplar Class 12 Chemistry Solutions Chapter 9 Coordination Compounds - Learning Outcome
  9. NCERT Exemplar Class 12 Chemistry Solutions
  10. 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 t2g to eg 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
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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

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

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Frequently Asked Questions (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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quartely question paper for mathematics cbse

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

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

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

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

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

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

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Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

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

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.

  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.

  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.

  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.

  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

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

I hope this information helps you.







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Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

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

Scholarship Details:

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

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

Hope you find this useful!

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Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

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

hope this helps.

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Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

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

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

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

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

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kumardurgesh1802 10 July,2024
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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
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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 3.8×107 J 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

 
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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
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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
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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 .
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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
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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.
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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.
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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
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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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