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Comprehension TypeDirection (Q. Nos. 16and 17) This section contains a paragraph, describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).PassageThe observed oxidation state of lanthanides either in solution or in insoluble compounds form tripositive cations. Based on the ionisation enthalpies and hydration energy factors, it is concluded that tripositive species is more stable than di or tetrapositive species in aqueous solution. In general, ions having half-filled (Tb4+ , Eu2+ , Gd3+) or completely filled 4f orbitals (Yb2+, Lu3+) or noble gas configuration (La3+, Ce4+) are stable. This is supported by their SRP values. Ions with + 4 oxidation state act as oxidising and ions having + 2 oxidation state act as reducing agents. Ions having same number of unpaired electrons in f-orbitals possess same colour.Q.Observe the following equations.Sm3+(aq) + e-→ Sm2+ (aq), E° = - 1.55 VEu3+(aq) + e-→ Eu2+ (aq), E° = - 0.43 VYb3+(aq) + e-→ Yb2+ (aq), E° = - 1.55VBased on the above equations, the correct reducing strength is in the order ofa)Sm2+ > Yb2+ > Eu2+b)Eu2+ > Yb2+ >Sm2+c)Eu2+ > Sm2+ > Yb2+d)Sm2+ > Eu2+ > Yb2+Correct answer is option 'A'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Comprehension TypeDirection (Q. Nos. 16and 17) This section contains a paragraph, describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).PassageThe observed oxidation state of lanthanides either in solution or in insoluble compounds form tripositive cations. Based on the ionisation enthalpies and hydration energy factors, it is concluded that tripositive species is more stable than di or tetrapositive species in aqueous solution. In general, ions having half-filled (Tb4+ , Eu2+ , Gd3+) or completely filled 4f orbitals (Yb2+, Lu3+) or noble gas configuration (La3+, Ce4+) are stable. This is supported by their SRP values. Ions with + 4 oxidation state act as oxidising and ions having + 2 oxidation state act as reducing agents. Ions having same number of unpaired electrons in f-orbitals possess same colour.Q.Observe the following equations.Sm3+(aq) + e-→ Sm2+ (aq), E° = - 1.55 VEu3+(aq) + e-→ Eu2+ (aq), E° = - 0.43 VYb3+(aq) + e-→ Yb2+ (aq), E° = - 1.55VBased on the above equations, the correct reducing strength is in the order ofa)Sm2+ > Yb2+ > Eu2+b)Eu2+ > Yb2+ >Sm2+c)Eu2+ > Sm2+ > Yb2+d)Sm2+ > Eu2+ > Yb2+Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Comprehension TypeDirection (Q. Nos. 16and 17) This section contains a paragraph, describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).PassageThe observed oxidation state of lanthanides either in solution or in insoluble compounds form tripositive cations. Based on the ionisation enthalpies and hydration energy factors, it is concluded that tripositive species is more stable than di or tetrapositive species in aqueous solution. In general, ions having half-filled (Tb4+ , Eu2+ , Gd3+) or completely filled 4f orbitals (Yb2+, Lu3+) or noble gas configuration (La3+, Ce4+) are stable. This is supported by their SRP values. Ions with + 4 oxidation state act as oxidising and ions having + 2 oxidation state act as reducing agents. Ions having same number of unpaired electrons in f-orbitals possess same colour.Q.Observe the following equations.Sm3+(aq) + e-→ Sm2+ (aq), E° = - 1.55 VEu3+(aq) + e-→ Eu2+ (aq), E° = - 0.43 VYb3+(aq) + e-→ Yb2+ (aq), E° = - 1.55VBased on the above equations, the correct reducing strength is in the order ofa)Sm2+ > Yb2+ > Eu2+b)Eu2+ > Yb2+ >Sm2+c)Eu2+ > Sm2+ > Yb2+d)Sm2+ > Eu2+ > Yb2+Correct answer is option 'A'. Can you explain this answer?.

Solutions for Comprehension TypeDirection (Q. Nos. 16and 17) This section contains a paragraph, describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).PassageThe observed oxidation state of lanthanides either in solution or in insoluble compounds form tripositive cations. Based on the ionisation enthalpies and hydration energy factors, it is concluded that tripositive species is more stable than di or tetrapositive species in aqueous solution. In general, ions having half-filled (Tb4+ , Eu2+ , Gd3+) or completely filled 4f orbitals (Yb2+, Lu3+) or noble gas configuration (La3+, Ce4+) are stable. This is supported by their SRP values. Ions with + 4 oxidation state act as oxidising and ions having + 2 oxidation state act as reducing agents. Ions having same number of unpaired electrons in f-orbitals possess same colour.Q.Observe the following equations.Sm3+(aq) + e-→ Sm2+ (aq), E° = - 1.55 VEu3+(aq) + e-→ Eu2+ (aq), E° = - 0.43 VYb3+(aq) + e-→ Yb2+ (aq), E° = - 1.55VBased on the above equations, the correct reducing strength is in the order ofa)Sm2+ > Yb2+ > Eu2+b)Eu2+ > Yb2+ >Sm2+c)Eu2+ > Sm2+ > Yb2+d)Sm2+ > Eu2+ > Yb2+Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12. Download more important topics, notes, lectures and mock test series for Class 12 Exam by signing up for free.

Here you can find the meaning of Comprehension TypeDirection (Q. Nos. 16and 17) This section contains a paragraph, describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).PassageThe observed oxidation state of lanthanides either in solution or in insoluble compounds form tripositive cations. Based on the ionisation enthalpies and hydration energy factors, it is concluded that tripositive species is more stable than di or tetrapositive species in aqueous solution. In general, ions having half-filled (Tb4+ , Eu2+ , Gd3+) or completely filled 4f orbitals (Yb2+, Lu3+) or noble gas configuration (La3+, Ce4+) are stable. This is supported by their SRP values. Ions with + 4 oxidation state act as oxidising and ions having + 2 oxidation state act as reducing agents. Ions having same number of unpaired electrons in f-orbitals possess same colour.Q.Observe the following equations.Sm3+(aq) + e-→ Sm2+ (aq), E° = - 1.55 VEu3+(aq) + e-→ Eu2+ (aq), E° = - 0.43 VYb3+(aq) + e-→ Yb2+ (aq), E° = - 1.55VBased on the above equations, the correct reducing strength is in the order ofa)Sm2+ > Yb2+ > Eu2+b)Eu2+ > Yb2+ >Sm2+c)Eu2+ > Sm2+ > Yb2+d)Sm2+ > Eu2+ > Yb2+Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Comprehension TypeDirection (Q. Nos. 16and 17) This section contains a paragraph, describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).PassageThe observed oxidation state of lanthanides either in solution or in insoluble compounds form tripositive cations. Based on the ionisation enthalpies and hydration energy factors, it is concluded that tripositive species is more stable than di or tetrapositive species in aqueous solution. In general, ions having half-filled (Tb4+ , Eu2+ , Gd3+) or completely filled 4f orbitals (Yb2+, Lu3+) or noble gas configuration (La3+, Ce4+) are stable. This is supported by their SRP values. Ions with + 4 oxidation state act as oxidising and ions having + 2 oxidation state act as reducing agents. Ions having same number of unpaired electrons in f-orbitals possess same colour.Q.Observe the following equations.Sm3+(aq) + e-→ Sm2+ (aq), E° = - 1.55 VEu3+(aq) + e-→ Eu2+ (aq), E° = - 0.43 VYb3+(aq) + e-→ Yb2+ (aq), E° = - 1.55VBased on the above equations, the correct reducing strength is in the order ofa)Sm2+ > Yb2+ > Eu2+b)Eu2+ > Yb2+ >Sm2+c)Eu2+ > Sm2+ > Yb2+d)Sm2+ > Eu2+ > Yb2+Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Comprehension TypeDirection (Q. Nos. 16and 17) This section contains a paragraph, describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).PassageThe observed oxidation state of lanthanides either in solution or in insoluble compounds form tripositive cations. Based on the ionisation enthalpies and hydration energy factors, it is concluded that tripositive species is more stable than di or tetrapositive species in aqueous solution. In general, ions having half-filled (Tb4+ , Eu2+ , Gd3+) or completely filled 4f orbitals (Yb2+, Lu3+) or noble gas configuration (La3+, Ce4+) are stable. This is supported by their SRP values. Ions with + 4 oxidation state act as oxidising and ions having + 2 oxidation state act as reducing agents. Ions having same number of unpaired electrons in f-orbitals possess same colour.Q.Observe the following equations.Sm3+(aq) + e-→ Sm2+ (aq), E° = - 1.55 VEu3+(aq) + e-→ Eu2+ (aq), E° = - 0.43 VYb3+(aq) + e-→ Yb2+ (aq), E° = - 1.55VBased on the above equations, the correct reducing strength is in the order ofa)Sm2+ > Yb2+ > Eu2+b)Eu2+ > Yb2+ >Sm2+c)Eu2+ > Sm2+ > Yb2+d)Sm2+ > Eu2+ > Yb2+Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Comprehension TypeDirection (Q. Nos. 16and 17) This section contains a paragraph, describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).PassageThe observed oxidation state of lanthanides either in solution or in insoluble compounds form tripositive cations. Based on the ionisation enthalpies and hydration energy factors, it is concluded that tripositive species is more stable than di or tetrapositive species in aqueous solution. In general, ions having half-filled (Tb4+ , Eu2+ , Gd3+) or completely filled 4f orbitals (Yb2+, Lu3+) or noble gas configuration (La3+, Ce4+) are stable. This is supported by their SRP values. Ions with + 4 oxidation state act as oxidising and ions having + 2 oxidation state act as reducing agents. Ions having same number of unpaired electrons in f-orbitals possess same colour.Q.Observe the following equations.Sm3+(aq) + e-→ Sm2+ (aq), E° = - 1.55 VEu3+(aq) + e-→ Eu2+ (aq), E° = - 0.43 VYb3+(aq) + e-→ Yb2+ (aq), E° = - 1.55VBased on the above equations, the correct reducing strength is in the order ofa)Sm2+ > Yb2+ > Eu2+b)Eu2+ > Yb2+ >Sm2+c)Eu2+ > Sm2+ > Yb2+d)Sm2+ > Eu2+ > Yb2+Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Comprehension TypeDirection (Q. Nos. 16and 17) This section contains a paragraph, describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).PassageThe observed oxidation state of lanthanides either in solution or in insoluble compounds form tripositive cations. Based on the ionisation enthalpies and hydration energy factors, it is concluded that tripositive species is more stable than di or tetrapositive species in aqueous solution. In general, ions having half-filled (Tb4+ , Eu2+ , Gd3+) or completely filled 4f orbitals (Yb2+, Lu3+) or noble gas configuration (La3+, Ce4+) are stable. This is supported by their SRP values. Ions with + 4 oxidation state act as oxidising and ions having + 2 oxidation state act as reducing agents. Ions having same number of unpaired electrons in f-orbitals possess same colour.Q.Observe the following equations.Sm3+(aq) + e-→ Sm2+ (aq), E° = - 1.55 VEu3+(aq) + e-→ Eu2+ (aq), E° = - 0.43 VYb3+(aq) + e-→ Yb2+ (aq), E° = - 1.55VBased on the above equations, the correct reducing strength is in the order ofa)Sm2+ > Yb2+ > Eu2+b)Eu2+ > Yb2+ >Sm2+c)Eu2+ > Sm2+ > Yb2+d)Sm2+ > Eu2+ > Yb2+Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Class 12 tests.