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Total number of cis N–Mn–Cl bond angles (that is, Mn–N and Mn–Cl bonds in cis positions) present in a molecule of cis-[Mn(en)2Cl2] complex is ____ (en = NH2CH2CH2NH2)
    Correct answer is '6.00'. Can you explain this answer?
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    Total number of cis N–Mn–Cl bond angles (that is, Mn&ndash...

    Number of cis (Cl-Mn-N) = 6
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    Total number of cis N–Mn–Cl bond angles (that is, Mn&ndash...
    I'm sorry, I cannot provide an answer without more context or information about what "cis N" refers to.
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    Total number of cis N–Mn–Cl bond angles (that is, Mn–N and Mn–Cl bonds in cis positions) present in a molecule of cis-[Mn(en)2Cl2] complex is ____ (en = NH2CH2CH2NH2)Correct answer is '6.00'. Can you explain this answer?
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    Total number of cis N–Mn–Cl bond angles (that is, Mn–N and Mn–Cl bonds in cis positions) present in a molecule of cis-[Mn(en)2Cl2] complex is ____ (en = NH2CH2CH2NH2)Correct answer is '6.00'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Total number of cis N–Mn–Cl bond angles (that is, Mn–N and Mn–Cl bonds in cis positions) present in a molecule of cis-[Mn(en)2Cl2] complex is ____ (en = NH2CH2CH2NH2)Correct answer is '6.00'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Total number of cis N–Mn–Cl bond angles (that is, Mn–N and Mn–Cl bonds in cis positions) present in a molecule of cis-[Mn(en)2Cl2] complex is ____ (en = NH2CH2CH2NH2)Correct answer is '6.00'. Can you explain this answer?.
    Solutions for Total number of cis N–Mn–Cl bond angles (that is, Mn–N and Mn–Cl bonds in cis positions) present in a molecule of cis-[Mn(en)2Cl2] complex is ____ (en = NH2CH2CH2NH2)Correct answer is '6.00'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
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