Shib Sagar posted an Question
August 24, 2020 • 21:33 pm 30 points
  • IIT JAM
  • Chemistry (CY)

Please solve these questions

ksmsm jsksks. jskssl > VO d) VO> TiO> MnO0> CaO 98.

2 Answer(s) Answer Now
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  • Suman Kumar

    Q 97 option D is correct

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    Ca ²+. is d⁰ configuration so zero CFSE. Mn²+ has also zero CFSE as it d5 configuration.

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    TM ion form complexes so we have to arrange in order of CFSE

  • Suman Kumar Best Answer

    Q 96 option C is correct

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    Flourine is a weak ligand

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    sir can you please explain it

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    See octahedral complexes has higher cfse so option B is eliminated.

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    Now CN- is stronger ligand than F-

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    sir in case of option a and c both r d8

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    if i consider sfl and wfl configaration is comes out e4t24

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    Ni(CN)4 ²- is a square planar complex

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    Square planar CFSE ke bare mai kya kehna. iski CFSE highest

  • Suman Kumar

    Q95 option B is correct

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    Co³+ d6 configuration so it will have very CFSE. So it normal spinel

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    Normal spinel have lamda value = 0

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    For inverse spinel lamda = 0.5

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    Priyanshu kumar best-answer

    96 ...Option C is correct.. Fluorine is a weak field ligand so it's CFSE value is less.

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    For 97 option D is correct

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    For ques 95 option B is correct lambda=0 for perfectly normal spinel

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    got this dear shib??

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    please ask if you have any doubt in this😊🙏

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    sir can you please explain 96

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    dear shib...F is a weak field ligand...and you know weak field ligand has less cfse value

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    CN is a strong field ligand

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