Payal posted an Question
July 11, 2020 • 18:25 pm 30 points
  • IIT JAM
  • Chemistry (CY)

The compound which does not show paramagnetism is: (a) [cu(nh,),jci, (c) 7. (b) agnh,),jci (d) no no

The compound which does not show paramagnetism is: (A) [Cu(NH,),JCI, (C) 7. (B) AgNH,),JCI (D) NO NO

2 Answer(s) Answer Now
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  • comment-profile-img>
    Priyanshu kumar

    see Meghna got this

    cropped6340560007341419580.jpg
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    got it...so 1 unpaired electron in 4p orbital...so paramagnetic.

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    please mark the answer if you got thisšŸ˜ŠšŸ™

  • comment-profile-img>
    Dinesh khalmaniya 1

    NO2 and NO have odd electrons and exit as dimer while Cu2+ have d8 configuration, d2sp3 hub so doesn't have lp so option A,C,D are incorrect

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    Ag+ have d10 electrons so doesn't have lp and is diamagnetic

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    if you have any doubt you can ask me I'll be happy to help you šŸ™šŸ˜Š

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    Option B is correct

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    yes, but 1 electron transfer to 4p orbital

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    check image sent by me

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    ok sir

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    please accept the answer

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    😊😊

  • comment-profile-img>
    Priyanshu kumar best-answer

    No2 is paramagnetic but its dimer is diamagnetic

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    please accept the answer if it helpsšŸ˜ŠšŸ™

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    check unpaired electrons...if it contains then it is para if not then dia...Ag + has d10 configuration...no unpaired electron so dia

  • Suman Kumar Best Answer

    Option B. [Ag(Nh3)2]Cl2

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    Cu²+ is d⁹ configuration. it has unpaired eletron. NO & NO2 too have unpaired eletron.

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    Ag¹+ is d¹⁰ so it is diamagnetic

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    I can't understand the underline part of it's solutions

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    the hybridization should beĀ dsp2. The Cu atom is in form ofĀ Cu+2Ā in the compound So, there would be a rearrangement of electrons inĀ Cu2+Ā because of theĀ NH3​ ligand (which is a strong one). And the last electron in the d-orbital would be out waiting for the N's electrons to fill up first. [becauseĀ NH3​ is a strong ligand and the electrons donated from it should have got a place first--this is only punctuation] So, the 4 electron pairs from N would be in one 3d, one 4s, & two 4p orbitals and in the third place of 4p orbital, theĀ eāˆ’Ā fromĀ 3dĀ would take place. So, you can say the hybridisation here would beĀ dsp2. Square planar with one unpaired electron.

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    I got it.Thank you sir

  • comment-profile-img>
    Dinesh khalmaniya 1

    Option B Ag+ have d10 configuration so doesn't have lone pair of electrons

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    please accept the ans if you got it šŸ™šŸ™

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    sir I can't understand the underlined part of it's solution.

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    wait I'm send you it's hybridisation

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