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

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

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

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Option B is correct

yes, but 1 electron transfer to 4p orbital

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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

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

Ag¹+ is d¹⁰ so it is diamagnetic

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

the hybridization should be dsp2. The Cu atom is in form of Cu+2 in the compound So, there would be a rearrangement of electrons in Cu2+ because of the 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. [because 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, the e− from 3d would take place. So, you can say the hybridisation here would be dsp2. Square planar with one unpaired electron.

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

wait I'm send you it's hybridisation