An unknown metal X forms diamagnetic square-planar [X(NH₃)₄]²⁺. Its atomic number is 46. For the same metal in oxidation state +4, assume octahedral coordination and that all NH₃ and halides are monod

Question

An unknown metal X forms diamagnetic square-planar [X(NH₃)₄]²⁺. Its atomic number is 46. For the same metal in oxidation state +4, assume octahedral coordination and that all NH₃ and halides are monodentate. Which statements are correct?

TestHub · Accepted Answer

Z = 46 identifies Pd; Pd(II) is d⁸. Pd(IV) with two ammines needs four inner halides, so [Pd(NH₃)₂Cl₄] is neutral. With four ammines only two halides fit inside, leaving two outside, not four. For total Cl₂BrI and two inner sites, pairs are ClCl, ClBr, ClI and BrI: four ionisation distributions before splitting into geometrical forms.

Chemistry - Coordination Chemistry Question with Solution | TestHub

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