Match the pair of complexes given in Column-I and the characteristic given in column-II. `{:("Column-I","Column-II"),((A)(NH_4)_2[NiCl_4] and (NH_(4))

Monjur Alahi

Match the pair of complexes given in Column-I and the characteristic given in column-II.
`{:("Column-I","Column-II"),((A)(NH_4)_2[NiCl_4] and (NH_(4))_(2)[NI(CN)_4],(p)"Both show same electrical conductance"),((B)CoCl_3. 6NH_3 and PtCl_4. 5NH_3,(q)"Both show same effective atomic number "),((C)[Pt(NH_3)_2Cl_2] and (NH_4)_2[PtCl_4],(r)"Both show same primary valencies "),((D )K_2[Fe(H_2O)_6] and K_4[FeCl_6], (s)"Both gives white participate with" AgNO_3 "solution"):}`


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Salim Ahmed Kawsar

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Correct Answer - A-p,q,r ; B-q,s ; C-q,r ; D-q,r
(A)`Ni^(2+)` ions in both complexes have same primary valencies, i.e, +2 Same number of ions i.e. 3 and so have same conductance. Both have same EAN i.e. 34. `Cl^(-)` ions in both complexes are in coordination sphere so no precipitate is obtained with `AgNO_3`.
(B)`overset(III)([Co(NH_3)_6]Cl_3) and overset(IV)([Pt(NH_3)_5Cl])Cl_3` have different primary valencies but same number of ions. So same electrical conductance . Do not have same effective atomic number as in both complexes metals have different atomic number. As there are three `Cl^-` ions out side the coordination sphere both will gives 3 moles of the precipitate of AgCl(white).
(C )`[Pt(NH_3)_2Cl_2] and (NH_4)_2 [PtCl_4]`. Pt is in +2 oxidation state so same primary valencies and same effective atomic number (84) and does not have same electrical conductance (former is neutral complex where as later one has 3 ions in aqueous solutons). No precipitate with `AgNO_3` as `Cl^-` ions in both complexes are in coordination sphere.
(D)Both have Fe in +2 oxidation state so same primary valencies but have different number of ions.so different electrical conductance (former has 3 ions where as later one has 5 ions in aqueous solution). Both complexes have same effective atomic number (i.e. 36).

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`Ag^(+)+hArr|Ag(NH_(3))_(2)|^(+), k_(1) = 6.8 xx 10^(-3)` `[Ag(NH_(3))]^(+) + NH_(3)hArr |Ag(NH_(3))_(2)|, k_(2) = 1.6xx10^(-3)` Then the formation co

Correct Answer - B As the reaction takes place in two stages `k = k_(1)xx k_(2) = (6.8 xx 10^(-3)) xx (1.6xx 10^(-3))` = `10.8 xx 10^(-6) = 1.08xx 10^(-5)`

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