JEE Advanced 2014 · Paper 2 · Chemistry Q33

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ChemistrySingle correct+3 / -1Question 33 of 60

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Paragraph For Question 33 and 34

An aqueous solution of metal ion M1 reacts separately with reagents Q and R in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion M2 always forms tetrahedral complexes with these reagents. Aqueous solution of M2 on reaction with reagent S gives white precipitate which dissolves in excess of S. The reactions are summarized in the scheme given below:

SCHEME: M1 with Q (excess) gives a tetrahedral complex and with R (excess) a square planar complex. M2 with Q (excess) or R (excess) gives tetrahedral complexes; M2 with S (stoichiometric amount) gives a white precipitate, which dissolves in excess S.

M1, Q and R, respectively are

\ce{Zn^{2+}}

, KCN and HCl

\ce{Ni^{2+}}

, HCl and KCN

\ce{Cd^{2+}}

, KCN and HCl

\ce{Co^{2+}}

, HCl, and KCN

Answer and worked solution

Official answer: B

M1 is $\ce{Ni^{2+}}$ ( $d^8$ ).

With excess HCl (Q): $\ce{[NiCl4]^{2-}}$ , weak-field, $sp^3$ : tetrahedral.
With excess KCN (R): $\ce{[Ni(CN)4]^{2-}}$ , strong-field, $dsp^2$ : square planar.

Other choices. $\ce{Zn^{2+}}$ and $\ce{Cd^{2+}}$ ( $d^{10}$ ) give only tetrahedral complexes with both reagents, so they fit M2, not M1. $\ce{Co^{2+}}$ does not form a square planar cyanide complex in excess cyanide.

(B) $\ce{Ni^{2+}}$ , HCl, KCN

Topics:Valence bond theory Crystal field theory

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JEE Advanced 2014 · Paper 2, Chemistry question 33