Coordination CompoundMedium
Question
Match List I with List II :
| List I (Complex/ion) | List II | ||
|---|---|---|---|
| A. | $$\left\lbrack Pt\left( {Cl}_{2} \right)\left( {NH}_{3} \right)_{2} \right\rbrack$$ | (I) | Octahedral |
| B. | $$\left\lbrack Co\left( {NH}_{3} \right)_{6} \right\rbrack{Cl}_{3}$$ | (II) | Trigonal bipyramidal |
| C. | $$\left\lbrack {NiCl}_{4} \right\rbrack^{2 -}$$ | (III) | Square planar |
| D | $$\left\lbrack Fe(CO)_{5} \right\rbrack$$ | (IV) | Tetrahedral |
Choose the correct Ans. from the options given below:
Options
A.A-III, B-IV, C-I, D-II
B.A-III, B-I, C-IV, D-II
C.A-IV, B-I, C-III, D-II
D.A-I, B-III, C-IV, D-II
Solution
Sol. [Key Concept: Matching reagents with trAns.formations]
- In $\left\lbrack Pt\left( {Cl}_{2} \right)\left( {NH}_{3} \right)_{2} \right\rbrack$; Pt has $dsp^{2}$ hybridisation, so shape of complex is square planar.
$- \ In\left\lbrack Co\left( {NH}_{3} \right)_{6} \right\rbrack{Cl}_{3};Co$ has $d^{2}sp^{3}$ hybridisation so shape of complex is octahedral.
- In $\lbrack NiCl\rbrack^{2 -}$; Ni has $sp^{3}$ hybridisation so shape of complex ion is tetrahedral.
- In $\left\lbrack Fe(CO)_{5} \right\rbrack$; shape of complex is trigonal bipyramidal.
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