Match List I with List II
Column I
- i. $[Pt(Cl_2)(NH_3)_2]$
- ii. $[Co(NH_3)_6]Cl_3$
- iii. $[NiCl_4]^{2-}$
- iv. $[Fe(CO)_5]$
Column II
- a. Octahedral
- b. Trigonal bipyramidal
- c. Square planar
- d. Tetrahedral
Correct answer: i → c, ii → a, iii → d, iv → b
Solution
$[Pt(Cl_2)(NH_3)_2]$: C.N. = 4, $Pt^{2+}$ – $5d^8$, all ligands considered as SFL, so hybridisation is $dsp^2$ – square planar
$[Co(NH_3)_6]Cl_3$: C.N. = 6, hybridisation $d^2sp^3$ – octahedral
$[NiCl_4]^{2-}$: C.N. = 4, $Ni^{2+}$, $Cl^-$ acts as WFL, no pairing, hence hybridisation is $sp^3$ – tetrahedral
$[Fe(CO)_5]$: C.N. = 5, hybridisation $dsp^3$ – trigonal bipyramidal
Correct match: i-c, ii-a, iii-d, iv-b.
Coordination Compounds in past papers
104 questions from this chapter have appeared across 27 exam years.
