Question

Match List-I with List-II. 

List-I	Complex / Ion
A	[CoCl₂(NH₃)₄]
B	[Co(NH₃)₆]Cl₃
C	[NiCl₄]²⁻
D	[Fe(CO)₅]

 

List-II	Shape / Geometry
I	Octahedral
II	Trigonal bipyramidal
III	Square planar
IV	Tetrahedral

• A. A-I, B-I, C-IV, D-II
• B. A-IV, B-I, C-III, D-II
• C. A-I, B-IV, C-II, D-III
• D. A-III, B-IV, C-I, D-II

Hint:

Coordination number \(6\) usually gives octahedral geometry, while \([Fe(CO)_5]\) is trigonal bipyramidal.

Answer 1

The Correct Option is A

Step 1: Understanding the Topic:
This problem deals with the geometry and hybridization of "Coordination Compounds." According to Valence Bond Theory (VBT), the spatial arrangement of ligands around a central metal ion depends on the metal's coordination number and its hybridization state. The hybridization, in turn, is influenced by whether the ligands are "strong field" (causing electron pairing) or "weak field" (leaving electrons unpaired). Mastering these shapes is fundamental to inorganic chemistry.
Step 2: Key Formulas and Approach:
The approach involves identifying the coordination number (CN) and the metal's electronic state:
CN 4: Tetrahedral ($sp^3$) or Square Planar ($dsp^2$).
CN 5: Trigonal Bipyramidal ($dsp^3$).
CN 6: Octahedral ($d^2sp^3$ or $sp^3d^2$).

Step 3: Detailed Explanation:

A. $[PtCl_2(NH_3)_2]$: Platinum is a heavy 5d metal in the $+2$ state. Coordination number is 4. For heavy metals like Pt and Pd, $CN=4$ almost always results in a Square Planar geometry regardless of ligand strength. A matches with III.
B. $[Co(NH_3)_6]Cl_3$: Cobalt is in the $+3$ state ($d^6$). With 6 ammonia ligands, it is an octahedral complex. $NH_3$ is a strong field ligand for $Co^{3+}$, forcing electron pairing and $d^2sp^3$ hybridization. B matches with I.
C. $[NiCl_4]^{2-$:} Nickel is in the $+2$ state ($d^8$). Chloride is a weak field ligand and cannot pair the electrons. Thus, it uses $sp^3$ hybridization, resulting in a Tetrahedral geometry. C matches with IV.
D. $[Fe(CO)_5]$: Iron is in the zero oxidation state. Coordination number is 5. Using $dsp^3$ hybridization, the shape is Trigonal bipyramidal. D matches with II.
Step 4: Final Answer:
The matching sequence is A-III, B-I, C-IV, D-II, which is option (B).