Question

Match List - I with List - II

Choose the correct answer from the options given below.

• A. a - ii, b – iii, c – iv, d - i
• B. a - ii, b - i, c - iii, d – iv
• C. a – iii, b – ii, c – iv, d – i
• D. a – i, b – iii, c – iv, d – ii

Hint:

For coordination number 4, the geometry depends on the ligand.

• \textbf{Strong field ligands} with d\(^8\) metals (like Ni\(^{2+}\)) cause pairing and lead to dsp\(^2\) (square planar) geometry.
• \textbf{Weak field ligands} with d\(^8\) metals do not cause pairing and lead to sp\(^3\) (tetrahedral) geometry.

For coordination number 5, the most common geometry is trigonal bipyramidal. For coordination number 6, the geometry is almost always octahedral.

Answer 1

The Correct Option is A

Step 1: Determine the geometry of each complex.
(a) [Co(NH\(_3\))\(_6\)]\(^{3+}\):

• Co atomic number = 27. Co\(^{3+}\) configuration: [Ar] 3d\(^6\).
• NH\(_3\) is a strong field ligand, so it will cause pairing of electrons. The 6 electrons will occupy three d-orbitals.
• Hybridization involves two empty 3d orbitals, one 4s, and three 4p orbitals, leading to d\(^2\)sp\(^3\) hybridization.
• The geometry for coordination number 6 is Octahedral.
• So, a \(\rightarrow\) ii.

(b) [NiCl\(_4\)]\(^{2-}\):

• Ni atomic number = 28. Ni\(^{2+}\) configuration: [Ar] 3d\(^8\).
• Cl\(^-\) is a weak field ligand. No pairing of electrons will occur.
• Hybridization involves one 4s and three 4p orbitals, as the 3d orbitals are not empty. This leads to sp\(^3\) hybridization.
• The geometry for sp\(^3\) hybridization is Tetrahedral.
• So, b \(\rightarrow\) iii.

(c) [Ni(CN)\(_4\)]\(^{2-}\):

• Ni\(^{2+}\) configuration: [Ar] 3d\(^8\).
• CN\(^-\) is a strong field ligand. It will force the two unpaired electrons in the 3d orbitals to pair up, leaving one 3d orbital empty.
• Hybridization involves the one empty 3d orbital, one 4s, and two 4p orbitals, leading to dsp\(^2\) hybridization.
• The geometry for dsp\(^2\) hybridization is Square planar.
• So, c \(\rightarrow\) iv.

(d) [Fe(CO)\(_5\)]:

• Fe atomic number = 26. Fe(0) configuration: [Ar] 3d\(^6\) 4s\(^2\).
• CO is a strong field ligand. In the complex, the 8 valence electrons of Fe rearrange and pair up in the 3d orbitals.
• The complex has coordination number 5. The hybridization is dsp\(^3\).
• The geometry for coordination number 5 (dsp\(^3\)) is Trigonal bipyramidal.
• So, d \(\rightarrow\) i.

Step 2: Compile the matches.
The correct matches are: a \(\rightarrow\) ii, b \(\rightarrow\) iii, c \(\rightarrow\) iv, d \(\rightarrow\) i.
Step 3: Final Answer.
This combination corresponds to option (A).