Match the species in List I with their geometry in List II
Choose the correct answer from the options given below:
Show Hint
When answering matching questions of this type, find the easiest or most distinct option first to eliminate incorrect alternatives quickly. For example, knowing that $\text{PCl}_5$ is Trigonal bipyramidal (A-III) instantly eliminates option (2). Subsequently, knowing that $\text{BF}_4^-$ is a classic tetrahedral species (C-I) helps you confidently select option (3) without needing to fully solve the transition metal complex configuration under high-pressure exam conditions!
Concept:
The geometry of a molecular species can be systematically determined using the Valence Shell Electron Pair Repulsion (VSEPR) theory and Coordination Chemistry models.
• Steric Number (S.N.): It is calculated as $\text{S.N.} = \frac{1}{2}[\text{Valence electrons of central atom} + \text{Number of monovalent atoms} - \text{Charge on cation} + \text{Charge on anion}]$.
• Hybridization and Geometry: The steric number indicates the hybridization and structural arrangement of electron pairs around the central atom.
• Coordination Complexes: For transition metal complexes, crystal field theory (CFT) and hybridization of $d$-orbitals dictate whether the complex is square planar ($dsp^2$) or tetrahedral ($sp^3$).
Step 1: Analyzing Species A ($\text{PCl}_5$)
Phosphorus (P) is the central atom belonging to Group 15, so it possesses 5 valence electrons. It is bonded to 5 monovalent chlorine atoms (Cl).
Using the steric number formula:
\[
\text{Steric Number} = \frac{5 + 5}{2} = 5
\]
A steric number of 5 implies $sp^3d$ hybridization. Since there are 5 bonded groups and 0 lone pairs on the central phosphorus atom, the molecular geometry matches the electronic geometry perfectly, which is
Trigonal bipyramidal.
Therefore,
A matches with III.
Step 2: Analyzing Species B ($\text{BrF}_5$)
Bromine (Br) is the central halogen atom belonging to Group 17, possessing 7 valence electrons. It is covalently bonded to 5 monovalent fluorine atoms (F).
Calculating the steric number:
\[
\text{Steric Number} = \frac{7 + 5}{2} = \frac{12}{2} = 6
\]
A steric number of 6 corresponds to $sp^3d^2$ hybridization, which defines an octahedral electronic arrangement.
Out of these 6 electron pairs, 5 are bonding pairs (associated with the F atoms) and 1 is a lone pair:
\[
\text{Number of lone pairs} = 6 - 5 = 1
\]
An octahedral geometry with one lone pair distorts to a
Square pyramidal geometry.
Therefore,
B matches with IV.
Step 3: Analyzing Species C ($\text{BF}_4^-$)
Boron (B) is the central atom belonging to Group 13, having 3 valence electrons. It is bonded to 4 fluorine atoms and carries a $-1$ anionic charge.
Calculating its steric number:
\[
\text{Steric Number} = \frac{3 + 4 - 0 + 1}{2} = \frac{8}{2} = 4
\]
A steric number of 4 corresponds to $sp^3$ hybridization. Since there are 4 sigma bonds and 0 lone pairs, the structural geometry of the molecule is perfectly
Tetrahedral.
Therefore,
C matches with I.
Step 4: Analyzing Species D ($[\text{Ni}(\text{CN})_4]^{2-}$)
This is a coordination complex where Nickel (Ni) is the central transition metal ion. Let us first determine the oxidation state of Ni:
\[
x + 4(-1) = -2 \implies x = +2
\]
Hence, we are dealing with a $\text{Ni}^{2+}$ ion. The ground state electronic configuration of neutral Nickel ($Z=28$) is $[\text{Ar}] 3d^8 4s^2$. For $\text{Ni}^{2+}$, the configuration becomes $[\text{Ar}] 3d^8 4s^0$.
Cyanide ($\text{CN}^-$) is a strong field ligand. According to Crystal Field Theory, a strong field ligand causes a pairing up of the electrons in the $3d$ orbitals:
\[
3d^8 \text{ (unpaired)} \xrightarrow{\text{pairing due to }\text{CN}^-} 3d^8 \text{ (completely paired up in 4 orbitals, leaving one } 3d \text{ orbital vacant)}
\]
The vacant $3d$ orbital, along with the $4s$ orbital and two $4p$ orbitals, undergo hybridization to form four $dsp^2$ hybrid orbitals. A coordination number of 4 with $dsp^2$ hybridization results in a
Square Planar geometry.
Therefore,
D matches with II.
Conclusion of Matching:
Combining all the deduced matches:
• A $\rightarrow$ III
• B $\rightarrow$ IV
• C $\rightarrow$ I
• D $\rightarrow$ II
This combination is exactly given in option (3).
