Match the species in List I with their geometry in List II
Choose the correct answer from the options given below:
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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!
Step 1: Set the method. For each species we find the steric number (or hybridisation) to get the geometry, using \(S.N. = \frac{1}{2}[\text{central valence electrons} + \text{monovalent atoms} - \text{cation charge} + \text{anion charge}]\). Step 2: Species A, \(PCl_5\). P has 5 valence electrons and 5 Cl atoms: \(S.N. = (5+5)/2 = 5\), so \(sp^3d\) with no lone pairs gives trigonal bipyramidal. So A goes with III. Step 3: Species B, \(BrF_5\). Br has 7 valence electrons and 5 F atoms: \(S.N. = (7+5)/2 = 6\), \(sp^3d^2\) with one lone pair, giving square pyramidal. So B goes with IV. Step 4: Species C, \(BF_4^-\). B has 3 valence electrons, 4 F atoms, and a \(-1\) charge: \(S.N. = (3+4+1)/2 = 4\), \(sp^3\) with no lone pairs, giving tetrahedral. So C goes with I. Step 5: Species D, \([Ni(CN)_4]^{2-}\). Ni is \(+2\) giving \(3d^8\); the strong-field \(CN^-\) pairs the electrons and frees one 3d orbital, giving \(dsp^2\) hybridisation and a square planar shape. So D goes with II. Step 6: Assemble the matches. A-III, B-IV, C-I, D-II, which is option (3). \[ \boxed{\text{A-III, B-IV, C-I, D-II}} \]
