Match the LIST-I with LIST-II A. PF5 I. dsp2 B. SF6 II. sp3d C. Ni(CO)4 III. sp3d2 D. [PtCl4]2- IV. sp3 Choose the correct answer from the options given below:

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q59 – JEE-MAIN 2025

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Home Chemistry Chemical BondingQuestion #249

Match the LIST-I with LIST-II

\begin{array}{lcl} LIST-I & LIST-II \\ A. P F_{5} & I. d s p^{2} \\ B. S F_{6} & II. s p^{3} d \\ C. N i (C O)_{4} & III. s p^{3} d^{2} \\ D. [P t C l_{4}]^{2 -} & IV. s p^{3} \end{array}

Choose the correct answer from the options given below:

Clue 1 of 3: Determine Steric Number

For main group elements, calculate the steric number (number of bond pairs + lone pairs) around the central atom to predict its geometry and hybridization.

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Ninja Strategy•Identify a Key Match First

By correctly identifying the hybridization of $P F_{5}$ as $s p^{3} d$ (II), all other options are immediately eliminated, leaving only the correct answer.

🧠 Full Solution Path

Step 1: Determine Hybridization for Main Group Compounds✦ Active

For $P F_{5}$ : Phosphorus (P) is the central atom with 5 valence electrons. It forms 5 single bonds with 5 Fluorine (F) atoms. There are no lone pairs. The steric number is $5 + 0 = 5$ , leading to $s p^{3} d$ hybridization (trigonal bipyramidal geometry). Thus, A $\to$ II.

For $S F_{6}$ : Sulfur (S) is the central atom with 6 valence electrons. It forms 6 single bonds with 6 Fluorine (F) atoms. There are no lone pairs. The steric number is $6 + 0 = 6$ , leading to $s p^{3} d^{2}$ hybridization (octahedral geometry). Thus, B $\to$ III.

Step 2: Determine Hybridization for Coordination Compounds○ Expand

For $N i (C O)_{4}$ : Nickel (Ni) is in the 0 oxidation state. Its electronic configuration is $[A r] 3 d^{8} 4 s^{2}$ . Carbonyl (CO) is a strong field ligand, causing pairing of electrons. The $4 s$ electrons are also pushed into the $3 d$ orbital, resulting in a $3 d^{10}$ configuration. The empty orbitals available for bonding are one $4 s$ and three $4 p$ orbitals, leading to $s p^{3}$ hybridization (tetrahedral geometry). Thus, C $\to$ IV.

For $[P t C l_{4}]^{2 -}$ : Platinum (Pt) is in the $+ 2$ oxidation state ( $x + 4 (- 1) = - 2 \Rightarrow x = + 2$ ). $P t^{2 +}$ has a $5 d^{8}$ electronic configuration. For 5d series elements, even weak field ligands like chloride ( $C l^{-}$ ) cause electron pairing. The $d^{8}$ configuration with pairing leads to one empty $d$ orbital, one $s$ orbital, and two $p$ orbitals for hybridization, resulting in $d s p^{2}$ hybridization (square planar geometry). Thus, D $\to$ I.

💡 Teacher's Secret Hint

Remember that ligand field effects (strong vs. weak field ligands) are crucial for determining hybridization in coordination compounds, especially for transition metals.

Step 3: Match the Options○ Expand

Combining the matches: A-II, B-III, C-IV, D-I. This corresponds to option 2.

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Chemistry Question 59 – JEE-MAIN 2025

Clue 1 of 3: Determine Steric Number

🧠 Full Solution Path