Question:
Match List I (species) with List II (hybridisation) and select the correct code given below
\[
\begin{array}{c c c}
List I & & List II
A.\; XeF_4 & & (i)\; dsp^2
B.\; H_2O & & (ii)\; sp^3
C.\; PCl_5 & & (iii)\; sp^3d^2
D.\; [Pt(NH_3)_4]^{2+} & & (iv)\; sp^3d
\end{array}
\]
Codes:
{c c c c c}
& A & B & C & D
Match List I (species) with List II (hybridisation) and select the correct code given below
\[
\begin{array}{c c c}
List I & & List II
A.\; XeF_4 & & (i)\; dsp^2
B.\; H_2O & & (ii)\; sp^3
C.\; PCl_5 & & (iii)\; sp^3d^2
D.\; [Pt(NH_3)_4]^{2+} & & (iv)\; sp^3d
\end{array}
\]
Codes:
{c c c c c}
& A & B & C & D
Show Hint
Square planar \(\Rightarrow\) \(dsp^2\), trigonal bipyramidal \(\Rightarrow\) \(sp^3d\), octahedral \(\Rightarrow\) \(sp^3d^2\).
Updated On: Apr 16, 2026
- & (iii) & (ii) & (iv) & (i)
- & (i) & (iv) & (ii) & (iii)
- & (iii) & (iv) & (ii) & (i)
- & (i) & (ii) & (iv) & (iii)
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Concept:
Hybridisation depends on number of electron pairs and geometry.
Step 1: XeF\(_4\).
Square planar \(\Rightarrow\) \[ sp^3d^2 \]
Step 2: H\(_2\)O.
Bent shape \(\Rightarrow\) \[ sp^3 \]
Step 3: PCl\(_5\).
Trigonal bipyramidal \(\Rightarrow\) \[ sp^3d \]
Step 4: [Pt(NH\(_3\))\(_4\)]\(^{2+}\).
Square planar (d\(^8\)) \(\Rightarrow\) \[ dsp^2 \] Final Matching: \[ A \to (iii),\quad B \to (ii),\quad C \to (iv),\quad D \to (i) \]
Step 1: XeF\(_4\).
Square planar \(\Rightarrow\) \[ sp^3d^2 \]
Step 2: H\(_2\)O.
Bent shape \(\Rightarrow\) \[ sp^3 \]
Step 3: PCl\(_5\).
Trigonal bipyramidal \(\Rightarrow\) \[ sp^3d \]
Step 4: [Pt(NH\(_3\))\(_4\)]\(^{2+}\).
Square planar (d\(^8\)) \(\Rightarrow\) \[ dsp^2 \] Final Matching: \[ A \to (iii),\quad B \to (ii),\quad C \to (iv),\quad D \to (i) \]
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