Question

Match the following

• A. A - II, B - I, C - III, D - IV
• B. A - I, B - II, C - III, D - IV
• C. A - III, B - IV, C - II, D - I
• D. A - IV, B - III, C - II, D - I

Hint:

Remember the basic shapes associated with common hybridizations. The number of hybrid orbitals corresponds to the number of sigma bonds and lone pairs around the central atom. \begin{itemize} \item sp: Linear \item sp$^2$: Trigonal planar \item sp$^3$: Tetrahedral \item dsp$^2$: Square planar \item sp$^3$d: Trigonal bipyramidal \item dsp$^3$: Trigonal bipyramidal (less common than sp$^3$d for this shape) \item sp$^3$d$^2$ or d$^2$sp$^3$: Octahedral \end{itemize}

Answer 1

The Correct Option is B

Step 1: Recall the relationship between hybridization and molecular shape.
The hybridization of the central atom in a molecule determines its electron geometry and, subsequently, its molecular shape (considering lone pairs).
Step 2: Match each hybridization with its corresponding shape. \begin{itemize} \item A. dsp$^2$: This hybridization involves one d, one s, and two p orbitals, resulting in a square planar geometry. Matches with I. Square planar. \item B. sp$^3$: This hybridization involves one s and three p orbitals, resulting in a tetrahedral geometry. Matches with II. Tetrahedral. \item C. d$^2$sp$^3$: This hybridization involves two d, one s, and three p orbitals, resulting in an octahedral geometry. Matches with III. Octahedral. \item D. sp$^3$d: This hybridization involves one s, three p, and one d orbital, resulting in a trigonal bipyramidal geometry. Matches with IV. Trigonal bipyramidal. \end{itemize}
Step 3: Combine the matches to find the correct option.
The correct matching is A - I, B - II, C - III, D - IV. Final Answer: The final answer is $\boxed{A - I, B - II, C - III, D - IV}$