Question

An element from the 1\(^\text{st}\) transition series and another element of the 3\(^\text{rd}\) transition series (same group) do not liberate \( \text{H}_2 \) gas from dilute acids like HCl. Both form halides. The hybridisation state of metal ion halide respectively are:

• A. Both \( \text{sp}^3 \)
• B. Both \( \text{dsp}^2 \)
• C. \( \text{sp}^3 \) and \( \text{dsp}^2 \)
• D. \( \text{dsp}^2 \) and \( \text{sp}^3 \)

Hint:

Transition elements can show different hybridisation states depending on their position in the periodic table and the availability of d-orbitals.

Answer 1

The Correct Option is C

Step 1: Characteristics of the elements.
The elements from the 1\(^\text{st}\) transition series have \( \text{sp}^3 \) hybridisation, while elements from the 3\(^\text{rd}\) transition series (e.g., \( \text{Cu}^{2+} \)) can have \( \text{dsp}^2 \) hybridisation due to the availability of d-orbitals for bonding. Step 2: Conclusion.
Therefore, the hybridisation states of the metal ions are \( \text{sp}^3 \) for the 1\(^\text{st}\) transition series element and \( \text{dsp}^2 \) for the 3\(^\text{rd}\) transition series element. Final Answer: \[ \boxed{\text{sp}^3 \text{ and } \text{dsp}^2} \]