The shape of $[Cu(NH_{3})_{4}]SO_{4}$ is:

Question

Accepted Answer

The shape of the complex $[Cu(NH_{3})_{4}]SO_{4}$ is square planar, as indicated by Option 1. Here is a detailed explanation supporting this:

1. **Justification for Square Planar (Option 1):**
   - The complex $[Cu(NH_{3})_{4}]SO_{4}$ involves a copper ion coordinated with four ammonia molecules. Copper in this complex is typically in the +2 oxidation state, which often leads to a d^9 electronic configuration.
   - In such cases, the Jahn-Teller effect is significant, which causes distortion in the geometry of the complex. This distortion often results in a square planar geometry for copper(II) complexes, especially when coordinated with strong field ligands like ammonia.
   - The square planar geometry minimizes electron repulsion and stabilizes the complex, making it the most favorable configuration for $[Cu(NH_{3})_{4}]^{2+}$.

2. **Explanation for Incorrectness of Other Options:**
   - **Pyramidal (Option 2):** A pyramidal shape is typically associated with molecules having a central atom with three bonds and one lone pair, such as in ammonia (NH₃). In the case of $[Cu(NH_{3})_{4}]SO_{4}$, the coordination number is four, and there are no lone pairs on the copper ion that would lead to a pyramidal shape.
   - **Octahedral (Option 3):** An octahedral geometry is common for complexes with a coordination number of six. Since $[Cu(NH_{3})_{4}]SO_{4}$ has a coordination number of four, it cannot adopt an octahedral shape.
   - **Tetrahedral (Option 4):** While tetrahedral geometry is possible for four-coordinate complexes, it is less common for copper(II) complexes with strong field ligands like ammonia. The square planar geometry is more stable due to the Jahn-Teller effect, as previously mentioned.

In conclusion, the square planar shape is the most appropriate and stable geometry for the $[Cu(NH_{3})_{4}]^{2+}$ complex, making Option 1 the correct answer.