Question

The shape of \[\left[ Cu{{\left( N{{H}_{3}} \right)}_{4}} \right]S{{O}_{4}}\] is:
a.Square Planar.
b.Pyramidal.
c.Octahedral
d.Tetrahedral.

Answer 1

Hint: We know that Copper is a d- block metal. The d block elements are found in the group \[3,4,5,6,7,8,910,11\] and \[12\] of the periodic table. These are also known as transition metals. The d orbital is filled with an electronic shell $n-1.$ There are a total of \[40\] d block elements.

Complete step by step solution:
The d block elements are small in size and generally have high electron positive density. They consist of d free orbitals to accept the free electrons from the ligand and hence, form complexes easily. Cuprammonium sulfate is a coordination compound. The coordination number is the number of the ions, atoms, or molecules attached to the central metal of the compound. The IUPAC name of Cuprammonium sulfate is tetraamminecopper \[\left( II \right)\] sulfate. Therefore, the structure of this complex is, \[\left[ Cu{{\left( N{{H}_{3}} \right)}_{4}} \right]S{{O}_{4}}\] . Now in this structure, there are four ammonia molecules directly attached with the copper. Therefore, the coordination number of copper in Cuprammonium sulfate is \[4.\]
The given compound is a coordination compound. To find the IUPAC name for the given compound, we need to follow some rules: Name the ions present in the compound in the order such that the name of the cation Is placed before the anion. When naming within the complex, before the name of the central atom, the name of the ligands is written. The names of the ligands must be written in alphabetic order.
Since \[-N{{H}_{3}}~\]is a strong field ligand, it will form square planar complex with \[C{{u}^{2+~}}\] with \[ds{{p}^{2}}~\] hybridization.

Therefore, the correct answer is option A.

Note:
Remember that Transition elements show a magnetic moment due to the presence of unpaired electrons to their d orbitals. With increasing the number of unpaired electrons, the spin magnetic moment value increases and vice versa. If there is no unpaired electron then it is diamagnetic.