Question

Which one of the following molecule/ion has square planar shape

• A. SF\(_4\)
• B. NH\(_4^+\)
• C. CH\(_2\)Cl\(_2\)
• D. CH\(_4\)
• E. XeF\(_4\)

Hint:

For square planar geometry, look for molecules with a central atom having 6 electron pairs (like XeF\(_4\), ICl\(_4^-\)).

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The shape of a molecule is determined by the arrangement of its electron pairs (both bonding and non-bonding) around the central atom, according to the Valence Shell Electron Pair Repulsion (VSEPR) theory. A square planar shape arises from an octahedral electron geometry where two positions are occupied by lone pairs. This corresponds to the VSEPR notation AX\(_4\)E\(_2\).
Step 2: Detailed Explanation:
Let's determine the shape for each option by finding the number of valence electrons, bonding pairs, and lone pairs on the central atom.

(A) SF\(_4\): Central atom is Sulfur (S, Group 16, 6 valence e\(^-\)). It is bonded to 4 Fluorine atoms.
Bonding pairs = 4
Lone pairs = (6 - 4) / 2 = 1
VSEPR type: AX\(_4\)E\(_1\). The shape is a see-saw.

(B) NH\(_4^+\): Central atom is Nitrogen (N, Group 15, 5 valence e\(^-\)). It has a +1 charge, so we subtract 1 electron. Total valence e\(^-\) = 5 - 1 = 4. It is bonded to 4 Hydrogen atoms.
Bonding pairs = 4
Lone pairs = (4 - 4) / 2 = 0
VSEPR type: AX\(_4\). The shape is tetrahedral.

(C) CH\(_2\)Cl\(_2\): Central atom is Carbon (C, Group 14, 4 valence e\(^-\)). It is bonded to 2 H and 2 Cl atoms.
Bonding pairs = 4
Lone pairs = (4 - 4) / 2 = 0
VSEPR type: AX\(_4\). The shape is tetrahedral.

(D) CH\(_4\): Central atom is Carbon (C, Group 14, 4 valence e\(^-\)). It is bonded to 4 H atoms.
Bonding pairs = 4
Lone pairs = (4 - 4) / 2 = 0
VSEPR type: AX\(_4\). The shape is tetrahedral.

(E) XeF\(_4\): Central atom is Xenon (Xe, Group 18, 8 valence e\(^-\)). It is bonded to 4 Fluorine atoms.
Bonding pairs = 4
Lone pairs = (8 - 4) / 2 = 2
VSEPR type: AX\(_4\)E\(_2\). The electron geometry is octahedral. The two lone pairs position themselves on opposite sides (axial positions) to minimize repulsion, forcing the four fluorine atoms into a plane around the central xenon atom. This results in a square planar shape.

Step 3: Final Answer:
The molecule with a square planar shape is XeF\(_4\). This corresponds to option (E).