Out of the following which will be see-saw shaped?

A. $XeO_2F_2$
B. $SF_4$
C. $BrF_3$
D. $XeO_4$

A and B only

The correct answer is Option (3) → A and B only.

To determine which of the given molecules are see-saw shaped, we need to analyze their molecular geometries using VSEPR (Valence Shell Electron Pair Repulsion) theory. The see-saw shape occurs when there are 5 regions of electron density (trigonal bipyramidal arrangement) around the central atom, with one lone pair, leading to a distorted shape.

Let us analyze each compound:

A. XeO₂F₂:

Xenon (Xe) has 8 valence electrons.

Oxygen and fluorine each contribute 2 and 1 electrons, respectively.

Total:  \( 8 + 2 \times 2 + 2 \times 1 = 12 \) electrons.

Xenon forms 4 bonds (2 with O and 2 with F) and has 1 lone pair.

Geometry: 5 regions of electron density (4 bonds + 1 lone pair) → see-saw shape.

B. SF₄:

Sulfur (S) has 6 valence electrons.

Fluorine contributes 4 electrons (one per bond).

Total: \( 6 + 4 \times 1 = 10 \) electrons.

Sulfur forms 4 bonds (with F) and has 1 lone pair.

Geometry: 5 regions of electron density (4 bonds + 1 lone pair) → see-saw shape.

C. BrF₃:

Bromine (Br) has 7 valence electrons.

Fluorine contributes 3 electrons (one per bond).

 Total: \( 7 + 3 = 10 \) electrons.

Bromine forms 3 bonds (with F) and has 2 lone pairs.

Geometry: 5 regions of electron density (3 bonds + 2 lone pairs) → T-shaped, not see-saw.

D. XeO₄:

Xenon (Xe) has 8 valence electrons.

Oxygen contributes 4 electrons (one per bond).

Total: \( 8 + 4 = 12 \) electrons.

Xenon forms 4 bonds (with O), with no lone pairs.

Geometry: 4 regions of electron density (4 bonds, no lone pairs) → tetrahedral, not see-saw.

Conclusion:

XeO₂F₂ and SF₄ both have a see-saw shape.

Therefore, the correct answer is A and B only.