Question

Give preparation, hybrididation and structure
of XeF4?​

Answer 1

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♦XeF4 consists of two lone pair electrons.

♦ Now if we follow the VSEPR theory, the bond pair electrons and lone pair electrons experience repulsion between them. ...

♦The lone pairs of Xenon lie in the perpendicular plane in an octahedral arrangement.

♦Therefore, XeF4 molecular geometry is square planar.

♦Name of the Molecule: Xenon Tetrafluoride

♦Geometry: Square Planar

♦Hybridization Type: sp2d2

♦Bond Angle: 90o or 180o

Answer 2

Explanation:

To Do :-

1. Preparation of XeF4
2. Hybridisation
3. Structure

1)

Preparation :-

• When Xenon reacts with fluoride (1:5) in 873 K and 7 bar pressure then it formed Xenon tetrafluoride.

Chemical equation :-

Xe(g) + 2F2(g)[1:5] → XeF4(s)

2)

Hybridisation of XeF4 is sp³d²

Calculation :-

H = 1/2×[V + M - C + A]

here,

V = valence electrons of central atom

M = number of monovalent atoms

C = charge of cation

A = charge of Anion

→ 1/2×[8 + 4]

→ 1/2 × 12

→ 6 (sp³d²)

3)

Structure :-

Refer to the attachment.