Q.Which of the following statements is not correct about XeF6?

XeF4 is square planar (sp3d2, two lone pairs axial); its symmetric shape cancels all bond dipoles, so its dipole moment is zero. Correct option (c).

Shapes and symmetry:

• XeF4: sp3d2, six electron pairs (4 bond + 2 lone), lone pairs occupy opposite axial positions, leaving a square planar arrangement of four Xe-F bonds. The dipoles cancel in pairs -> net dipole = 0.
• XeO3: pyramidal (one lone pair) -> polar, non-zero dipole. …

Statements (a), (b), (c) about XeF6 are correct; statement (d) is wrong because hydrolysis of XeF6 gives oxyfluorides/XeO3 and HF, not Xe + HF + O2. Correct answer (the not-correct statement) = (d).

Checking each:

• (a) Oxidation number of Xe in XeF6 is +6 (6 F, each -1). Correct.
• (b) Xe has 7 electron pairs (6 bond + 1 lone), so hybridisation is sp3d3. Correct.
• (c) The lone pair distorts the octahedron, giving a distorted (capped) octahedral shape. Correct. …

Xe in XeF4 has 6 electron domains (4 bonding + 2 lone pairs), sp3d2 hybridised. The two lone pairs occupy axial positions, leaving the four F atoms in a square plane.

Count electron pairs on Xe in XeF4:

• Xe has 8 valence electrons; 4 are used to bond 4 F atoms, leaving 2 lone pairs.
• Total domains = 4 bond pairs + 2 lone pairs = 6 -> sp3d2 hybridisation -> octahedral electron geometry. …

XeO3's geometry follows from Xe being sp³ hybridised with one lone pair occupying the fourth position — the same electron arrangement as NH3, giving a pyramidal shape.

Xenon trioxide, XeO3, has Xenon in the +6 oxidation state (formed by careful hydrolysis of XeF6). Around the central Xe atom there are 3 Xe=O bond pairs plus 1 lone pair, giving 4 electron domains total → sp³ hybridisation. Since one of the four sp³ positions is occupied by a lone pair (rather than a bonded atom), the molecular shape (ignoring the lone pair) is pyramidal (trigonal pyramidal), analogous to the shape of NH3 — with Xe at the apex, above th …

Xenon forms the largest number of compounds among the noble gases — option (D).

Noble gases are largely inert, but reactivity increases down the group as ionisation enthalpy falls (the outer electrons become easier to involve in bonding). Xenon, having a relatively low ionisation enthalpy and large size, readily reacts with the most electronegative elements (fluorine and oxygen) to form many compounds such as XeF2, XeF4, XeF6, XeO3 and XeOF4. Krypton forms only a few (e.g. …

$XeF_4$ has $sp^3d^2$ hybridisation with 2 lone pairs on Xe, giving an octahedral electron geometry but a square planar molecular shape.

Xenon in $XeF_4$ has 8 valence electrons; 4 are used to form 4 Xe–F bonds, leaving 2 lone pairs. This gives 6 electron domains around Xe (4 bond pairs + 2 lone pairs), so the hybridisation is $sp^3d^2$ and the parent electronic geometry is octahedral. The two lone pairs occupy positions opposite each other (axial) to …

Xe in XeO3 has 3 bond pairs and 1 lone pair (sp3 hybridisation), giving a pyramidal molecular shape (tetrahedral electron geometry).

In $\text{XeO}_3$, the central xenon atom is bonded to three oxygen atoms (via three Xe=O bonds) and also carries one lone pair of electrons. This gives xenon a steric number of 4 (3 bond pairs + 1 lone pair), corresponding to $sp^3$ hybridisation and a tetrahedral electron-pair geometry.

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