The type of hybridisation and number of lone pairs of electrons on $Xe$ in $XeO{F_4},$ respectively, are :

JEE CHEMISTRY MCQ: P-BLOCK ELEMENTS

1 Medium

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Analyzing +4 | -1

The type of hybridisation and number of lone pairs
of electrons on

Xe

XeO{F_4},

respectively, are :

View answer and explanation

Correct option: C

s{p^3}{d^2}

and 1

Option C (

s{p^3}{d^2}

and 1) is correct because

XeOF_4

has a square pyramidal geometry with

s{p^3}{d^2}

hybridization and one lone pair on Xe.

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17^{\text {th }}

{O}_2

{O}_2

\mathrm{C}{\mathrm{N}^ - }

{\mathrm{N}_2}

\mathrm{C}{\mathrm{N}^ - },\mathrm{\;}{\mathrm{N}_2}

2 {Fe}^{3+}+{SO}_2+2 {H}_2 {O} \rightarrow 2 {Fe}^{2+}+{SO}_4^{2-}+4 {H}^{+}

{H}_2 {~S}_2 {O}_8

{S}={O}

{S}-{OH}

{O}-

{O}

{NH}_3

{H}_2 {SO}_4

{H}_2 {SO}_4