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Compounds of Xenon

SbF5 reacts w...

Question

$S b F_{5}$ reacts with $X e F_{4}$ to form an adduct. The shapes of cation and anion in the adduct are respectively:

A

square planar, trigonal bipyramidal

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B

T-shaped, octahedral

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C

square pyramidal, octahedral

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D

square planar, octahedral

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Solution

The correct option is B T-shaped, octahedral
$X e F_{4}$ or xenon tetrafluoride reacts with $S b F_{5}$ and forms adduct. $X e F_{4} + S b F_{5} \to [X e F_{3}]^{+} [S b F_{6}]^{-}$
Thus, the cation is T-shaped and the anion is octahedral.
Hence, the correct option is b.

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Similar questions

S b F_{5}

X e F_{4}

X e F_{6}

to form ionic compounds

{[X e F_{3}]}_{3}^{+} {[S b F_{6}]}_{6}^{-}

{[X e F_{5}]}_{5}^{+} {[S b F_{6}]}_{6}^{-}

then molecular shape of

{[X e F_{3}]}_{3}^{+}

{[X e F_{5}]}_{5}^{+}

ion respectively.

S b F_{5}

X e F_{4}

X e F_{6}

to form ionic compounds

[X e F_{3}^{+}] [{S b F_{6}}^{-}]

[X e F_{5}^{+}] [{S b F_{6}}^{-}]

. The geometry of

X e F_{3}^{+}

[X e F_{5}^{+}]

ion respectively is:

S b F_{5}

X e F_{4}

X e F_{6}

to form ionic compounds

[X e F_{3}^{+}] [S b F_{6}]^{-} a n d [X e F_{5}^{+}] [S b F_{6}^{-}] .

The geometry of

X e F_{3}^{+}

X e F_{5}^{+}

ion, respectively, is :

B F_{3}

, the B - F bond length is 1.30 A. When

B F_{3}

is allowed to react with

M e_{3} N

, it forms an adduct,

M e_{3} N

\to

B F_{3}

. The bond length of B - F in the adduct is:

B F_{3}

forms an adduct with

N H_{3}

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