What atomic or hybrid orbitals make up the sigma bond between br and f in bromine trifluoride, brf3?

1 answer:

8 0

<span>Draw the Lewis structure. Bromine has 3 bonds and two lone pairs for a trigonal bipyramidal electron geometry and an sp^3d hybridization. Fluorine is peripheral it does not require hybridization, but we often consider it to be hybridized too - it has 1 bond and 3 lone pairs for sp^3 hybridization. So the sigma bonds come from an overlap of an sp^3d orbital on Br with an sp^3 orbital on F. If you don't consider the F to be hybridized the overlap would have to be to a p orbital on the F</span>

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VladimirAG [237]

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3 years ago

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Match each substance with the correct designation for the equation HSO3- + CH3NH2 <=> SO32- + CH3NH3+ HSO3- CH3NH2 SO32- C

Zanzabum

Answer:

$HSO_3^-$ : conjugate acid of $SO_3^{2-}$

$CH_3NH_2$ : conjugate base of $CH_3NH_3^+$

$SO_3^{2-}$ : conjugate base of $HSO_3^-$

$CH_3NH_3^+$ : conjugate acid of $CH_3NH_2$

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

$HSO_3^-+CH_3NH_2\rightleftharpoons SO_3^{2-}+CH_3NH_3^+$

Here in forward reaction $CH_3NH_2$ is accepting a proton, thus it is considered as a base and after accepting a proton, it forms $CH_3NH_3^+$ which is a conjugate acid.

And $HSO_3^-$ is losing a proton, thus it is considered as an acid and after loosing a proton, it forms $SO_3^{2-}$ which is a conjugate base.

Similarly in the backward reaction, $CH_3NH_3^+$ is loosing a proton, thus it is considered as a acid and after loosing a proton, it forms $CH_3NH_2$ which is a conjugate base.

And $SO_3^{2-}$ is accepting a proton, thus it is considered as a base and after accepting a proton, it forms $HSO_3^{-}$ which is a conjugate acid.