
Assertion:The acidic strength sequence $BM{{e}_{3}} < B{{(MeO)}_{3}}$
Reason: The acidic strength can also be explained by considering the existing back bonding.
(A) Assertion is true, Reason is true, and Reason is the correct explanation for Assertion
(B) Assertion is true, Reason is true, and Reason is not the correct explanation for Assertion
(C) Assertion is true, Reason is false
(D) Assertion is false, Reason is true
Answer
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Hint: An atom in a compound has a lone pair of electrons and another atom has a vacant orbital placed adjacent to each other involving back bonding which makes the compound to be stable as this compound satisfies the octet rule. This bonding is effective only when the size of the valence shell matches.
Complete answer:
Given Assertion,
The acidic strength sequence $BM{{e}_{3}} < B{{(MeO)}_{3}}$
From the above statement, both triethyl borane and trimethoxy borane form 3 covalent bonds with boron in each compound.
Boron has 3 valence electrons and gains 3 more electrons from the covalent bonds and it has a total of 6 electrons. Because of the 6 valence electrons, both compounds do not fulfill the octet rule and boron becomes electron-deficient which acts as a Lewis acid.
So, trimethyl borane is more acidic than trimethoxy borane. i.e, $BM{{e}_{3}}>B{{(MeO)}_{3}}$
Hence, the Assertion is false.
Since triethyl borane is more acidic than trimethoxy borane due to the involvement of back-bonding of boron with oxygen.
Hence, Assertion is false, Reason is true
So, the correct answer is option C.
Note:
The back bonding of any acidic compound is inversely proportional to its acidic strength. So, in case of acidic strength, boron compounds have to know about back bonding. The compound which satisfies the octet rule in $\pi $ back bonding electrons move from the atomic orbital of the atom to ${{\pi }^{*}}$ the atomic orbital of another atom.
Complete answer:
Given Assertion,
The acidic strength sequence $BM{{e}_{3}} < B{{(MeO)}_{3}}$
From the above statement, both triethyl borane and trimethoxy borane form 3 covalent bonds with boron in each compound.
Boron has 3 valence electrons and gains 3 more electrons from the covalent bonds and it has a total of 6 electrons. Because of the 6 valence electrons, both compounds do not fulfill the octet rule and boron becomes electron-deficient which acts as a Lewis acid.
So, trimethyl borane is more acidic than trimethoxy borane. i.e, $BM{{e}_{3}}>B{{(MeO)}_{3}}$
Hence, the Assertion is false.
Since triethyl borane is more acidic than trimethoxy borane due to the involvement of back-bonding of boron with oxygen.
Hence, Assertion is false, Reason is true
So, the correct answer is option C.
Note:
The back bonding of any acidic compound is inversely proportional to its acidic strength. So, in case of acidic strength, boron compounds have to know about back bonding. The compound which satisfies the octet rule in $\pi $ back bonding electrons move from the atomic orbital of the atom to ${{\pi }^{*}}$ the atomic orbital of another atom.
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