
The correct Lewis acid order for boron halides is:
A)${\text{BB}}{{\text{r}}_{\text{3}}}{\text{ > BC}}{{\text{l}}_{\text{3}}}{\text{ > B}}{{\text{I}}_{\text{3}}}{\text{ > B}}{{\text{F}}_{\text{3}}}$
B)${\text{B}}{{\text{I}}_{\text{3}}}{\text{ > B}}{{\text{F}}_{\text{3}}}{\text{ > BB}}{{\text{r}}_{\text{3}}}{\text{ > BC}}{{\text{l}}_{\text{3}}}$
C)${\text{B}}{{\text{F}}_{\text{3}}}{\text{ > BC}}{{\text{l}}_{\text{3}}}{\text{ > BB}}{{\text{r}}_{\text{3}}}{\text{ > B}}{{\text{I}}_{\text{3}}}$
D)${\text{B}}{{\text{I}}_{\text{3}}}{\text{ > BB}}{{\text{r}}_{\text{3}}}{\text{ > BC}}{{\text{l}}_{\text{3}}}{\text{ > B}}{{\text{F}}_{\text{3}}}$
Answer
126.9k+ views
Hint:To solve this question, it is required to have knowledge about Lewis acidity and back bonding. Lewis acidity of a compound is defined as its tendency to accept electrons. It is prominent in compounds which have vacant orbitals to accept electrons. Back bonding occurs when there is a transfer of electrons from the electron rich species to anti-bonding pi orbital of the other species attached to it. It occurs when there is a lone pair on the Lewis base and a vacant p-orbital on the Lewis acid species.
Complete step by step answer:
Boron is an electron deficient species. Its electronic configuration is ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{1}}}$. It donates the three electrons in its valence shell to each halogen atom to form ${\text{B}}{{\text{X}}_{\text{3}}}$ . This means that boron has three vacant p-orbitals where it can accept new electrons and thus acts as Lewis acid.
In the case of ${\text{B}}{{\text{F}}_{\text{3}}}$ the small size of boron and fluorine means that the atoms are really close to each other, where fluorine is the electron rich species with lone pairs and boron is the electron deficient species with an empty p-orbital. As a result, the fluorine atom donates a lone pair to the anti-bonding p-orbital of boron. So, the ${\text{B - F}}$ bond acquires partial double bond character and also contains a filled p-orbital. Thus, it does not act as a Lewis acid due to the absence of vacant p-orbital and inability to accept extra electrons.
In ${\text{BC}}{{\text{l}}_3}$ this effect is not so prominent as in ${\text{B}}{{\text{F}}_{\text{3}}}$ due to the increase in size in chlorine atom. The ${\text{2p\pi - 2p\pi }}$ back-bonding effect is more prominent than ${\text{2p\pi - 3p\pi }}$ back bonding effect. Due to the absence of back-bonding, ${\text{BC}}{{\text{l}}_3}$ shows Lewis acid character better than ${\text{B}}{{\text{F}}_{\text{3}}}$ . Similarly, this trend is followed for the other halogens as well with Lewis acidic character increasing down the group as back-bonding becomes less and less prominent. Thus, the correct order of Lewis acid character of boron halides will be : ${\text{B}}{{\text{I}}_{\text{3}}}{\text{ > BB}}{{\text{r}}_{\text{3}}}{\text{ > BC}}{{\text{l}}_{\text{3}}}{\text{ > B}}{{\text{F}}_{\text{3}}}$
$\therefore $ The correct option is option D, i.e. ${\text{B}}{{\text{I}}_{\text{3}}}{\text{ > BB}}{{\text{r}}_{\text{3}}}{\text{ > BC}}{{\text{l}}_{\text{3}}}{\text{ > B}}{{\text{F}}_{\text{3}}}$.
Note: In the questions involving halogens, we generally apply the concept of electronegativity. In this case, if we did so then fluorine being most electronegative would pull electron density towards itself and boron fluoride would be a better Lewis acid. But, the small size of fluorine atom and the small size of boron atoms causes back-bonding to occur and instead causes boron fluoride to have the least Lewis acidic character.
Complete step by step answer:
Boron is an electron deficient species. Its electronic configuration is ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{1}}}$. It donates the three electrons in its valence shell to each halogen atom to form ${\text{B}}{{\text{X}}_{\text{3}}}$ . This means that boron has three vacant p-orbitals where it can accept new electrons and thus acts as Lewis acid.
In the case of ${\text{B}}{{\text{F}}_{\text{3}}}$ the small size of boron and fluorine means that the atoms are really close to each other, where fluorine is the electron rich species with lone pairs and boron is the electron deficient species with an empty p-orbital. As a result, the fluorine atom donates a lone pair to the anti-bonding p-orbital of boron. So, the ${\text{B - F}}$ bond acquires partial double bond character and also contains a filled p-orbital. Thus, it does not act as a Lewis acid due to the absence of vacant p-orbital and inability to accept extra electrons.
In ${\text{BC}}{{\text{l}}_3}$ this effect is not so prominent as in ${\text{B}}{{\text{F}}_{\text{3}}}$ due to the increase in size in chlorine atom. The ${\text{2p\pi - 2p\pi }}$ back-bonding effect is more prominent than ${\text{2p\pi - 3p\pi }}$ back bonding effect. Due to the absence of back-bonding, ${\text{BC}}{{\text{l}}_3}$ shows Lewis acid character better than ${\text{B}}{{\text{F}}_{\text{3}}}$ . Similarly, this trend is followed for the other halogens as well with Lewis acidic character increasing down the group as back-bonding becomes less and less prominent. Thus, the correct order of Lewis acid character of boron halides will be : ${\text{B}}{{\text{I}}_{\text{3}}}{\text{ > BB}}{{\text{r}}_{\text{3}}}{\text{ > BC}}{{\text{l}}_{\text{3}}}{\text{ > B}}{{\text{F}}_{\text{3}}}$
$\therefore $ The correct option is option D, i.e. ${\text{B}}{{\text{I}}_{\text{3}}}{\text{ > BB}}{{\text{r}}_{\text{3}}}{\text{ > BC}}{{\text{l}}_{\text{3}}}{\text{ > B}}{{\text{F}}_{\text{3}}}$.
Note: In the questions involving halogens, we generally apply the concept of electronegativity. In this case, if we did so then fluorine being most electronegative would pull electron density towards itself and boron fluoride would be a better Lewis acid. But, the small size of fluorine atom and the small size of boron atoms causes back-bonding to occur and instead causes boron fluoride to have the least Lewis acidic character.
Recently Updated Pages
JEE Main 2023 (January 30th Shift 2) Maths Question Paper with Answer Key

Know The Difference Between Fluid And Liquid

JEE Main 2022 (July 25th Shift 2) Physics Question Paper with Answer Key

Classification of Elements and Periodicity in Properties Chapter For JEE Main Chemistry

JEE Main 2023 (January 25th Shift 1) Maths Question Paper with Answer Key

JEE Main 2023 (January 24th Shift 2) Chemistry Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

JEE Main Login 2045: Step-by-Step Instructions and Details

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

JEE Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics

NCERT Solutions for Class 11 Chemistry Chapter 8 Organic Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 6 Equilibrium

NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrocarbons
