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CHXI11:THE P-BLOCK ELEMENTS

316045 The strongest Lewis acid among boron halides is

1

{BBr}_{3}

2

{BCl}_{3}

3

{BI}_{3}

4

{BF}_{3}

Explanation:

Generally, as the electronegativity increases, acidic character increases, but in case of boron halides, ${BI}_{3}$ is most acidic, i.e. strongest Lewis acid.This is because there is back donation of electrons from halogen atom to boron in ${BX}_{3} (X = F, Cl, Br, I)$ This makes the ${BX}_{3}$ less electron deficient. The tendency of back donation decreases as the size of halogen atom increases (due to large energy difference).Hence, due to large size of $I, {BI}_{3}$ is most electron deficient and thus, the strongest Lewis acid.

CHXI11:THE P-BLOCK ELEMENTS

316046 In $B F_{3}$ , the $B - F$ bond length is $1 .30 \overset{o}{A}$ , when ${BF}_{3}$ is allowed to be treated with ${Me}_{3} N$ , it forms an adduct, ${Me}_{3} N \to {BF}_{3}$ , The bond length of $B$ $F$ in the adduct is:

1 Greater than

1 .30 \overset{o}{A}

2 Smaller than

1 .30 \overset{o}{A}

3 Equal to

1 .30 \overset{o}{A}

4 None of these

CHXI11:THE P-BLOCK ELEMENTS

316047 The power of halides of Boron to act as Lewis acids decreases in the order of

1

{BF}_{3} > {BCl}_{3} > {BBr}_{3}

2

{BBr}_{3} > {BCl}_{3} > {BF}_{3}

3

{BCl}_{3} > {BF}_{3} > {BBr}_{3}

4

{BCl}_{3} > {BBr}_{3} > {BF}_{3}

CHXI11:THE P-BLOCK ELEMENTS

316048 In $B F_{3}$

1 B-F bond has some double bond character and this bond is delocalised

2 All the B-F bonds are single covalent in nature

3 Bond energy and bond-length of B-F bond indicate its single bond character

4 All the bonds are ionic

CHXI11:THE P-BLOCK ELEMENTS

316049 The correct statement is

1

B I_{3}

is the weakest Lewis acid among the boron halides

2 there is minimum

p π -  p π

back bonding in

{BF}_{3}

3

{BF}_{3}

is the strongest Lewis acid among the other boron halides

4 there is maximum

p π -  p π

back bonding in

{BF}_{3}

.

Explanation:

There is maximum $p π - p π$ back bonding in ${BF}_{3}$ due to identical size of $2p$ orbitals of $B$ and $F$ .

KCET - 2015

CHXI11:THE P-BLOCK ELEMENTS

316045 The strongest Lewis acid among boron halides is

1

{BBr}_{3}

2

{BCl}_{3}

3

{BI}_{3}

4

{BF}_{3}

Explanation:

Generally, as the electronegativity increases, acidic character increases, but in case of boron halides, ${BI}_{3}$ is most acidic, i.e. strongest Lewis acid.This is because there is back donation of electrons from halogen atom to boron in ${BX}_{3} (X = F, Cl, Br, I)$ This makes the ${BX}_{3}$ less electron deficient. The tendency of back donation decreases as the size of halogen atom increases (due to large energy difference).Hence, due to large size of $I, {BI}_{3}$ is most electron deficient and thus, the strongest Lewis acid.

CHXI11:THE P-BLOCK ELEMENTS

316046 In $B F_{3}$ , the $B - F$ bond length is $1 .30 \overset{o}{A}$ , when ${BF}_{3}$ is allowed to be treated with ${Me}_{3} N$ , it forms an adduct, ${Me}_{3} N \to {BF}_{3}$ , The bond length of $B$ $F$ in the adduct is:

1 Greater than

1 .30 \overset{o}{A}

2 Smaller than

1 .30 \overset{o}{A}

3 Equal to

1 .30 \overset{o}{A}

4 None of these

CHXI11:THE P-BLOCK ELEMENTS

316047 The power of halides of Boron to act as Lewis acids decreases in the order of

1

{BF}_{3} > {BCl}_{3} > {BBr}_{3}

2

{BBr}_{3} > {BCl}_{3} > {BF}_{3}

3

{BCl}_{3} > {BF}_{3} > {BBr}_{3}

4

{BCl}_{3} > {BBr}_{3} > {BF}_{3}

CHXI11:THE P-BLOCK ELEMENTS

316048 In $B F_{3}$

1 B-F bond has some double bond character and this bond is delocalised

2 All the B-F bonds are single covalent in nature

3 Bond energy and bond-length of B-F bond indicate its single bond character

4 All the bonds are ionic

CHXI11:THE P-BLOCK ELEMENTS

316049 The correct statement is

1

B I_{3}

is the weakest Lewis acid among the boron halides

2 there is minimum

p π -  p π

back bonding in

{BF}_{3}

3

{BF}_{3}

is the strongest Lewis acid among the other boron halides

4 there is maximum

p π -  p π

back bonding in

{BF}_{3}

.

Explanation:

There is maximum $p π - p π$ back bonding in ${BF}_{3}$ due to identical size of $2p$ orbitals of $B$ and $F$ .

KCET - 2015

CHXI11:THE P-BLOCK ELEMENTS

316045 The strongest Lewis acid among boron halides is

1

{BBr}_{3}

2

{BCl}_{3}

3

{BI}_{3}

4

{BF}_{3}

Explanation:

Generally, as the electronegativity increases, acidic character increases, but in case of boron halides, ${BI}_{3}$ is most acidic, i.e. strongest Lewis acid.This is because there is back donation of electrons from halogen atom to boron in ${BX}_{3} (X = F, Cl, Br, I)$ This makes the ${BX}_{3}$ less electron deficient. The tendency of back donation decreases as the size of halogen atom increases (due to large energy difference).Hence, due to large size of $I, {BI}_{3}$ is most electron deficient and thus, the strongest Lewis acid.

CHXI11:THE P-BLOCK ELEMENTS

316046 In $B F_{3}$ , the $B - F$ bond length is $1 .30 \overset{o}{A}$ , when ${BF}_{3}$ is allowed to be treated with ${Me}_{3} N$ , it forms an adduct, ${Me}_{3} N \to {BF}_{3}$ , The bond length of $B$ $F$ in the adduct is:

1 Greater than

1 .30 \overset{o}{A}

2 Smaller than

1 .30 \overset{o}{A}

3 Equal to

1 .30 \overset{o}{A}

4 None of these

CHXI11:THE P-BLOCK ELEMENTS

316047 The power of halides of Boron to act as Lewis acids decreases in the order of

1

{BF}_{3} > {BCl}_{3} > {BBr}_{3}

2

{BBr}_{3} > {BCl}_{3} > {BF}_{3}

3

{BCl}_{3} > {BF}_{3} > {BBr}_{3}

4

{BCl}_{3} > {BBr}_{3} > {BF}_{3}

CHXI11:THE P-BLOCK ELEMENTS

316048 In $B F_{3}$

1 B-F bond has some double bond character and this bond is delocalised

2 All the B-F bonds are single covalent in nature

3 Bond energy and bond-length of B-F bond indicate its single bond character

4 All the bonds are ionic

CHXI11:THE P-BLOCK ELEMENTS

316049 The correct statement is

1

B I_{3}

is the weakest Lewis acid among the boron halides

2 there is minimum

p π -  p π

back bonding in

{BF}_{3}

3

{BF}_{3}

is the strongest Lewis acid among the other boron halides

4 there is maximum

p π -  p π

back bonding in

{BF}_{3}

.

Explanation:

There is maximum $p π - p π$ back bonding in ${BF}_{3}$ due to identical size of $2p$ orbitals of $B$ and $F$ .

KCET - 2015

CHXI11:THE P-BLOCK ELEMENTS

316045 The strongest Lewis acid among boron halides is

1

{BBr}_{3}

2

{BCl}_{3}

3

{BI}_{3}

4

{BF}_{3}

Explanation:

Generally, as the electronegativity increases, acidic character increases, but in case of boron halides, ${BI}_{3}$ is most acidic, i.e. strongest Lewis acid.This is because there is back donation of electrons from halogen atom to boron in ${BX}_{3} (X = F, Cl, Br, I)$ This makes the ${BX}_{3}$ less electron deficient. The tendency of back donation decreases as the size of halogen atom increases (due to large energy difference).Hence, due to large size of $I, {BI}_{3}$ is most electron deficient and thus, the strongest Lewis acid.

CHXI11:THE P-BLOCK ELEMENTS

316046 In $B F_{3}$ , the $B - F$ bond length is $1 .30 \overset{o}{A}$ , when ${BF}_{3}$ is allowed to be treated with ${Me}_{3} N$ , it forms an adduct, ${Me}_{3} N \to {BF}_{3}$ , The bond length of $B$ $F$ in the adduct is:

1 Greater than

1 .30 \overset{o}{A}

2 Smaller than

1 .30 \overset{o}{A}

3 Equal to

1 .30 \overset{o}{A}

4 None of these

CHXI11:THE P-BLOCK ELEMENTS

316047 The power of halides of Boron to act as Lewis acids decreases in the order of

1

{BF}_{3} > {BCl}_{3} > {BBr}_{3}

2

{BBr}_{3} > {BCl}_{3} > {BF}_{3}

3

{BCl}_{3} > {BF}_{3} > {BBr}_{3}

4

{BCl}_{3} > {BBr}_{3} > {BF}_{3}

CHXI11:THE P-BLOCK ELEMENTS

316048 In $B F_{3}$

1 B-F bond has some double bond character and this bond is delocalised

2 All the B-F bonds are single covalent in nature

3 Bond energy and bond-length of B-F bond indicate its single bond character

4 All the bonds are ionic

CHXI11:THE P-BLOCK ELEMENTS

316049 The correct statement is

1

B I_{3}

is the weakest Lewis acid among the boron halides

2 there is minimum

p π -  p π

back bonding in

{BF}_{3}

3

{BF}_{3}

is the strongest Lewis acid among the other boron halides

4 there is maximum

p π -  p π

back bonding in

{BF}_{3}

.

Explanation:

There is maximum $p π - p π$ back bonding in ${BF}_{3}$ due to identical size of $2p$ orbitals of $B$ and $F$ .

KCET - 2015

CHXI11:THE P-BLOCK ELEMENTS

316045 The strongest Lewis acid among boron halides is

1

{BBr}_{3}

2

{BCl}_{3}

3

{BI}_{3}

4

{BF}_{3}

Explanation:

Generally, as the electronegativity increases, acidic character increases, but in case of boron halides, ${BI}_{3}$ is most acidic, i.e. strongest Lewis acid.This is because there is back donation of electrons from halogen atom to boron in ${BX}_{3} (X = F, Cl, Br, I)$ This makes the ${BX}_{3}$ less electron deficient. The tendency of back donation decreases as the size of halogen atom increases (due to large energy difference).Hence, due to large size of $I, {BI}_{3}$ is most electron deficient and thus, the strongest Lewis acid.

CHXI11:THE P-BLOCK ELEMENTS

316046 In $B F_{3}$ , the $B - F$ bond length is $1 .30 \overset{o}{A}$ , when ${BF}_{3}$ is allowed to be treated with ${Me}_{3} N$ , it forms an adduct, ${Me}_{3} N \to {BF}_{3}$ , The bond length of $B$ $F$ in the adduct is:

1 Greater than

1 .30 \overset{o}{A}

2 Smaller than

1 .30 \overset{o}{A}

3 Equal to

1 .30 \overset{o}{A}

4 None of these

CHXI11:THE P-BLOCK ELEMENTS

316047 The power of halides of Boron to act as Lewis acids decreases in the order of

1

{BF}_{3} > {BCl}_{3} > {BBr}_{3}

2

{BBr}_{3} > {BCl}_{3} > {BF}_{3}

3

{BCl}_{3} > {BF}_{3} > {BBr}_{3}

4

{BCl}_{3} > {BBr}_{3} > {BF}_{3}

CHXI11:THE P-BLOCK ELEMENTS

316048 In $B F_{3}$

1 B-F bond has some double bond character and this bond is delocalised

2 All the B-F bonds are single covalent in nature

3 Bond energy and bond-length of B-F bond indicate its single bond character

4 All the bonds are ionic

CHXI11:THE P-BLOCK ELEMENTS

316049 The correct statement is

1

B I_{3}

is the weakest Lewis acid among the boron halides

2 there is minimum

p π -  p π

back bonding in

{BF}_{3}

3

{BF}_{3}

is the strongest Lewis acid among the other boron halides

4 there is maximum

p π -  p π

back bonding in

{BF}_{3}

.

Explanation:

There is maximum $p π - p π$ back bonding in ${BF}_{3}$ due to identical size of $2p$ orbitals of $B$ and $F$ .

KCET - 2015

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