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Above order is correct for
\[{\text{B}}{{\text{F}}_{\text{3}}}{\text{ < BC}}{{\text{l}}_{\text{3}}}{\text{ < BB}}{{\text{r}}_{\text{3}}}{\text{ < B}}{{\text{I}}_{\text{3}}}\]
A) Bond angle
B) Electron density on B
C) Extent of back bonding
D) None

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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Hint: In modern periodic table totally \[118\]elements. Boron is one of the elements in the periodic table. The atomic number of boron is\[5\]. The mass number of boron is\[10.8\]. Boron is present in the second period and \[13th\] group in the periodic table. Fluorine, chlorine, bromine and iodine are halogens. Halogens are present in the seventeen group periodic table. The symbols of fluorine, chlorine, bromine and iodine are \[{\text{f,Cl}}{\text{.BrandI}}\]respectively.

Complete answer:
We must remember that the bond angle means the angle difference between one bond from central atom to one surrounding atom and the other bond from central atom to second surrounding atom.
We also need to know that the bond angle of boron halide depends on the size of the surrounding atom. The atomic size of the halides plays a vital role for bond angle in boron halides. The atomic size of the halogen is directly proportional to the bond angle.
The increasing order of the atomic size of the halogen is given below,
\[{\text{F < Cl < Br < I}}\]
Hence, the increasing order of the bond angle of boron trihalide is,
\[{\text{B}}{{\text{F}}_{\text{3}}}{\text{ < BC}}{{\text{l}}_{\text{3}}}{\text{ < BB}}{{\text{r}}_{\text{3}}}{\text{ < B}}{{\text{I}}_{\text{3}}}\].
Electron density on B
Here electron density on boron in all boron trihalides are the same. Because all are halogen hence, electron density surrounding the boron is equal only.
Extent of back bonding
Here back bonding in boron trihalide is dependent on the electronegativity of the bonded atom. In that case we seen this concept means \[{\text{B}}{{\text{F}}_{\text{3}}}{\text{ > BC}}{{\text{l}}_{\text{3}}}{\text{ > BB}}{{\text{r}}_{\text{3}}}{\text{ > B}}{{\text{I}}_{\text{3}}}\]
According to the above discussion
Option A is correct, the order of bond order is \[{\text{B}}{{\text{F}}_{\text{3}}}{\text{ < BC}}{{\text{l}}_{\text{3}}}{\text{ < BB}}{{\text{r}}_{\text{3}}}{\text{ < B}}{{\text{I}}_{\text{3}}}\]
Option B is not correct, the order of electron density on B is \[{\text{B}}{{\text{F}}_{\text{3}}}{\text{ = BC}}{{\text{l}}_{\text{3}}}{\text{ = BB}}{{\text{r}}_{\text{3}}}{\text{ = B}}{{\text{I}}_{\text{3}}}\]
Option C is also not correct, the order of back bonding is \[{\text{B}}{{\text{F}}_{\text{3}}}{\text{ > BC}}{{\text{l}}_{\text{3}}}{\text{ > BB}}{{\text{r}}_{\text{3}}}{\text{ > B}}{{\text{I}}_{\text{3}}}\]
Option D is not applicable, because option A is correct.

Hence, option A is correct order for increasing order of
\[{\text{B}}{{\text{F}}_{\text{3}}}{\text{ < BC}}{{\text{l}}_{\text{3}}}{\text{ < BB}}{{\text{r}}_{\text{3}}}{\text{ < B}}{{\text{I}}_{\text{3}}}\]


Note:
We have to remember that the halogen is the most electron affinity group in the periodic table. Halogen is also the most electronegativity group in the periodic table. The ionisation energy of the halogen is very less compared to other groups. In halogen from top to bottom, electron affinity decreases. Chlorine is the highest electron affinity species in the periodic table, because the size of the fluorine is very small compared to chlorine. But, the electronegativity value of fluorine is high compared to the electronegativity value of the chlorine.
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