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

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Last updated date: 01st Mar 2024
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IVSAT 2024
Answer
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Hint: Electronegativity on a periodic table increases as we move from left to right and from bottom to top. And bond angle and acidic character of any molecule is inversely proportional to the electronegativity of the atom i.e. lesser the electronegativity, then more the atom is acidic and has more bond angle.

Complete step by step solution:
We can get the answer by referring to the electronegativity of the halides in the given question.
We know that the electronegativity of an element decreases down the group and increases across a period i.e. from left to right. So, the electronegativity of the halogens decreases from fluorine to iodine.
Hence, according to this $B{{F}_{3}}$ is more electronegative than $B{{I}_{3}}$ and so it should be the most Lewis acidic due to the electron-withdrawing nature of fluorine atoms. But the order is just the opposite due to back-bonding. Back bonding means a type of bonding that occurs between atoms in a compound in which one atom has a lone pair of electrons and the other has vacant orbital. It has a pi-bonding character.
So, back-bonding due to the lone pair of electrons of halogen atoms increases the electron density on $B$ centre, thereby reducing its Lewis acid character. Thus, this back-bonding is more in case of $B{{F}_{3}}$ due to similar size of orbitals of boron and fluorine and it decreases down the group due to increase in the size of the halogen atoms thereby increasing the bond angle of the boron trihalides.
Thus, we can see that,
The order of electronegativity, back-bonding nature decreases down the group but the bond angle and Lewis acidic, electron density and bond angle nature of the trihalides increases down the group.

Hence, the correct option is A.

Note: Effective overlapping of orbitals among the trihalides decreases due to the large size of p orbitals of halides thereby decreasing the electron deficiency of boron atom in its trihalides and bond length of the compound. Remember bond length is inversely proportional to the bond angle. The order will be like $1p-2p>3p-2p>2p-2p>4p-2p$ starting from boron iodide followed by boron bromide, boron chloride and boron fluoride. This is the decreasing order of Lewis acid character which is similar to the order of bond angle.