Question

Valency of $ Boron $ in $ B{F_3} $ molecules is ________
A) Zero
B) One
C) two
D) three

Answer 1

Hint: Group- $ 13 $ have the following elements- $ Boron,Alu\min ium,Gallium,Indium,Thallium $ and the most common oxidation state among them is $ + 3 $ . They are also known as p-block elements. These metals mostly have non-metallic character.

Complete step by step solution:
 $ Boron $ , is the 1st member of Group- $ 13 $ . It belongs to the p- block family.
The elements of the p-block family are non-metals as they have high electronegativities, their size increases as we move down the group so the value of ionization energy decreases.
Non-metals are responsible for the formation of acidic hydroxides.
In the halogen family, $ Fluorine $ is the 1st member and also it is the most electronegative atom in the entire periodic table.
The compound is covalent in nature. It is a colourless gas with boiling point $ - {101^\circ }C $ . It can be prepared by
 $ {B_2}{O_3} + 3Ca{F_2} + conc.3{H_2}S{O_4} \to 2B{F_3} + 3CaS{O_4} + 3{H_2}O $
Also there is another method-
 $ {B_2}{O_3} + 6N{H_4}B{F_4} \to 8B{F_3} + 6N{H_3} + 3{H_2}O $
This compound has a trigonal planar structure, with bond angle $ {120^\circ } $ which is predicted by $ VSEPR $ theory. The bond lengths are $ 1.30A^\circ $ which is shorter than the sum of covalent radii. The bond energy is very high $ 646KJmo{l^{ - 1}} $ which is due to the strong $ p\pi - p\pi $ bond which induces a double bond like character.
Also halogens do not show a positive oxidation state due to high electronegativities.
Hence the oxidation state of $ Fluorine $ here will be $ - 1 $ .
Now calculating the oxidation state $ Boron $ in $ B{F_3} $ , let’s say it is “a”
 $ \begin{gathered}
  a + 3 \times ( - 1) = 0 \\
   \Rightarrow a - 3 = 0 \\
   \Rightarrow a = + 3 \\
\end{gathered} $
The electronic configuration of $ Boron $ in $ + 3 $ oxidation state $ 1{s^2}2{s^0}2{p^0} $
 $ \therefore $ $ B{F_3} $ Has three bond pairs and thus the Valency is three.
Correct option is (D).

Note:
 $ Boron $ also forms trihalides with chlorine, bromine and iodine. $ BC{l_3} $ , $ BB{r_3} $ are liquid unlike $ B{F_3} $ whereas $ B{I_3} $ is a solid compound. $ Boron $ Has electron deficiency in the trihalides thus it can be used as Lewis acids in many chemical reactions.