question_answer

\[B{{I}_{3}}\] is symmetrical planar molecule, all the \[(B-I)\] bonds lie at \[120{}^\circ \] of each other. The distance between the I atoms is\[3.54\overset{o}{\mathop{A}}\,\]. The radius of covalently bonded I atom is\[1.33\overset{o}{\mathop{A}}\,\]. Covalent radius of boron atom is  

A)  \[0.6\overset{o}{\mathop{A}}\,\]           

B)        \[0.7\overset{o}{\mathop{A}}\,\]             

C)   \[0.8\overset{o}{\mathop{A}}\,\]                 

D)       \[0.9\overset{o}{\mathop{A}}\,\]

Correct Answer: B

Solution :

[b] \[x={{d}_{IB}}=\frac{1.77}{\sin 60{}^\circ }=2.04\overset{o}{\mathop{A}}\,\] \[{{r}_{B}}=(2.04\overset{o}{\mathop{A}}\,)-(1.33\overset{o}{\mathop{A}}\,)=0.71\overset{o}{\mathop{A}}\,\]