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Standard XI

Chemistry

Covalent Radius

The distance ...

Question

The distance between any two $I$ atoms in $B I_{3}$ is found to be $3.46 o A$ . What is the covalent radius of boron. If covalent radius of $I$ atom is $1.33 o A$ ? (Assume all bonds are single bonds in $B I_{3}$ )
(Given, $\sqrt{3} = 1.73, \sqrt{3} .99 = 1.99$ )

0.66 o A

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1.77 o A

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1.2 o A

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1.32 o A

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Solution

The correct option is A $0.66 o A$
The hybridisation of $B I_{3}$ is $s p^{2}$
So, it has trigonal planar shape.
$∴ ∠ B I B = 120^{o}$
Given,
The distance between two $I$ atoms is $3.46 A^{o}$ .
Covalent radius of $I$ atom is $= 1.33 o A$

Since,
$B - I$ bond length are same, thus $∠ B I I = 30^{o}$

If we draw a perpendicular $B x$ , then length of $I x$ would be $1.73 o A$
Now,
$t a n 30^{o} = \frac{B x}{I x}$
$\Rightarrow \frac{1}{\sqrt{3}} = \frac{B x}{1.73}$
$\Rightarrow B x = \frac{1.73}{1.73} = 1$
Again,
$B I^{2} = B x^{2} + I x^{2}$
$= 1^{2} + (1.73)^{2}$
$= 3.99$
$\Rightarrow B I = 1.99$
Let, the covalent radius of $B = Z$
$∴$
According to the question,
$1.33 + Z = 1.99$
$\Rightarrow Z = 0.66 o A$

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The distance between any two I atoms in BI3 is found to be 3.46 oA. What is the covalent radius of boron. If covalent radius of I atom is 1.33 oA? (Assume all bonds are single bonds in BI3) (Given, √3=1.73, √3.99=1.99)

The distance between any two $I$ atoms in $B I_{3}$ is found to be $3.46 o A$ . What is the covalent radius of boron. If covalent radius of $I$ atom is $1.33 o A$ ? (Assume all bonds are single bonds in $B I_{3}$ )
(Given, $\sqrt{3} = 1.73, \sqrt{3} .99 = 1.99$ )