The X -X bond length is 1 -A and Y - Y bond length is 1-54 -A- If electronegativities of -X- and -Y- are 3-0 and 2-0 respectively- the X - Y bond length is likely to be

Atomic radius of nbsp; X = Bond length2= 12 = 0.5 ∘A Atomic radius of Y = 1.542 = 0.77 ∘A Hence, using Stevenson equation, nbsp;X Y bond length = 0.5 + 0. ...

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Standard XI

Chemistry

Shape and Geometry Using VSEPR

The X -X bond...

Question

The X -X bond length is 1 $\circ A$ and Y - Y bond length is $1.54 \circ A$ . If electronegativities of 'X' and 'Y' are 3.0 and 2.0 respectively, the X - Y bond length is likely to be

1.08 \circ A

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1.18 \circ A

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1.28 \circ A

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0.08 \circ A

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Solution

The correct option is B $1.18 \circ A$
Atomic radius of
$X = \frac{Bond length}{2} = \frac{1}{2} = 0.5 \circ A$
Atomic radius of $Y = \frac{1.54}{2} = 0.77 \circ A$
Hence,
using Stevenson equation, X - Y bond length = 0.5 + 0.77 - 0.09 $\times$ (3-2) $= 1.18 \circ A$

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Standard XI Chemistry

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The X -X bond length is 1 ∘A and Y - Y bond length is 1.54 ∘A. If electronegativities of 'X' and 'Y' are 3.0 and 2.0 respectively, the X - Y bond length is likely to be

The correct option is B 1.18∘AAtomic radius of X=Bond length2=12=0.5∘A Atomic radius of Y=1.542=0.77∘A Hence, using Stevenson equation, X - Y bond length = 0.5 + 0.77 - 0.09 × (3-2) =1.18∘A

The X -X bond length is 1 $\circ A$ and Y - Y bond length is $1.54 \circ A$ . If electronegativities of 'X' and 'Y' are 3.0 and 2.0 respectively, the X - Y bond length is likely to be

The correct option is B $1.18 \circ A$
Atomic radius of
$X = \frac{Bond length}{2} = \frac{1}{2} = 0.5 \circ A$
Atomic radius of $Y = \frac{1.54}{2} = 0.77 \circ A$
Hence,
using Stevenson equation, X - Y bond length = 0.5 + 0.77 - 0.09 $\times$ (3-2) $= 1.18 \circ A$