# The X - X bond length is 1.5 : AA and Y-Y bond length is 2.54 : AA. If electronegativities of 'X' and 'Y' are 3.5 and 2.0 respectively, the X-Y bond length is likely to be (A) 2.02 : AA (B) 1.44 : AA (C) 2.88 : AA (D) 1.88 : AA

## Question

The $X$ - $X$ bond length is $1.5 : \AA$ and $Y-Y$ bond length is $2.54 : \AA$. If electronegativities of '$X$' and '$Y$' are 3.5 and 2.0 respectively, the $X-Y$ bond length is likely to be

(A) $2.02 : \AA$ (B) $1.44 : \AA$ (C) $2.88 : \AA$ (D) $1.88 : \AA$

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## Answer

The correct option is **D: $1.88 : \AA$**.

First, calculate the **atomic radius** of $X$ and $Y$.

For $X$: $$ \text{Atomic radius of } X = \frac{\text{Bond length}}{2} = \frac{1.5}{2} = 0.75 : \AA $$

For $Y$: $$ \text{Atomic radius of } Y = \frac{2.54}{2} = 1.27 : \AA $$

Next, apply the **Stevenson equation** to estimate the $X-Y$ bond length. The equation is:
$$
X-Y \text{ bond length} = r_X + r_Y - 0.09 \times (\chi_X - \chi_Y)
$$

where $r_X$ and $r_Y$ are the atomic radii, and $\chi_X$ and $\chi_Y$ are the electronegativities of $X$ and $Y$ respectively. Using the given values: $$ X-Y \text{ bond length} = 0.75 + 1.27 - 0.09 \times (3.5 - 2) $$

Calculate the difference in electronegativity: $$ 3.5 - 2 = 1.5 $$

Now, multiply this by 0.09: $$ 0.09 \times 1.5 = 0.135 $$

Finally, subtract this value from the sum of the atomic radii: $$ X-Y \text{ bond length} = 0.75 + 1.27 - 0.135 = 1.885 : \AA $$

Which rounds to approximately: $$ 1.88 : \AA $$

Thus, the most probable $X-Y$ bond length is **1.88 Å**.