Question

Nishit Bhandari went to meet his friend Rohit, where he saw that his friend was doing the study of a particular chemistry book. But he could not find the theoretical value of bond-length in H-F but he found that ${{r}_{H}}\text{ and }{{\text{r}}_{F}}$ are $0.37\text{ }{{\text{A}}^{\circ }}\text{ and 0}\text{.72}{{\text{A}}^{\circ }}$ respectively and electronegativity of F and H 4.0 and 2.1 respectively. What is the bond length of H-F bond?
[A] $1.09{{A}^{\circ }}$
[B] $1.784{{A}^{\circ }}$
[C] $0.92{{A}^{\circ }}$
[D] $0.46{{A}^{\circ }}$

Answer 1

Hint: As we can see that hydrogen and fluorine are non-identical atoms. Therefore we cannot add their ionic radii to get the bond length. We can use the formula \[{{d}_{X-Y}}={{r}^{+}}+{{r}^{-}}-0.09\left| {{E}^{-}}-{{E}^{+}} \right|\] to find the bond length of H-F bond.

Complete answer:
In the question, we have the value of ionic radii of hydrogen and fluorine.
We know that the bond length is the length of the bond between the two participating atoms and it depends on the factors like electronegativity and size of the atoms.
Bond length of a triple bond is generally less than that of a double bond and a single bond is generally the longest among the both.
In the question, the ionic radii and the electronegativity of hydrogen and fluorine atom is given to us. Bond length is the sum of the ionic radii of the two atoms if the atoms are identical but here, as the two atoms i.e. hydrogen and fluorine are now identical therefore simply adding their ionic radii will not give us the correct solution.
Her, we can us the formula-
     \[{{d}_{X-Y}}={{r}^{+}}+{{r}^{-}}-0.09\left| {{E}^{-}}-{{E}^{+}} \right|\]
Here, \[{{d}_{X-Y}}\] is the bond length of the X-Y bond.
\[{{r}^{+}}\] is the ionic radii of the cation and \[{{r}^{-}}\] is the ionic radii of the anion.
\[{{E}^{-}}\] is the electronegativity of the anion and \[{{E}^{+}}\]is the electronegativity of the cation.
For the given question we can write that, \[{{d}_{H-F}}={{r}_{H}}+{{r}_{F}}-0.09\left| {{E}_{F}}-{{E}_{H}} \right|\]
Putting the values of the ionic radii and electronegativity, we will get-
     \[\begin{align}
  & {{d}_{H-F}}=0.37\text{ }{{\text{A}}^{\circ }}\text{+0}\text{.72}{{\text{A}}^{\circ }}-0.09\left| 4-2.1 \right| \\
 & \Rightarrow {{d}_{H-F}}=1.09-0.09\left( 1.9 \right)=1.09-0.171 \\
 & \Rightarrow {{d}_{H-F}}=0.92{{\text{A}}^{\circ }} \\
\end{align}\]
We can see from the above calculation that the bond length of H-F is $0.92{{A}^{\circ }}$.

Therefore, the correct answer is option [C] $0.92{{A}^{\circ }}$.

Note: Generally, the bold length is the sum of covalent radii of the two atoms. Bond length is inversely proportional to bond strength. In ionic molecules, bond strength depends upon the charge and size of the atoms. In covalent molecules, the bold strength depends on the electronegativity and the bond length.