Question

The bond length$(pm)$ of ${{F}_{2}},{{H}_{2}},C{{l}_{2}}$ and ${{I}_{2}}$ , respectively is:
A) $144,74,199,267$
B) $74,144,199,267$
C) $74,267,199,144$
D) $144,74,267,199$

Answer 1

Hint: In chemistry, the atoms bond with each other to form molecules. Different atoms have different bond order, bond length. Bond length is defined as the distance between the center of two atoms that are bonded with each other covalently. We can determine the number of bonded electrons with the help of the length of the bond.

Complete step-by-step answer:
Atomic size is also known as atomic radius which means the distance from the center of an atom to the boundary of the surrounding shells of electrons. There are many types of atomic radius:
Vander Waal radius: It is defined as the half the distance between the center of two atoms that are not bound to the same molecule.
I) Ionic radius: It is the distance between the center of an ion to the outermost shell of an ion.
II) Covalent radius: It is defined as the half of the distance between the center of the two similar atoms that are bonded together by a covalent bond.

When the force of attraction between the two bonded atoms increases, the atoms become smaller and its bond length decreases, but as we decrease the force of attraction between the two bonded atoms, the size of the atoms increase and their bond length also increases. So, the bond length is directly proportional to atomic size.

In this question, as we can see that fluorine, bromine and iodine are from the ${{17}^{th}}$ group and as we down the group, the size of an atom increases due to the increase in the number of shells. So, here we can say that iodine shows the largest size. Therefore, the bond length of iodine will be more than that of fluorine and bromine. In the given options, the maximum bond length is $267$ , so the bond length of iodine must be equal to $267$

Hydrogen atoms are the smallest among all the elements as it consists of one electron and one proton, therefore the bond length of hydrogen will be the minimum one, that is, $74$. And thereby, the bond length of fluorine must be $144$ , and the bond length of chlorine must be $199$
The bond length order of the given elements can be given as:
${{I}_{2}}>C{{l}_{2}}>{{F}_{2}}>{{H}_{2}}$

Therefore, the correct option is (A).
Note: In this question, we have discussed that the atomic size is directly proportional to the bond length. Among the given options, iodine is the largest atom, so its bond length will be maximum, and bond length of hydrogen will be minimum.

Bond length is inversely proportional to bond order. So, if we increase the bond order, it will result in stronger bonds because the force of attraction increases between the atoms. Shorter the bond, the more will be the force of attraction.