Question

Out of $N{O^ + }$ , $NO_2^ -$ , $NO_3^ -$ which has the longest NO bond and which has the shortest NO bond?

Answer 1

Hint : The distance between the nuclei of two atoms chemically bound in a molecule is called bond length. Bond order of covalent bond is equal to the total number of electron pairs covalently bonded between the two atoms in a molecule. By drawing the lewis structure of the molecule and counting the total number of electron pairs between two atoms in a molecule, bond order is calculated.

Complete Step By Step Answer:
Bond length is equal to the sum of the covalent radii of two atoms. Bond length is inversely proportional to bond order in covalent molecules. Bond order is calculated for the given compounds.
The bond order of $N{O^ + }$ is $\dfrac{{3{\text{ electrons}}}}{{1{\text{ bonding groups}}}}$ $= 3$
The bond order of $NO_2^ -$ is $\dfrac{{3{\text{ electrons}}}}{{{\text{2 bonding groups}}}}$ $= \dfrac{3}{2}$ $= 1.5$
The bond order of $NO_3^ -$ is $\dfrac{{{\text{4 electrons}}}}{{{\text{3 bonding groups}}}} = \dfrac{4}{3} \approx 1.33$
We know that bond length is inversely proportional to the bond order. That means if the bond is stronger then it is also shorter.
Therefore the order of bond lengths is:
$NO_3^ - > NO_2^ - > N{O^ + }$ .

Note :
Bond length is directly proportional to the atomic radii of atoms involved in the reaction. So, we can say that the periodic trends of bond length are similar to that of atomic radii. The measure of strength of bond is given by bond energy. It is also called bond enthalpy. Bond energy is directly proportional to the bond to the bond order which means multiple bonds have higher bond energies. Bond energy is inversely proportional to the bond length that means shorter bonds have less bond energies. The difference of electronegativities of atoms involved in chemical bonds also affect the bond energy.