Question

The correct order regarding bond length:
(A) $N_2{H}_2<N_2{H}_4$ ( $N-N$ bond length)
(B) $NO>N{O}^{+}$ ( $N-O$ bond length)
(C) $C_2{H}_4>K[PtC{l}_3(C_2{H}_4)]$ ( $C-C$ bond length)
(D) $PO{F}_3>POC{l}_3$ ( $P-O$ bond length)

Answer 1

Hint:The bond length is the average distance between the nuclei of two atoms bonded in a molecule. We determine the bond length of a molecule by the number of bonded electrons i.e. bond order. The higher the bond order, the stronger will be the pull between two atoms and hence the shorter will be the bond length of the molecule.

Complete step-by-step answer:We know that bond length is inversely proportional to bond order which means the bond length will become shorter if a greater number of electrons are participating in bond formation. Also, in a bond present between two identical atoms, the half of the bond distance is equal to the covalent radius.
We will now compare each of the options separately:
In the first option we see that in $N_2{H}_2$ a double bond is present between the nitrogen atoms and a single bond is present between the nitrogen atoms in $N_2{H}_4$ hence double bond will be shorter than the single bonds as double bond is stronger than single bond so more electrons will be present and will be held by a strong force of attraction. Thus, the statement,$N_2{H}_2<N_2{H}_4$ is not valid.
In the second option in $NO$ a double bond is present and in $N{O}^{+}$ a triple bond is present between the nitrogen and oxygen atom. Hence as we know that the stronger the bond will be, the shorter will be the bond as a greater number of electrons are participating in the bond formation. Thus, this statement i.e. $NO>N{O}^{+}$ is absolutely correct.
In the third option the carbon-carbon bond in $K[PtC{l}_3(C_2{H}_4)]$ is longer than the bond present in $C_2{H}_4$ as in $K[PtC{l}_3(C_2{H}_4)]$ there exists some $d\mathrm{\Pi }-p\mathrm{\Pi }$ back bonding from the platinum atom to ethylene. Hence the third option is incorrect.
In the fourth option we see that the phosphorus and oxygen bond will be longer when bonded with the chlorine atom as chlorine is less electronegative than fluorine and we know that the bond length becomes shorter if bonded with an electronegative atom as the electronegative atom will strongly attract electrons towards itself. Therefore, the fourth statement is also wrong.

Hence the only correct answer is Option 2. i.e. $NO>N{O}^{+}$.

Note:A double bond is composed of one single sigma bond and one pi bond, the first bond i.e. the sigma bond is already holding the atoms together. The pi bond increases the strength of the bond and therefore draws the atoms closer to each other. This results in the sigma-pi bond strengthening as orbital overlap increases.