Answer
Verified
38.1k+ views
Hint:When the atomic orbitals overlap with each other in the region where density of electrons is high, the molecular orbitals are formed. Overlap of the atomic orbitals determines the efficiency of the interaction between the atomic orbitals. Energy of bonding molecular orbitals is less than the nonbonding molecular orbitals.
Complete step by step solution:
The bond order is half of the difference between the total numbers of the bonding electrons
and antibonding electrons in the given molecule.
Formula to calculate Bond order: \[Bond\,Order=\dfrac{1}{2}\left( \begin{align}
& Electron\,in\,bonding\,molecular\,orbital- \\
& Electron\,in\,anti\,bonding\,molecular\,orbital \\
\end{align} \right)\]
Bond order of \[{{F}_{2}}\]:\[\begin{align}
& {{F}_{2}}=\sigma {{(1s)}^{2}}\sigma *{{(1s)}^{2}}\sigma {{(2s)}^{2}}\sigma *{{(2s)}^{2}}\sigma {{(2{{p}_{z}})}^{2}}\pi {{(2{{p}_{x}})}^{2}}\pi {{(2{{p}_{y}})}^{2}}\pi *{{(2{{p}_{x}})}^{2}}\pi *{{(2{{p}_{y}})}^{2}} \\
& Bond\,Order=\dfrac{10-8}{2}=1 \\
\end{align}\]
Similarly, we can calculate the bond order of \[O_{2}^{2-}\] by writing its molecular orbital configuration which is 1.
Bond order of \[{{N}_{2}}\]:
\[\begin{align}
& {{N}_{2}}=\sigma {{(1s)}^{2}}\sigma *{{(1s)}^{2}}\sigma {{(2s)}^{2}}\sigma *{{(2s)}^{2}}\sigma {{(2{{p}_{x}})}^{2}}\sigma {{(2{{p}_{y}})}^{2}}\sigma {{(2{{p}_{z}})}^{2}} \\
& Bond\,Order=\dfrac{10-4}{2}=3 \\
\end{align}\]
Similarly, we can calculate for CO and \[N{{O}^{+}}\], by again by writing its molecular orbital configuration which is 3.
So, the assertion is correct.
The bond order of a molecule is directly proportional to the stability of that molecule. With increasing bond order, the bond length is decreased. Consequently, the amount of energy to dissociate the shorter bond is higher than that of the larger bond. Hence, the bond order of a molecule is more stable.
So, the reason is correct.
Since, nothing is stated in the assertion about stability, the reason becomes unrelated to it. Therefore, the correct option is (b).
Note: If the bond order is low, there will be less attraction between electrons and this causes the atoms to be held together more loosely. As a result, they dissociate easily, meaning they are less stable.
Complete step by step solution:
The bond order is half of the difference between the total numbers of the bonding electrons
and antibonding electrons in the given molecule.
Formula to calculate Bond order: \[Bond\,Order=\dfrac{1}{2}\left( \begin{align}
& Electron\,in\,bonding\,molecular\,orbital- \\
& Electron\,in\,anti\,bonding\,molecular\,orbital \\
\end{align} \right)\]
Bond order of \[{{F}_{2}}\]:\[\begin{align}
& {{F}_{2}}=\sigma {{(1s)}^{2}}\sigma *{{(1s)}^{2}}\sigma {{(2s)}^{2}}\sigma *{{(2s)}^{2}}\sigma {{(2{{p}_{z}})}^{2}}\pi {{(2{{p}_{x}})}^{2}}\pi {{(2{{p}_{y}})}^{2}}\pi *{{(2{{p}_{x}})}^{2}}\pi *{{(2{{p}_{y}})}^{2}} \\
& Bond\,Order=\dfrac{10-8}{2}=1 \\
\end{align}\]
Similarly, we can calculate the bond order of \[O_{2}^{2-}\] by writing its molecular orbital configuration which is 1.
Bond order of \[{{N}_{2}}\]:
\[\begin{align}
& {{N}_{2}}=\sigma {{(1s)}^{2}}\sigma *{{(1s)}^{2}}\sigma {{(2s)}^{2}}\sigma *{{(2s)}^{2}}\sigma {{(2{{p}_{x}})}^{2}}\sigma {{(2{{p}_{y}})}^{2}}\sigma {{(2{{p}_{z}})}^{2}} \\
& Bond\,Order=\dfrac{10-4}{2}=3 \\
\end{align}\]
Similarly, we can calculate for CO and \[N{{O}^{+}}\], by again by writing its molecular orbital configuration which is 3.
So, the assertion is correct.
The bond order of a molecule is directly proportional to the stability of that molecule. With increasing bond order, the bond length is decreased. Consequently, the amount of energy to dissociate the shorter bond is higher than that of the larger bond. Hence, the bond order of a molecule is more stable.
So, the reason is correct.
Since, nothing is stated in the assertion about stability, the reason becomes unrelated to it. Therefore, the correct option is (b).
Note: If the bond order is low, there will be less attraction between electrons and this causes the atoms to be held together more loosely. As a result, they dissociate easily, meaning they are less stable.
Recently Updated Pages
silver wire has diameter 04mm and resistivity 16 times class 12 physics JEE_Main
A parallel plate capacitor has a capacitance C When class 12 physics JEE_Main
A series combination of n1 capacitors each of value class 12 physics JEE_Main
When propyne is treated with aqueous H2SO4 in presence class 12 chemistry JEE_Main
Which of the following is not true in the case of motion class 12 physics JEE_Main
The length of a potentiometer wire is 10m The distance class 12 physics JEE_MAIN
Other Pages
A point charge q placed at the point A is A In stable class 12 physics JEE_Main
Oxidation state of S in H2S2O8 is A 6 B 7 C +8 D 0 class 12 chemistry JEE_Main
The mole fraction of the solute in a 1 molal aqueous class 11 chemistry JEE_Main
The reaction of Zinc with dilute and concentrated nitric class 12 chemistry JEE_Main
How many grams of concentrated nitric acid solution class 11 chemistry JEE_Main
P Q and R long parallel straight wires in air carrying class 12 physics JEE_Main