The correct order of bond dissociation energy | vedclass

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Question

(C) The bond order is directly proportional to the bond dissociation energy. Calculating the bond order using Molecular Orbital Theory: For $N_2$ ($14

Vedclass · Accepted Answer

$N_2 > O_2 > O_2^-$

Vedclass · Answer

(C) The bond order is directly proportional to the bond dissociation energy. Calculating the bond order using Molecular Orbital Theory: For $N_2$ ($14$ electrons): Bond order = $(10-4)/2 = 3$. For $O_2$ ($16$ electrons): Bond order = $(10-6)/2 = 2$. For $O_2^-$ ($17$ electrons): Bond order = $(10-7)/2 = 1.5$. Since the bond order follows the order $N_2 (3) > O_2 (2) > O_2^- (1.5)$,the bond dissociation energy follows the same order: $N_2 > O_2 > O_2^-$.

The correct order of bond dissociation energy among $N_2, O_2, O_2^-$ is shown in which of the following arrangements?

Similar Questions

Thermal decomposition of $AgNO_3$ produces two paramagnetic gases. The total number of electrons present in the antibonding molecular orbitals of the gas that has the higher number of unpaired electrons is. . . . .

Bonding in which of the following diatomic molecule$(s)$ becomes stronger,on the basis of $M.O.$ Theory,by removal of an electron?
$(A)$ $NO$
$(B)$ $N_2$
$(C)$ $O_2$
$(D)$ $C_2$
$(E)$ $B_2$
Choose the most appropriate answer from the options given below.

How many among the given species have a bond order of $0.5$?
$H_2^{+}, He_2^{+}, He_2^{-}, B_2^{+}, F_2^{-}, Be_2^{2-}$

Which molecule has the highest bond order?

Bond order is an inverse measure of

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