When N_2 goes to N_2^+, the N-N bond distance | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) According to Molecular Orbital Theory, bond order is given by $\frac{1}{2}(N_b - N_a)$.For $N_2$ ($14$ electrons), the bond order is $3$. When it

Vedclass · Accepted Answer

Increase, decrease

Vedclass · Answer

(B) According to Molecular Orbital Theory, bond order is given by $\frac{1}{2}(N_b - N_a)$.For $N_2$ ($14$ electrons), the bond order is $3$. When it forms $N_2^+$, an electron is removed from a bonding molecular orbital, so the bond order decreases to $2.5$. Since bond distance is inversely proportional to bond order, the $N-N$ bond distance increases.For $O_2$ ($16$ electrons), the bond order is $2$. When it forms $O_2^+$, an electron is removed from an antibonding molecular orbital ($\pi^*$), so the bond order increases to $2.5$. Consequently, the $O-O$ bond distance decreases.

When $N_2$ goes to $N_2^+$, the $N-N$ bond distance ..... and when $O_2$ goes to $O_2^+$, the $O-O$ bond distance .......

Similar Questions

What is the total number of electrons present in bonding and antibonding molecular orbitals respectively in an $F_2$ molecule according to molecular orbital $(MO)$ theory?

Provide the energy level diagram obtained by the overlapping of $2p_{z}$ orbitals and show the diagram of the resulting molecular orbitals.

In which of the following pairs of molecules/ions,both the species are not likely to exist?

Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.

The number of antibonding electron pairs in $O_{2}^{2-}$ molecular ion on the basis of molecular orbital theory is (Atomic number of $O$ is $8$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module