${N_2}$ and ${O_2}$ are converted into monoanions $N_2^ - $ and $O_2^ - $ respectively, which of the following statements is wrong
${N_2}$ and ${O_2}$ are converted into monoanions $N_2^ - $ and $O_2^ - $ respectively, which of the following statements is wrong
- [AIPMT 1997]
- A
In ${N_2},$ the $N - N$ bond weakens
- B
In ${O_2},$ the $O - O$ bond order increases
- C
In ${O_2},$ bond length increases
- D
$N_2^ - $ becomes diamagnetic
Similar Questions
Which of the following is paramagnetic ?
Which of the following is paramagnetic ?
The correct order of bound orders of $C _{2}^{2-}, N _{2}^{2-}$ and $O _{2}^{2-}$ is, respectively.
The correct order of bound orders of $C _{2}^{2-}, N _{2}^{2-}$ and $O _{2}^{2-}$ is, respectively.
- [JEE MAIN 2022]
$O_2^{2 - }$is the symbol of .. ion
$O_2^{2 - }$is the symbol of .. ion
Which one is paramagnetic from the following
Which one is paramagnetic from the following
- [AIPMT 1995]
Match $List-I$ with $List-II$.
$List-I$
$List-II$
$(A)$ $\Psi_{ MO }=\Psi_{ A }-\Psi_{ B }$
$(I)$ Dipole moment
$(B)$ $\mu=Q \times I$
$(II)$ Bonding molecular orbital
$(C)$ $\frac{N_{b}-N_{a}}{2}$
$(III)$ Anti-bonding molecualr orbital
$(D)$ $\Psi_{ MO }=\Psi_{ A }+\Psi_{ B }$
$(IV)$ Bond order
Match $List-I$ with $List-II$.
| $List-I$ | $List-II$ |
| $(A)$ $\Psi_{ MO }=\Psi_{ A }-\Psi_{ B }$ | $(I)$ Dipole moment |
| $(B)$ $\mu=Q \times I$ | $(II)$ Bonding molecular orbital |
| $(C)$ $\frac{N_{b}-N_{a}}{2}$ | $(III)$ Anti-bonding molecualr orbital |
| $(D)$ $\Psi_{ MO }=\Psi_{ A }+\Psi_{ B }$ | $(IV)$ Bond order |
- [JEE MAIN 2022]