The species,which has the bond order same as | vedclass

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

Question

(B) The total number of electrons in $F_2$ is $18$. According to Molecular Orbital Theory,the electronic configuration of $F_2$ is $\sigma 1s^2, \sigm

Vedclass · Accepted Answer

$O_2^{2-}$

Vedclass · Answer

(B) The total number of electrons in $F_2$ is $18$. According to Molecular Orbital Theory,the electronic configuration of $F_2$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. Bond order $= \frac{1}{2} (N_b - N_a) = \frac{1}{2} (10 - 8) = 1$. Now,checking the options: $O_2^+$ has $15$ electrons,bond order $= 2.5$. $O_2^{2-}$ has $18$ electrons,bond order $= \frac{1}{2} (10 - 8) = 1$. $O_2$ has $16$ electrons,bond order $= 2$. $N_2^+$ has $13$ electrons,bond order $= 2.5$. Thus,$O_2^{2-}$ has the same bond order as $F_2$.

The species,which has the bond order same as that of $F_2$ molecule is

Similar Questions

Match the following molecules/ions in List-$I$ with their respective bond orders in List-$II$:
| List-$I$ (Molecules/ions) | List-$II$ (Bond order) |
| :--- | :--- |
| $A. N_2^+$ | $I. 1.0$ |
| $B. CO$ | $II. 1.5$ |
| $C. O_2$ | $III. 2.0$ |
| $D. O_2^-$ | $IV. 2.5$ |
| | $V. 3.0$ |

The total number of species from the following in which one unpaired electron is present,is . . . . . . $N_2, O_2, C_2^{-}, O_2^{-}, O_2^{2-}, H_2^{+}, CN^{-}, He_2^{+}$

The ground state electronic configuration of $CO$ molecule is

The magnetic property of $NO_2$ is

During the formation of molecular orbitals from atomic orbitals,the electron density .....

Vedclass Test Series

Exam Paper Generator

Online Exam Module