Consider the following statements.I. In the c | vedclass

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

Question

(C) The electronic configuration of $O_2$ ($16$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi

Vedclass · Accepted Answer

$III$ & $IV$ only

Vedclass · Answer

(C) The electronic configuration of $O_2$ ($16$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Bond order = $(10-6)/2 = 2$. It is paramagnetic.The electronic configuration of $O_2^{2+}$ ($14$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. Bond order = $(10-4)/2 = 3$. It is diamagnetic.Statement $I$ is incorrect because bond order increases from $2$ to $3$.Statement $II$ is incorrect because $O_2$ is paramagnetic and $O_2^{2+}$ is diamagnetic.Statement $III$ is correct because bond order increases,so bond length decreases.For $O_2^{2-}$ ($18$ electrons),bond order = $(10-8)/2 = 1$. For $B_2$ ($10$ electrons),bond order = $(6-4)/2 = 1$.Statement $IV$ is correct because both have a bond order of $1$.

Similar Questions

In $O_2^-$,$O_2$,and $O_2^{2-}$ molecular species,the total number of antibonding electrons respectively are

According to molecular orbital theory,antibonding molecular orbitals of $O_2$ contain

The set of species having only fractional bond order values is

If the magnetic moment of a dioxygen species is $1.73 \; B.M$,it may be

The sum of the bond orders of $O_2^{2+}, O_2^{2-}, O_2^{+}, O_2^{-}, O_2$ and the sum of the unpaired electrons present in them respectively are

Vedclass Test Series

Exam Paper Generator

Online Exam Module