Which of the following is paramagnetic?

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(A) Paramagnetism is due to the presence of unpaired electrons.In $O_2[AsF_6]$,the $O_2^+$ ion has a bond order of $2.5$ and contains one unpaired ele

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$O_2[AsF_6]$

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(A) Paramagnetism is due to the presence of unpaired electrons.In $O_2[AsF_6]$,the $O_2^+$ ion has a bond order of $2.5$ and contains one unpaired electron in its molecular orbitals,making it paramagnetic.$Al_2O_3$ is an ionic solid where $Al^{3+}$ ($[Ne]$ configuration) and $O^{2-}$ ($[Ne]$ configuration) have no unpaired electrons,making it diamagnetic.For $C_2$,the molecular orbital configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2, \pi 2p_y^2$. All electrons are paired,so it is diamagnetic.For $Be_2$,the configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2$. All electrons are paired,so it is diamagnetic.

List-$I$	List-$II$
$(a)$ $Ne_2$	$(i)$ $1$
$(b)$ $N_2$	$(ii)$ $2$
$(c)$ $F_2$	$(iii)$ $0$
$(d)$ $O_2$	$(iv)$ $3$

Similar Questions

Consider the ions/molecule $O_{2}^{+}, O_{2}, O_{2}^{-}, O_{2}^{2-}$. For increasing bond order,the correct option is ..... .

Match List-$I$ with List-$II$ based on the bond order of the molecules.

List-$I$ List-$II$

$(a)$ $Ne_2$ $(i)$ $1$

$(b)$ $N_2$ $(ii)$ $2$

$(c)$ $F_2$ $(iii)$ $0$

$(d)$ $O_2$ $(iv)$ $3$

Choose the correct answer from the options given below:

The correct sequence of bond order is

Observe the following molecules / ions $H_2, N_2, O_2, N_2^+, O_2^+, O_2^-, F_2$. Identify the correct statement.

The bond order in $NO$ is $2.5$ while that in $NO^{+}$ is $3$. Which of the following statements is true for these two species?

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