Among the following,the paramagnetic compound | vedclass

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(D) Paramagnetism is due to the presence of unpaired electrons.$1$. $Na_2O_2$ contains the peroxide ion $O_2^{2-}$. Its molecular orbital configuratio

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$KO_2$

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(D) Paramagnetism is due to the presence of unpaired electrons.$1$. $Na_2O_2$ contains the peroxide ion $O_2^{2-}$. Its molecular orbital configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. It has $0$ unpaired electrons,so it is diamagnetic.$2$. $O_3$ (Ozone) is diamagnetic as all electrons are paired.$3$. $N_2O$ is diamagnetic as all electrons are paired.$4$. $KO_2$ contains the superoxide ion $O_2^-$. Its molecular orbital configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. It has $1$ unpaired electron,making it paramagnetic.Therefore,$KO_2$ is the paramagnetic compound. The correct option is $(D)$.

Among the following,the paramagnetic compound 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) |
| :--- | :--- |
| $A. N_2^+$ | $I. 1.0$ |
| $B. CO$ | $II. 1.5$ |
| $C. O_2$ | $III. 2.0$ |
| $D. O_2^-$ | $IV. 2.5$ |
| | $V. 3.0$ |

Which amongst the following is an incorrect statement?

Bond order is an inverse measure of

Find the odd-electron molecules from the following:
$(i) C_2, (ii) H_2, (iii) SCl_2, (iv) NO, (v) NO_2$

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

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