The bond order and magnetic property of acety | vedclass

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(A) The acetylide ion is $C_2^{2-}$.Total number of electrons in $C_2^{2-} = (6 	imes 2) + 2 = 14 \ e^-$.According to Molecular Orbital Theory,the el

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$NO^{+}$

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(A) The acetylide ion is $C_2^{2-}$.Total number of electrons in $C_2^{2-} = (6 	imes 2) + 2 = 14 \ e^-$.According to Molecular Orbital Theory,the electronic configuration of $C_2^{2-}$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$.Bond order $= \frac{1}{2} (N_b - N_a) = \frac{1}{2} (10 - 4) = 3$.Since all electrons are paired,it is diamagnetic.Now,check $NO^{+}$: Total electrons $= 7 + 8 - 1 = 14 \ e^-$.$NO^{+}$ also has a bond order of $3$ and is diamagnetic.Therefore,the properties match $NO^{+}$.

List-$I$	List-$II$
$(1)$ ${\rm{NO}}$	$(A)$ $1.5$
$(2)$ ${\rm{CO}}$	$(B)$ $2.0$
$(3)$ ${\rm{O}}_2^-$	$(C)$ $2.5$
$(4)$ ${\rm{O}}_2$	$(D)$ $3.0$

The bond order and magnetic property of acetylide ion are same as that of $........$.

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Arrange the following species in the correct order of their stabilities: $N_2^{-}, C_2, Ne_2, O_2^{2-}$

Match the species given in List-$I$ with their bond orders given in List-$II$.

List-$I$ List-$II$

$(1)$ ${\rm{NO}}$ $(A)$ $1.5$

$(2)$ ${\rm{CO}}$ $(B)$ $2.0$

$(3)$ ${\rm{O}}_2^-$ $(C)$ $2.5$

$(4)$ ${\rm{O}}_2$ $(D)$ $3.0$

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