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4.Chemical Bonding and Molecular Structure
hard
If Hund's Rule does not hold good, then which of the following pairs is diamagnetic
A
$B_2, O_2^+$
B
$C_2, O_2$
C
$O_2^-, N_2$
D
$N_2, NO$
Solution
$B_2 \rightarrow$ Total $e^- \Rightarrow 10$
$\sigma s^2 \sigma^* 1 s^2 \sigma_2 s^2 \sigma^*2 s^2 \pi 2 P_x^2=\pi 2 P_y{ }^0$.
diamagnetic
$O _2 \rightarrow$ Total $e^- \rightarrow 16$
$\sigma 1s^2 \sigma^*1 s^2 \quad \sigma 2 s^2 \quad \sigma^*2 s^2 \sigma 2 P_z^2 \,\,\pi 2 P_x^2=\pi 2 P_y^2\, \pi^*2 P_{x^2}=\pi 2 P_y{ }^0 \quad$ diagmagnetic
$N _2 \rightarrow$ diamagnetic.
$C _2 \rightarrow$ Total $e^- \Rightarrow 12$
$\sigma 1s^2 \sigma^*1s^2 \sigma 2 s^2 \sigma^*2 s^2 \pi 2 P_x^2=\pi 2 P_y 2$
diamagnetic
Standard 11
Chemistry
Similar Questions
Match $List-I$ with $List-II$.
| $List-I$ | $List-II$ |
| $(A)$ $\Psi_{ MO }=\Psi_{ A }-\Psi_{ B }$ | $(I)$ Dipole moment |
| $(B)$ $\mu=Q \times I$ | $(II)$ Bonding molecular orbital |
| $(C)$ $\frac{N_{b}-N_{a}}{2}$ | $(III)$ Anti-bonding molecualr orbital |
| $(D)$ $\Psi_{ MO }=\Psi_{ A }+\Psi_{ B }$ | $(IV)$ Bond order |
