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(A) For each transformation,we calculate the bond order $(BO)$ and magnetic behavior:$(a)$ $C^{+}_2$ ($7$ electrons,$\sigma 1s^2, \sigma^* 1s^2, \sigm

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$C^{+}_2 	o C_2$

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(A) For each transformation,we calculate the bond order $(BO)$ and magnetic behavior:$(a)$ $C^{+}_2$ ($7$ electrons,$\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \pi 2p^1$,$BO = (5-2)/2 = 1.5$,paramagnetic) $	o C_2$ ($8$ electrons,$\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \pi 2p^2$,$BO = (6-2)/2 = 2.0$,diamagnetic). Here,$BO$ increases and magnetic behavior changes.$(b)$ $NO^{+}$ ($14$ electrons,$BO = 3.0$,diamagnetic) $	o NO$ ($15$ electrons,$BO = 2.5$,paramagnetic). Here,$BO$ decreases.$(c)$ $O_2$ ($16$ electrons,$BO = 2.0$,paramagnetic) $	o O^{+}_2$ ($15$ electrons,$BO = 2.5$,paramagnetic). Here,$BO$ increases but magnetic behavior remains paramagnetic.$(d)$ $N_2$ ($14$ electrons,$BO = 3.0$,diamagnetic) $	o N^{+}_2$ ($13$ electrons,$BO = 2.5$,paramagnetic). Here,$BO$ decreases.

List-$I$	List-$II$
$A$. $\Psi_{MO} = \Psi_{A} - \Psi_{B}$	$I$. Dipole moment
$B$. $\mu = Q \times r$	$II$. Bonding molecular orbital
$C$. $\frac{N_{b} - N_{a}}{2}$	$III$. Anti-bonding molecular orbital
$D$. $\Psi_{MO} = \Psi_{A} + \Psi_{B}$	$IV$. Bond order

In which of the following transformations,the bond order has increased and the magnetic behaviour has changed?

Similar Questions

Among the following,the molecule expected to be stabilized by anion formation is

Match List-$I$ with List-$II$.

List-$I$ List-$II$

$A$. $\Psi_{MO} = \Psi_{A} - \Psi_{B}$ $I$. Dipole moment

$B$. $\mu = Q \times r$ $II$. Bonding molecular orbital

$C$. $\frac{N_{b} - N_{a}}{2}$ $III$. Anti-bonding molecular orbital

$D$. $\Psi_{MO} = \Psi_{A} + \Psi_{B}$ $IV$. Bond order

Which of the following best describes the given molecular orbital diagram?

Use molecular orbital theory to explain why the $Be_{2}$ molecule does not exist.

The molecular orbital configuration of a diatomic molecule is
$\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_x^2 \pi 2p_y^2 \pi 2p_z^2$
Its bond order is

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