Which of the following is the correct relatio | vedclass

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(C) The bond order of $N_2$ is $3.0$ (electronic configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sig

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Bond dissociation energy of $N_2 > $ Bond dissociation energy of $N_2^+$

Vedclass · Answer

(C) The bond order of $N_2$ is $3.0$ (electronic configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$).The bond order of $N_2^+$ is $2.5$ (electronic configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^1$).Since bond dissociation energy is directly proportional to bond order,the bond dissociation energy of $N_2$ is greater than that of $N_2^+$.

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 is the correct relationship between the bond dissociation energy of $N_2$ and $N_2^+$?

Similar Questions

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

List-$I$ List-$II$

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

$B$. $\mu = Q \times r$ $II$. 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

Given below are two statements:
Statement $(I)$: The correct sequence of bond lengths in the following species is:
$O_2^+ < O_2 < O_2^- < O_2^{2-}$
Statement $(II)$: The correct sequence of number of unpaired electrons in the following species is:
$O_2 > O_2^+ > O_2^- > O_2^{2-}$

State whether the bond order increases or decreases when $O_2$ changes to $O_2^+$ and when $N_2$ changes to $N_2^+$.

Molecular orbital theory was developed mainly by

In dinitrogen tetroxide,$N_2O_4$:

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