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(C) The atomic number of fluorine $(F)$ is $9$. Therefore, an $F_2$ molecule has $9 	imes 2 = 18$ electrons.The molecular orbital configuration for $

Vedclass · Accepted Answer

Bonding - $10$, Antibonding - $8$

Vedclass · Answer

(C) The atomic number of fluorine $(F)$ is $9$. Therefore, an $F_2$ molecule has $9 	imes 2 = 18$ electrons.The molecular orbital configuration for $F_2$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^2, \pi^* 2p_y^2$.Bonding electrons are those in orbitals without an asterisk $(*)$. Total bonding electrons = $2 (\sigma 1s) + 2 (\sigma 2s) + 2 (\sigma 2p_z) + 2 (\pi 2p_x) + 2 (\pi 2p_y) = 10$.Antibonding electrons are those in orbitals with an asterisk $(*)$. Total antibonding electrons = $2 (\sigma^* 1s) + 2 (\sigma^* 2s) + 2 (\pi^* 2p_x) + 2 (\pi^* 2p_y) = 8$.Thus, bonding electrons are $10$ and antibonding electrons are $8$.

List-$I$ (Molecule / ion)	List-$II$ (Bond order)
$1$. $NO$	$i$. $1.5$
$2$. $CO$	$ii$. $2.0$
$3$. $O_2^{-}$	$iii$. $2.5$
$4$. $O_2$	$iv$. $3.0$

What is the total number of electrons present in bonding and antibonding molecular orbitals respectively in an $F_2$ molecule according to molecular orbital $(MO)$ theory?

Similar Questions

What is the number of electrons present in antibonding orbitals of $N_2$ molecule according to molecular orbital theory?

Match List-$I$ with List-$II$ and select the correct option:
List-$I$ (Molecule / ion) List-$II$ (Bond order)
$1$. $NO$ $i$. $1.5$
$2$. $CO$ $ii$. $2.0$
$3$. $O_2^{-}$ $iii$. $2.5$
$4$. $O_2$ $iv$. $3.0$

The electronic configuration of the $N_2$ molecule in its ground state can be written as: $KK[\sigma 2s^2 \sigma^* 2s^2 \pi 2p_x^2 \pi 2p_y^2 \sigma 2p_z^2]$. What is the bond order of the $N_2$ molecule?

The correct order of energies of molecular orbitals of $N_2$ molecule is

During the change of $O_2$ to $O_2^-$ ion,the electron adds to which one of the following orbitals?

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