Which of the following pictorial diagrams mos | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) $\pi^*$ antibonding molecular orbital is formed by the out-of-phase overlap of two atomic orbitals (e.g.,$p_x$ or $p_y$ orbitals). This results in

Vedclass · Accepted Answer

Vedclass · Answer

(C) $\pi^*$ antibonding molecular orbital is formed by the out-of-phase overlap of two atomic orbitals (e.g.,$p_x$ or $p_y$ orbitals). This results in the presence of a nodal plane between the two nuclei where the electron probability density is zero, in addition to the nodal plane containing the internuclear axis. In the $\pi^*$ orbital, the lobes of the atomic orbitals have opposite signs (represented by shading) facing each other, leading to destructive interference. Diagram $C$ correctly depicts this out-of-phase combination, showing the nodal plane between the nuclei.

Which of the following pictorial diagrams most correctly represents the $\pi^*$ ($\pi$-antibonding) molecular orbital between two atoms if the internuclear axis is taken to be in the $z$-direction?

Similar Questions

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:
$O_2 > O_2^+ > O_2^- > O_2^{2-}$

Between $1s-1s$ and $1s-2s$ combinations,which one does not form a molecular orbital? Why?

Distinguish between bonding molecular orbitals and antibonding molecular orbitals.

How many of the following species have the bond order $2$?
$C_2, B_2^{2-}, N_2^{2+}, CN^{+}, NO^{-}, O_2, C_2^{+}$

Which of the following species,$O_2^-$ or $O_2^{2-}$,has a higher bond order?

Vedclass Test Series

Exam Paper Generator

Online Exam Module