The correct set of symbols for the molecular orbitals given below is:

  • A
    $I. = \sigma^*, II. = \sigma, III. = \pi^*, IV. = \pi$
  • B
    $I. = \sigma^*, II. = \pi, III. = \pi^*, IV. = \sigma$
  • C
    $I. = \pi^*, II. = \sigma, III. = \sigma^*, IV. = \pi$
  • D
    $I. = \pi, II. = \sigma^*, III. = \sigma, IV. = \pi^*$

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Similar Questions

Assertion : Fluorine molecule has bond order one.
Reason : The number of electrons in the antibonding molecular orbitals is two less than that in bonding molecular orbitals.

Given below are two statements:
Statement $I$: $A$ $\pi$ bonding $MO$ has lower electron density above and below the inter-nuclear axis.
Statement $II$: The $\pi^*$ antibonding $MO$ has a node between the nuclei.
In the light of the above statements,choose the most appropriate answer from the options given below:

The energy of $\sigma 2p_z$ molecular orbital is greater than $\pi 2p_x$ and $\pi 2p_y$ molecular orbitals in nitrogen molecule. Write the complete sequence of energy levels in the increasing order of energy in the molecule. Compare the relative stability and the magnetic behaviour of the following species : $N_2, N_2^+, N_2^-, N_2^{2+}$

Calculate the bond order for $H_2^+$,$He_2^+$,and $He_2^{2+}$.

Which of the following statements is correct?

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