All Exams Test series for 1 year @ ₹349 only
Question

The highest occupied molecular orbital for $Ne_2$ is

This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
$\sigma^*_{2p}$

Understanding Ne2 Molecular Orbitals

To determine the Highest Occupied Molecular Orbital (HOMO) for $Ne_2$, we need to analyze its molecular orbital (MO) configuration.

Neon Atomic Configuration

Neon (Ne) is element number 10. Its electron configuration is $1s^2 2s^2 2p^6$. For molecular orbital formation, we focus on the valence electrons, which are in the 2s and 2p subshells.

Ne2 Molecular Orbitals

A $Ne_2$ molecule is formed from two Neon atoms, resulting in a total of $2 \times 8 = 16$ electrons. The molecular orbitals are formed by combining the atomic orbitals. For elements like Neon, the energy order of molecular orbitals derived from 2s and 2p atomic orbitals is typically:

$\sigma_{2s}, \sigma^*_{2s}, \pi_{2p}, \sigma_{2p}, \pi^*_{2p}, \sigma^*_{2p}$

Note: The $\pi_{2p}$ orbitals are doubly degenerate, and the $\pi^*_{2p}$ orbitals are also doubly degenerate.

Filling Molecular Orbitals

We fill these molecular orbitals with the 16 total electrons according to the Pauli Exclusion Principle and Hund's rule:

  • $\sigma_{2s}$: 2 electrons
  • $\sigma^*_{2s}$: 2 electrons
  • $\pi_{2p}$: 4 electrons (2 in each degenerate orbital)
  • $\sigma_{2p}$: 2 electrons
  • $\pi^*_{2p}$: 4 electrons (2 in each degenerate orbital)
  • $\sigma^*_{2p}$: 2 electrons

The complete electron configuration is: $\sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p}^4 \sigma_{2p}^2 \pi^*_{2p}^4 \sigma^*_{2p}^2$.

Identifying the HOMO

The Highest Occupied Molecular Orbital (HOMO) is the last molecular orbital filled with electrons. In the $Ne_2$ configuration, the last filled orbital is $\sigma^*_{2p}$.

Therefore, the HOMO for $Ne_2$ is $\sigma^*_{2p}$.

Was this answer helpful?

Similar Questions

  1. Match List I with List II :

    List I (Transition metal/compound)List II (Catalytic Role)
    A. $V_2O_5$I. Preparation of ammonia from $N_2/H_2$ mixture
    B. FeII. Polymerisation of alkynes
    C. $PdCl_2$III. Preparation of $H_2SO_4$ from $SO_2$
    D. Ni complexIV. Oxidation of ethyne to ethanal

    Choose the correct answer :

  2. Identify the incorrect statement from the following :
  3. Identify the correct statement about $ClF_3$ from the following options :

  4. The correct formal charges on oxygen atoms numbered 2, 1 and 3 respectively in ozone are :

  5. Match the species in List I with their geometry in List II
    List IList II
    A. $PCl_5$I. Tetrahedral
    B. $BrF_5$II. Square Planar
    C. $BF_4^-$III. Trigonal bipyramidal
    D. $[Ni(CN)_4]^{2-}$IV. Square pyramidal

    Choose the correct answer from the options given below:
  6. Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R.
    Assertion A : The first ionization enthalpy of O is lower than that of N and F.
    Reason R : The loss of an electron from O leads to stable half-filled p orbital.
    In light of the above statements, choose the most appropriate answer from the options given below:
  7. According to crystal field theory, the correct order of ligands with respect to their decreasing order of field strength is
  8. Given below are two statements:
    Statement-I : $[Fe(ox)_3]^{3-}$ is chiral.
    Statement-II : $trans-[Cr(H_2O)_2(ox)_2]^-$ is chiral.
    (Given : $oxH_2 = HOOC-COOH$)
    In light of the above statements, choose the most appropriate answer from the options given below:
  9. Given below are two statements:
    Statement-I : Heating NaCl with concentrated $H_2SO_4$ and $MnO_2$ results in oxidation of Mn.
    Statement-II : Heating NaI with concentrated $H_2SO_4$ and $MnO_2$ results in reduction of Mn.
    In light of the above statements, choose the most appropriate answer from the options given below:
  10. The green paramagnetic species formed by heating $KMnO_4$ at $513\text{ K}$ is

Important Questions from Inorganic Chemistry

  1. A chromium complex with a formula $\text{CrCl}_3\cdot 6\text{H}_2\text{O}$ has a spin only magnetic moment value of $3.87 \text{ BM}$ and its solution conductivity corresponds to $1 : 2$ electrolyte. $2.75 \text{ g}$ of the complex solution was initially passed through a cation exchanger. The solution obtained after the process was reacted with excess of $\text{AgNO}_3$. The amount of $\text{AgCl}$ formed in the above process is _____ g. (Nearest integer)
    [Given: Molar mass in $\text{g mol}^{-1}$ $\text{Cr} : 52; \text{Cl}: 35.5, \text{Ag}: 108, \text{O}: 16, \text{H}: 1$]
  2. Consider the following reactions :
    $\text{NaCl} + \text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 \rightarrow \text{A} + \text{KHSO}_4 + \text{NaHSO}_4 + \text{H}_2\text{O}$
    $\text{A} + \text{NaOH} \rightarrow \text{B} + \text{NaCl} + \text{H}_2\text{O}$
    $\text{B} + \text{H}_2\text{SO}_4 + \text{H}_2\text{O}_2 \rightarrow \text{C} + \text{Na}_2\text{SO}_4 + \text{H}_2\text{O}$
    In the product 'C', 'X' is the number of $\text{O}_2^{2-}$ units, 'Y' is the total number oxygen atoms present and 'Z' is the oxidation state of Cr. The value of X + Y + Z is ____________.
  3. Identify the metal ions among $\text{Co}^{2+}, \text{Ni}^{2+}, \text{Fe}^{2+}, \text{V}^{3+}$ and $\text{Ti}^{2+}$ having a spin-only magnetic moment value more than $3.0\text{ BM}$. The sum of unpaired electrons present in the high spin octahedral complexes formed by those metal ions is __________.
  4. The crystal field splitting energy of $[\text{Co(oxalate)}_3]^{3-}$ complex is '$n$' times that of the $[\text{Cr(oxalate)}_3]^{3-}$ complex. Here '$n$' is ___________. (Assume $\Delta_o >> P$)

  5. Which of the following is the structure of pyrosulphuric acid?
Need Expert Advice?
More Questions from NEET

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App