Q.Which one from O2^- and O2^2- may exhibit paramagnetism?

Using molecular orbital (MO) theory, bond order = 1/2 (bonding electrons - antibonding electrons). A bond order of zero means no net bond forms, so the molecule does not exist.

For He2, each He atom contributes 2 electrons, so He2 has 4 electrons total. The MO filling order for these light diatomics is sigma1s, sigma*1s, sigma2s, ...

He2 electron configuration: (sigma1s)^2 (sigma*1s)^2

Bonding electrons = 2, antibonding electrons = 2

Bond order = 1/2 (2 - 2) = 0

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This quantity is the bond order, defined as half the difference between electrons in bonding and antibonding molecular orbitals.

In Molecular Orbital Theory (MOT), atomic orbitals combine to form bonding molecular orbitals (lower energy, stabilizing) and antibonding molecular orbitals (higher energy, destabilizing). The bond order of a molecule is defined as:

Bond order = 1/2 (Nb − Na) …

O2 has 2 unpaired electrons (paramagnetic).

Using molecular orbital theory for O2 (16 electrons), the last two electrons go one each into the two degenerate π* (antibonding) orbitals (π2px and π2py) with parallel spins (Hund's rule). This leaves …

Filling He2's molecular orbitals gives 2 bonding and 2 antibonding electrons; since bond order = (bonding - antibonding)/2, this comes out to exactly zero, which is precisely why the reason correctly explains the assertion.

Each He atom contributes 2 electrons, so $He_2$ (hypothetical) would have 4 electrons total. Filling molecular orbitals in order of energy:

$\sigma 1s^2\ \sigma^*1s^2$

• Bonding electrons ($N_b$) = 2 (in $\sigma 1s$)
• Antibonding electrons ($N_a$) = 2 (in $\sigma^*1s$)

Bond order formula:

$\text{Bond order} = \frac{N_b - N_a}{2} = \frac{2-2}{2} = 0$

Assertion: "The bond order of helium molecule is zero." — TRUE, as shown above.

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The term describing the number of bonds between two bonded atoms is bond order, matching option (d).

Bond order is defined as the number of bonds (single, double or triple) between two atoms in a molecule. For example, bond order is 1 for a single bond, 2 for a double bond, and 3 for a triple bond -- so 'No. of Bonds' directly c …

O2 is paramagnetic because of two unpaired electrons in pi* orbitals.

Molecular orbital theory (MOT) explains magnetic behaviour. For O2 (16 electrons) the configuration ends as ...(pi2px2 pi2py2)(pi2px1 pi2py1): the last two electrons occupy the two degenerate pi* antibonding orbitals singly (Hund's rule), giving 2 unpaired electrons — hence paramagnetism, one of MOT's classic triumphs. In N2 the pi* orbitals are empty and all electrons are p …

N2 has bond order 3.

Bond order = (number of bonding electrons - number of antibonding electrons)/2. For N2 (14 electrons) the MO filling gives 10 bonding and 4 antibonding electrons, so bond order = (10 - 4)/2 = 3, consistent with the N triple-bond N structure. By comparison, O2 has bond order 2, and F2 and Cl2 have bond o …

Bond order = (bonding electrons - antibonding electrons) / 2; for He2, the bonding and antibonding electrons exactly cancel out.

Each He atom has 2 electrons, so He2 (if it formed) would have 4 electrons total.

Using molecular orbital (MO) theory, these 4 electrons fill the molecular orbitals in order of increasing energy: sigma1s (2 electrons) then sigma1s (2 electrons) -- i.e. configuration sigma1s2 sigma1s2.

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LCAO (Linear Combination of Atomic Orbitals) is the mathematical method used in Molecular Orbital Theory to combine atomic orbitals into bonding and antibonding molecular orbitals.

In Molecular Orbital Theory, atomic orbitals of similar energy and matching symmetry on different atoms are combined (added or subtracted) to give new molecular orbitals that belong to the molecule as a whole. This method is called the Linear Combination of Atomic Orbitals (LCAO): adding two …

N2, NO+ and CN- are all isoelectronic (14 electrons) with bond order 3; NO- has one extra electron in an antibonding orbital, giving bond order 2.5 - the odd one out.

A note on the printed stem

As transcribed, the question asks for the species with bond order different from 'CO2'. However, CO2 is a triatomic molecule (two separate C=O double bonds, each of bond order 2) and does not fit naturally into a comparison with the diatomic/pseudo-diatomic species given as options (NO-, NO+, CN-, N2). This is a well-known standard question comparing bond orders of the isoelectronic 14-electron species N2, CO, CN-, NO+ (all bond order 3) against NO- (bond order 2.5) - so the intended reference species is almost certainly 'CO' (carbon monoxide), not 'CO2'. The solution below uses this standard interpretation.

Molecular orbital bond order for each species

Using the standard total-electron-count approach for these second-period diatomics (isoelectronic species share the same MO filling pattern and hence the same bond order):

• CO: 14 total electrons, same bonding pattern as N2, bond order = 3.
• N2: 14 electrons, bond order = 3.
• CN-: C (6e) + N (7e) + 1 extra electron (negative charge) = 14 electrons, isoelectronic with N2/CO, bond order = 3.
• NO+: N (7e) + O (8e) - 1 electron (positive charge) = 14 electrons, isoelectronic with N2/CO, bond order = 3. …