Molecular orbital theory Questions in English

(N/A) $LCAO$ stands for Linear Combination of Atomic Orbitals.
Atomic orbitals and $\psi$: According to wave mechanics,atomic orbitals are expressed by wave functions $(\psi)$,which represent the amplitude of electron waves. These are obtained from the solution of the $Schrodinger$ wave equation.
Molecular orbitals and $LCAO$: The $Schrodinger$ wave equation cannot be solved exactly for any system containing more than one electron. Since molecular orbitals are one-electron wave functions for molecules,they are difficult to obtain directly from the $Schrodinger$ equation.
To overcome this,an approximate method known as $LCAO$ is used. It assumes that molecular orbitals can be formed by the linear combination (addition or subtraction) of atomic orbitals of the atoms involved in bonding.

(N/A) $LCAO$ Method of molecular orbitals: Molecular orbitals are not obtained directly by the $Schrodinger$ wave equation,but they can be obtained by the $LCAO$ method.
$LCAO$ method for Hydrogen Molecule $(H_2)$:
- Hydrogen is a homonuclear diatomic molecule. Consider the hydrogen molecule $(H_2)$ consisting of two atoms $H_A$ and $H_B$.
- Mathematically,the formation of molecular orbitals may be described by the linear combination of atomic orbitals,which can take place by the addition and subtraction of wave functions of individual atomic orbitals as shown below:
$\psi_{MO} = \psi_A + \psi_B$ (Bonding)
$\psi^*_{MO} = \psi_A - \psi_B$ (Antibonding)
Bonding molecular orbital $(\psi_{MO})$ e.g.,$\sigma$: The molecular orbital $\sigma$ formed by the addition of atomic orbitals is called the bonding molecular orbital. Here,for $\sigma$ type molecular orbital,$\psi_{MO} = \sigma(H_2) = \psi_A + \psi_B$.
Antibonding molecular orbital $(\psi^*_{MO})$ e.g.,$\sigma^*$: The molecular orbital $\sigma^*$ formed by the subtraction of atomic orbitals $(\psi_A$ and $\psi_B)$ is called the antibonding molecular orbital. Here,for $\sigma^*$ type antibonding molecular orbital,$\psi^*_{MO}(H_2) = \sigma^*(H_2) = \psi_A - \psi_B$.

(N/A) $LCAO$ Method of molecular orbitals: Molecular orbitals are not obtained directly by the Schrodinger wave equation,but they can be obtained by the $LCAO$ (Linear Combination of Atomic Orbitals) method.
$LCAO$ method for Hydrogen Molecule $(H_{2})$:
- Hydrogen is a homonuclear diatomic molecule. Consider the hydrogen molecule $(H_{2})$ consisting of two atoms $H_{A}$ and $H_{B}$.
- Mathematically,the formation of molecular orbitals may be described by the linear combination of atomic orbitals,which can take place by addition or subtraction of the wave functions of individual atomic orbitals as shown below:
$\psi_{MO} = \psi_{A} + \psi_{B} \quad \text{or} \quad \psi^{*}_{MO} = \psi_{A} - \psi_{B}$
- Bonding molecular orbital $(\psi_{MO})$,e.g.,$\sigma$: The molecular orbital $\sigma$ formed by the addition of atomic orbitals is called the bonding molecular orbital. Here,$\psi_{MO} = \sigma(H_{2}) = \psi_{A} + \psi_{B}$.
- Antibonding molecular orbital $(\psi^{*}_{MO})$,e.g.,$\sigma^{*}$: The molecular orbital $\sigma^{*}$ formed by the subtraction of atomic orbitals $(\psi_{A}$ and $\psi_{B})$ is called the antibonding molecular orbital. Here,$\psi^{*}_{MO}(H_{2}) = \sigma^{*}_{(H_{2})} = \psi_{A} - \psi_{B}$.
- The energy level diagram shows that the bonding orbital has lower energy than the atomic orbitals,while the antibonding orbital has higher energy.

(N/A)

Bonding Molecular Orbitals $(BMO)$	Antibonding Molecular Orbitals $(ABMO)$
It is abbreviated as $BMO$. Its wave function is expressed by $\psi_{MO} = \psi_{A} + \psi_{B}$.	It is abbreviated as $ABMO$. Its wave function is expressed by $\psi_{MO}^{*} = \psi_{A} - \psi_{B}$.
They are formed by the additive effect of atomic orbitals.	They are formed by the subtractive effect of atomic orbitals.
Formation involves constructive interference of electron waves,reinforcing each other.	Formation involves destructive interference of electron waves,canceling each other.
Electron density is concentrated between the nuclei,reducing internuclear repulsion.	Electron density is concentrated away from the space between the nuclei.
No nodal plane is present between the nuclei.	$A$ nodal plane (where electron density is zero) exists between the nuclei.
Electrons in $BMO$ stabilize the molecule.	Electrons in $ABMO$ destabilize the molecule.
$BMO$ possesses lower energy than the combining atomic orbitals.	$ABMO$ possesses higher energy than the combining atomic orbitals.
$BMO$ is stable. Examples: $\sigma, \pi$.	$ABMO$ is unstable. Examples: $\sigma^{*}, \pi^{*}$.

(N/A) The linear combination of atomic orbitals $(LCAO)$ to form molecular orbitals takes place only if the following conditions are satisfied:
$1$. The combining atomic orbitals must have the same or nearly the same energy. For example,a $1s$ orbital can combine with another $1s$ orbital but not with a $2s$ orbital because the energy of the $2s$ orbital is significantly higher than that of the $1s$ orbital.
$2$. The combining atomic orbitals must have the same symmetry about the molecular axis. By convention,the $z$-axis is taken as the molecular axis. Atomic orbitals with similar energy will not combine if they do not possess the same symmetry. For example,a $2p_z$ orbital of one atom can combine with a $2p_z$ orbital of another atom,but not with $2p_x$ or $2p_y$ orbitals due to their different symmetries.
$3$. The combining atomic orbitals must overlap to the maximum extent. The greater the extent of overlap,the greater will be the electron density between the nuclei of the molecular orbital.

(N/A) Molecular orbitals of diatomic molecules are designated as $\sigma$ (sigma),$\pi$ (pi),$\delta$ (delta),etc.
$1$. $\sigma$ (sigma) molecular orbitals: These are symmetrical around the bond axis. For example,$\sigma_{1s}$ and $\sigma_{1s}^{*}$ are formed by the linear combination of $1s$ orbitals. Similarly,if the internuclear axis is the $Z$-direction,the linear combination of $2p_{z}$ orbitals produces $\sigma(2p_{z})$ and $\sigma^{*}(2p_{z})$ orbitals. All these are symmetrical around the bond axis.
$2$. $\pi$ (pi) molecular orbitals: These are not symmetrical around the bond axis. Molecular orbitals obtained from $2p_{x}$ and $2p_{y}$ orbitals are labeled as $\pi$ and $\pi^{*}$. $A$ $\pi$ bonding $MO$ has larger electron density above and below the internuclear axis,while the $\pi^{*}$ antibonding $MO$ has a node between the nuclei.

(N/A) The $1s$ atomic orbitals on two atoms (e.g.,hydrogen) combine to form two molecular orbitals designated as $\sigma 1s$ and $\sigma^{*} 1s$.
$\sigma 1s$ is the bonding molecular orbital $(BMO)$ and $\sigma^{*} 1s$ is the antibonding molecular orbital $(ABMO)$.
The energy order is: Energy of $\sigma 1s < $ Energy of atomic orbital $1s  < $ Energy of $\sigma^{*} 1s$.
The sum of the energies of the molecular orbitals is equal to the sum of the energies of the two $1s$ atomic orbitals.
The energy diagram and the formation of these orbitals are shown below:
$AO =$ Atomic orbitals,$MO =$ Molecular orbitals
$BMO$ is formed by the constructive interference (addition) of atomic orbitals,while $ABMO$ is formed by the destructive interference (subtraction) of atomic orbitals.

(N/A) The $1s$ atomic orbitals on two hydrogen atoms overlap to form two molecular orbitals,designated as $\sigma 1s$ and $\sigma^{*} 1s$.
$\sigma 1s$ is the bonding molecular orbital $(BMO)$ and $\sigma^{*} 1s$ is the antibonding molecular orbital $(ABMO)$.
The energy order is: Energy of $\sigma 1s < $ Energy of atomic orbital $1s < $ Energy of $\sigma^{*} 1s$.
The sum of the energies of the two molecular orbitals is equal to the sum of the energies of the two $1s$ atomic orbitals.
The energy level diagram for the formation of $H_{2}$ from two $H$ atoms is shown below.
$AO =$ Atomic orbitals,$MO =$ Molecular orbitals,$BMO =$ Bonding molecular orbital,$ABMO =$ Antibonding molecular orbital.
The formation of these molecular orbitals by the linear combination of atomic orbitals $(LCAO)$ is represented as:
$(i)$ $ABMO$ $(\sigma^{*} 1s)$: Formed by the subtractive overlap of two $1s$ orbitals.
(ii) $BMO$ $(\sigma 1s)$: Formed by the additive overlap of two $1s$ orbitals.

(N/A) According to the Linear Combination of Atomic Orbitals $(LCAO)$ method,two $2p_{z}$ atomic orbitals overlap to form two molecular orbitals ($MO$s): a bonding molecular orbital $(BMO)$ and an antibonding molecular orbital $(ABMO)$.
$1$. Energy Level Diagram:
- The $2p_{z}$ atomic orbitals $(AO)$ combine to form a lower energy bonding orbital $\sigma 2p_{z}$ and a higher energy antibonding orbital $\sigma^{*} 2p_{z}$.
- Energy order: $\sigma 2p_{z} < 2p_{z} < \sigma^{*} 2p_{z}$.
$2$. Orbital Diagrams:
- Bonding $(\sigma 2p_{z})$: Formed by constructive interference (addition) of wave functions,resulting in increased electron density between the nuclei.
- Antibonding $(\sigma^{*} 2p_{z})$: Formed by destructive interference (subtraction) of wave functions,resulting in a nodal plane between the nuclei where electron density is zero.

(N/A) According to the Linear Combination of Atomic Orbitals $(LCAO)$ method,the overlapping of two $2p_x$ atomic orbitals $(AO)$ results in the formation of two molecular orbitals $(MO)$: a bonding molecular orbital $(BMO)$ denoted as $\pi 2p_x$ and an antibonding molecular orbital $(ABMO)$ denoted as $\pi^* 2p_x$.
Energy order: $\pi 2p_x < 2p_x < \pi^* 2p_x$.
In the $\pi$ type $BMO$,the $(+)$ and $(-)$ phases of the wave functions overlap constructively,leading to increased electron density between the two nuclei.
In the $\pi^*$ type $ABMO$,the phases overlap destructively,resulting in a vertical nodal plane between the two nuclei where the electron density is zero.

Molecular orbitals formed by $LCAO$:

$AO$	Combination of atomic orbitals $(LCAO)$	Molecular orbital $(MO)$
$2s$	$\psi(2s) + \psi(2s)$,$\psi(2s) - \psi(2s)$	$BMO: \sigma(2s)$,$ABMO: \sigma^{*}(2s)$
$2p_{z}$	$\psi(2p_{z}) + \psi(2p_{z})$,$\psi(2p_{z}) - \psi(2p_{z})$	$BMO: \sigma(2p_{z})$,$ABMO: \sigma^{*}(2p_{z})$
$2p_{x}$	$\psi(2p_{x}) + \psi(2p_{x})$,$\psi(2p_{x}) - \psi(2p_{x})$	$BMO: \pi(2p_{x})$,$ABMO: \pi^{*}(2p_{x})$
$2p_{y}$	$\psi(2p_{y}) + \psi(2p_{y})$,$\psi(2p_{y}) - \psi(2p_{y})$	$BMO: \pi(2p_{y})$,$ABMO: \pi^{*}(2p_{y})$

Energy order of orbitals:
The increasing order of energy of $MO$ for $Li_{2}, Be_{2}, B_{2}, C_{2}, N_{2}$ is:
$\sigma 1s < \sigma^{*} 1s < \sigma 2s < \sigma^{*} 2s < (\pi 2p_{x} = \pi 2p_{y}) < \sigma 2p_{z} < (\pi^{*} 2p_{x} = \pi^{*} 2p_{y}) < \sigma^{*} 2p_{z}$
The increasing order of energy of $MO$ for $O_{2}$ and $F_{2}$ is:
$\sigma 1s < \sigma^{*} 1s < \sigma 2s < \sigma^{*} 2s < \sigma 2p_{z} < (\pi 2p_{x} = \pi 2p_{y}) < (\pi^{*} 2p_{x} = \pi^{*} 2p_{y}) < \sigma^{*} 2p_{z}$

(D) The electronic configuration of a molecule in $MO$ theory provides the following information:
$1$. Stability of molecules: If $N_{b} > N_{a}$,the molecule is stable. If $N_{b} < N_{a}$,the molecule is unstable. Stability is directly proportional to bond order.
$2$. Bond order: Calculated as $BO = \frac{1}{2} (N_{b} - N_{a})$.
$3$. Nature of bond: Integral bond order values of $1$,$2$,and $3$ correspond to single,double,or triple bonds,respectively.
$4$. Bond length: Bond length decreases as bond order increases.
$5$. Magnetic nature: If all molecular orbitals are doubly occupied,the substance is diamagnetic. If one or more molecular orbitals are singly occupied,it is paramagnetic (e.g.,$O_{2}$).
Where,$N_{a} =$ number of electrons in antibonding orbitals and $N_{b} =$ number of electrons in bonding orbitals.

Electron configuration: It is formed by the combination of two hydrogen atoms $(1s^1)$. The total number of electrons in the $H_2$ molecule is $2$.
In the $H_2$ molecule,the molecular orbitals are $\sigma_{1s}$ and $\sigma_{1s}^*$.
Therefore,the electron configuration of the $H_2$ molecule is $(\sigma_{1s})^2 (\sigma_{1s}^*)^0$.
Bond order: The number of electrons in the bonding molecular orbital $(BMO)$ $\sigma_{1s} = 2$ and in the antibonding molecular orbital $(ABMO)$ $\sigma_{1s}^* = 0$.
Therefore,Bond order $= \frac{N_b - N_a}{2} = \frac{2 - 0}{2} = 1$.
This means two $H$ atoms are bonded together by a single covalent bond.
Magnetic property: In the $H_2$ molecule,there are no unpaired electrons; all electrons are paired. Therefore,the $H_2$ molecule is diamagnetic.

Bond order according to Lewis theory: The bond order is defined as the number of chemical bonds (shared electron pairs) between two atoms in a molecule.
Examples: The bond order in $H_2, F_2, Cl_2,$ and $HCl$ is $1$. In $O_2$,the bond order is $2$,and in $N_2$,it is $3$,because the number of shared electron pairs between the two atoms is $2$ and $3$ respectively.
Bond order according to Molecular Orbital $(MO)$ theory: The formula is $\text{Bond Order} = \frac{1}{2}(N_b - N_a)$.
Where:
$N_b = \text{Total number of bonding electrons in } BMO$.
$N_a = \text{Total number of antibonding electrons in } ABMO$.
Examples using $MO$ theory:
$1.$ For $H_2$: $\text{Bond Order} = \frac{1}{2}(2 - 0) = 1$.
$2.$ For $F_2, Cl_2, Br_2$: $\text{Bond Order} = \frac{1}{2}(10 - 8) = 1$.
$3.$ For $O_2$: $\text{Bond Order} = \frac{1}{2}(10 - 6) = 2$.
$4.$ For $N_2$: $\text{Bond Order} = \frac{1}{2}(10 - 4) = 3$.

(N/A) Bond order is defined as one half of the difference between the number of electrons in bonding molecular orbitals $(N_{b})$ and the number of electrons in antibonding molecular orbitals $(N_{a})$.
The formula is: $\text{Bond Order} = \frac{1}{2} (N_{b} - N_{a})$.
Significance:
$1$. $A$ positive bond order $(N_{b} > N_{a})$ indicates a stable molecule.
$2$. $A$ negative or zero bond order $(N_{b} \leq N_{a})$ indicates an unstable molecule or that the bond is not possible.

(N/A) The atomic number of $Be$ is $4$,so its electronic configuration is $1s^{2} 2s^{2}$.
In a $Be_{2}$ molecule,there are a total of $8$ electrons.
The molecular orbital configuration of $Be_{2}$ is $(\sigma_{1s})^{2} (\sigma_{1s}^{*})^{2} (\sigma_{2s})^{2} (\sigma_{2s}^{*})^{2}$.
Here,the number of bonding electrons $(N_{b})$ is $4$ and the number of antibonding electrons $(N_{a})$ is $4$.
The bond order $(BO)$ is calculated as: $BO = \frac{1}{2}(N_{b} - N_{a}) = \frac{1}{2}(4 - 4) = 0$.
Since the bond order of $Be_{2}$ is $0$,the molecule is unstable and does not exist.

For $N_2$: Total electrons $= 14$. Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2$. Bond Order $(BO) = \frac{1}{2}(10 - 4) = 3$.
For $O_2$: Total electrons $= 16$. Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$. $BO = \frac{1}{2}(10 - 6) = 2$.
For $O_2^+$: Total electrons $= 15$. Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1$. $BO = \frac{1}{2}(10 - 5) = 2.5$.
For $O_2^-$: Total electrons $= 17$. Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^1$. $BO = \frac{1}{2}(10 - 7) = 1.5$.

The stability of a species is directly proportional to its bond order $(BO)$. The bond order is calculated using the formula: $BO = \frac{1}{2}(N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
$1$. $O_2$ $(16 \ e^-)$: Configuration: $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi^* 2p_x)^1(\pi^* 2p_y)^1$. $BO = \frac{1}{2}(10 - 6) = 2.0$. It has unpaired electrons,so it is paramagnetic.
$2$. $O_2^+$ $(15 \ e^-)$: Configuration: $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi^* 2p_x)^1$. $BO = \frac{1}{2}(10 - 5) = 2.5$. It has an unpaired electron,so it is paramagnetic.
$3$. $O_2^-$ $(17 \ e^-)$: Configuration: $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi^* 2p_x)^2(\pi^* 2p_y)^1$. $BO = \frac{1}{2}(10 - 7) = 1.5$. It has an unpaired electron,so it is paramagnetic.
$4$. $O_2^{2-}$ $(18 \ e^-)$: Configuration: $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi^* 2p_x)^2(\pi^* 2p_y)^2$. $BO = \frac{1}{2}(10 - 8) = 1.0$. All electrons are paired,so it is diamagnetic.
Stability order: $O_2^+ (2.5) > O_2 (2.0) > O_2^- (1.5) > O_2^{2-} (1.0)$.

(N/A) $He$ $(Z=2)$,so the total number of electrons in $He_{2} = 4$.
Electron configuration in $MO$ for $He_{2}$: $(\sigma_{1s})^{2}(\sigma_{1s}^{*})^{2}$.
All electrons are paired in $He_{2}$,so it is diamagnetic.
Bond order $= \frac{1}{2}(N_{b} - N_{a}) = \frac{1}{2}(2 - 2) = 0$.
Since the bond order in $He_{2}$ is zero,it is unstable and does not exist.
The $MO$ energy diagram of $He_{2}$ is shown below.

(N/A) $He$ $(Z=2)$,so the total number of electrons in $He_{2}$ is $4$.
The molecular orbital $(MO)$ electronic configuration of $He_{2}$ is: $(\sigma_{1s})^{2} (\sigma_{1s}^{*})^{2}$.
All electrons are paired in $He_{2}$,so it is diamagnetic.
Bond order $= \frac{1}{2} (N_{b} - N_{a}) = \frac{1}{2} (2 - 2) = 0$.
Since the bond order of $He_{2}$ is $0$,the molecule is unstable and does not exist.
The $MO$ energy diagram of $He_{2}$ is shown below.

(N/A) $Li$ $(Z=3)$,so the total number of electrons in $Li_{2} = 6$.
Electron configuration in $MO$ for $Li_{2}$:
$(\sigma 1s)^{2} (\sigma^{*} 1s)^{2} (\sigma 2s)^{2}$ or $KK (\sigma 2s)^{2}$.
Magnetic Property: All electrons are paired,so it is diamagnetic.
Bond Order Calculation:
$N_{b} = 4$ (bonding electrons),$N_{a} = 2$ (antibonding electrons).
Bond order $= \frac{1}{2} (N_{b} - N_{a}) = \frac{1}{2} (4 - 2) = 1$.
Since the bond order is $1$,$Li_{2}$ has a single bond and is stable.
Note: $KK$ represents the core electrons $[He_{2}] = (\sigma 1s)^{2} (\sigma^{*} 1s)^{2}$.
The energy diagram for the $Li_{2}$ molecule is shown below:

(N/A) $Be$ $(Z=4)$. The total number of electrons in $Be_{2}$ is $8$.
Electron configuration in $MO$ for $Be_{2}$:
$(\sigma 1s)^{2}(\sigma^{*} 1s)^{2}(\sigma 2s)^{2}(\sigma^{*} 2s)^{2}$ $OR$ $KK(\sigma 2s)^{2}(\sigma^{*} 2s)^{2}$.
Magnetic Property: All electrons are paired,so it is diamagnetic.
Bond order $(BO) = \frac{1}{2} (N_{b} - N_{a}) = \frac{1}{2} (4 - 4) = 0$.
Since the bond order is zero,$Be_{2}$ is unstable and does not exist.
The energy diagram for the $Be_{2}$ molecule is shown below.

(N/A) $B_{2}$ $(Z=5)$ has the atomic configuration $1s^{2} 2s^{2} 2p^{1}$. Thus, the total number of electrons in $B_{2}$ is $10$.
The molecular orbital $(MO)$ electron configuration for $B_{2}$ is: $KK(\sigma_{2s})^{2}(\sigma^{*}_{2s})^{2}(\pi 2p_{x})^{1}(\pi 2p_{y})^{1}$. Note that for $B_{2}$, the energy of $\pi 2p$ orbitals is lower than that of the $\sigma 2p_{z}$ orbital due to $s-p$ mixing.
Bond order $= \frac{1}{2} (N_{b} - N_{a}) = \frac{1}{2} (6 - 4) = 1$.
Since there are two unpaired electrons in the $\pi 2p$ orbitals, the $B_{2}$ molecule is paramagnetic.
It possesses a single bond, which makes it stable in the gas phase.
The bond length is $159 \ pm$ and the bond dissociation energy is $290 \ kJ \ mol^{-1}$.
The energy level diagram is provided below.

(N/A) $C_2$ $(Z=6)$ has an atomic configuration of $1s^2 2s^2 2p^2$. Thus,the total number of electrons in $C_2$ is $12$.
Molecular orbital $(MO)$ electron configuration for $C_2$:
$(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2$ or $KK(\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2$.
Bond order $= \frac{1}{2} (N_b - N_a) = \frac{1}{2} (8 - 4) = 2$.
Magnetic Property: All electrons are paired,therefore,it is diamagnetic.
Note: In the $C_2$ molecule,both bonds in the double bond are $\pi$-bonds.
Energy level diagram for $C_2$ molecule:

(N/A) For $N_{2}$ $(Z=7)$: The electronic configuration of $N$ is $1s^{2} 2s^{2} 2p^{3}$.
Total electrons in $N_{2} = 14$.
The molecular orbital $(MO)$ configuration for $N_{2}$ is:
$KK(\sigma 2s)^{2} (\sigma^{*} 2s)^{2} (\pi 2p_{x})^{2} = (\pi 2p_{y})^{2} (\sigma 2p_{z})^{2}$
Bond order $(BO)$ $= \frac{1}{2} (N_{b} - N_{a}) = \frac{1}{2} (10 - 4) = 3$.
This indicates a triple bond in $N_{2}$.
Magnetic Property: Since all electrons are paired,$N_{2}$ is diamagnetic.
The energy diagram is provided below.

(N/A) $O_2$ $(Z=8)$ has the configuration $1s^2 2s^2 2p^4$. Total electrons in $O_2 = 16$.
Molecular orbital $(MO)$ configuration for $O_2$:
$KK(\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$
Bond order $= \frac{1}{2} (N_b - N_a) = \frac{1}{2} (10 - 6) = 2$ (Double bond in $O_2$).
Since there are two unpaired electrons in the $\pi^* 2p_x$ and $\pi^* 2p_y$ orbitals,the molecule is paramagnetic.
The energy diagram for the $O_2$ molecule is provided below:

(N/A) The atomic number of fluorine $(F)$ is $Z=9$,and its electronic configuration is $1s^2 2s^2 2p^5$.
There are $7$ valence electrons in each $F$ atom,so the total number of electrons in the $F_2$ molecule is $14$.
The molecular orbital $(MO)$ configuration for $F_2$ is: $KK(\sigma_{2s})^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi^* 2p_x)^2(\pi^* 2p_y)^2$.
Bond order $= \frac{1}{2} (N_b - N_a) = \frac{1}{2} (10 - 8) = 1$.
This indicates a single bond between the two fluorine atoms $(F-F)$.
Magnetic property: Since all electrons are paired,the $F_2$ molecule is diamagnetic.
The energy level diagram is provided below:

(N/A) $Ne_{2}$ $(Z=10)$: $1s^{2} 2s^{2} 2p^{6}$.
Electron configuration in $MO$ for $Ne_{2}$:
$KK(\sigma 2s)^{2} (\sigma^{*} 2s)^{2} (\sigma 2p_{z})^{2} (\pi 2p_{x})^{2} (\pi 2p_{y})^{2} (\pi^{*} 2p_{x})^{2} (\pi^{*} 2p_{y})^{2} (\sigma^{*} 2p_{z})^{2}$
Bond order $= \frac{1}{2} (N_{b} - N_{a}) = \frac{1}{2} (10 - 10) = 0$
Since the bond order is $0$,the molecule is unstable and does not exist. The energy level diagram for $Ne_{2}$ is shown below:

(N/A) The molecular orbital $(MO)$ occupancy,bond order,magnetic properties,and valence electron configurations for the homonuclear diatomic molecules $B_2, C_2, N_2, O_2, F_2,$ and $Ne_2$ are summarized in the following table:
| Molecule | Valence Electron Configuration | Bond Order | Magnetic Property |
| :--- | :--- | :--- | :--- |
| $B_2$ | $(\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p})^2$ | $1$ | Paramagnetic |
| $C_2$ | $(\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p})^4$ | $2$ | Diamagnetic |
| $N_2$ | $(\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p})^4 (\sigma_{2p})^2$ | $3$ | Diamagnetic |
| $O_2$ | $(\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi^*_{2p})^2$ | $2$ | Paramagnetic |
| $F_2$ | $(\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi^*_{2p})^4$ | $1$ | Diamagnetic |
| $Ne_2$ | $(\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi^*_{2p})^4 (\sigma^*_{2p})^2$ | $0$ | Diamagnetic |

(N/A) According to molecular orbital theory,the electronic configurations are as follows:
$O_2^{+}: (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x^2, \pi 2p_y^2) (\pi^* 2p_x^1)$
Bond order of $O_2^{+} = \frac{10-5}{2} = 2.5$
$O_2^{2-}: (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x^2, \pi 2p_y^2) (\pi^* 2p_x^2, \pi^* 2p_y^2)$
Bond order of $O_2^{2-} = \frac{10-8}{2} = 1.0$
Since bond energy is directly proportional to bond order,$O_2^{+}$ has higher bond energy than $O_2^{2-}$.
$O_2^{+}$ is paramagnetic due to the presence of one unpaired electron,whereas $O_2^{2-}$ is diamagnetic as all electrons are paired.

(N/A) The electronic configuration of $N$ atom $(Z=7)$ is $1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$. Total number of electrons in $N_2$ is $14$. The increasing order of energy levels for $N_2$ is: $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \pi 2p_x = \pi 2p_y < \sigma 2p_z < \pi^* 2p_x = \pi^* 2p_y < \sigma^* 2p_z$.
$(i)$ $N_2$ ($14$ $e^-$): Configuration: $(\sigma 1s)^2, (\sigma^* 1s)^2, (\sigma 2s)^2, (\sigma^* 2s)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\sigma 2p_z)^2$. Bond Order $(BO)$ = $\frac{1}{2}(10-4) = 3$. Diamagnetic (no unpaired electrons).
$(ii)$ $N_2^+$ ($13$ $e^-$): Configuration: $(\sigma 1s)^2, (\sigma^* 1s)^2, (\sigma 2s)^2, (\sigma^* 2s)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\sigma 2p_z)^1$. $BO = \frac{1}{2}(9-4) = 2.5$. Paramagnetic (one unpaired electron).
$(iii)$ $N_2^-$ ($15$ $e^-$): Configuration: $(\sigma 1s)^2, (\sigma^* 1s)^2, (\sigma 2s)^2, (\sigma^* 2s)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\sigma 2p_z)^2, (\pi^* 2p_x)^1$. $BO = \frac{1}{2}(10-5) = 2.5$. Paramagnetic (one unpaired electron).
$(iv)$ $N_2^{2+}$ ($12$ $e^-$): Configuration: $(\sigma 1s)^2, (\sigma^* 1s)^2, (\sigma 2s)^2, (\sigma^* 2s)^2, (\pi 2p_x)^2, (\pi 2p_y)^2$. $BO = \frac{1}{2}(8-4) = 2$. Diamagnetic (no unpaired electrons).
Stability order based on $BO$: $N_2 > N_2^+ \approx N_2^- > N_2^{2+}$.

(N/A) According to molecular orbital theory,the electronic configurations and bond order of $N_{2}$,$N_{2}^{+}$,$O_{2}$,and $O_{2}^{+}$ species are as follows:
$N_{2} (14 \ e^{-}) = \sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, (\pi 2p_{x}^{2} \approx \pi 2p_{y}^{2}), \sigma 2p_{z}^{2}$
Bond order $= \frac{1}{2} (N_{b} - N_{a}) = \frac{1}{2} (10 - 4) = 3$
$N_{2}^{+} (13 \ e^{-}) = \sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, (\pi 2p_{x}^{2} \approx \pi 2p_{y}^{2}), \sigma 2p_{z}^{1}$
Bond order $= \frac{1}{2} (9 - 4) = 2.5$
$O_{2} (16 \ e^{-}) = \sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \sigma 2p_{z}^{2}, (\pi 2p_{x}^{2} \approx \pi 2p_{y}^{2}), (\pi^{*} 2p_{x}^{1} \approx \pi^{*} 2p_{y}^{1})$
Bond order $= \frac{1}{2} (10 - 6) = 2$
$O_{2}^{+} (15 \ e^{-}) = \sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \sigma 2p_{z}^{2}, (\pi 2p_{x}^{2} \approx \pi 2p_{y}^{2}), \pi^{*} 2p_{x}^{1}$
Bond order $= \frac{1}{2} (10 - 5) = 2.5$
$(A)$ For $N_{2} \to N_{2}^{+} + e^{-}$,the bond order changes from $3$ to $2.5$. Thus,the bond order decreases.
$(B)$ For $O_{2} \to O_{2}^{+} + e^{-}$,the bond order changes from $2$ to $2.5$. Thus,the bond order increases.

(N/A) Formation of $N_{2}$ molecule:
Electronic configuration of $N$-atom: ${ }_{7} N = 1s^{2}, 2s^{2}, 2p_{x}^{1}, 2p_{y}^{1}, 2p_{z}^{1}$
$N_{2}$ molecule: $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \pi 2p_{x}^{2}, \pi 2p_{y}^{2}, \sigma 2p_{z}^{2}$
Bond order = $\frac{1}{2} (N_{b} - N_{a}) = \frac{1}{2} (10 - 4) = 3$. $A$ bond order of $3$ indicates a triple bond.
Formation of $F_{2}$ molecule:
Electronic configuration of $F$-atom: ${ }_{9} F = 1s^{2}, 2s^{2}, 2p_{x}^{2}, 2p_{y}^{2}, 2p_{z}^{1}$
$F_{2}$ molecule: $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \sigma 2p_{z}^{2}, \pi 2p_{x}^{2} = \pi 2p_{y}^{2}, \pi^{*} 2p_{x}^{2} = \pi^{*} 2p_{y}^{2}$
Bond order = $\frac{1}{2} (10 - 8) = 1$. $A$ bond order of $1$ indicates a single bond.
Formation of $Ne_{2}$ molecule:
Electronic configuration of $Ne$-atom: ${ }_{10} Ne = 1s^{2}, 2s^{2}, 2p_{x}^{2}, 2p_{y}^{2}, 2p_{z}^{2}$
$Ne_{2}$ molecule: $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \sigma 2p_{z}^{2}, \pi 2p_{x}^{2} = \pi 2p_{y}^{2}, \pi^{*} 2p_{x}^{2} = \pi^{*} 2p_{y}^{2}, \sigma^{*} 2p_{z}^{2}$
Bond order = $\frac{1}{2} (10 - 10) = 0$. $A$ bond order of $0$ indicates no bond exists.

(A) According to Molecular Orbital Theory $(MOT)$,when two oxygen atoms (each with $8$ electrons,$1s^2 2s^2 2p^4$) combine to form $O_2$ ($16$ electrons),they form $10$ molecular orbitals (MOs) in total.
The electronic configuration of $O_2$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1, \sigma^* 2p_z^0$.
Counting these: $\sigma 1s, \sigma^* 1s, \sigma 2s, \sigma^* 2s, \sigma 2p_z, \pi 2p_x, \pi 2p_y, \pi^* 2p_x, \pi^* 2p_y, \sigma^* 2p_z$ equals $10$ MOs.
Therefore,statement $(A)$ is correct.

(D) The number of nodal planes in molecular orbitals are as follows:
$1$. For $\sigma^{*} 2s$: It has $1$ nodal plane (perpendicular to the internuclear axis).
$2$. For $\sigma^{*} 2p_{z}$: It has $1$ nodal plane (perpendicular to the internuclear axis).
$3$. For $\pi^{*} 2p_{x}$: It has $2$ nodal planes (one is the molecular plane and the other is perpendicular to the internuclear axis).
$4$. For $\pi^{*} 2p_{y}$: It has $2$ nodal planes (one is the molecular plane and the other is perpendicular to the internuclear axis).
Wait,let us re-evaluate the nodal planes based on standard molecular orbital theory:
- $\sigma^{*} 2s$ has $1$ nodal plane.
- $\sigma^{*} 2p_{z}$ has $1$ nodal plane.
- $\pi^{*} 2p_{x}$ has $2$ nodal planes.
- $\pi^{*} 2p_{y}$ has $2$ nodal planes.
Actually,both $\pi^{*} 2p_{x}$ and $\pi^{*} 2p_{y}$ have $2$ nodal planes. However,in many textbook contexts for this specific question,$\pi^{*} 2p_{y}$ is often cited as the example for $2$ nodal planes. Given the options,both $C$ and $D$ are technically correct,but $D$ is the standard answer provided in this context.

(B) The bond order is calculated using the formula: $\text{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
For $O_2^-$: Total electrons = $8 + 8 + 1 = 17$. Configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. Bond order = $\frac{10 - 7}{2} = 1.5$.
For $N_2^+$: Total electrons = $7 + 7 - 1 = 13$. Configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^1$. Bond order = $\frac{9 - 4}{2} = 2.5$.
For $O_2^+$: Total electrons = $15$. Bond order = $\frac{10 - 5}{2} = 2.5$.
For $N_2^-$: Total electrons = $15$. Bond order = $\frac{10 - 5}{2} = 2.5$.
Comparing the options,$O_2^+$ and $N_2^-$ both have a bond order of $2.5$.

(A, B, D) The energy order of molecular orbitals for molecules with $14$ or fewer electrons (like $N_2$) is: $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \pi 2p_x \approx \pi 2p_y < \sigma 2p_z < \pi^* 2p_x \approx \pi^* 2p_y < \sigma^* 2p_z$.
In this sequence,the $\pi 2p_x$ and $\pi 2p_y$ orbitals fill before the $\sigma 2p_z$ orbital.
For molecules with more than $14$ electrons (like $O_2, F_2, Ne_2$),the energy order is: $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < \pi 2p_x \approx \pi 2p_y < \pi^* 2p_x \approx \pi^* 2p_y < \sigma^* 2p_z$.
In this case,the $\pi 2p_x$ and $\pi 2p_y$ orbitals fill after the $\sigma 2p_z$ orbital.
Among the given options,$O_2$ ($16$ electrons),$Ne_2$ ($20$ electrons),and $F_2$ ($18$ electrons) follow this order. However,usually,such questions look for the specific case where the $\sigma 2p_z$ is lower in energy than $\pi 2p$ orbitals. Since $O_2, F_2, Ne_2$ all follow this,the question implies identifying the group. Given the standard options,$O_2, F_2, Ne_2$ all satisfy the condition.

(N/A) Molecular orbitals are formed by the Linear Combination of Atomic Orbitals $(LCAO)$.
The bonding molecular orbital is formed by the addition of atomic wave functions: $\psi_{MO} = \psi_A + \psi_B$.
The antibonding molecular orbital is formed by the subtraction of atomic wave functions: $\psi_{MO}^* = \psi_A - \psi_B$.

(B) The $1s-2s$ combination does not form a molecular orbital because the energy difference between these two orbitals is significantly large.

(B) In diatomic molecules from $Li_2$ to $N_2$,the energy of the $\sigma_{2p_z}$ orbital is higher than the $\pi_{2p_x}$ and $\pi_{2p_y}$ orbitals due to $s-p$ mixing.
In molecules from $O_2$ to $Ne_2$,the energy of the $\sigma_{2p_z}$ orbital is lower than the $\pi_{2p_x}$ and $\pi_{2p_y}$ orbitals because $s-p$ mixing is negligible.

(B) The stability of a molecule is directly proportional to its bond order.
Bond order calculations:
$N_2 (14 \ e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. Bond order = $(10-4)/2 = 3.0$.
$N_2^+ (13 \ e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^1$. Bond order = $(9-4)/2 = 2.5$.
$N_2^- (15 \ e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2, \pi^* 2p_x^1$. Bond order = $(10-5)/2 = 2.5$.
$N_2^{2+} (12 \ e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2$. Bond order = $(8-4)/2 = 2.0$.
Comparing bond orders: $N_2 (3.0) > N_2^+ (2.5) = N_2^- (2.5) > N_2^{2+} (2.0)$.
Thus,the stability order is $N_2 > N_2^+ = N_2^- > N_2^{2+}$.

(A) Bond dissociation energy is directly proportional to the bond order.
Calculate the bond order for each species:
$O_2$ ($16$ electrons): Bond order = $(10-6)/2 = 2.0$
$O_2^+$ ($15$ electrons): Bond order = $(10-5)/2 = 2.5$
$O_2^-$ ($17$ electrons): Bond order = $(10-7)/2 = 1.5$
$O_2^{2-}$ ($18$ electrons): Bond order = $(10-8)/2 = 1.0$
Since bond order follows the order $O_2^{2-} (1.0) < O_2^- (1.5) < O_2 (2.0) < O_2^+ (2.5)$,the bond dissociation energy follows the same order.
Therefore,the correct order is $O_2^{2-} < O_2^- < O_2 < O_2^+$.

(B) The electronic configuration of $Ne$ $(Z=10)$ is $1s^2 2s^2 2p^6$.
For $Ne_2$ molecule,the total number of electrons is $20$.
The molecular orbital configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2, \sigma^* 2p_z^2$.
Bond order = $\frac{1}{2} (N_b - N_a) = \frac{1}{2} (10 - 10) = 0$.
Since the bond order is $0$,the $Ne_2$ molecule is unstable and does not exist.

(A) The bond order is calculated using the formula: $Bond \ Order = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.

Species	Electronic Configuration	Bond Order
$H_2^+$	$(\sigma 1s)^1$	$(1-0)/2 = 0.5$
$He_2^+$	$(\sigma 1s)^2 (\sigma^* 1s)^1$	$(2-1)/2 = 0.5$
$He_2^{2+}$	$(\sigma 1s)^2$	$(2-0)/2 = 1$

(N/A) The bond order is calculated using the formula: $\text{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.

Species	Bond Order
$NO$	$2.5$
$NO^+$	$3.0$
$CN$	$2.5$
$CN^-$	$3.0$
$CO$	$3.0$

(D) To determine the bond order,we calculate the total number of valence electrons for each species:

Species	Total Valence Electrons
$NO^+$	$5 + 6 - 1 = 10$
$CN^-$	$4 + 5 + 1 = 10$
$CO$	$4 + 6 = 10$
$N_2$	$5 + 5 = 10$

Since all these species are isoelectronic (having $10$ valence electrons),they follow the same molecular orbital configuration,resulting in a bond order of $3.0$.

(A) The bond order is calculated using the formula: $\text{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.

Species	Electronic Configuration	Bond Order
$H_2^+$	$\sigma 1s^1$	$(1-0)/2 = 0.5$
$He_2^-$	$\sigma 1s^2, \sigma^* 1s^1$	$(2-1)/2 = 0.5$
$He_2^{2-}$	$\sigma 1s^2, \sigma^* 1s^2$	$(2-2)/2 = 0$

Thus,$H_2^+$ and $He_2^-$ have the same bond order of $0.5$.

(A) According to Molecular Orbital Theory $(MOT)$,the electronic configuration of $O_2$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$.
For $O_2^-$,the bond order is $\frac{10 - 7}{2} = 1.5$.
For $O_2^{2-}$,the bond order is $\frac{10 - 8}{2} = 1.0$.
Therefore,$O_2^-$ has a higher bond order than $O_2^{2-}$.

(A) According to Molecular Orbital Theory $(MOT)$,the electronic configuration of the species is as follows:
$1$. $O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. It has $2$ unpaired electrons,so it is paramagnetic.
$2$. $O_2^-$ ($17$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. It has $1$ unpaired electron,so it is paramagnetic.
$3$. $O_2^{2-}$ ($18$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. It has no unpaired electrons,so it is diamagnetic.
Therefore,$O_2$ and $O_2^-$ are paramagnetic.

(C) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as: $\text{Bond Order} = \frac{1}{2} (N_b - N_a)$.

Species	Bond Order
$O_2$	$2.0$
$O_2^+$	$2.5$
$O_2^-$	$1.5$

Comparing the values,$O_2^-$ has the lowest bond order of $1.5$.