Molecular orbital theory Questions in English

(A) The total number of electrons in $O_2^+$ is $15$. The molecular orbital configuration is $KK(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi^* 2p_x)^1$.
Bond order $= \frac{1}{2}(N_b - N_a) = \frac{1}{2}(10 - 5) = 2.5$.
The total number of electrons in $N_2^+$ is $13$. The molecular orbital configuration is $KK(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x)^2(\pi 2p_y)^1$.
Bond order $= \frac{1}{2}(9 - 4) = 2.5$.
Thus,the bond order of $O_2^+$ is the same as that of $N_2^+$.

(A) The electronic configuration of $O_2$ ($16$ electrons) is:
$(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$.
The number of bonding electrons $(N_b)$ $= 10$.
The number of antibonding electrons $(N_a)$ $= 6$.
Bond order $= \frac{1}{2} [N_b - N_a] = \frac{1}{2} [10 - 6] = 2$.

(A) The bond length is inversely proportional to the bond order.
Bond order is calculated as: $Bond \; Order = \frac{N_b - N_a}{2}$.
For $O_2$ $(16 \; e^-)$: $Bond \; Order = \frac{10 - 6}{2} = 2.0$.
For $O_2^+$ $(15 \; e^-)$: $Bond \; Order = \frac{10 - 5}{2} = 2.5$.
For $O_2^-$ $(17 \; e^-)$: $Bond \; Order = \frac{10 - 7}{2} = 1.5$.
The bond order sequence is $O_2^+ (2.5) > O_2 (2.0) > O_2^- (1.5)$.
Since bond length is inversely proportional to bond order,the bond length order is $O_2^- > O_2 > O_2^+$.

(B) The molecular orbital configuration for $O_2$ is $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x^2 \equiv \pi 2p_y^2)(\pi^* 2p_x^1 \equiv \pi^* 2p_y^1)$.
Bond order of $O_2 = \frac{10-6}{2} = 2.0$. It has two unpaired electrons,so it is paramagnetic.
The molecular orbital configuration for $O_2^+$ is $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x^2 \equiv \pi 2p_y^2)(\pi^* 2p_x^1 \equiv \pi^* 2p_y^0)$.
Bond order of $O_2^+ = \frac{10-5}{2} = 2.5$.
Since $O_2^+$ has one unpaired electron,it is paramagnetic.
Comparing the values,$2.5 > 2.0$,so the bond order of $O_2^+ > O_2$.

(B) The bond length is inversely proportional to the bond order.
Since the bond order of $NO^{+}$ $(3)$ is greater than the bond order of $NO$ $(2.5)$,the bond length of $NO^{+}$ will be shorter than that of $NO$.
Therefore,the bond length in $NO$ is greater than in $NO^{+}$.

(D) The electronic configuration of $O_2$ is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$,which contains two unpaired electrons,making it paramagnetic.
$B_2$ has the configuration $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^1 (\pi 2p_y)^1$,which contains two unpaired electrons,making it paramagnetic.
$N_2^{+}$ has $13$ electrons with the configuration $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^1$,which contains one unpaired electron,making it paramagnetic.
Since all the given species are paramagnetic,the correct option is $D$.

(C) The bond order of $NO^{+}$,$NO$,and $NO^{-}$ are $3$,$2.5$,and $2$ respectively.
Since bond energy is directly proportional to bond order,the order of bond energies is $NO^{+} > NO > NO^{-}$.

(A) Paramagnetic property arises due to the presence of unpaired electrons.
$B_2$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1 = \pi 2p_y^1$ (contains $2$ unpaired electrons).
$C_2$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2, \pi 2p_y^2$ (no unpaired electrons).
$N_2$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_y^2, \pi 2p_z^2, \sigma 2p_x^2$ (no unpaired electrons).
$F_2$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_x^2, \pi 2p_y^2, \pi 2p_z^2, \pi^* 2p_y^2, \pi^* 2p_z^2$ (no unpaired electrons).
Therefore,only $B_2$ has unpaired electrons and exhibits paramagnetism.

(B) According to Molecular Orbital Theory $(MOT)$,the electronic configuration of $O_2$ ($16$ electrons) is:
$O_2: (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$.
Due to the presence of two unpaired electrons in the antibonding $\pi^*$ orbitals,$O_2$ is paramagnetic.
$H_2$,$N_2$,and $CO$ have all paired electrons and are diamagnetic.

(A) The molecular orbital electronic configurations are as follows:
$O_2^- (17 \ e^-) = \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^2, \pi^* 2p_y^1$. Number of antibonding electrons = $2+2+2+1 = 7$.
$O_2 (16 \ e^-) = \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^1, \pi^* 2p_y^1$. Number of antibonding electrons = $2+2+1+1 = 6$.
$O_2^{2-} (18 \ e^-) = \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^2, \pi^* 2p_y^2$. Number of antibonding electrons = $2+2+2+2 = 8$.
Therefore,the number of antibonding electrons are $7, 6,$ and $8$ respectively.

(C) Species with unpaired electrons are paramagnetic.
$O_2$ has $2$ unpaired electrons in its antibonding $\pi^*$ orbitals.
$O_2^{1+}$ has $1$ unpaired electron.
$O_2^{-}$ has $1$ unpaired electron.
$O_2^{2-}$ (peroxide ion) has $18$ electrons,and its molecular orbital configuration is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^2$. Since all electrons are paired,it is diamagnetic.

(A) According to Molecular Orbital Theory $(MOT)$,the electronic configuration of the species are as follows:
$O_2$ ($16$ 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.
$O_2^+$ ($15$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. It has $1$ unpaired electron.
$O_2^-$ ($17$ 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.
$O_2^{2-}$ ($18$ 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 $0$ unpaired electrons.
Therefore,$O_2$ has the maximum number of unpaired electrons.

(C) To determine the magnetic behavior,we use Molecular Orbital Theory $(MOT)$:
$N_2$ ($14$ electrons): $KK \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p_x)^2 \pi(2p_y)^2 \sigma(2p_z)^2$. All electrons are paired,so it is diamagnetic.
$C_2$ ($12$ electrons): $KK \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p_x)^2 \pi(2p_y)^2$. All electrons are paired,so it is diamagnetic.
$N_2^+$ ($13$ electrons): $KK \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p_x)^2 \pi(2p_y)^2 \sigma(2p_z)^1$. There is one unpaired electron in the $\sigma(2p_z)$ orbital,so it is paramagnetic.
$O_2^{2-}$ ($18$ electrons): $KK \sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x)^2 \pi(2p_y)^2 \pi^*(2p_x)^2 \pi^*(2p_y)^2$. All electrons are paired,so it is diamagnetic.
Therefore,the correct option is $C$.

(C) The molecular orbital configuration of the $O_2$ molecule ($16$ electrons) is:
$[\sigma (1s)^2, \sigma^*(1s)^2, \sigma (2s)^2, \sigma^*(2s)^2, \sigma (2p_z)^2, \pi (2p_x)^2, \pi (2p_y)^2, \pi^*(2p_x)^1, \pi^*(2p_y)^1]$
Counting the electrons in the filled orbitals:
- $\sigma (1s)^2$: $2$ electrons ($1$ pair)
- $\sigma^*(1s)^2$: $2$ electrons ($1$ pair)
- $\sigma (2s)^2$: $2$ electrons ($1$ pair)
- $\sigma^*(2s)^2$: $2$ electrons ($1$ pair)
- $\sigma (2p_z)^2$: $2$ electrons ($1$ pair)
- $\pi (2p_x)^2$: $2$ electrons ($1$ pair)
- $\pi (2p_y)^2$: $2$ electrons ($1$ pair)
Total number of pairs = $7$.
Total number of paired electrons = $7 \times 2 = 14$.

(B) According to Molecular Orbital Theory, bond order is given by $\frac{1}{2}(N_b - N_a)$.
For $N_2$ ($14$ electrons), the bond order is $3$. When it forms $N_2^+$, an electron is removed from a bonding molecular orbital, so the bond order decreases to $2.5$. Since bond distance is inversely proportional to bond order, the $N-N$ bond distance increases.
For $O_2$ ($16$ electrons), the bond order is $2$. When it forms $O_2^+$, an electron is removed from an antibonding molecular orbital ($\pi^*$), so the bond order increases to $2.5$. Consequently, the $O-O$ bond distance decreases.

(A) The total number of electrons in each species is as follows:
$CN^{-} = 6 + 7 + 1 = 14$ electrons.
$CO = 6 + 8 = 14$ electrons.
$NO^{+} = 7 + 8 - 1 = 14$ electrons.
Since all species have $14$ electrons,they are isoelectronic.
Using the molecular orbital theory,the bond order is calculated as:
$B.O. = \frac{1}{2} [N_b - N_a] = \frac{1}{2} [10 - 4] = 3$.
Thus,they all have a bond order of $3$ and are isoelectronic.

(D) $H^-$ is the most stable species among the given options because it has a completely filled $1s^2$ electronic configuration,which is similar to the stable noble gas configuration of Helium $(He)$.

(B) three-electron bond is a type of chemical bond where three electrons are shared between two atoms.
Nitric oxide $(NO)$ has a total of $11$ valence electrons ($5$ from $N$ and $6$ from $O$). Its Lewis structure shows a double bond between $N$ and $O$ with an unpaired electron on the nitrogen atom. This unpaired electron effectively participates in a $3$-electron bond character in its molecular orbital description.
Among the given options,$NO$ (Nitric oxide) is the molecule that contains this characteristic $3$-electron bond.

(A) According to Molecular Orbital Theory $(MOT)$, the electronic configuration of $O_2$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$.
Since the oxygen molecule contains two unpaired electrons in the antibonding molecular orbitals ($\pi^* 2p_x$ and $\pi^* 2p_y$), it exhibits paramagnetism.

(C) The electronic configurations are as follows:
$1$. $KO_2$ contains the superoxide ion $O_2^-$. The molecular orbital configuration of $O_2^-$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. It has one unpaired electron.
$2$. $NO_2^-$ has $5 + 6(2) + 1 = 18$ valence electrons. All electrons are paired.
$3$. $BaO_2$ contains the peroxide ion $O_2^{2-}$. The configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. All electrons are paired.
$4$. $NO_2^+$ has $5 + 6(2) - 1 = 16$ valence electrons. All electrons are paired.
Therefore,only $KO_2$ contains an unpaired electron.

(D) Paramagnetism is a property of substances that are attracted by a magnetic field. This behavior arises due to the presence of one or more unpaired electrons in the atoms,molecules,or ions. Therefore,the correct option is $(D)$.

(D) The molecular formula of dinitrogen tetroxide is $(N_2O_4)$.
In $(N_2O_4)$,each nitrogen atom is bonded to two oxygen atoms and one nitrogen atom.
The total number of valence electrons is $(2 \times 5) + (4 \times 6) = 10 + 24 = 34$.
Since all electrons are paired in the structure of $(N_2O_4)$,there are no unpaired electrons.
Substances with no unpaired electrons are diamagnetic.
Therefore,$(N_2O_4)$ has no unpaired electrons and is diamagnetic.

(A) The total number of electrons in $O_2^+$ is $15$.
The molecular orbital configuration is: $KK \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 = \pi 2p_y^2 \pi^* 2p_x^1$.
Bond order is calculated as: $\text{Bond Order} = \frac{N_b - N_a}{2} = \frac{10 - 5}{2} = 2.5$.
Since it contains one unpaired electron,it is paramagnetic.
The bond order of $O_2^+$ $(2.5)$ is higher than that of $O_2$ $(2.0)$,so $O_2^+$ is more stable than $O_2$.

(B) Degenerate orbitals are those that have the same energy level.
In homonuclear diatomic molecules,the molecular orbitals $\pi 2p_x$ and $\pi 2p_y$ have identical energy levels due to the symmetry of the $p_x$ and $p_y$ orbitals along the internuclear axis.
Therefore,the set $(\pi 2p_x, \pi 2p_y)$ is degenerate.

(A) The bond length is inversely proportional to the bond order.
Mathematically,$\text{Bond Length} \propto \frac{1}{\text{Bond Order}}$.
Therefore,a higher bond order corresponds to a shorter bond length.
Comparing the given values: $2.5 > 2.0 > 1.5 > 0.5$.
Thus,the bond order of $2.5$ corresponds to the shortest bond length.

(C) According to the Molecular Orbital Theory,the bond order of a molecule is calculated as: $Bond \ Order = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
For $He_2$,the total number of electrons is $4$.
The electronic configuration is: $\sigma_{1s}^2 \sigma_{1s}^{*2}$.
Bond order $= \frac{2 - 2}{2} = 0$.
$A$ bond order of $0$ indicates that the molecule is unstable and does not exist.

(C) 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.
$1. O_2$ ($16$ electrons): Configuration is $KK(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi^* 2p_x)^1(\pi^* 2p_y)^1$. Bond order $= \frac{10 - 6}{2} = 2$.
$2. O_2^-$ ($17$ electrons): Configuration adds one electron to $\pi^* 2p_y$. Bond order $= \frac{10 - 7}{2} = 1.5$.
$3. O_2^+$ ($15$ electrons): Configuration removes one electron from $\pi^* 2p_y$. Bond order $= \frac{10 - 5}{2} = 2.5$.
$4. O_2^{2-}$ ($18$ electrons): Configuration adds two electrons to $\pi^* 2p_x$ and $\pi^* 2p_y$. Bond order $= \frac{10 - 8}{2} = 1$.
Comparing the values,$O_2^+$ has the highest bond order of $2.5$.

(B) To determine paramagnetism,we look for the presence of unpaired electrons using Molecular Orbital Theory $(MOT)$.
$1$. $NO$ has $15$ electrons: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1$. It has one unpaired electron,so it is paramagnetic.
$2$. $S^{2-}$ has $18$ electrons $(16 + 2)$: $1s^2 2s^2 2p^6 3s^2 3p^6$. All electrons are paired,so it is diamagnetic.
$3$. $O_2^-$ has $17$ 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 one unpaired electron,so it is paramagnetic.
$4$. $N_2^-$ has $15$ electrons: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2 (\pi^* 2p_x)^1$. It has one unpaired electron,so it is paramagnetic.
Thus,$S^{2-}$ is the only species that is not paramagnetic (it is diamagnetic).

(C) Molecules or ions that contain one or more unpaired electrons are attracted by an external magnetic field and are said to be paramagnetic.

(B) The $NO$ molecule has $15$ electrons. According to Molecular Orbital Theory,its electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$.
Since it contains one unpaired electron in the $\pi^*$ orbital,it exhibits paramagnetic behavior.

(A) The bond order is calculated using the formula: $\text{Bond Order} = \frac{1}{2}(N_b - N_a)$.
Here,$N_b$ is the number of electrons in bonding molecular orbitals and $N_a$ is the number of electrons in antibonding molecular orbitals.
From the given configuration: $N_b = 2 + 2 + 2 + 2 = 8$ (excluding $KK$ shell) and $N_a = 2$.
Therefore,$\text{Bond Order} = \frac{1}{2}(8 - 2) = \frac{6}{2} = 3$.

(B) According to Molecular Orbital Theory,the total number of electrons in $CO$ is $14$ $(6 + 8 = 14)$.
The electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$.
Since all electrons are paired,$CO$ is diamagnetic.

(C) The bond order of $N_2$ is $3.0$ (electronic configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$).
The bond order of $N_2^+$ is $2.5$ (electronic configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^1$).
Since bond dissociation energy is directly proportional to bond order,the bond dissociation energy of $N_2$ is greater than that of $N_2^+$.

(A) The total number of electrons in $O_2^-$ is $8 + 8 + 1 = 17$.
According to Molecular Orbital Theory,the electronic configuration is: $KK \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 = \pi 2p_y^2 \pi^* 2p_x^2 = \pi^* 2p_y^1$.
Number of bonding electrons $(N_b)$ = $8$.
Number of antibonding electrons $(N_a)$ = $7$.
Bond Order $(B.O.)$ = $\frac{1}{2}(N_b - N_a) = \frac{8 - 7}{2} = 0.5$.

(A) The total number of electrons in $N_2^+$ is $7 + 7 - 1 = 13$.
According to the Molecular Orbital Theory for molecules with $14$ or fewer electrons,the energy order is $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \pi 2p_x = \pi 2p_y < \sigma 2p_z < \pi^* 2p_x = \pi^* 2p_y < \sigma^* 2p_z$.
Filling $13$ electrons into these orbitals: $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \pi 2p_x^2 \pi 2p_y^2 \sigma 2p_z^1$.

(A) The antibonding molecular orbital $\pi^* 2p_x$ has two nodal planes perpendicular to each other. One nodal plane is the molecular plane (containing the internuclear axis),and the other is perpendicular to the molecular plane.

(A) The total number of electrons in $O_2^{2-}$ is $18$.
The molecular orbital configuration is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$.
Counting the electrons in the atomic orbitals that do not participate in bonding (the inner shell $1s$ and $2s$ orbitals,and the lone pairs in the valence shell),we look at the non-bonding electrons.
Specifically,the electrons in the $1s, 2s$ orbitals and the lone pairs in the $O$ atoms contribute to the non-bonding character.
In $O_2^{2-}$,there are $16$ non-bonding electrons,which corresponds to $8$ non-bonding electron pairs.

(D) The bond order is inversely proportional to the bond length,i.e.,$\text{Bond Order} \propto \frac{1}{\text{Bond Length}}$.
Since the bond order of $NO$ $(2.5)$ is less than the bond order of $NO^+$ $(3.0)$,the bond length of $NO$ must be greater than the bond length of $NO^+$.
Therefore,the bond length of $NO > NO^+$.

(A) The total number of electrons in $CN^-$ is $6 + 7 + 1 = 14$.
The total number of electrons in $CO$ is $6 + 8 = 14$.
The total number of electrons in $NO^+$ is $7 + 8 - 1 = 14$.
Since all these species have $14$ electrons,they are isoelectronic.
According to Molecular Orbital Theory,the bond order for a $14$-electron species is calculated as: $\text{Bond Order} = \frac{1}{2}(N_b - N_a) = \frac{1}{2}(10 - 4) = 3$.
Therefore,all these species have a bond order of $3$ and are isoelectronic.

(A) According to Molecular Orbital Theory $(MOT)$:
$NO$ has a total of $15$ electrons ($7$ from $N$ and $8$ from $O$).
The electronic configuration is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$.
Since there is one electron in the antibonding $\pi^* 2p_x$ orbital,$NO$ is paramagnetic and contains one unpaired electron.
$CO$ ($14$ electrons) and $CN^{-}$ ($14$ electrons) are diamagnetic (zero unpaired electrons).
$O_2$ ($16$ electrons) has two unpaired electrons in the $\pi^* 2p$ orbitals.

(C) The total number of electrons in $O_2^{2-}$ is $8 + 8 + 2 = 18$.
According to the Molecular Orbital Theory,the electronic configuration of $O_2^{2-}$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$.
Number of bonding electrons $(N_b)$ = $10$.
Number of antibonding electrons $(N_a)$ = $8$.
Bond Order = $\frac{1}{2} (N_b - N_a) = \frac{1}{2} (10 - 8) = \frac{2}{2} = 1$.

(B) The molecular orbital configuration of $O_2^+$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$.
Since the last electron enters the antibonding $\pi^*$ orbital,the electron lost during the ionization of $O_2^+$ to $O_2^{2+}$ is removed from the antibonding $\pi-$ orbital.

(B) The bond order of a species can be determined by its total number of electrons using Molecular Orbital Theory.
For $CN^{-}$: Total electrons = $6 (C) + 7 (N) + 1 = 14$. Bond order = $\frac{1}{2}(10 - 4) = 3$.
For $CO$: Total electrons = $6 (C) + 8 (O) = 14$. Bond order = $\frac{1}{2}(10 - 4) = 3$.
Since both $CN^{-}$ and $CO$ have $14$ electrons,they have the same bond order of $3$.

(D) Stability is directly proportional to the bond order of the species.
$1$. Calculate the total number of electrons and bond order $(B.O.)$ for each species:
- $O_2^{+}$ $(15 \ e^{-})$: $B.O. = 2.5$
- $O_2$ $(16 \ e^{-})$: $B.O. = 2.0$
- $O_2^{-}$ $(17 \ e^{-})$: $B.O. = 1.5$
- $O_2^{2-}$ $(18 \ e^{-})$: $B.O. = 1.0$
$2$. Since stability $\propto B.O.$,the decreasing order of stability is:
$O_2^{+} > O_2 > O_2^{-} > O_2^{2-}$

(B) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as: $\text{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
For $O_{2}^{-}$ ($17$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. Bond order = $\frac{10 - 7}{2} = 1.5$.
For $O_{2}^{+}$ ($15$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Bond order = $\frac{10 - 5}{2} = 2.5$.
For $O_{2}^{2+}$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. Bond order = $\frac{10 - 4}{2} = 3.0$.
Thus,the increasing order of bond order is $O_{2}^{-} < O_{2}^{+} < O_{2}^{2+}$.

(A) According to Molecular Orbital Theory $(MOT)$:
Number of electrons:
$O_2^- = 17$,$O_2 = 16$,$O_2^+ = 15$.
Bond order calculation:
Bond Order = $\frac{1}{2} (N_b - N_a)$.
For $O_2^-$: Bond order = $1.5$.
For $O_2$: Bond order = $2.0$.
For $O_2^+$: Bond order = $2.5$.
Thus,the correct order of bond order is $O_2^- < O_2 < O_2^+$.

(D) $CN^{-}$,$CO$,and $NO^{+}$ are isoelectronic with $14$ electrons each,and there are no unpaired electrons in the Molecular Orbital $(MO)$ configuration of these species.
Therefore,these are diamagnetic.
$O_{2}^{-}$ has $17$ electrons and is paramagnetic due to the presence of one unpaired electron in its antibonding $\pi^{*}$ orbital.
$(CN^{-}, CO, NO^{+}) \; 14 \; e^{-} = \sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \sigma 2p_{z}^{2}, \pi 2p_{x}^{2} \approx \pi 2p_{y}^{2}$ (No unpaired electron,diamagnetic).
$O_{2}^{-} = \sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \sigma 2p_{z}^{2}, \pi 2p_{x}^{2} \approx \pi 2p_{y}^{2}, \pi^{*} 2p_{x}^{2} \approx \pi^{*} 2p_{y}^{1}$ (One unpaired electron,paramagnetic).

(A) The bond order is calculated using the formula: $Bond \ order = \frac{N_b - N_a}{2}$,where $N_b$ is the number of electrons in bonding molecular orbitals and $N_a$ is the number of electrons in antibonding molecular orbitals.
For $CO$ ($6+8=14$ electrons): Bond order $= \frac{10-4}{2} = 3$.
For $NO^{+}$ ($7+8-1=14$ electrons): Bond order $= \frac{10-4}{2} = 3$.
Since both $CO$ and $NO^{+}$ have $14$ electrons,they have the same bond order of $3$.
For other options:
$NO^{-}$ ($16$ electrons,$BO=2$),$CN^{-}$ ($14$ electrons,$BO=3$).
$O_2$ ($16$ electrons,$BO=2$),$N_2$ ($14$ electrons,$BO=3$).
$O_2$ ($16$ electrons,$BO=2$),$B_2$ ($10$ electrons,$BO=1$).