Molecular orbital theory Questions in English

(C) Bond dissociation energy $(E)$ is directly proportional to the bond order,while bond length $(R)$ is inversely proportional to the bond order.
The bond orders are: $N_2$ (triple bond,$B.O. = 3$),$O_2$ (double bond,$B.O. = 2$),and $F_2$ (single bond,$B.O. = 1$).
Therefore,the order of bond dissociation energy is: $E(N_2) > E(O_2) > E(F_2)$.
The order of bond length is: $R(F_2) > R(O_2) > R(N_2)$.

(A) According to Molecular Orbital Theory,for molecules with $Z > 7$ (like $O_2$ and $F_2$),the energy of the $\sigma_{2p_z}$ orbital is lower than the energy of the $\pi_{2p_x}$ and $\pi_{2p_y}$ orbitals.
The correct energy order for $F_2$ is: $\sigma_{2p_z} < \pi_{2p_x} = \pi_{2p_y} < \pi_{2p_x}^* = \pi_{2p_y}^* < \sigma_{2p_z}^*$.

(C) Molecular orbital configuration:
$(i)$ $B_2$ ($10$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1 \approx \pi 2p_y^1$. It has two unpaired electrons,hence it is paramagnetic.
$(ii)$ $N_2$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 \approx \pi 2p_y^2, \sigma 2p_z^2$. It has no unpaired electrons,hence it is diamagnetic.
$(iii)$ $O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 \approx \pi 2p_y^2, \pi^* 2p_x^1 \approx \pi^* 2p_y^1$. It has two unpaired electrons,hence it is paramagnetic.
$(iv)$ $C_2$ ($12$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 \approx \pi 2p_y^2$. It has no unpaired electrons,hence it is diamagnetic.
Therefore,$N_2$ and $C_2$ are diamagnetic.

(A) The bond order is a measure of the number of chemical bonds between a pair of atoms.
It is directly proportional to the bond dissociation energy and the stability of the molecule.
However,it is inversely proportional to the bond length.
As the bond order increases,the atoms are pulled closer together,resulting in a shorter bond length.
Therefore,$Bond \ Order \propto \frac{1}{Bond \ Length}$.

(C) The formation of molecular orbitals is described by the Linear Combination of Atomic Orbitals $(LCAO)$ method.
Bonding molecular orbitals are formed by the constructive interference of atomic orbitals,which corresponds to the addition of their wave functions.
Therefore,the wave function for a bonding molecular orbital is given by $\psi_{bonding} = \psi_A + \psi_B$.
Conversely,antibonding molecular orbitals are formed by the destructive interference,represented by the subtraction of wave functions,$\psi_{antibonding} = \psi_A - \psi_B$.

(D) According to molecular orbital theory:
$I$. $\sigma^*$-orbitals are formed by the subtraction of two wave functions,$\psi_A$ and $\psi_B$,therefore $\sigma^* = \psi_A - \psi_B$. This is correct.
$II$. $\sigma$-orbitals are formed by the constructive interference of two orbitals in the same phase,and thus,they do not have a nodal plane between the two nuclei. This statement is incorrect.
$III$. The combination of two $1s$ atomic orbitals gives two molecular orbitals,one of which is of lower energy,called the $\sigma$ bonding orbital,and the other is of higher energy,called the $\sigma^*$ anti-bonding orbital. This is correct.
Hence,$(I)$ and $(III)$ are the correct statements,and option $(D)$ is the correct answer.

(C) Bond order can be calculated by using the formula,$\text{B.O.} = \frac{1}{2} [N_b - N_a]$.
Molecular orbital electronic configuration of $B_2 = (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x^1 = \pi 2p_y^1)$.
Bond order of $B_2 = \frac{1}{2} [6 - 4] = 1$.
Molecular orbital electronic configuration of $He_2 = (\sigma 1s)^2 (\sigma^* 1s)^2$.
Bond order of $He_2 = \frac{1}{2} [2 - 2] = 0$.
Molecular orbital electronic configuration of $N_2 = (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x^2 = \pi 2p_y^2) (\sigma 2p_z)^2$.
Bond order of $N_2 = \frac{1}{2} [10 - 4] = 3$.
Molecular orbital electronic configuration of $C_2 = (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x^2 = \pi 2p_y^2)$.
Bond order of $C_2 = \frac{1}{2} [8 - 4] = 2$.
Therefore,the increasing order of bond order values is $He_2 (0) < B_2 (1) < C_2 (2) < N_2 (3)$.

(C) The electronic configuration of $O_2$ ($16$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Bond order = $(10-6)/2 = 2$. It is paramagnetic.
The electronic configuration of $O_2^{2+}$ ($14$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. Bond order = $(10-4)/2 = 3$. It is diamagnetic.
Statement $I$ is incorrect because bond order increases from $2$ to $3$.
Statement $II$ is incorrect because $O_2$ is paramagnetic and $O_2^{2+}$ is diamagnetic.
Statement $III$ is correct because bond order increases,so bond length decreases.
For $O_2^{2-}$ ($18$ electrons),bond order = $(10-8)/2 = 1$. For $B_2$ ($10$ electrons),bond order = $(6-4)/2 = 1$.
Statement $IV$ is correct because both have a bond order of $1$.

(D) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as $BO = \frac{1}{2}(N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
For $O_2$ ($16$ electrons): Bond order = $2.0$.
For $O_2^{+}$ ($15$ electrons): Bond order = $2.5$.
For $O_2^{-}$ ($17$ electrons): Bond order = $1.5$.
For $O_2^{2+}$ ($14$ electrons): Bond order = $3.0$.
Sum of bond orders = $2.0 + 2.5 + 1.5 + 3.0 = 9.0$.

(A) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as $\frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
For $O_2$ ($16$ electrons): Bond order = $2.0$.
For $O_2^{-}$ ($17$ electrons): Bond order = $1.5$.
For $O_2^{2-}$ ($18$ electrons): Bond order = $1.0$.
Sum of bond orders of $O_2^{-}$ and $O_2^{2-}$ = $1.5 + 1.0 = 2.5$. Thus,$x = 2.5$.
For $O_2^{2+}$ ($14$ electrons): Bond order = $3.0$.
We need to express the bond order of $O_2^{2+}$ $(3.0)$ in terms of $x$ $(2.5)$.
$3.0 = k \times 2.5 \implies k = \frac{3.0}{2.5} = 1.2$.
Therefore,the bond order of $O_2^{2+}$ is $1.2x$.

(B) The bond order is calculated as: $\text{Bond order} = \frac{1}{2} [N_b - N_a]$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of anti-bonding electrons.
For $O_2$ ($16$ electrons): $\text{Bond order} = \frac{1}{2} [10 - 6] = 2.0$.
For $O_2^{-}$ ($17$ electrons): $\text{Bond order} = \frac{1}{2} [10 - 7] = 1.5$.
For $O_2^{+}$ ($15$ electrons): $\text{Bond order} = \frac{1}{2} [10 - 5] = 2.5$.
For $O_2^{2-}$ ($18$ electrons): $\text{Bond order} = \frac{1}{2} [10 - 8] = 1.0$.
For $O_2^{2+}$ ($14$ electrons): $\text{Bond order} = \frac{1}{2} [10 - 4] = 3.0$.
Calculating the differences:
$(A)$ $|2.5 - 1.5| = 1.0$
$(B)$ $|3.0 - 1.0| = 2.0$
$(C)$ $|3.0 - 2.0| = 1.0$
$(D)$ $|3.0 - 2.5| = 0.5$
The maximum difference is $2.0$ for the pair $O_2^{2-}$ and $O_2^{2+}$.

(D) The electronic configuration of $C_2$ ($12$ electrons) is: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2$.
Bond order $= \frac{1}{2}(N_b - N_a) = \frac{1}{2}(8 - 4) = 2$.
Since all electrons are paired,$C_2$ is diamagnetic.
$C_2^{-}$ has $13$ electrons,configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^1$. It has one unpaired electron,so it is paramagnetic.
In $C_2$,the two highest occupied molecular orbitals are $\pi$ orbitals,meaning $C_2$ consists of two $\pi$ bonds.
Therefore,the statement that $C_2$ consists of $1 \sigma$ and $1 \pi$ bond is incorrect.

(B) $\text{B.O.} = \frac{N_b - N_a}{2}$
For $O_2^{+}$: $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 \pi 2p_y^2 \pi^* 2p_x^1$
$\Rightarrow \text{B.O.} = \frac{10 - 5}{2} = 2.5 = x$
For $O_2^{-}$: $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 \pi 2p_y^2 \pi^* 2p_x^2 \pi^* 2p_y^1$
$\Rightarrow \text{B.O.} = \frac{10 - 7}{2} = 1.5$
For $O_2^{2+}$: $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 \pi 2p_y^2$
$\Rightarrow \text{B.O.} = \frac{10 - 4}{2} = 3$
Now,express $1.5$ and $3$ in terms of $x = 2.5$:
$\frac{1.5}{2.5} x = \frac{3}{5} x$
$\frac{3}{2.5} x = \frac{6}{5} x$
Thus,the bond orders are $\frac{3}{5} x$ and $\frac{6}{5} x$.

(C) The bond order is determined by the total number of electrons in the species.
$CN^{-}$ has $6 + 7 + 1 = 14$ electrons.
$N_2^{+}$ has $7 + 7 - 1 = 13$ electrons.
$O_2^{2-}$ has $8 + 8 + 2 = 18$ electrons.
$NO^{+}$ has $7 + 8 - 1 = 14$ electrons.
Since $CN^{-}$ and $NO^{+}$ both have $14$ electrons,they have the same bond order of $3$.

(A) The molecular orbital configuration for $C_2$ ($12$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2$.
In $C_2$,the bond order is $\frac{1}{2}(N_b - N_a) = \frac{1}{2}(8 - 4) = 2$.
Since the last electrons enter the $\pi 2p$ orbitals,both bonds in $C_2$ are $\pi$-bonds.
Therefore,$C_2$ is the molecule that contains only $\pi$-bonds.

(C) The bond order can be determined by 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.
$1. C_2$ ($12$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2, \pi 2p_y^2$. $N_b = 8, N_a = 4$. $B.O. = \frac{8-4}{2} = 2$.
$2. B_2^{2-}$ ($12$ electrons): Same configuration as $C_2$. $B.O. = 2$.
$3. N_2^{2+}$ ($12$ electrons): Same configuration as $C_2$. $B.O. = 2$.
$4. CN^{+}$ ($12$ electrons): Same configuration as $C_2$. $B.O. = 2$.
$5. NO^{-}$ ($12$ electrons): Same configuration as $C_2$. $B.O. = 2$.
$6. O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^1, \pi^* 2p_y^1$. $N_b = 10, N_a = 6$. $B.O. = \frac{10-6}{2} = 2$.
$7. C_2^{+}$ ($11$ electrons): $B.O. = \frac{7-4}{2} = 1.5$.
Thus,the species with bond order $2$ are $C_2, B_2^{2-}, N_2^{2+}, CN^{+}, NO^{-}, O_2$. Total $6$ species.

(B) To determine the bond order and magnetic character,we use Molecular Orbital Theory $(MOT)$. The total number of electrons for each species is:
$O_2$ $(16 \ e^-)$: Bond order = $\frac{10-6}{2} = 2$,Paramagnetic.
$O_2^{+}$ $(15 \ e^-)$: Bond order = $\frac{10-5}{2} = 2.5$,Paramagnetic.
$O_2^{-}$ $(17 \ e^-)$: Bond order = $\frac{10-7}{2} = 1.5$,Paramagnetic.
$O_2^{2-}$ $(18 \ e^-)$: Bond order = $\frac{10-8}{2} = 1$,Diamagnetic.
$(A)$ The species with the highest bond order is $O_2^{+}$ (Bond order = $2.5$).
$(B)$ The species with diamagnetic character is $O_2^{2-}$ (all electrons are paired).
Therefore,the pair is $(O_2^{+}, O_2^{2-})$.

(B) According to Molecular Orbital Theory,a species does not exist if its Bond Order $(B.O.)$ is $0$.
$B.O. = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
$1$. $H_2^{+} (1e^-): \sigma 1s^1, B.O. = 0.5$ (Exists)
$2$. $He_2^{2+} (2e^-): \sigma 1s^2, B.O. = 1$ (Exists)
$3$. $Li_2^{2-} (8e^-): \sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2, B.O. = \frac{1}{2}(4-4) = 0$ (Does not exist)
$4$. $Ne_2 (20e^-): \sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2pz^2 \pi 2px^2 \pi 2py^2 \pi^* 2px^2 \pi^* 2py^2 \sigma^* 2pz^2, B.O. = \frac{1}{2}(10-10) = 0$ (Does not exist)
$5$. $Be_2^{-} (9e^-): \sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2pz^1, B.O. = 0.5$ (Exists)
$6$. $He_2 (4e^-): \sigma 1s^2 \sigma^* 1s^2, B.O. = \frac{1}{2}(2-2) = 0$ (Does not exist)
Thus,$Li_2^{2-}, Ne_2,$ and $He_2$ do not exist.

(B) According to molecular orbital theory,the bond order is calculated as:
$Bond \ order = \frac{(\text{Number of } e^{-} \text{ in bonding MO}) - (\text{Number of } e^{-} \text{ in antibonding MO})}{2}$
Calculating bond orders for the given species:
$O_2^{+} (15 \ e^{-}): \sigma 1s^2, \sigma^{*} 1s^2, \sigma 2s^2, \sigma^{*} 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^{*} 2p_x^1$. $B.O. = \frac{10-5}{2} = 2.5$
$O_2 (16 \ e^{-}): \sigma 1s^2, \sigma^{*} 1s^2, \sigma 2s^2, \sigma^{*} 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^{*} 2p_x^1 = \pi^{*} 2p_y^1$. $B.O. = \frac{10-6}{2} = 2.0$
$O_2^{-} (17 \ e^{-}): \sigma 1s^2, \sigma^{*} 1s^2, \sigma 2s^2, \sigma^{*} 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^{*} 2p_x^2 = \pi^{*} 2p_y^1$. $B.O. = \frac{10-7}{2} = 1.5$
$O_2^{2-} (18 \ e^{-}): \sigma 1s^2, \sigma^{*} 1s^2, \sigma 2s^2, \sigma^{*} 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^{*} 2p_x^2 = \pi^{*} 2p_y^2$. $B.O. = \frac{10-8}{2} = 1.0$
Comparing the values: $2.5 > 2.0 > 1.5 > 1.0$.
Thus,the correct sequence is $O_2^{+} > O_2 > O_2^{-} > O_2^{2-}$.

(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 anti-bonding electrons.
$1$. For $He_2$ ($4$ electrons): Configuration is $\sigma 1s^2 \sigma^* 1s^2$. $BO = \frac{2-2}{2} = 0$.
$2$. For $He_2^+$ ($3$ electrons): Configuration is $\sigma 1s^2 \sigma^* 1s^1$. $BO = \frac{2-1}{2} = 0.5$.
$3$. For $O_2$ ($16$ electrons): Configuration is $KK \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 = \pi 2p_y^2 \pi^* 2p_x^1 = \pi^* 2p_y^1$. $BO = \frac{10-6}{2} = 2$.
$4$. For $O_2^+$ ($15$ electrons): Configuration is $KK \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 = \pi 2p_y^2 \pi^* 2p_x^1$. $BO = \frac{10-5}{2} = 2.5$.
Thus,the bond orders are $0, 0.5, 2$ and $2.5$ respectively.

(B) According to 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 $Be_2$ ($Z=4$,total $8$ electrons),the electronic configuration is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2$.
Here,$N_b = 4$ and $N_a = 4$.
$Bond \ Order = \frac{1}{2} (4 - 4) = 0$.
Since the bond order is $0$,the $Be_2$ molecule is unstable and does not exist.

(D) If all the electrons are paired,the molecule is diamagnetic,while if one or more electrons in a molecule are unpaired,the species is paramagnetic.
$O_2^{2-}$ is not paramagnetic because in $O_2^{2-}$ all the electrons are paired.
Configuration of $O_2^{2-}$ (Total electrons $= 18$): $(\sigma 1s)^2, (\sigma^* 1s)^2, (\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\pi^* 2p_x)^2, (\pi^* 2p_y)^2$.
Since all electrons are paired,$O_2^{2-}$ is diamagnetic.

(C) Based on the molecular orbital theory diagrams:
$(i)$ Represents an antibonding pi orbital,denoted as $\pi^*$.
(ii) Represents a bonding sigma orbital,denoted as $\sigma$.
(iii) Represents an antibonding sigma orbital,denoted as $\sigma^*$.
(iv) Represents a bonding pi orbital,denoted as $\pi$.
Therefore,the correct set is $I. = \pi^*, II. = \sigma, III. = \sigma^*, IV. = \pi$.

(D) Bond length is inversely proportional to bond order.
The bond order for these species is calculated using Molecular Orbital Theory:
Bond order of $O_2^+ = 2.5$
Bond order of $O_2 = 2.0$
Bond order of $O_2^- = 1.5$
Bond order of $O_2^{2-} = 1.0$
Since bond length $\propto \frac{1}{\text{bond order}}$,the order of bond length is $O_2^{2-} (IV) > O_2^- (III) > O_2 (I) > O_2^+ (II)$.
Thus,the correct sequence is $IV > III > I > II$.

(B) According to molecular orbital theory,species with all paired electrons in their electronic configuration are diamagnetic in nature.
$He_2^{+} \rightarrow$ Total $3$ electrons,configuration $\sigma 1s^2, \sigma^* 1s^1$. It contains one unpaired electron and is paramagnetic.
$H_2 \rightarrow$ Total $2$ electrons,configuration $\sigma 1s^2$. All electrons are paired,hence it is diamagnetic.
$H_2^{+} \rightarrow$ Total $1$ electron,configuration $\sigma 1s^1$. It contains one unpaired electron and is paramagnetic.
$H_2^{-} \rightarrow$ Total $3$ electrons,configuration $\sigma 1s^2, \sigma^* 1s^1$. It contains one unpaired electron and is paramagnetic.
$He \rightarrow$ Total $2$ electrons,configuration $\sigma 1s^2$. All electrons are paired,hence it is diamagnetic.
Therefore,$H_2$ and $He$ are diamagnetic. The total count is $2$.

(B) 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
$N_2^{-}$	$2.5$
$NO^{-}$	$2$
$C_2^{-}$	$2.5$
$N_2^{+}$	$2.5$
$C_2^{2-}$	$3$
$CN^{+}$	$2$

From the table,the ions with a bond order of $2.5$ are $N_2^{-}$,$C_2^{-}$,and $N_2^{+}$.
Therefore,there are $3$ such ions.

(A) The bond order is calculated using the formula: $\text{Bond order} = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
$A. N_2^+$ ($13$ electrons): $\sigma 1s^2 < \sigma^* 1s^2 < \sigma 2s^2 < \sigma^* 2s^2 < \pi 2p_x^2 = \pi 2p_y^2 < \sigma 2p_z^1$. Bond order = $\frac{9-4}{2} = 2.5$ $(IV)$.
$B. CO$ ($14$ 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 = $\frac{10-4}{2} = 3.0$ $(V)$.
$C. O_2$ ($16$ electrons): $\sigma 1s^2 < \sigma^* 1s^2 < \sigma 2s^2 < \sigma^* 2s^2 < \sigma 2p_z^2 < \pi 2p_x^2 = \pi 2p_y^2 < \pi^* 2p_x^1 = \pi^* 2p_y^1$. Bond order = $\frac{10-6}{2} = 2.0$ $(III)$.
$D. O_2^-$ ($17$ 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 = $\frac{10-7}{2} = 1.5$ $(II)$.
Therefore,the correct match is $A-IV, B-V, C-III, D-II$.

(D) The molecular orbital electronic configuration of $Be_2$ is: $\sigma 1s^2 < \sigma^* 1s^2 < \sigma 2s^2 < \sigma^* 2s^2$.
The molecular orbital electronic configuration of $C_2$ is: $\sigma 1s^2 < \sigma^* 1s^2 < \sigma 2s^2 < \sigma^* 2s^2 < [\pi 2p_{x}^2 = \pi 2p_{y}^2]$.
The molecular orbital electronic configuration of $N_2$ is: $\sigma 1s^2 < \sigma^* 1s^2 < \sigma 2s^2 < \sigma^* 2s^2 < [\pi 2p_{x}^2 = \pi 2p_{y}^2] < \sigma 2p_{z}^2$.
The molecular orbital electronic configuration of $O_2$ is: $\sigma 1s^2 < \sigma^* 1s^2 < \sigma 2s^2 < \sigma^* 2s^2 < \sigma 2p_{z}^2 < [\pi 2p_{x}^2 = \pi 2p_{y}^2] < [\pi^* 2p_{x}^1 = \pi^* 2p_{y}^1]$.
Molecules having unpaired electrons are paramagnetic,and those without unpaired electrons are diamagnetic.
Since $O_2$ has $2$ unpaired electrons in its antibonding molecular orbitals,it is paramagnetic. Conversely,$Be_2$,$C_2$,and $N_2$ have no unpaired electrons,making them diamagnetic.

(A) According to Molecular Orbital Theory $(MOT)$:
Bond Order $(B.O.)$ = $\frac{1}{2} (N_b - N_a)$
For $Li_2$ ($6$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2$. $B.O. = \frac{1}{2} (4 - 2) = 1.0$ $(iii)$.
For $N_2$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. $B.O. = \frac{1}{2} (10 - 4) = 3$ $(i)$.
For $Be_2$ ($8$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2$. $B.O. = \frac{1}{2} (4 - 4) = 0$ $(iv)$.
For $O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. $B.O. = \frac{1}{2} (10 - 6) = 2$ $(v)$.
Therefore,the correct match is $A-iii, B-i, C-iv, D-v$.

(A) To group the species by bond order,we calculate the bond order $(B.O.)$ for each using Molecular Orbital Theory:
$1$. $O_2^{2-}$ ($18$ electrons): $B.O. = \frac{1}{2}(10-8) = 1$
$2$. $Li_2$ ($6$ electrons): $B.O. = \frac{1}{2}(4-2) = 1$
$3$. $F_2$ ($18$ electrons): $B.O. = \frac{1}{2}(10-8) = 1$
$4$. $He_2^{2+}$ ($2$ electrons): $B.O. = \frac{1}{2}(2-0) = 1$
$5$. $N_2$ ($14$ electrons): $B.O. = \frac{1}{2}(10-4) = 3$
$6$. $O_2^{2+}$ ($14$ electrons): $B.O. = \frac{1}{2}(10-4) = 3$
Thus,the species with $B.O. = 1$ are $(O_2^{2-}, Li_2, F_2, He_2^{2+})$ and the species with $B.O. = 3$ are $(N_2, O_2^{2+})$.

(A) The molecular orbital electronic configuration of $He_2$ is $\sigma 1s^2, \sigma^* 1s^2$.
Bond order $= \frac{(\text{electrons in bonding MO}) - (\text{electrons in antibonding MO})}{2}$.
For $He_2$,bond order $= \frac{2-2}{2} = 0$.
Since the bond order is $0$,the $He_2$ molecule does not exist.
For $He_2^{+}$,the molecular orbital electronic configuration is $\sigma 1s^2, \sigma^* 1s^1$.
Bond order $= \frac{2-1}{2} = 1/2$.

(A) The electronic configuration of $N_2$ ($14$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$.
Bond order is calculated as: $\text{Bond order} = \frac{N_b - N_a}{2} = \frac{10 - 4}{2} = 3$.

(C) The molecular orbital electronic configuration of $O_2$ ($16$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^1, \pi^* 2p_y^1$.
Bonding molecular orbitals are those without an asterisk $(*)$.
The bonding orbitals are $\sigma 1s, \sigma 2s, \sigma 2p_z, \pi 2p_x, \pi 2p_y$.
Each of these $5$ orbitals contains $2$ electrons,totaling $10$ bonding electrons.
Therefore,the number of bonding electron pairs is $10 / 2 = 5$.

(D) $KO_2$ is a superoxide compound consisting of $K^{+}$ and $O_2^{-}$ ions.
The $K^{+}$ ion has a noble gas configuration and is diamagnetic.
The superoxide ion $O_2^{-}$ has $17$ electrons.
According to molecular orbital theory,the electronic configuration of $O_2^{-}$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$.
Due to the presence of one unpaired electron in the $\pi^* 2p$ orbital,$O_2^{-}$ is paramagnetic.

(C) An odd-electron molecule is a molecule that contains an odd number of valence electrons.
$1$. $C_2$: Total valence electrons = $4 + 4 = 8$ (even).
$2$. $H_2$: Total valence electrons = $1 + 1 = 2$ (even).
$3$. $SCl_2$: Total valence electrons = $6 + 7(2) = 20$ (even).
$4$. $NO$: Total valence electrons = $5 + 6 = 11$ (odd).
$5$. $NO_2$: Total valence electrons = $5 + 6(2) = 17$ (odd).
Therefore,$NO$ and $NO_2$ are odd-electron molecules.

(A) The total number of electrons in $HeH^{+}$ is $2 + 1 - 1 = 2$.
The molecular orbital electronic configuration is $\sigma_{1s}^2$.
Bond Order $= \frac{1}{2} [N_b - N_a] = \frac{1}{2} [2 - 0] = 1$.

(B) The bond order is calculated using the formula: $B.O. = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
For $He_{2}$ $(4 \ e^{-})$: Configuration is $\sigma(1s)^{2}, \sigma^{*}(1s)^{2}$. $B.O. = \frac{2-2}{2} = 0$.
For $He_{2}^{+}$ $(3 \ e^{-})$: Configuration is $\sigma(1s)^{2}, \sigma^{*}(1s)^{1}$. $B.O. = \frac{2-1}{2} = 0.5$.
For $He_{2}^{2+}$ $(2 \ e^{-})$: Configuration is $\sigma(1s)^{2}$. $B.O. = \frac{2-0}{2} = 1$.
Thus,the bond orders are $0, 0.5, 1$ respectively.

(C) The molecule is $O_2$.
$1$. $O_2$ has two unpaired electrons in its antibonding $\pi^*{2p}$ orbitals,which corresponds to a $2$-electron magnetic moment (paramagnetic).
$2$. The one-electron reduced species is the superoxide ion $(O_2^-)$ and the two-electron reduced species is the peroxide ion $(O_2^{2-})$. Both can act as oxidizing and reducing agents.
$3$. When $O_2$ is one-electron oxidized,it forms the dioxygenyl cation $(O_2^+)$. In $O_2$,the bond order is $2.0$ (electrons in antibonding orbitals). Removing one electron from the antibonding orbital increases the bond order to $2.5$,which causes the bond length to decrease.

(B, C) The correct statements for the peroxide ion $(O_{2}^{2-})$ are that it is diamagnetic and it has a bond order of $1$.
Electronic configuration of $O_{2}^{2-}$ is: $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \sigma 2p_{z}^{2}, (\pi 2p_{x}^{2} = \pi 2p_{y}^{2}), (\pi^{*} 2p_{x}^{2} = \pi^{*} 2p_{y}^{2})$.
Bond order $= \frac{1}{2} (\text{Number of bonding electrons} - \text{Number of anti-bonding electrons})$.
$BO = \frac{10 - 8}{2} = \frac{2}{2} = 1$.
Since the number of unpaired electrons is $0$,$O_{2}^{2-}$ is diamagnetic in nature.

(B) In a heteronuclear diatomic molecule $AB$,if $A$ is more electronegative than $B$,the bonding molecular orbital is closer in energy to the atomic orbital of the more electronegative atom $A$.
Consequently,the electron density in the bonding molecular orbital is concentrated more towards the more electronegative atom $A$.
Thus,the bonding molecular orbital resembles the character of $A$ more than that of $B$.

(A) The $CO$ molecule has a total of $14$ electrons ($6$ from $C$ and $8$ from $O$).
According to Molecular Orbital Theory,for molecules with $14$ or fewer electrons,the energy order of molecular orbitals 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 the $14$ electrons into these orbitals:
$1\sigma^2, 2\sigma^2, 3\sigma^2, 4\sigma^2, 1\pi^4, 5\sigma^2$.
In many textbooks,the notation is simplified as $1\sigma^2 2\sigma^2 1\pi^4 3\sigma^2$ where $1\sigma$ is $\sigma 1s$,$2\sigma$ is $\sigma^* 1s$,$3\sigma$ is $\sigma 2s$,$4\sigma$ is $\sigma^* 2s$,$1\pi$ is $\pi 2p$,and $5\sigma$ is $\sigma 2p_z$.
Thus,the correct configuration is $1\sigma^2 2\sigma^2 3\sigma^2 4\sigma^2 1\pi^4 5\sigma^2$,which corresponds to option $A$ when simplified.

(B) The total number of electrons in $B_2$ is $5 + 5 = 10 \ e^-$.
According to Molecular Orbital Theory,the electronic configuration is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1, \pi 2p_y^1$.
Due to the presence of $2$ unpaired electrons in the $\pi 2p$ bonding molecular orbitals,$B_2$ exhibits paramagnetic behaviour.

(B) Paramagnetic substances contain one or more unpaired electrons.
According to Molecular Orbital Theory $(MOT)$:
$N_2$ (total $14$ electrons): $({\sigma}1s)^2 ({\sigma}^*1s)^2 ({\sigma}2s)^2 ({\sigma}^*2s)^2 ({\pi}2p_x)^2 ({\pi}2p_y)^2 ({\sigma}2p_z)^2$. All electrons are paired,so it is diamagnetic.
$NO$ (total $15$ 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 in the antibonding ${\pi}^*$ orbital,so it is paramagnetic.
$CO$ (total $14$ electrons): Similar to $N_2$,it is diamagnetic.
$O_3$ (total $24$ electrons): It is a diamagnetic molecule.
Therefore,$NO$ is the correct answer.

(C) The molecular orbital configuration of $CO$ ($14$ electrons) is: $\sigma(1s)^2, \sigma^*(1s)^2, \sigma(2s)^2, \sigma^*(2s)^2, \pi(2p_x)^2 = \pi(2p_y)^2, \sigma(2p_z)^2$.
Number of bonding electrons $(N_b)$ = $10$.
Number of antibonding electrons $(N_a)$ = $4$.
Bond Order = $\frac{N_b - N_a}{2} = \frac{10 - 4}{2} = 3$.

(A) According to molecular orbital theory,the electronic configuration for $B_{2}$ ($10$ electrons) is: $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \pi 2p_{x}^{1} = \pi 2p_{y}^{1}$.
Since it contains two unpaired electrons,$B_{2}$ is paramagnetic.
The electronic configuration for $C_{2}$ ($12$ electrons) is: $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \pi 2p_{x}^{2} = \pi 2p_{y}^{2}$.
Since all electrons are paired,$C_{2}$ is diamagnetic.

(C) To determine the bond order and magnetic nature,we use the Molecular Orbital Theory $(MOT)$ configuration:

Species	Bond Order and Magnetic Nature
$O_{2}^{+}$ ($15$ electrons)	$2.5$,Paramagnetic
$O_{2}^{-}$ ($17$ electrons)	$1.5$,Paramagnetic
$N_{2}^{+}$ ($13$ electrons)	$2.5$,Paramagnetic
$N_{2}^{-}$ ($15$ electrons)	$2.5$,Paramagnetic
$N_{2}^{2-}$ ($16$ electrons)	$2.0$,Paramagnetic
$N_{2}$ ($14$ electrons)	$3.0$,Diamagnetic

Comparing the options:
$O_{2}^{+}$ has a bond order of $2.5$ and is paramagnetic.
$N_{2}^{-}$ has a bond order of $2.5$ and is paramagnetic.
Therefore,the pair $(O_{2}^{+}, N_{2}^{-})$ has the same bond order and magnetic character.

(C) $\pi^*$ antibonding molecular orbital is formed by the out-of-phase overlap of two atomic orbitals (e.g.,$p_x$ or $p_y$ orbitals).
This results in the presence of a nodal plane between the two nuclei where the electron probability density is zero,in addition to the nodal plane containing the internuclear axis.
In the $\pi^*$ orbital,the lobes of the atomic orbitals have opposite signs (represented by shading) facing each other,leading to destructive interference.
Diagram $C$ correctly depicts this out-of-phase combination,showing the nodal plane between the nuclei.