Molecular orbital theory Questions in English

(C) For $O_2 \rightarrow O_2^+$,the bond order changes from $2.0$ to $2.5$. Thus,the bond order increases.
For $N_2 \rightarrow N_2^+$,the bond order changes from $3.0$ to $2.5$. Thus,the bond order decreases.

(B) The electronic configuration of $O_2$ is $KK(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^1(\pi_{2p_y}^*)^1$.
When an electron is added to form $O_2^-$,it enters the lowest energy vacant antibonding molecular orbital,which is $\pi_{2p_x}^*$ (or $\pi_{2p_y}^*$).
Thus,the electron is added to the $\pi^*$ molecular orbital.

(C) The electronic configuration of the $O_2$ molecule ($16$ electrons) is: $\sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \sigma_{2p_z}^2, \pi_{2p_x}^2 = \pi_{2p_y}^2, \pi_{2p_x}^{*1} = \pi_{2p_y}^{*1}$.
Antibonding molecular orbitals are those denoted with an asterisk $(*)$.
These are: $\sigma_{1s}^{*2}, \sigma_{2s}^{*2}, \pi_{2p_x}^{*1}, \pi_{2p_y}^{*1}$.
Total number of electrons in antibonding orbitals = $2 + 2 + 1 + 1 = 6$.

(N/A) $(i)$ $F. Hund$ and $R.S. Mulliken$
$(ii)$ Bonding molecular orbitals
$(iii)$ Electronic configuration of the molecule
$(iv)$ Bond order

(C) The bond order is calculated using the formula: $\text{Bond Order} = \frac{N_b - N_a}{2}$.
$(1)$ ${\rm{NO}}$ ($15$ 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 = $\frac{10 - 5}{2} = 2.5$.
$(2)$ ${\rm{CO}}$ ($14$ electrons): Bond order = $3.0$.
$(3)$ ${\rm{O}}_2^-$ ($17$ electrons): Bond order = $\frac{10 - 7}{2} = 1.5$.
$(4)$ ${\rm{O}}_2$ ($16$ electrons): Bond order = $\frac{10 - 6}{2} = 2.0$.
Therefore,the correct matching is: $(1-C, 2-D, 3-A, 4-B)$.

(A) The bond strength is directly proportional to the bond order of the species.
Total electrons in $NO^{2+}$ = $7 + 8 - 2 = 13$. Bond order = $\frac{1}{2}(10 - 3) = 3.5$ (Correction: $NO^{2+}$ has $13$ electrons,configuration $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^1$. Bond order = $\frac{1}{2}(8 - 5) = 1.5$).
Total electrons in $NO^{+}$ = $7 + 8 - 1 = 14$. Bond order = $\frac{1}{2}(10 - 4) = 3$.
Total electrons in $NO$ = $7 + 8 = 15$. Bond order = $\frac{1}{2}(10 - 5) = 2.5$.
Total electrons in $NO^{-}$ = $7 + 8 + 1 = 16$. Bond order = $\frac{1}{2}(10 - 6) = 2$.
Comparing the bond orders: $NO^{+} (3) > NO (2.5) > NO^{-} (2) > NO^{2+} (1.5)$.
Therefore,$NO^{2+}$ has the minimum bond strength.

(B) For $He_{2}$,the total number of electrons is $4$.
The molecular orbital configuration is $\sigma_{1s}^{2} \sigma_{1s}^{*2}$.
Bond Order $(B.O.) = \frac{1}{2} [N_{b} - N_{a}] = \frac{1}{2} [2 - 2] = 0$.
Since the bond order is $0$,the molecule $He_{2}$ does not exist.

(C) The electronic configuration of $O_{2}$ ($16$ electrons) is: $\sigma 1s^{2} \sigma^{*} 1s^{2} \sigma 2s^{2} \sigma^{*} 2s^{2} \sigma 2p_{z}^{2} \pi 2p_{x}^{2} = \pi 2p_{y}^{2} \pi^{*} 2p_{x}^{1} = \pi^{*} 2p_{y}^{1}$.
When an electron is added to $O_{2}$ to form $O_{2}^{-}$,the incoming electron enters the lowest energy vacant or half-filled molecular orbital,which is the $\pi^{*}$ antibonding orbital.
Thus,the electron enters the $\pi^{*} 2p_{x}$ or $\pi^{*} 2p_{y}$ orbital.
Therefore,option $(C)$ is correct.

(B) According to Molecular Orbital Theory,the bond order is calculated as $\text{Bond Order} = \frac{N_b - N_a}{2}$.
For a diatomic molecule with $15$ electrons (like $NO$),the molecular orbital configuration is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$.
Number of bonding electrons $(N_b)$ = $2+2+2+2+2 = 10$.
Number of antibonding electrons $(N_a)$ = $2+2+1 = 5$.
$\text{Bond Order} = \frac{10 - 5}{2} = \frac{5}{2} = 2.5$.
Thus,the total number of electrons in $AX$ is $15$.

(C) .

Chemical Species	Bond Order
$He_{2}^{+}$	$0.5$
$He_{2}^{-}$	$0.5$
$Be_{2}$	$0$
$O_{2}^{2-}$	$1$

According to $M.O.T.$,if the bond order of a chemical species is zero,then that chemical species does not exist.

(A) The bond order $(B.O.)$ 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.
$(a) \ Ne_2$ $(20 \ e^-)$: Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2, \sigma^* 2p_z^2$. $B.O. = \frac{10-10}{2} = 0$.
$(b) \ N_2$ $(14 \ e^-)$: $B.O. = \frac{10-4}{2} = 3$.
$(c) \ F_2$ $(18 \ e^-)$: $B.O. = \frac{10-8}{2} = 1$.
$(d) \ O_2$ $(16 \ e^-)$: $B.O. = \frac{10-6}{2} = 2$.
Thus,the correct matching is: $(a)$ $\rightarrow (iii), (b)$ $\rightarrow (iv), (c)$ $\rightarrow (i), (d)$ $\rightarrow (ii)$.

(A) The electronic configuration of $O_{2}^{-}$ ($17$ electrons) is: $(\sigma_{1s})^{2}(\sigma_{1s}^{*})^{2}(\sigma_{2s})^{2}(\sigma_{2s}^{*})^{2}(\sigma_{2p_{z}})^{2}(\pi_{2p_{x}}^{2} = \pi_{2p_{y}}^{2})(\pi_{2p_{x}}^{*2} = \pi_{2p_{y}}^{*1})$.
$\text{Bond order} = \frac{N_b - N_a}{2} = \frac{10 - 7}{2} = 1.5$.
Since there is one unpaired electron in the $\pi^{*}$ orbital,the ion is paramagnetic.

(B) The total number of electrons in $O_{2}^{2-}$ is $8 + 8 + 2 = 18$.
The molecular orbital configuration is: $\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \sigma_{2p_{z}}^{2} (\pi 2p_{x}^{2} = \pi 2p_{y}^{2}) (\pi_{2p_{x}}^{*2} = \pi_{2p_{y}}^{*2})$.
Since all molecular orbitals are completely filled,there are $0$ unpaired electrons.

(C) The total number of electrons in $B_{2}^{+}$ is $5 + 5 - 1 = 9$.
The molecular orbital configuration of $B_{2}^{+}$ is $\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \pi_{2py}^{1}$.
There is $1$ unpaired electron $(n=1)$.
The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$.
$\mu = \sqrt{1(1+2)} = \sqrt{3} \ BM$.
Given $\sqrt{3} = 1.73$,so $\mu = 1.73 \ BM = 173 \times 10^{-2} \ BM$.

(A) 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 $O_{2}^{+} (15 \ e^-)$: $\text{Bond Order} = \frac{10 - 5}{2} = 2.5$
For $O_{2} (16 \ e^-)$: $\text{Bond Order} = \frac{10 - 6}{2} = 2.0$
For $O_{2}^{-} (17 \ e^-)$: $\text{Bond Order} = \frac{10 - 7}{2} = 1.5$
For $O_{2}^{2-} (18 \ e^-)$: $\text{Bond Order} = \frac{10 - 8}{2} = 1.0$
Thus,the correct sequence is $O_{2}^{+} > O_{2} > O_{2}^{-} > O_{2}^{2-}$.

(A) The total number of electrons in $CO$ is $6 + 8 = 14$. According to Molecular Orbital Theory,the bond order of a $14$-electron species is $\frac{10-4}{2} = 3$.
The total number of electrons in $NO^{\oplus}$ is $7 + 8 - 1 = 14$. Similarly,the bond order of $NO^{\oplus}$ is $\frac{10-4}{2} = 3$.
The difference between the bond orders is $|3 - 3| = 0$.
Given that the difference is $\frac{x}{2}$,we have $\frac{x}{2} = 0$,which implies $x = 0$.

(B) The total number of electrons in $O_{2}^{2-}$ is $8 + 8 + 2 = 18$.
The molecular orbital configuration of $O_{2}^{2-}$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^2, \pi^* 2p_y^2$.
The bonding molecular orbitals are $\sigma 1s, \sigma 2s, \sigma 2p_z, \pi 2p_x, \text{ and } \pi 2p_y$.
The number of electrons in these bonding orbitals are $2, 2, 2, 2, \text{ and } 2$ respectively.
Total bonding electrons = $2 + 2 + 2 + 2 + 2 = 10$.

(N/A) Metals conduct electricity through the movement of electrons. In metals,atomic orbitals overlap to form molecular orbitals that are so close in energy that they form a band.
If this band is partially filled or overlaps with a higher energy unoccupied conduction band,electrons can flow easily under an applied electric field,resulting in electrical conductivity.
Metals conduct electricity in both solid and molten states.

(C) The electronic configuration of $O_{2}^{+}$ is $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \sigma 2p_{z}^{2}, \pi 2p_{x}^{2} = \pi 2p_{y}^{2}, \pi^{*} 2p_{x}^{1}$.
Since $O_{2}^{+}$ ion has $15$ electrons,it contains one unpaired electron in the $\pi^{*}$ orbital.
Therefore,it is paramagnetic in nature,making the statement in option $C$ incorrect.

(B) To determine the bond order,we use the Molecular Orbital Theory $(MOT)$ configuration:
$1$. For $C_{2}^{2-}$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. Bond order = $(10-4)/2 = 3$.
$2$. For $N_{2}^{2-}$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Bond order = $(10-6)/2 = 2$.
$3$. For $O_{2}^{2-}$ ($18$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. Bond order = $(10-8)/2 = 1$.
Thus,the order of bond orders is $O_{2}^{2-} (1) < N_{2}^{2-} (2) < C_{2}^{2-} (3)$.

(A) According to $M.O.$ Theory,bond strength is directly proportional to bond order $(B.O.)$.
If an electron is removed from an antibonding molecular orbital $(M.O.)$,the bond order increases,making the bond stronger.
Let us analyze the molecules:
$1$. $NO$ ($15$ $e^-$): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Removing an electron from $\pi^* 2p_x$ increases $B.O.$ from $2.5$ to $3.0$.
$2$. $N_2$ ($14$ $e^-$): Removing an electron from a bonding orbital $(\sigma 2p_z)$ decreases $B.O.$ from $3.0$ to $2.5$.
$3$. $O_2$ ($16$ $e^-$): Configuration has electrons in $\pi^* 2p_x$ and $\pi^* 2p_y$. Removing an electron from $\pi^*$ increases $B.O.$ from $2.0$ to $2.5$.
$4$. $C_2$ ($12$ $e^-$): Removing an electron from a bonding orbital $(\pi 2p)$ decreases $B.O.$ from $2.0$ to $1.5$.
$5$. $B_2$ ($10$ $e^-$): Removing an electron from a bonding orbital $(\pi 2p)$ decreases $B.O.$ from $1.0$ to $0.5$.
Thus,the bond becomes stronger for $NO$ and $O_2$.

(A) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as: $\text{Bond Order} = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of electrons in bonding molecular orbitals and $N_a$ is the number of electrons in antibonding molecular orbitals.

Species	Bond Order
$O_{2}^{2-}$	$1.0$
$O_{2}^{-}$	$1.5$
$O_{2}$	$2.0$
$O_{2}^{+}$	$2.5$

The increasing order of bond order is: $O_{2}^{2-} < O_{2}^{-} < O_{2} < O_{2}^{+}$.

(C) To determine diamagnetism,we check the number of electrons in each species:
$N_{2}$ ($14$ $e^-$): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$ (All paired,Diamagnetic)
$N_{2}^{+}$ ($13$ $e^-$): Paramagnetic
$N_{2}^{-}$ ($15$ $e^-$): Paramagnetic
$N_{2}^{2-}$ ($16$ $e^-$): Paramagnetic
$O_{2}$ ($16$ $e^-$): Paramagnetic
$O_{2}^{+}$ ($15$ $e^-$): Paramagnetic
$O_{2}^{-}$ ($17$ $e^-$): Paramagnetic
$O_{2}^{2-}$ ($18$ $e^-$): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$ (All paired,Diamagnetic)
Thus,the species showing diamagnetism are $N_{2}$ and $O_{2}^{2-}$.
The total number of diamagnetic species is $2$.

(C) . $\Psi_{MO} = \Psi_{A} - \Psi_{B}$ represents the Anti-bonding molecular orbital $(III)$.
$B$. $\mu = Q \times r$ is the formula for Dipole moment $(I)$.
$C$. $\frac{N_{b} - N_{a}}{2}$ is the formula for Bond order $(IV)$.
$D$. $\Psi_{MO} = \Psi_{A} + \Psi_{B}$ represents the Bonding molecular orbital $(II)$.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(B) The bond order for each species is calculated as follows:
$CN^{-}$ ($14$ electrons): Bond order $= \frac{10-4}{2} = 3$
$NO^{+}$ ($14$ electrons): Bond order $= \frac{10-4}{2} = 3$
$O_{2}$ ($16$ electrons): Bond order $= \frac{10-6}{2} = 2$
$O_{2}^{+}$ ($15$ electrons): Bond order $= \frac{10-5}{2} = 2.5$
$O_{2}^{2+}$ ($14$ electrons): Bond order $= \frac{10-4}{2} = 3$
Thus,$CN^{-}, NO^{+},$ and $O_{2}^{2+}$ have an identical bond order of $3$.
The total number of such species is $3$.

(D) To determine the paramagnetic nature,we use Molecular Orbital Theory $(MOT)$ to find the number of unpaired electrons:
$1. B_{2} (10 \ e^-): (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^1 (\pi 2p_y)^1$ ($2$ unpaired electrons,paramagnetic)
$2. Li_{2} (6 \ e^-): (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2$ ($0$ unpaired electrons,diamagnetic)
$3. C_{2} (12 \ e^-): (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2$ ($0$ unpaired electrons,diamagnetic)
$4. C_{2}^{-} (13 \ e^-): (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^1$ ($1$ unpaired electron,paramagnetic)
$5. O_{2}^{2-} (18 \ e^-): (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^2$ ($0$ unpaired electrons,diamagnetic)
$6. O_{2}^{+} (15 \ e^-): (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1$ ($1$ unpaired electron,paramagnetic)
$7. He_{2}^{+} (3 \ e^-): (\sigma 1s)^2 (\sigma^* 1s)^1$ ($1$ unpaired electron,paramagnetic)
Thus,the paramagnetic species are $B_{2}, C_{2}^{-}, O_{2}^{+},$ and $He_{2}^{+}$. The total count is $4$.

(D) The electronic configuration of the $C_{2}$ molecule is $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \pi 2p_{y}^{2} = \pi 2p_{x}^{2}$.
According to the molecular orbital theory for $C_{2}$,the next available molecular orbital after the $\pi 2p$ orbitals is the $\sigma 2p_{z}$ orbital.
Therefore,if an extra electron is added to the $C_{2}$ molecule,it will occupy the $\sigma 2p_{z}$ molecular orbital.
Thus,the correct option is $D$.

(D) To be diamagnetic,no unpaired electron must be present.
Write the molecular orbital electronic configuration of each species.
For all options given,the configuration $\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \sigma_{2p_z}^{2} \pi_{2p_x}^{2} = \pi_{2p_y}^{2}$ is common.
After that:
$O_{2}^{+}: \pi_{2p_x}^{*1}$ (one unpaired electron,paramagnetic)
$O_{2}^{-}: \pi_{2p_x}^{*2} = \pi_{2p_y}^{*1}$ (one unpaired electron,paramagnetic)
$O_{2}: \pi_{2p_x}^{*1} = \pi_{2p_y}^{*1}$ (two unpaired electrons,paramagnetic)
$O_{2}^{2-}: \pi_{2p_x}^{*2} = \pi_{2p_y}^{*2}$ (no unpaired electrons,diamagnetic).
Therefore,$O_{2}^{2-}$ is the diamagnetic species.

(C)
The bond order $(BO)$ of a species is calculated as $BO = \frac{N_b - N_a}{2}$,where $N_b$ is the number of electrons in bonding orbitals and $N_a$ is the number of electrons in antibonding orbitals.
$(a)$ $O_2$: Electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^1, \pi^* 2p_y^1$. $BO = \frac{10 - 6}{2} = 2.0$.
$(b)$ $F_2$: Electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^2, \pi^* 2p_y^2$. $BO = \frac{10 - 8}{2} = 1.0$.
$(c)$ $O_2^{+}$: Electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^1$. $BO = \frac{10 - 5}{2} = 2.5$.
$(d)$ $F_2^{-}$: Electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^2, \pi^* 2p_y^2, \sigma^* 2p_z^1$. $BO = \frac{10 - 9}{2} = 0.5$.
Thus,$O_2^{+}$ has the highest bond order.

(D)
Total number of electrons in $O_2^{2-} = 18$.
The molecular orbital 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 is calculated using the formula: $BO = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
Here,$N_b = 10$ and $N_a = 8$.
Therefore,$BO = \frac{10 - 8}{2} = 1$.

(C) Isoelectronic species are those species which have the same number of electrons.
The total number of electrons in $CO$ is $6 + 8 = 14$.
$(A)$ Total number of electrons in $O_2^{+}$ is $8 + 8 - 1 = 15$.
$(B)$ Total number of electrons in $O_2^{-}$ is $8 + 8 + 1 = 17$.
$(C)$ Total number of electrons in $CN^{-}$ is $6 + 7 + 1 = 14$.
$(D)$ Total number of electrons in $N_2^{+}$ is $7 + 7 - 1 = 13$.
Thus,$CN^{-}$ is isoelectronic with $CO$.

(C) The bond order is calculated as $\text{B.O.} = \frac{1}{2}(N_b - N_a)$,where $N_b$ is the number of electrons in bonding orbitals and $N_a$ is the number of electrons in antibonding orbitals.
$(A)$ $CO$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2$. $\text{B.O.} = \frac{1}{2}(10 - 4) = 3$.
$O_2^{2-}$ ($18$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^2, \pi^* 2p_y^2$. $\text{B.O.} = \frac{1}{2}(10 - 8) = 1$.
$(B)$ $O_2^{-}$ ($17$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^2, \pi^* 2p_y^1$. $\text{B.O.} = \frac{1}{2}(10 - 7) = 1.5$.
$CO$ has $\text{B.O.} = 3$.
$(C)$ $O_2^{2-}$ has $\text{B.O.} = 1$.
$B_2$ ($10$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1, \pi 2p_y^1$. $\text{B.O.} = \frac{1}{2}(6 - 4) = 1$.
$(D)$ $CO$ has $\text{B.O.} = 3$.
$N_2^{+}$ ($13$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2, \pi 2p_y^2, \sigma 2p_z^1$. $\text{B.O.} = \frac{1}{2}(9 - 4) = 2.5$.
Since $O_2^{2-}$ and $B_2$ both have a bond order of $1$,option $(C)$ is correct.

(D)
Diamagnetic species are those which have all electrons paired in their molecular orbitals.
$(i)$ $NO$: Total electrons $= 7+8=15$. It has an unpaired electron in the $\pi^* 2p_x$ orbital,so it is paramagnetic.
$(ii)$ $NO_2$: Total electrons $= 7+8+8=23$. It has an odd number of electrons,so it is paramagnetic.
$(iii)$ $O_2$: Total electrons $= 16$. According to Molecular Orbital Theory,it has two unpaired electrons in the $\pi^* 2p_x$ and $\pi^* 2p_y$ orbitals,so it is paramagnetic.
$(iv)$ $CO_2$: Total electrons $= 6+8+8=22$. The structure is $O=C=O$. All electrons are paired in bonding and lone pair orbitals,making it diamagnetic.

(A) $N_2$ $(14 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. $HOMO$ is $\sigma 2p_z$,which has $0$ unpaired electrons.
$N_2^{+}$ $(13 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^1$. $HOMO$ is $\sigma 2p_z$,which has $1$ unpaired electron.
$O_2$ $(16 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. $HOMO$ is $\pi^* 2p$,which has $2$ unpaired electrons.
$O_2^{+}$ $(15 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. $HOMO$ is $\pi^* 2p$,which has $1$ unpaired electron.
Thus,the number of unpaired electrons is $0, 1, 2, 1$ respectively.

(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 electrons in bonding orbitals and $N_a$ is the number of electrons in antibonding orbitals.
$1$. For $O_2^{2-}$ ($18$ electrons): The configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. Bond order = $\frac{1}{2} (10 - 8) = 1$.
$2$. For $CO$ ($14$ electrons): The configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. Bond order = $\frac{1}{2} (10 - 4) = 3$.
$3$. For $NO^{+}$ ($14$ electrons): Being isoelectronic with $CO$,it also has a bond order of $3$.

(A) The acetylide ion is $C_2^{2-}$.
Total number of electrons in $C_2^{2-} = (6 \times 2) + 2 = 14 \ e^-$.
According to Molecular Orbital Theory,the electronic configuration of $C_2^{2-}$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$.
Bond order $= \frac{1}{2} (N_b - N_a) = \frac{1}{2} (10 - 4) = 3$.
Since all electrons are paired,it is diamagnetic.
Now,check $NO^{+}$: Total electrons $= 7 + 8 - 1 = 14 \ e^-$.
$NO^{+}$ also has a bond order of $3$ and is diamagnetic.
Therefore,the properties match $NO^{+}$.

(B) The electronic configuration of $NO$ ($15$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. It has one unpaired electron,so it is paramagnetic. Bond order $= (10-5)/2 = 2.5$.
After losing one electron,$NO^{+}$ ($14$ electrons) has the configuration $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. It has no unpaired electrons,so it is diamagnetic. Bond order $= (10-4)/2 = 3$.
Thus,in the process $NO \rightarrow NO^{+}$,the bond order increases from $2.5$ to $3$ and the character changes from paramagnetic to diamagnetic.

(B) According to Molecular Orbital Theory,for homonuclear diatomic molecules like $B_2, C_2,$ and $N_2$,the $2s-2p$ mixing causes the $\sigma 2p_z$ orbital to have higher energy than the $\pi 2p_x$ and $\pi 2p_y$ orbitals.
Therefore,the correct increasing order of energy for $N_2$ is $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$.

(A) The bond order of $CO$ is $3$ because it has a triple bond $(C \equiv O)$.
The bond order of $NO^{+}$ is $3$ because it is isoelectronic with $CO$ and has a triple bond $(N \equiv O^{+})$.
The sum of the bond orders is $3 + 3 = 6$.

(D) To determine the number of species that are paramagnetic and have a bond order of $1$,we analyze each species using Molecular Orbital Theory:

$Species$	$Magnetic \ behaviour$	$Bond \ order$
$H_2$	Diamagnetic	$1$
$He_2^{+}$	Paramagnetic	$0.5$
$O_2^{+}$	Paramagnetic	$2.5$
$N_2^{2-}$	Paramagnetic	$2$
$O_2^{2-}$	Diamagnetic	$1$
$F_2$	Diamagnetic	$1$
$Ne_2^{+}$	Paramagnetic	$0.5$
$B_2$	Paramagnetic	$1$

From the table,only $B_2$ is both paramagnetic and has a bond order of $1$.
Therefore,the total number of such species is $1$.

(A) In a diatomic molecule,the $2s$ atomic orbitals combine to form one bonding molecular orbital $(\sigma 2s)$ and one antibonding molecular orbital $(\sigma^* 2s)$.
The $2p$ atomic orbitals combine to form three bonding molecular orbitals $(\sigma 2p_z, \pi 2p_x, \pi 2p_y)$ and three antibonding molecular orbitals $(\sigma^* 2p_z, \pi^* 2p_x, \pi^* 2p_y)$.
Total number of antibonding molecular orbitals $= 1 (\text{from } 2s) + 3 (\text{from } 2p) = 4$.

(A) According to Molecular Orbital Theory, the number of molecular orbitals formed is equal to the number of atomic orbitals combined.
$1.$ From $2s$ atomic orbitals: $2$ atomic orbitals ($2s$ from each atom) combine to form $2$ molecular orbitals ($\sigma 2s$ and $\sigma^* 2s$).
$2.$ From $2p$ atomic orbitals: $6$ atomic orbitals ($2p_x, 2p_y, 2p_z$ from each atom) combine to form $6$ molecular orbitals $(\sigma 2p_z, \sigma^* 2p_z, \pi 2p_x, \pi 2p_y, \pi^* 2p_x, \pi^* 2p_y)$.
Total molecular orbitals = $2 + 6 = 8$.

(B) The conditions for the linear combination of atomic orbitals $(LCAO)$ are:
$1$. The combining atomic orbitals must have comparable energies.
$2$. The combining atomic orbitals must have the same symmetry about the molecular axis.
$3$. The combining atomic orbitals must overlap to the maximum extent.
Based on these conditions,statements $A$ and $C$ are correct. Therefore,the correct option is $B$.

(C) Statement $I$ is false because a $\pi$ bonding molecular orbital $(MO)$ has higher electron density above and below the inter-nuclear axis.
Statement $II$ is true because the $\pi^*$ antibonding molecular orbital has a nodal plane perpendicular to the inter-nuclear axis between the two nuclei,where the electron density is zero.
Therefore,Statement $I$ is false but Statement $II$ is true.

(B) According to $M.O.T.$ (Molecular Orbital Theory):
$O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Unpaired electrons $= 2$.
$O_2^{-}$ ($17$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. Unpaired electrons $= 1$.
$NO$ ($15$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Unpaired electrons $= 1$.
$CN^{-}$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. Unpaired electrons $= 0$.
$O_2^{2-}$ ($18$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. Unpaired electrons $= 0$.
Thus,the species with one unpaired electron are $O_2^{-}$ and $NO$. The total number is $2$.

(B) Antibonding molecular orbitals $(\sigma^*)$ are formed by the destructive interference of atomic orbitals.
The wave function for the antibonding molecular orbital is given by the subtraction of the wave functions of the participating atomic orbitals.
Therefore,$\sigma^* = \psi_A - \psi_B$.

(B) The bond order is calculated using the formula: $\text{B.O.} = \frac{N_b - N_a}{2}$.
$C_2$ $(12 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2$. $\text{B.O.} = \frac{8-4}{2} = 2$.
$O_2$ $(16 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. $\text{B.O.} = \frac{10-6}{2} = 2$.
$Be_2$ $(8 \ e^-)$: $\text{B.O.} = 0$.
$Li_2$ $(6 \ e^-)$: $\text{B.O.} = 1$.
$Ne_2$ $(20 \ e^-)$: $\text{B.O.} = 0$.
$N_2$ $(14 \ e^-)$: $\text{B.O.} = 3$.
$He_2$ $(4 \ e^-)$: $\text{B.O.} = 0$.
Thus,only $C_2$ and $O_2$ have a bond order of $2$. The total count is $2$.

(C) To determine the number of unpaired electrons,we use Molecular Orbital Theory $(MOT)$:
$N_2$: $(14 \ e^-) \rightarrow \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$ ($0$ unpaired $e^-$)
$O_2$: $(16 \ e^-) \rightarrow \dots, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$ ($2$ unpaired $e^-$)
$C_2^{-}$: $(13 \ e^-) \rightarrow \dots, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^1$ ($1$ unpaired $e^-$)
$O_2^{-}$: $(17 \ e^-) \rightarrow \dots, \pi^* 2p_x^2 = \pi^* 2p_y^1$ ($1$ unpaired $e^-$)
$O_2^{2-}$: $(18 \ e^-) \rightarrow \dots, \pi^* 2p_x^2 = \pi^* 2p_y^2$ ($0$ unpaired $e^-$)
$H_2^{+}$: $(1 \ e^-) \rightarrow \sigma 1s^1$ ($1$ unpaired $e^-$)
$CN^{-}$: $(14 \ e^-) \rightarrow \dots, \sigma 2p_z^2$ ($0$ unpaired $e^-$)
$He_2^{+}$: $(3 \ e^-) \rightarrow \sigma 1s^2, \sigma^* 1s^1$ ($1$ unpaired $e^-$)
Species with one unpaired electron are $C_2^{-}, O_2^{-}, H_2^{+}, He_2^{+}$.
Total count = $4$.