Molecular orbital theory Questions in English

(D) $(A) \, \text{The M.O. configuration for } C_2 \text{ is } \sigma_{1s}^2 < \sigma_{1s}^* {}^2 < \sigma_{2s}^2 < \sigma_{2s}^* {}^2 < \pi_{2p_y}^2 = \pi_{2p_z}^2 < \sigma_{2p_x}. \text{ Here, } \pi_{2p_y}^2 = \pi_{2p_z}^2 \text{ is the } HOMO \text{ and } \sigma_{2p_x} \text{ is the } LUMO. \text{ Thus, statement } (A) \text{ is correct.}$
$(B) \, \text{In } C_2, \text{ the four valence electrons occupy the two } \pi \text{ molecular orbitals } (\pi_{2p_y} \text{ and } \pi_{2p_z}), \text{ resulting in two } \pi \text{ bonds. Thus, statement } (B) \text{ is correct.}$
$(C) \, \text{The } C_2^{2-} \text{ ion has } 14 \text{ electrons. Its configuration is } \sigma_{1s}^2 < \sigma_{1s}^* {}^2 < \sigma_{2s}^2 < \sigma_{2s}^* {}^2 < \pi_{2p_y}^2 = \pi_{2p_z}^2 < \sigma_{2p_x}^2. \text{ The bond order is } \frac{1}{2}(10-4) = 3, \text{ consisting of one } \sigma \text{ and two } \pi \text{ bonds. Thus, statement } (C) \text{ is correct.}$
$(D) \, \text{Since all statements are correct, the correct option is } (D).$

(C) Stability of a molecule is directly proportional to its Bond Order $(B.O.)$.
For $N_2$ ($14$ electrons): $B.O. = 3$. For $N_2^+$ ($13$ electrons): $B.O. = 2.5$. For $N_2^-$ ($15$ electrons): $B.O. = 2.5$. Since $N_2^+$ and $N_2^-$ have the same $B.O.$,stability is determined by the number of antibonding electrons. $N_2^+$ is more stable than $N_2^-$. Thus,$N_2 > N_2^+ > N_2^-$ is correct.
For $O_2$ ($16$ electrons): $B.O. = 2$. For $O_2^+$ ($15$ electrons): $B.O. = 2.5$. For $O_2^{+2}$ ($14$ electrons): $B.O. = 3$. Thus,$O_2^{+2} > O_2^+ > O_2$ is correct.
For $O_2^{-2}$ ($18$ electrons): $B.O. = 1$. Thus,$O_2^{+2} > O_2 > O_2^{-2}$ is correct.
Option $C$ states $N_2^+ > N_2 > N_2^-$,which is incorrect because $N_2$ has the highest $B.O.$ $(3)$ and is the most stable.

(A) The electronic configuration of $O_2$ ($16$ electrons) is: $KK (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2pz)^2 (\pi 2px)^2 (\pi 2py)^2 (\pi^* 2px)^1 (\pi^* 2py)^1$. It has $2$ unpaired electrons.
The electronic configuration of $O_2^+$ ($15$ electrons) is: $KK (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2pz)^2 (\pi 2px)^2 (\pi 2py)^2 (\pi^* 2px)^1$. It has $1$ unpaired electron.
The electronic configuration of $O_2^-$ ($17$ electrons) is: $KK (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2pz)^2 (\pi 2px)^2 (\pi 2py)^2 (\pi^* 2px)^2 (\pi^* 2py)^1$. It has $1$ unpaired electron.
The electronic configuration of $O_2^{2-}$ ($18$ electrons) is: $KK (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2pz)^2 (\pi 2px)^2 (\pi 2py)^2 (\pi^* 2px)^2 (\pi^* 2py)^2$. It has $0$ unpaired electrons.
Thus,$O_2$ has the maximum number of unpaired electrons.

(C) nodal plane is a plane where the probability of finding an electron is zero.
$1$. $\sigma _{2s}$ molecular orbital has zero nodal planes perpendicular to the internuclear axis.
$2$. $\pi _{2p_y}$ bonding molecular orbital has one nodal plane (the molecular plane itself).
$3$. $\pi ^* _{2p_y}$ antibonding molecular orbital has two nodal planes: one is the molecular plane,and the other is a plane perpendicular to the internuclear axis passing between the nuclei.
$4$. $\sigma ^* _{2p_z}$ antibonding molecular orbital has one nodal plane perpendicular to the internuclear axis.
Thus,the $\pi ^* _{2p_y}$ antibonding molecular orbital has two nodal planes.

(B) Bond length is inversely proportional to bond order. Higher bond order results in a shorter bond length.
In $N_2H_4$ $(H_2N-NH_2)$,the $N-N$ bond is a single bond (bond order = $1$).
In $N_2O_4$ $(O_2N-NO_2)$,the $N-N$ bond is a single bond (bond order = $1$).
In $N_2O$ $(N \equiv N^+-O^-)$,the $N-N$ bond is a triple bond (bond order = $3$).
In $N_2$ $(N \equiv N)$,the $N-N$ bond is a triple bond (bond order = $3$).
Comparing $N_2$ and $N_2O$,the $N-N$ bond in $N_2$ is a pure triple bond,whereas in $N_2O$,the resonance structures result in a bond order slightly less than $3$. Therefore,$N_2$ has the shortest $N-N$ bond length.

(C) To determine the presence of unpaired electrons,we analyze the electronic structure of each species:
$1$. $KO_2$: This 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.
$2$. $AlO_2^-$: The aluminate ion has a closed-shell structure with no unpaired electrons.
$3$. $BaO_2$: This 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.
$4$. $NO_2^+$: The nitronium ion has $16$ valence electrons $(5 + 6 \times 2 - 1 = 16)$. It is isoelectronic with $CO_2$ and has no unpaired electrons.
Therefore,only $KO_2$ contains an unpaired electron.

(A) nodal plane is defined as a region in space where the probability of finding an electron is zero.
Therefore,during the formation of molecular orbitals,the electron density is zero in the nodal plane.

(D) Let us analyze the changes in bond order $(BO)$ and number of unpaired electrons $(n)$:
$1$. $N_2 (14e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. $BO = 3, n = 0$.
$N_2^{-2} (16e^-): \dots \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. $BO = 2, n = 2$. (Both decrease/increase differently)
$2$. $B_2 (10e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1 = \pi 2p_y^1$. $BO = 1, n = 2$.
$B_2^{-} (11e^-): \dots \pi 2p_x^2 = \pi 2p_y^1$. $BO = 1.5, n = 1$.
$3$. $O_2^{-} (17e^-): \dots \sigma 2p_z^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. $BO = 1.5, n = 1$.
$O_2^{2-} (18e^-): \dots \pi^* 2p_x^2 = \pi^* 2p_y^2$. $BO = 1, n = 0$.
$4$. $O_2^{-} (17e^-): \dots \sigma 2p_z^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. $BO = 1.5, n = 1$.
$O_2 (16e^-): \dots \sigma 2p_z^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. $BO = 2, n = 2$.
In the change $O_2^{-} \to O_2$,the bond order increases from $1.5$ to $2$ and the number of unpaired electrons increases from $1$ to $2$.

(D) Bond order $= 1/2 \times (N_B - N_{AB})$.
For $N_2$ ($14$ electrons): $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p_x)^2(\pi 2p_y)^2(\sigma 2p_z)^2$. Bond order $= 1/2 \times (10-4) = 3$.
For $N_2^+$ ($13$ 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 $= 1/2 \times (9-4) = 2.5$. Since bond order decreases,the $N-N$ bond weakens.
For $O_2$ ($16$ electrons): $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi^* 2p_x)^1(\pi^* 2p_y)^1$. Bond order $= 1/2 \times (10-6) = 2$.
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 $= 1/2 \times (10-5) = 2.5$. Since bond order increases,bond length decreases and paramagnetism decreases (from $2$ unpaired electrons to $1$).
$N_2^+$ has $1$ unpaired electron,so it is paramagnetic,not diamagnetic. Thus,statement $D$ is incorrect.

(D) nodal plane is a plane where the probability of finding an electron is zero.
For molecular orbitals,the number of nodal planes is related to the symmetry and the number of nodes.
- The $ \sigma_{1s} $ and $ \sigma^*_{2s} $ orbitals have $0$ and $1$ nodal plane respectively.
- The $ \sigma_{2p_z} $ orbital has $0$ nodal planes perpendicular to the internuclear axis.
- The $ \pi^*_{2p_x} $ antibonding molecular orbital is formed by the out-of-phase overlap of $ p_x $ orbitals. It possesses one nodal plane containing the internuclear axis and another nodal plane perpendicular to the internuclear axis.
Thus,the $ \pi^*_{2p_x} $ orbital has two nodal planes.

(C) Let us calculate the number of unpaired electrons and bond order for each pair:
$1$. $CO$ ($14$ electrons): Bond order = $3$,unpaired electrons = $0$. $CN^{-}$ ($14$ electrons): Bond order = $3$,unpaired electrons = $0$. (Same bond order,same unpaired electrons).
$2$. $O_2^{+}$ ($15$ electrons): Bond order = $2.5$,unpaired electrons = $1$. $O_2$ ($16$ electrons): Bond order = $2$,unpaired electrons = $2$. (Different bond order,different unpaired electrons).
$3$. $O_2$ ($16$ electrons): Bond order = $2$,unpaired electrons = $2$. $B_2$ ($10$ electrons): Bond order = $1$,unpaired electrons = $2$. (Different bond order,same unpaired electrons).
$4$. $NO^{+}$ ($14$ electrons): Bond order = $3$,unpaired electrons = $0$. $N_2$ ($14$ electrons): Bond order = $3$,unpaired electrons = $0$. (Same bond order,same unpaired electrons).
Thus,the pair $O_2$ and $B_2$ have the same number of unpaired electrons $(2)$ but different bond orders ($2$ and $1$ respectively).

(D) To determine the $O-O$ bond distance,we look at the bond order of the species:
$1$. $O_2$: Bond order = $2.0$ (Electronic configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$).
$2$. $O_2^-$ (Superoxide): Bond order = $1.5$.
$3$. $O_2^{2-}$ (Peroxide): Bond order = $1.0$.
Since bond distance is inversely proportional to bond order,the species with the highest bond order will have the minimum bond distance.
$O_2$ has the highest bond order $(2.0)$,therefore it has the shortest $O-O$ bond distance.

(D) The electronic configuration of $C_2$ ($12$ electrons) is $KK \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p_x)^2 \pi(2p_y)^2$. The bond order is $(6-2)/2 = 2$,consisting of two $\pi$ bonds.
The electronic configuration of $C_2^{2-}$ ($14$ electrons) is $KK \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p_x)^2 \pi(2p_y)^2 \sigma(2p_z)^2$. The bond order is $(8-2)/2 = 3$,consisting of two $\pi$ bonds and one $\sigma$ bond.
Comparing the two,the bond order increases from $2$ to $3$,and the number of $\sigma$ bonds increases from $0$ to $1$. However,the number of $\pi$ bonds remains $2$ in both cases. Therefore,the statement that the number of $\pi$ bonds increases is incorrect.

(B) The $B_2$ molecule is formed by the combination of two Boron atoms.
The atomic number of Boron is $5$,and its electronic configuration is $1s^2 \, 2s^2 \, 2p^1$.
According to Molecular Orbital Theory,the molecular orbital configuration for $B_2$ ($10$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1 = \pi 2p_y^1$.
Number of bonding electrons $(N_b)$ = $6$.
Number of antibonding electrons $(N_a)$ = $4$.
Bond order = $\frac{1}{2}(N_b - N_a) = \frac{1}{2}(6 - 4) = \frac{2}{2} = 1$.

(C) In the $ClO_4^-$ ion,the central chlorine atom is $sp^3$ hybridized.
It forms four $Cl-O$ bonds.
The $\pi$ bonds are formed by the lateral overlap of the filled $2p$ orbitals of oxygen with the empty $3d$ orbitals of chlorine.
Therefore,the $\pi$ bond is formed by $p\pi-d\pi$ overlapping,specifically between the $2p$ orbital of oxygen and the $3d$ orbital of chlorine.

(A) According to Molecular Orbital Theory $(M.O.T)$,the electronic configuration of oxygen species is as follows:
$1$. For $O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. The number of antibonding electrons is $2 + 2 + 1 + 1 = 6$.
$2$. For $O_2^-$ ($17$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. The number of antibonding electrons is $2 + 2 + 2 + 1 = 7$.
$3$. For $O_2^{2-}$ ($18$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. The number of antibonding electrons is $2 + 2 + 2 + 2 = 8$.
Thus,the number of antibonding electrons in $O_2^-$,$O_2$,and $O_2^{2-}$ are $7, 6, 8$ respectively.

(D) The bond order is calculated using the Molecular Orbital Theory $(MOT)$ formula: $\text{Bond Order} = \frac{N_b - N_a}{2}$.
For $O_2^+ (15 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Bond order = $\frac{10 - 5}{2} = 2.5$.
For $O_2 (16 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Bond order = $\frac{10 - 6}{2} = 2.0$.
For $O_2^- (17 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. Bond order = $\frac{10 - 7}{2} = 1.5$.
Thus,the correct sequence is $O_2^+ > O_2 > O_2^-$.

(C) According to Molecular Orbital Theory $(MOT)$:
The electronic configuration of the peroxide ion $(O_2^{2-})$ is: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2pz}^2 \pi_{2px}^2 \pi_{2py}^2 \pi_{2px}^{*2} \pi_{2py}^{*2}$.
When $O_2^{2-}$ is converted to $O_2^-$,one electron is removed from the highest occupied molecular orbital,which is the antibonding $\pi^*$ orbital.
Therefore,the electron is removed from $\pi_{2py}^*$ (or $\pi_{2px}^*$).

(A) In $N(SiH_3)_3$ (trisilylamine),the lone pair on the $N$ atom is donated into the empty $d$-orbitals of the $Si$ atoms,which is known as $p\pi-d\pi$ back-bonding.
This back-bonding gives the $N-Si$ bond partial double bond character,making it shorter than a standard $N-Si$ single bond.
Due to this back-bonding,the $N$ atom adopts $sp^2$ hybridization,making the molecule planar.
Since the lone pair is involved in back-bonding,it is not available for donation,making $N(SiH_3)_3$ a much weaker base than $NH_3$.
Option $A$ is incorrect because the central $N$ atom is $sp^2$ hybridized and the $Si$ atoms are $sp^3$ hybridized,but the statement implies a single central atom has two hybridizations,which is not the case; however,in the context of the molecule's structure,the statement that it is associated with two types of hybridization is technically true for the molecule as a whole,but the question asks for the incorrect statement. Upon review,all statements $B, C, D$ are correct. Statement $A$ is often considered misleading or incorrect in specific contexts because the hybridization is distinct for different atoms.

(C) In $N(SiH_3)_3$,the nitrogen atom has a lone pair of electrons.
Silicon has vacant $3d$ orbitals.
The lone pair on the nitrogen atom is donated into the vacant $3d$ orbital of silicon,forming a $p\pi-d\pi$ back bond.
This back bonding changes the hybridization of nitrogen from $sp^3$ to $sp^2$,making the molecule planar (trigonal planar).
While $BF_3$ is also planar,it is planar due to its $sp^2$ hybridization and the nature of its bonding,but $N(SiH_3)_3$ is a classic example of a molecule that becomes planar specifically due to back bonding where it would otherwise be pyramidal like $N(CH_3)_3$.

(A) The bond order is calculated using the formula: $Bond \ order = \frac{1}{2} \times (N_b - N_a)$, where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
(1) For $N_2^+$ ($13$ electrons): $Bond \ order = \frac{1}{2} \times (9 - 4) = 2.5$.
For $O_2^+$ ($15$ electrons): $Bond \ order = \frac{1}{2} \times (10 - 5) = 2.5$.
Since both have a bond order of $2.5$, this pair is correct.
(2) For $F_2$ ($18$ electrons): $Bond \ order = 1$.
For $Ne_2$ ($20$ electrons): $Bond \ order = 0$.
(3) For $O_2$ ($16$ electrons): $Bond \ order = 2$.
For $B_2$ ($10$ electrons): $Bond \ order = 1$.
(4) For $C_2$ ($12$ electrons): $Bond \ order = 2$.
For $N_2$ ($14$ electrons): $Bond \ order = 3$.

(B) To determine the presence of unpaired electrons in antibonding molecular orbitals,we write the $MO$ configurations:
$C_2$ ($12$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2$. (No unpaired electrons).
$O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. (Has $2$ unpaired electrons in antibonding $\pi^*$ orbitals).
$He_2^+$ ($3$ electrons): $\sigma 1s^2, \sigma^* 1s^1$. (Has $1$ unpaired electron in antibonding $\sigma^*$ orbital).
Thus,the pair $O_2$ and $He_2^+$ both contain unpaired electrons in their antibonding molecular orbitals.

(A) $N_2$ and $CN^-$ are isoelectronic,both having $14$ electrons.
$N_2$ is a neutral molecule with a very strong triple bond $(N \equiv N)$ and high bond dissociation energy $(941 \ kJ \ mol^{-1})$,making it chemically inert.
$CN^-$ is an anion with a negative charge,which makes it more reactive as a nucleophile compared to the neutral $N_2$ molecule.
Therefore,the inertness of $N_2$ compared to $CN^-$ is primarily due to its higher bond energy and stability as a neutral species.

(C) Isoelectronic species are those that have the same number of electrons.
For $CO$: $6 + 8 = 14$ electrons.
For $NO^{+}$: $7 + 8 - 1 = 14$ electrons.
For $CN^{-}$: $6 + 7 + 1 = 14$ electrons.
For $C_2^{2-}$: $(6 \times 2) + 2 = 14$ electrons.
Since all these species contain $14$ electrons,they are isoelectronic.

(A) Bond length is inversely proportional to bond order.
Bond order for $O_2$ is $2.0$.
Bond order for $O_3$ is $1.5$ (due to resonance).
Bond order for $O_2^{2-}$ (peroxide ion) is $1.0$.
Since bond length $\propto \frac{1}{\text{bond order}}$,the order of bond length is $O_2 < O_3 < O_2^{2-}$.

(D) The electronic configuration of $N_2$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. Bond order = $(10-4)/2 = 3$.
In $N_2^+$,one electron is removed from the bonding molecular orbital $(\sigma 2p_z)$,so the configuration becomes $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^1$. Bond order = $(9-4)/2 = 2.5$. Since the bond order decreases,the $N-N$ bond weakens and $N_2^+$ is paramagnetic.
For $O_2$,the configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Bond order = $(10-6)/2 = 2$.
In $O_2^+$,one electron is removed from the antibonding molecular orbital $(\pi^*)$,so the configuration becomes $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Bond order = $(10-5)/2 = 2.5$. The bond order increases,and paramagnetism decreases as the number of unpaired electrons reduces from $2$ to $1$.
Therefore,the statement that $N_2^+$ becomes diamagnetic is wrong.

(C) 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.
$1$. For $CN^{-}$ ($14$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. $B.O. = \frac{10 - 4}{2} = 3$.
$2$. For $O_2^-$ ($17$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. $B.O. = \frac{10 - 7}{2} = 1.5$.
$3$. For $NO^{+}$ ($14$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. $B.O. = \frac{10 - 4}{2} = 3$.
$4$. For $CN^{+}$ ($12$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2$. $B.O. = \frac{8 - 4}{2} = 2$.
Thus,$CN^{-}$ and $NO^{+}$ both have a bond order of $3$.

(D) The total number of electrons in a $B_2$ molecule is $10$.
The molecular orbital 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 two unpaired electrons in the $\pi 2p$ orbitals,the $B_2$ molecule is paramagnetic.
The highest occupied molecular orbital $(HOMO)$ is of $\pi$ type,not $\sigma$ type.
Therefore,both the Assertion and the Reason are incorrect.

(A) The $MO$ electronic configuration of the $F_2$ molecule ($18$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$.
Number of bonding electrons $(N_b)$ $= 10$.
Number of antibonding electrons $(N_a)$ $= 8$.
Bond order $= \frac{N_b - N_a}{2} = \frac{10 - 8}{2} = 1$.
Since $N_b - N_a = 2$,the number of electrons in antibonding orbitals is indeed two less than in bonding orbitals.
Thus,both Assertion and Reason are correct,and the Reason is the correct explanation for the Assertion.

(C) According to $M.O.T.$,the electronic configuration of the $C_2$ molecule is:
$\sigma 1s^2 < \sigma^* 1s^2 < \sigma 2s^2 < \sigma^* 2s^2 < \pi 2p_x^2 = \pi 2p_y^2$
In the $C_2$ molecule,the bond order is $2$,and both bonds are $\pi$ bonds due to the presence of electrons only in the $\pi 2p$ orbitals.

(D) According to Molecular Orbital Theory $(MOT)$,the electronic configuration of $O_2$ ($16$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$.
Since $O_2$ contains two unpaired electrons in the $\pi^*$ antibonding orbitals,it exhibits paramagnetic behavior.
In contrast,$N_2$ ($14$ electrons),$H_2$ ($2$ electrons),and $Li_2$ ($6$ electrons) have all paired electrons,making them diamagnetic.

(B) The bond order of a diatomic species can be 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 species with total electrons up to $14$,the configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$.
$1$. $CN^{+}$ ($12$ electrons): $\text{Bond Order} = \frac{8-4}{2} = 2.0$
$2$. $CN$ ($13$ electrons): $\text{Bond Order} = \frac{9-4}{2} = 2.5$
$3$. $CN^{-}$ ($14$ electrons): $\text{Bond Order} = \frac{10-4}{2} = 3.0$
$4$. $NO$ ($15$ electrons): $\text{Bond Order} = \frac{10-5}{2} = 2.5$
Comparing the values,$CN^{-}$ has the highest bond order of $3.0$.

(A) According to Molecular Orbital Theory $(MOT)$:
The total number of electrons in $CN^{-}$ is $6 + 7 + 1 = 14$.
The electronic configuration is: $\sigma_{1s}^{2}, \sigma_{1s}^{*2}, \sigma_{2s}^{2}, \sigma_{2s}^{*2}, \pi_{2p_{x}}^{2} = \pi_{2p_{y}}^{2}, \sigma_{2p_{z}}^{2}$.
Bond order $= \frac{1}{2}(N_b - N_a) = \frac{1}{2}(10 - 4) = 3$.
Since all electrons are paired,$CN^{-}$ is diamagnetic.

(D) In the complex $[Ni(CO)_4]$,the synergistic bonding involving $d \pi - p \pi$ back-bonding leads to a decrease in the $Ni-C$ bond length and an increase in the $C-O$ bond length.
This phenomenon,along with the magnetic properties and the nature of the metal-ligand bond,is best explained by $Molecular \ Orbital \ Theory$ $(MOT)$.

(A) The magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n+2)} \; B.M$,where $n$ is the number of unpaired electrons. For $\mu = 1.73 \; B.M$,we have $n = 1$.

Species	Unpaired electrons $(n)$
$O_{2}^{+}$	$1$
$O_{2}^{-}$	$1$
$O_{2}$	$2$

Since both $O_{2}^{+}$ and $O_{2}^{-}$ have $1$ unpaired electron,they both exhibit a magnetic moment of $1.73 \; B.M$.

(B) Bond strength represents the extent of bonding between two atoms forming a molecule.
According to molecular orbital theory,the bond strength is directly proportional to the bond order.
Therefore,a higher bond order indicates a stronger bond between the atoms.

(N/A) The following conditions must be satisfied by atomic orbitals to form molecular orbitals:
$(a)$ The combining atomic orbitals must have the same or nearly the same energy. This means that in a homonuclear molecule,the $1s$ atomic orbital of an atom can combine with the $1s$ atomic orbital of another atom,and not with the $2s$ orbital.
$(b)$ The combining atomic orbitals must have proper orientations to ensure that the overlap is maximum.
$(c)$ The extent of overlapping should be large.

(N/A) The electronic configuration of Beryllium $(Be)$ is $1s^{2} \, 2s^{2}$.
The molecular orbital electronic configuration for the $Be_{2}$ molecule is:
$\sigma_{1s}^2 \, \sigma_{1s}^{*2} \, \sigma_{2s}^2 \, \sigma_{2s}^{*2}$.
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 anti-bonding molecular orbitals.
For $Be_{2}$,$N_{b} = 4$ and $N_{a} = 4$.
$\therefore \text{Bond Order} = \frac{1}{2}(4 - 4) = 0$.
$A$ bond order of $0$ indicates that the molecule is unstable and does not exist.

(A) The stability of a species is directly proportional to its bond order. The bond order is calculated as $\frac{1}{2}(N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of anti-bonding electrons.
$1$. For $O_2$ ($16$ electrons): Configuration is $KK [\sigma(2s)]^2 [\sigma^*(2s)]^2 [\sigma(2p_z)]^2 [\pi(2p_x)]^2 [\pi(2p_y)]^2 [\pi^*(2p_x)]^1 [\pi^*(2p_y)]^1$. Bond order $= \frac{1}{2}(10 - 6) = 2$. It is paramagnetic due to two unpaired electrons.
$2$. For $O_2^+$ ($15$ electrons): Configuration is $KK [\sigma(2s)]^2 [\sigma^*(2s)]^2 [\sigma(2p_z)]^2 [\pi(2p_x)]^2 [\pi(2p_y)]^2 [\pi^*(2p_x)]^1$. Bond order $= \frac{1}{2}(10 - 5) = 2.5$. It is paramagnetic due to one unpaired electron.
$3$. For $O_2^-$ ($17$ electrons): Configuration is $KK [\sigma(2s)]^2 [\sigma^*(2s)]^2 [\sigma(2p_z)]^2 [\pi(2p_x)]^2 [\pi(2p_y)]^2 [\pi^*(2p_x)]^2 [\pi^*(2p_y)]^1$. Bond order $= \frac{1}{2}(10 - 7) = 1.5$. It is paramagnetic due to one unpaired electron.
$4$. For $O_2^{2-}$ ($18$ electrons): Configuration is $KK [\sigma(2s)]^2 [\sigma^*(2s)]^2 [\sigma(2p_z)]^2 [\pi(2p_x)]^2 [\pi(2p_y)]^2 [\pi^*(2p_x)]^2 [\pi^*(2p_y)]^2$. Bond order $= \frac{1}{2}(10 - 8) = 1$. It is diamagnetic as all electrons are paired.
Stability order: $O_2^+ > O_2 > O_2^- > O_2^{2-}$.

(N/A) The plus $(+)$ and minus $(-)$ signs in orbital representations correspond to the phase of the electron wave function $(\psi)$. Their significance includes:
$(i)$ Determining whether constructive or destructive interference (overlapping) occurs between orbitals.
$(ii)$ Indicating whether the energy of the system increases (repulsion) or decreases (attraction) upon interaction.
$(iii)$ Predicting whether a chemical bond will form or not.
$(iv)$ Providing information regarding bond formation,molecular geometry,and bond angles.

Bond order is defined as one half of the difference between the number of electrons present in the bonding and anti-bonding orbitals of a molecule.
If $N_{b}$ is the number of electrons in bonding orbitals and $N_{a}$ is the number of electrons in anti-bonding orbitals,then:
Bond order $= \frac{1}{2} (N_{b} - N_{a})$
$1$. For $N_{2}$ ($14$ electrons):
Configuration: $[\sigma(1s)]^2 [\sigma^{*}(1s)]^2 [\sigma(2s)]^2 [\sigma^{*}(2s)]^2 [\pi(2p_{x})]^2 [\pi(2p_{y})]^2 [\sigma(2p_{z})]^2$
$N_{b} = 10, N_{a} = 4$
Bond order $= \frac{1}{2} (10 - 4) = 3$
$2$. For $O_{2}$ ($16$ electrons):
Configuration: $[\sigma(1s)]^2 [\sigma^{*}(1s)]^2 [\sigma(2s)]^2 [\sigma^{*}(2s)]^2 [\sigma(2p_{z})]^2 [\pi(2p_{x})]^2 [\pi(2p_{y})]^2 [\pi^{*}(2p_{x})]^1 [\pi^{*}(2p_{y})]^1$
$N_{b} = 10, N_{a} = 6$
Bond order $= \frac{1}{2} (10 - 6) = 2$
$3$. For $O_{2}^{+}$ ($15$ electrons):
Configuration: $KK [\sigma(2s)]^2 [\sigma^{*}(2s)]^2 [\sigma(2p_{z})]^2 [\pi(2p_{x})]^2 [\pi(2p_{y})]^2 [\pi^{*}(2p_{x})]^1$
$N_{b} = 8, N_{a} = 3$ (excluding $KK$ shell)
Bond order $= \frac{1}{2} (8 - 3) = 2.5$
$4$. For $O_{2}^{-}$ ($17$ electrons):
Configuration: $KK [\sigma(2s)]^2 [\sigma^{*}(2s)]^2 [\sigma(2p_{z})]^2 [\pi(2p_{x})]^2 [\pi(2p_{y})]^2 [\pi^{*}(2p_{x})]^2 [\pi^{*}(2p_{y})]^1$
$N_{b} = 8, N_{a} = 5$ (excluding $KK$ shell)
Bond order $= \frac{1}{2} (8 - 5) = 1.5$

(B) The compound $KO_{2}$ consists of $K^{+}$ and $O_{2}^{-}$ ions.
The superoxide ion $O_{2}^{-}$ has $13$ valence electrons.
According to molecular orbital theory,the electronic configuration of $O_{2}^{-}$ is $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \sigma 2p_{z}^{2}, \pi 2p_{x}^{2} = \pi 2p_{y}^{2}, \pi^{*} 2p_{x}^{2} = \pi^{*} 2p_{y}^{1}$.
The presence of one unpaired electron in the $\pi^{*} 2p$ molecular orbital makes $O_{2}^{-}$ paramagnetic.

(N/A) Isoelectronic species are those that contain the same total number of electrons. According to the Molecular Orbital Theory,if two or more diatomic molecules or ions are isoelectronic,they possess the same bond order.
Example-$1$: $F_{2}$ and $O_{2}^{2-}$ both contain $18$ electrons. Their bond order is $1$.
Example-$2$: $N_{2}$,$CO$,and $NO^{+}$ are isoelectronic,each containing $14$ electrons. They all exhibit a triple bond.

Molecule/ion	$N_{2}, CO, NO^{+}$
Bond structure	$N \equiv N, C \equiv O, N \equiv O^{+}$

(N/A) The bond order between two atoms is directly proportional to the bond strength. As the bond order increases,the bond enthalpy increases,the bond length decreases,and the stability of the molecule increases.
$\text{Bond Strength} \propto \text{Bond order} \propto \frac{1}{\text{Bond length}} \propto \text{Stability of molecule}$
This relationship is illustrated by the following examples:

Bond	Bond order	Bond length	Strength
$C-C$	$1$	$154 \text{ pm}$	Minimum
$C=C$	$2$	$133 \text{ pm}$	Intermediate
$C \equiv C$	$3$	$120 \text{ pm}$	Maximum
$C-N$	$1$	$143 \text{ pm}$	Minimum
$C=N$	$2$	$138 \text{ pm}$	Intermediate
$C \equiv N$	$3$	$116 \text{ pm}$	Maximum

(A) According to Molecular Orbital Theory $(MOT)$,the electronic configuration of the $O_2$ molecule ($16$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$.
Since there are two unpaired electrons in the $\pi^*$ antibonding orbitals,the $O_2$ molecule is paramagnetic.
The bond order is calculated as: $\text{Bond Order} = \frac{N_b - N_a}{2} = \frac{10 - 6}{2} = 2$.

(N/A) $1$. Positive (bonding) overlapping: In this type of overlapping,the overlapping lobes of the atomic orbitals have the same sign (phase). This leads to an increase in electron density between the two nuclei,resulting in the formation of a bonding molecular orbital (e.g.,$s+s$ or $p_z+p_z$ with same phase).
$2$. Negative (antibonding) overlapping: In this type of overlapping,the overlapping lobes of the atomic orbitals have opposite signs (phases). This leads to a decrease in electron density between the two nuclei,resulting in the formation of an antibonding molecular orbital (e.g.,$s-s$ or $p_z-p_z$ with opposite phase).

(B) According to Molecular Orbital Theory $(MOT)$,the electronic configuration of the dioxygen molecule $(O_2)$ 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 molecule contains two unpaired electrons in the antibonding $\pi^*$ molecular orbitals,it exhibits paramagnetic behavior.

(A) The number of electrons in $H_2^+$ is calculated as $(1 + 1 - 1) = 1$.
The number of electrons in $H_2$ is $(1 + 1) = 2$.
The number of electrons in $O_2^+$ is calculated as $(8 + 8 - 1) = 15$.

(N/A) Molecular Orbital $(MO)$ theory was developed by $F. Hund$ and $R.S. Mulliken$ in $1932$. The features of this theory are as follows:
$(i)$ The electrons in a molecule are present in various molecular orbitals,just as electrons in atoms are present in various atomic orbitals.
$(ii)$ Atomic orbitals of comparable energies and proper symmetry combine to form molecular orbitals.
$(iii)$ While an electron in an atomic orbital is influenced by one nucleus,in a molecular orbital,it is influenced by two or more nuclei depending upon the number of atoms in the molecule. Thus,an atomic orbital is monocentric,while a molecular orbital is polycentric.
$(iv)$ The number of molecular orbitals formed is equal to the number of combining atomic orbitals. When two atomic orbitals combine,two molecular orbitals are formed: one is known as a bonding molecular orbital,and the other is called an antibonding molecular orbital.
$(v)$ The bonding molecular orbital has lower energy and hence greater stability than the corresponding antibonding molecular orbital.
$(vi)$ Just as the electron probability distribution around a nucleus in an atom is given by an atomic orbital,the electron probability distribution around a group of nuclei in a molecule is given by a molecular orbital.
$(vii)$ Molecular orbitals,like atomic orbitals,are filled in accordance with the $Aufbau$ principle,obeying the $Pauli$ exclusion principle and $Hund$'s rule.