Molecular orbital theory Questions in English

(B) The bond order is calculated using the formula: $\text{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
For $O_{2}^{+}$ (total $15$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Bond order $= \frac{10 - 5}{2} = 2.5$.
For $O_{2}^{-}$ (total $17$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. Bond order $= \frac{10 - 7}{2} = 1.5$.
For $O_{2}^{2-}$ (total $18$ electrons): Bond order $= \frac{10 - 8}{2} = 1.0$.
For $O_{2}$ (total $16$ electrons): Bond order $= \frac{10 - 6}{2} = 2.0$.
Thus,$O_{2}^{-}$ has a bond order of $1.5$.

(A) To find the bond order,we use 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.
$1$. For $O_2^+$: Total electrons = $15$. Configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Bond order = $\frac{1}{2}(10 - 5) = 2.5$.
$2$. For $NO^+$: Total electrons = $14$. Configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. Bond order = $\frac{1}{2}(10 - 4) = 3.0$.
$3$. For $O_2^{2-}$: Total electrons = $18$. Bond order = $1.0$.
$4$. For $B_2$: Total electrons = $10$. Bond order = $1.0$.
Since both $O_2^{2-}$ and $B_2$ have a bond order of $1.0$,they form the correct pair.

(A) 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 $O_2$ gains an electron to form $O_2^-$,the incoming electron enters the next available orbital,which is the antibonding $\pi^*$ orbital.
Thus,the configuration of $O_2^-$ becomes $KK \sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x)^2 \pi(2p_y)^2 \pi^*(2p_x)^2 \pi^*(2p_y)^1$.

(D) The bond order is calculated using the formula: $Bond \ Order = \frac{1}{2} (N_b - N_a)$.
For $He_2^+$ ($3$ electrons): $\sigma 1s^2, \sigma^* 1s^1$,$BO = \frac{2-1}{2} = 0.5$.
For $O_2^-$ ($17$ electrons): $BO = \frac{10-7}{2} = 1.5$.
For $NO$ ($15$ electrons): $BO = \frac{10-5}{2} = 2.5$.
For $C_2^{2-}$ ($14$ electrons): $BO = \frac{10-4}{2} = 3.0$.
Thus,the increasing order of bond order is: $He_2^+ < O_2^- < NO < C_2^{2-}$.

(A) The bond order is inversely proportional to the bond length.
Bond order of $O_2^+ = \frac{10-5}{2} = 2.5$
Bond order of $O_2 = \frac{10-6}{2} = 2.0$
Bond order of $O_2^- = \frac{10-7}{2} = 1.5$
Bond order of $O_2^{2-} = \frac{10-8}{2} = 1.0$
Since $O_2^+$ has the highest bond order $(2.5)$,it has the minimum bond length.

(C) The electronic configuration of $O_2^+$ ($13$ electrons) is: $KK \sigma 2s^2 \sigma^* 2s^2 \sigma 2pz^2 (\pi 2px^2 = \pi 2py^2) (\pi^* 2px^1)$. It has one unpaired electron,so it is paramagnetic.
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 two unpaired electrons,so it is paramagnetic.
Since both $O_2^+$ and $O_2$ contain unpaired electrons in their $\pi^*$ antibonding molecular orbitals,both are paramagnetic.

(B) The species $Be_2$ does not exist under normal conditions.
The electronic configuration of $Be_2$ ($Z=4$,total $8$ electrons) is: $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2$.
The bond order is calculated as: $\text{Bond order} = \frac{N_b - N_a}{2} = \frac{4 - 4}{2} = 0$.
Since the bond order is $0$,the molecule is unstable and does not exist.

(A) The molecular orbital configuration for $N_2$ ($14$ 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 $(B.O.)$ $= \frac{10-4}{2} = 3$.
For $N_2^-$ ($15$ electrons): $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p_x)^2(\pi 2p_y)^2(\sigma 2p_z)^2(\pi^* 2p_x)^1$.
$B.O. = \frac{10-5}{2} = 2.5$.
For $N_2^{2-}$ ($16$ electrons): $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p_x)^2(\pi 2p_y)^2(\sigma 2p_z)^2(\pi^* 2p_x)^1(\pi^* 2p_y)^1$.
$B.O. = \frac{10-6}{2} = 2$.
Comparing the bond orders: $2 < 2.5 < 3$.
Therefore,the increasing order is $N_2^{2-} < N_2^- < N_2$.

(B) Isoelectronic species are those that possess the same number of electrons.
For $NO^{+}$: $7 + 8 - 1 = 14$ electrons.
For $C_2^{2-}$: $6 \times 2 + 2 = 14$ electrons.
For $CN^{-}$: $6 + 7 + 1 = 14$ electrons.
For $N_2$: $7 \times 2 = 14$ electrons.
Since $NO^{+}, C_2^{2-}, CN^{-}$,and $N_2$ each have $14$ electrons,they are isoelectronic.

(C) According to Molecular Orbital Theory $(MOT)$,we determine the presence of unpaired electrons by filling the molecular orbitals:
$1$. For $N_2^+$ ($13$ electrons): $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^1$. It has $1$ unpaired electron.
$2$. For $O_2$ ($16$ electrons): $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*1} \pi_{2p_y}^{*1}$. It has $2$ unpaired electrons.
$3$. For $O_2^{2-}$ ($18$ electrons): $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*2} \pi_{2p_y}^{*2}$. All electrons are paired.
$4$. For $B_2$ ($10$ electrons): $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p_x}^1 \pi_{2p_y}^1$. It has $2$ unpaired electrons.
Therefore,$O_2^{2-}$ is the only species without unpaired electrons.

(B) To determine the magnetic behavior,we analyze the molecular orbital configuration of each species.
$1$. $NO$ 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 is paramagnetic due to the unpaired electron.
$2$. $O_2^{2-}$ has $18$ electrons. Its 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.
$3$. $O_2^{+}$ has $15$ electrons and is paramagnetic.
$4$. $O_2$ has $16$ electrons and is paramagnetic due to two unpaired electrons in the $\pi^*$ orbitals.

(C) The bond order $(BO)$ 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 antibonding electrons.
For $NO$ ($15$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. $BO = \frac{1}{2}(10 - 5) = 2.5$. It is paramagnetic.
For $NO^+$ ($14$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. $BO = \frac{1}{2}(10 - 4) = 3.0$. It is diamagnetic.
In the process $NO \rightarrow NO^+$,the electron is removed from the antibonding $\pi^*$ orbital,which increases the bond order from $2.5$ to $3.0$ and changes the magnetic behavior from paramagnetic to diamagnetic.

(A) The bond order of a diatomic species depends on the total number of electrons in the molecule or ion.
Species with the same number of electrons are isoelectronic and generally possess the same bond order.
Calculating the total number of electrons for each species:
$CN^{-}$: $6 + 7 + 1 = 14$ electrons.
$NO^{+}$: $7 + 8 - 1 = 14$ electrons.
$CN^{+}$: $6 + 7 - 1 = 12$ electrons.
$O_2^{-}$: $8 + 8 + 1 = 17$ electrons.
Since $CN^{-}$ and $NO^{+}$ both have $14$ electrons,they have the same bond order (which is $3$).

(D) According to $MO$ theory,bond order is inversely proportional to bond length.
$O_2^{2+}$ has a bond order of $3.0$.
$O_2^+$ has a bond order of $2.5$.
$O_2^-$ has a bond order of $1.5$.
$O_2^{2-}$ has a bond order of $1.0$.
Since $O_2^{2+}$ has the highest bond order,it has the shortest bond length.

(A) According to Molecular Orbital Theory $(MOT)$,a molecule is diamagnetic if all its electrons are paired.
For $C_2$ ($12$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2$. All electrons are paired,so it is diamagnetic.
For $F_2$ ($18$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. All electrons are paired,so it is diamagnetic.
For $O_2$ ($16$ electrons): The configuration ends with $\pi^* 2p_x^1 = \pi^* 2p_y^1$,having $2$ unpaired electrons,making it paramagnetic.
For $S_2$ ($32$ electrons): Similar to $O_2$,it has $2$ unpaired electrons in the $\pi^*$ orbitals,making it paramagnetic.
Note: Both $C_2$ and $F_2$ are diamagnetic. However,in standard textbook contexts for this specific question,$C_2$ is frequently cited as the primary example of a molecule with a unique bonding structure exhibiting diamagnetism due to the $\pi$ bonding.

(C) According to Molecular Orbital Theory,species with a bond order of $0$ or less do not exist.
For $H_{2}^{2+}$: Total electrons $= 1 + 1 - 2 = 0$. Electronic configuration $= \sigma 1s^{0}$. Bond order $= 0$.
For $He_{2}$: Total electrons $= 2 + 2 = 4$. Electronic configuration $= \sigma 1s^{2}, \sigma^* 1s^{2}$. Bond order $= \frac{2 - 2}{2} = 0$.
Since both $H_{2}^{2+}$ and $He_{2}$ have a bond order of $0$,they are not likely to exist.

(B) The stability of a molecule is directly proportional to its bond order. If bond orders are equal, the species with more electrons in antibonding orbitals is less stable.
$Li_2$ ($6$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2$. Bond order $= \frac{4 - 2}{2} = 1.0$.
$Li_2^+$ ($5$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^1$. Bond order $= \frac{3 - 2}{2} = 0.5$.
$Li_2^-$ ($7$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^1$. Bond order $= \frac{4 - 3}{2} = 0.5$.
Comparing $Li_2^+$ and $Li_2^-$, both have a bond order of $0.5$, but $Li_2^-$ has more electrons in the antibonding orbital $(\sigma^* 2s^1)$, making it less stable than $Li_2^+$.
Therefore, the stability order is $Li_2^- < Li_2^+ < Li_2$.

(B) Paramagnetism is due to the presence of unpaired electrons.
$1.$ $NO$: Total electrons = $15$. It has an odd number of electrons,so it must have at least one unpaired electron,making it paramagnetic.
$2.$ $CO$: Total electrons = $14$. All electrons are paired in its molecular orbitals,so it is diamagnetic (not paramagnetic).
$3.$ $O_2$: Total electrons = $16$. According to Molecular Orbital $(MO)$ theory,it has two unpaired electrons in its antibonding $\pi^* 2p_x$ and $\pi^* 2p_y$ orbitals,so it is paramagnetic.
$4.$ $B_2$: Total electrons = $10$. It has two unpaired electrons in its $\pi 2p_x$ and $\pi 2p_y$ orbitals,so it is paramagnetic.

(C) The bond order of a molecule is defined by 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.
If two molecules have the same bond order,then $\frac{N_{b1} - N_{a1}}{2} = \frac{N_{b2} - N_{a2}}{2}$.
This implies that $N_{b1} - N_{a1} = N_{b2} - N_{a2}$.
Therefore,the difference between the number of bonding and antibonding electrons must be the same for both molecules.
Option $(C)$ correctly states this relationship.

(D) The bond strength is directly proportional to the Bond Order $(B.O.)$.
$(a) \ O_2^- (B.O. = 1.5), O_2 (B.O. = 2.0)$. Here,the second species is stronger.
$(b) \ N_2 (B.O. = 3.0), N_2^+ (B.O. = 2.5)$. Here,the first species is stronger.
$(c) \ NO^+ (B.O. = 3.0), NO^- (B.O. = 2.0)$. Here,the first species is stronger.
Therefore,the pairs where the first species has a stronger bond are $(b)$ and $(c)$.

(A) According to Molecular Orbital Theory $(MOT)$,for homonuclear and heteronuclear diatomic molecules with $14$ or fewer electrons,the order of energy is $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \pi 2px = \pi 2py < \sigma 2pz < \pi^* 2px = \pi^* 2py < \sigma^* 2pz$.
For molecules with more than $14$ electrons,the order is $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2pz < \pi 2px = \pi 2py < \pi^* 2px = \pi^* 2py < \sigma^* 2pz$.
$CO$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2px^2, \pi 2py^2, \sigma 2pz^2$.
$CN^-$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2px^2, \pi 2py^2, \sigma 2pz^2$.
$NO$ ($15$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2pz^2, \pi 2px^2, \pi 2py^2, \pi^* 2px^1$.
$N_2^+$ ($13$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2px^2, \pi 2py^2, \sigma 2pz^1$.
All given options represent correct configurations based on the $MOT$ energy order.

(B) Total number of electrons in $OF = 8$ (from $O$) $+ 9$ (from $F$) $= 17$ electrons.
The molecular orbital configuration for $17$ electrons is: $\sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \sigma_{2p_z}^2, \pi_{2p_x}^2 = \pi_{2p_y}^2, \pi_{2p_x}^{*2} = \pi_{2p_y}^{*1}$.
Number of bonding electrons $(N_b)$ $= 10$.
Number of anti-bonding electrons $(N_a)$ $= 7$.
Bond order $= \frac{1}{2}(N_b - N_a) = \frac{1}{2}(10 - 7) = \frac{3}{2} = 1.5$.

(A) Stability is directly proportional to bond order.
Bond order $= \frac{1}{2} \times (N_B - N_{AB})$.
$1$. $N_2$ (valence electrons $= 10$): $(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p)^4(\sigma 2p)^2$. Bond order $= \frac{1}{2} \times (8 - 2) = 3$.
$2$. $N_2^+$ (valence electrons $= 9$): $(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p)^4(\sigma 2p)^1$. Bond order $= \frac{1}{2} \times (7 - 2) = 2.5$.
$3$. $N_2^-$ (valence electrons $= 11$): $(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p)^4(\sigma 2p)^2(\pi^* 2p)^1$. Bond order $= \frac{1}{2} \times (8 - 3) = 2.5$.
$4$. $N_2^{2-}$ (valence electrons $= 12$): $(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p)^4(\sigma 2p)^2(\pi^* 2p)^2$. Bond order $= \frac{1}{2} \times (8 - 4) = 2$.
Although $N_2^+$ and $N_2^-$ have the same bond order,$N_2^-$ has an electron in an antibonding orbital,which decreases its stability compared to $N_2^+$.
Thus,the stability order is $N_2 > N_2^+ > N_2^- > N_2^{2-}$.
Therefore,the correct option is $A$.

(B) To determine the change in bond order and magnetic nature,we analyze the molecular orbital configurations:
$1$. For $O_2 (16e^-)$: Bond order = $2.0$,Paramagnetic. For $O_2^+ (15e^-)$: Bond order = $2.5$,Paramagnetic. (Both change).
$2$. For $N_2^+ (13e^-)$: Bond order = $2.5$,Paramagnetic. For $N_2^- (15e^-)$: Bond order = $2.5$,Paramagnetic. (Bond order and magnetic nature remain unchanged).
$3$. For $NO (15e^-)$: Bond order = $2.5$,Paramagnetic. For $NO^+ (14e^-)$: Bond order = $3.0$,Diamagnetic. (Both change).
$4$. For $CO (14e^-)$: Bond order = $3.0$,Diamagnetic. For $CO^+ (13e^-)$: Bond order = $2.5$,Paramagnetic. (Both change).
Thus,the process $N_2^+ \to N_2^-$ maintains both properties.

(C) To determine paramagnetism,we look for the presence of unpaired electrons.
$1$. $O_2$: According to Molecular Orbital Theory,the electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. It has $2$ unpaired electrons,so it is paramagnetic.
$2$. $Na_2O_2$: Contains the peroxide ion $O_2^{2-}$. The configuration is $\sigma^* 2p_x^2 = \sigma^* 2p_y^2$. All electrons are paired,so it is diamagnetic.
$3$. $BaO_2$: Contains the peroxide ion $O_2^{2-}$,which is diamagnetic.
$4$. $KO_3$: Contains the ozonide ion $O_3^-$. The $O_3^-$ ion has $19$ valence electrons. Molecular orbital calculations show it has an unpaired electron in the antibonding $\pi^*$ orbital,making it paramagnetic.
Therefore,$O_2$ and $KO_3$ are both paramagnetic.

(B) To determine the bond length,we first calculate the bond order for each species.
$1$. For $CO$: The bond order is $3.0$.
$2$. For $CO^{+}$: The bond order is $2.5$ (as one electron is removed from the antibonding orbital of $CO$).
$3$. For $C_2O_{4}^{2-}$ (oxalate ion): The $C-O$ bond has partial double bond character due to resonance,with an average bond order of $1.5$.
Since bond length is inversely proportional to bond order $(Bond \ Length \propto \frac{1}{Bond \ Order})$,the order of bond length is: $C_2O_{4}^{2-} (1.5) > CO^{+} (2.5) > CO (3.0)$.
Therefore,the correct order is $C_2O_{4}^{2-} > CO^{+} > CO$.

(B) The compound $(SiH_3)_3N$ is planar and less basic than $(CH_3)_3N$.
In $(SiH_3)_3N$,the lone pair of electrons on the $N$ atom is donated into the vacant $d$-orbitals of the $Si$ atoms,resulting in $p\pi - d\pi$ back bonding.
This delocalization of the lone pair makes the molecule planar and significantly reduces its availability for protonation,making it less basic than $(CH_3)_3N$.
In $(CH_3)_3N$,carbon lacks vacant $d$-orbitals,so no such back bonding occurs,and the lone pair remains localized on the $N$ atom.

(A) The bond order of oxygen species is determined by the number of electrons in the molecular orbitals:
$(I)$ $Na_2O_2$ contains the peroxide ion $O_2^{2-}$. The number of electrons is $18$. Bond order $= \frac{1}{2}(10 - 8) = 1.0$.
$(II)$ $O_2[AsF_6]$ contains the dioxygenyl cation $O_2^+$. The number of electrons is $15$. Bond order $= \frac{1}{2}(10 - 5) = 2.5$.
$(III)$ $CsO_2$ contains the superoxide ion $O_2^-$. The number of electrons is $17$. Bond order $= \frac{1}{2}(10 - 7) = 1.5$.
$(IV)$ $O_2$ is the neutral dioxygen molecule. The number of electrons is $16$. Bond order $= \frac{1}{2}(10 - 6) = 2.0$.
Comparing the bond orders: $2.5 (II) > 2.0 (IV) > 1.5 (III) > 1.0 (I)$.
Thus,the correct decreasing order is $(II) > (IV) > (III) > (I)$.

(C) In $O_2$ ($16$ electrons),the molecular orbital configuration is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$.
$sp$ mixing is generally significant for molecules with $14$ or fewer electrons (like $B_2, C_2, N_2$).
For $O_2$,the energy of $2s$ and $2p$ orbitals is sufficiently different that $sp$ mixing is negligible.
Even if $sp$ mixing were considered,the magnetic nature of $O_2$ (paramagnetic due to two unpaired electrons in $\pi^*$ orbitals) remains unchanged because the filling order of the higher energy orbitals is not affected by $sp$ mixing in this case.

(A) $HOMO$ stands for Highest Occupied Molecular Orbital.
For the $N_2$ molecule,the electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$.
The $HOMO$ is the $\sigma 2p_z$ orbital.
This orbital is formed by the head-on overlap of two $2p_z$ orbitals with constructive interference (same signs overlapping),which is represented by the combination of two $p$ orbitals where the lobes of the same sign face each other.

(A) Paramagnetism is due to the presence of unpaired electrons.
In $O_2[AsF_6]$,the $O_2^+$ ion has a bond order of $2.5$ and contains one unpaired electron in its molecular orbitals,making it paramagnetic.
$Al_2O_3$ is an ionic solid where $Al^{3+}$ ($[Ne]$ configuration) and $O^{2-}$ ($[Ne]$ configuration) have no unpaired electrons,making it diamagnetic.
For $C_2$,the molecular orbital configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2, \pi 2p_y^2$. All electrons are paired,so it is diamagnetic.
For $Be_2$,the configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2$. All electrons are paired,so it is diamagnetic.

(B) The $\sigma _{2s}^*$ molecular orbital is an antibonding molecular orbital formed by the out-of-phase combination of two $2s$ atomic orbitals.
It possesses a nodal plane perpendicular to the internuclear axis,which is a characteristic feature of $p$ orbitals (specifically $p_x$ or $p_y$ orbitals depending on the internuclear axis).
Therefore,the nodal structure of the $\sigma _{2s}^*$ orbital is similar to that of a $p$ orbital.

(D) In molecular orbital theory,the $\pi$ bond is formed by the lateral (sideways) overlap of atomic orbitals.
Option $A$ correctly shows the lateral overlap of two $p$-orbitals forming a $\pi_{2p}$ bond.
Options $B$ and $C$ represent the same lateral overlap of $d$-orbitals to form a $\pi_{3d}$ bond.
Option $D$ shows an incorrect representation for the antibonding $\pi^*_{3d}$ orbital,as the phase relationship and nodal planes do not correspond to the standard antibonding $\pi$ orbital structure for $d$-orbitals.

(D) The potential energy $(P.E.)$ curve for a molecule depends on its bond order and bond dissociation energy.
For $H_2$,the bond order is $1$ and it is a stable molecule with a deep potential energy well.
For $H_2^+$,the bond order is $0.5$,making it less stable than $H_2$,so it has a shallower potential energy well compared to $H_2$.
Looking at the provided graph,the curve labeled $a$ corresponds to $H_2$,$b$ corresponds to $H_2^+$,and $c$ corresponds to $H_2^-$.
However,the graph shows $H_2$ at $a$,$H_2^+$ at $b$,and $H_2^-$ at $c$. The stability order is $H_2 > H_2^+ \approx H_2^-$. The graph correctly depicts the relative depths of the potential energy wells for these species. Therefore,all are correctly matched.

(B) The total number of electrons in a $C_2$ molecule is $12$ $(6 + 6)$.
According to Molecular Orbital Theory,the electronic configuration of $C_2$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2$.
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.
Here,$N_b = 8$ and $N_a = 4$.
$\text{Bond Order} = \frac{8 - 4}{2} = \frac{4}{2} = 2$.

(A) In the $O_2$ molecule,if the $x-axis$ is considered the internuclear (molecular) axis,then the $p_x$ orbitals undergo head-on overlap to form a $\sigma-$ bond.
The $p_y$ and $p_z$ orbitals undergo side-on overlap to form $\pi-$ bonds.
For a $\pi-$ bond formed by $p_z$ orbitals,the electron density is concentrated above and below the molecular axis $(x-axis)$.
The nodal plane is the plane where the probability of finding an electron is zero.
For a $\pi_{p_z}$ bond,the nodal plane contains the molecular axis $(x-axis)$ and is perpendicular to the plane of the orbital lobes.
Since the $p_z$ orbitals lie in the $xz-plane$,the nodal plane for the $\pi_{p_z}$ bond is the $xy-plane$.

(B) The electronic configuration of $N_2$ ($14$ electrons) is: $KK \sigma 2s^2 \sigma^* 2s^2 \pi 2p_x^2 = \pi 2p_y^2 \sigma 2p_z^2$.
When $N_2$ is converted to $N_2^+$,one electron is removed from the highest occupied molecular orbital $(HOMO)$.
In $N_2$,the $HOMO$ is the $\sigma 2p_z$ orbital.
Therefore,the electron is lost from the $\sigma 2p_z$ molecular orbital.

(B) molecule or ion is paramagnetic if it contains at least one unpaired electron.
$NO$ $(15 \ e^-)$ and $NO_2$ $(23 \ e^-)$ are odd-electron species and are paramagnetic.
$O_2^+$ $(15 \ e^-)$ has one unpaired electron in the $\pi^* 2p$ orbital,making it paramagnetic.
$O_2^-$ $(17 \ e^-)$ has one unpaired electron in the $\pi^* 2p$ orbital,making it paramagnetic.
Therefore,the set $O_2^+, O_2^-, NO, NO_2$ contains only paramagnetic species.

(D) In $Cl_2O_7$,the $Cl$ atom is $sp^3$ hybridized.
The $\pi$-bonds are formed by the back-bonding mechanism involving the overlap of the empty $3d$ orbitals of $Cl$ and the filled $2p$ orbitals of $O$.
Therefore,the $\pi$-bond is formed by $3d-2p$ overlapping.

(C) The molecular orbital ${\pi 2p_x}$ is a Bonding Molecular Orbital $(B.M.O.)$ which has $1$ nodal plane.
Antibonding molecular orbitals like ${\sigma ^* 2p_z}$ have higher energy than bonding orbitals and possess nodal planes perpendicular to the internuclear axis.
Among the given options,${\sigma ^* 2p_z}$ is an antibonding orbital with a nodal plane,making it higher in energy than the bonding orbitals.

(B) According to Molecular Orbital Theory $(M.O.T.)$,the bond order is inversely proportional to the bond length.
For $F_2^+$: Total electrons = $17$,Bond Order = $(10-7)/2 = 1.5$.
For $F_2$: Total electrons = $18$,Bond Order = $(10-8)/2 = 1.0$.
For $F_2^-$: Total electrons = $19$,Bond Order = $(10-9)/2 = 0.5$.
Since the bond order follows the order $F_2^+ > F_2 > F_2^-$,the bond length follows the order $F_2^+ < F_2 < F_2^-$.

(D) nodal plane is a plane where the probability of finding an electron is zero.
For the antibonding molecular orbital $ \pi_{2px}^* $,there are two nodal planes: one is the nodal plane of the atomic $ p_x $ orbitals (the $ yz $-plane),and the other is the nodal plane perpendicular to the internuclear axis passing through the center of the bond (due to the antibonding nature).
Thus,$ \pi_{2px}^* $ has two nodal planes.

(C) nodal plane is defined as a region in space where the probability of finding an electron is zero.

(A) To determine the bond order and magnetic property,we use Molecular Orbital Theory $(MOT)$:
$1$. For $O_2^+$ ($15$ electrons): Configuration is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1$. Bond order = $\frac{10-5}{2} = 2.5$. It has one unpaired electron,so it is paramagnetic.
$2$. For $B_2$ ($10$ electrons): Configuration is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^1 (\pi 2p_y)^1$. Bond order = $\frac{6-4}{2} = 1.0$. It is paramagnetic but has an integer bond order.
$3$. For $O_2$ ($16$ electrons): Bond order = $\frac{10-6}{2} = 2.0$. It is paramagnetic but has an integer bond order.
$4$. For $N_2$ ($14$ electrons): Bond order = $\frac{10-4}{2} = 3.0$. It is diamagnetic.
Thus,$O_2^+$ is both paramagnetic and has a fractional bond order.

(A) For $C_2^+ \to C_2$:
$C_2^+$ has $9$ electrons,configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p^1$. Bond order = $(6-4)/2 = 1.0$. It is paramagnetic.
$C_2$ has $12$ electrons,configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p^4$. Bond order = $(8-4)/2 = 2.0$. It is diamagnetic.
Since the bond order increased $(1.0 \to 2.0)$ and magnetic behavior changed (paramagnetic to diamagnetic),this is the correct transformation.

(A) To determine the magnetic nature,we calculate the total number of electrons and use Molecular Orbital Theory $(MOT)$.
$NO^{-}$ has $16$ electrons. The electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$.
Since it contains two unpaired electrons in the $\pi^*$ orbitals,it is paramagnetic.
$O_2^{2-}$ has $18$ electrons and is diamagnetic.
$CN^{-}$ has $14$ electrons and is diamagnetic.
$CO$ has $14$ electrons and is diamagnetic.

(C) The electronic configuration of the peroxide ion $(O_2^{2-})$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$.
$1.$ The $\pi^*$ antibonding molecular orbitals are completely filled with $4$ electrons,so statement $A$ is correct.
$2.$ Since all electrons are paired,the ion is diamagnetic,so statement $B$ is correct.
$3.$ Bond order = $\frac{1}{2} (N_b - N_a) = \frac{1}{2} (10 - 8) = 1$. Thus,statement $C$ is correct.
$4.$ The total number of electrons in $O_2^{2-}$ is $8 + 8 + 2 = 18$,while Neon has $10$ electrons. Thus,statement $D$ is incorrect.
Therefore,statements $A, B,$ and $C$ are correct.

(B) According to Molecular Orbital Theory $(M.O.T.)$,the electronic configuration of $B_2$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1, \pi 2p_y^1$.
In $B_2$,the bond order is $1$,and the bond is formed by the overlap of $2p$ orbitals,which is a $\pi$-bond.
Since $B_2$ has two unpaired electrons in the $\pi 2p$ orbitals,it is paramagnetic.
$C_2$ has a bond order of $2$ (both $\pi$-bonds) but is diamagnetic.
${O_2}^-$ has both $\sigma$ and $\pi$ bonds and is paramagnetic.
$N_2$ has one $\sigma$ and two $\pi$ bonds and is diamagnetic.
Therefore,$B_2$ is the correct species.

(B) The bond length is inversely proportional to the bond order $(B.O.)$.
$1$. For $O_2$ (dioxygen molecule): The bond order is $2.0$.
$2$. For $Na_2O_2$ (contains peroxide ion,$O_2^{2-}$): The bond order is $1.0$.
$3$. For $KO_2$ (contains superoxide ion,$O_2^-$): The bond order is $1.5$.
Comparing the bond orders: $2.0 (O_2) > 1.5 (O_2^-) > 1.0 (O_2^{2-})$.
Since bond length is inversely proportional to bond order,the species with the lowest bond order will have the maximum bond length.
Therefore,$O_2^{2-}$ (in $Na_2O_2$) has the maximum bond length.