Molecular orbital theory Questions in English

(A) The molecular orbital configuration for $O_2$ $(16 \ e^-)$ is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^1(\pi_{2p_y}^*)^1$.
Number of electrons in $(\pi^*)$ of $O_2 = 2$.
For $O_2^{+}$ $(15 \ e^-)$,one electron is removed from the $(\pi^*)$ orbital,so number of electrons in $(\pi^*)$ of $O_2^{+} = 1$.
For $O_2^{-}$ $(17 \ e^-)$,one electron is added to the $(\pi^*)$ orbital,so number of electrons in $(\pi^*)$ of $O_2^{-} = 3$.
Total number of electrons in $(\pi^*)$ orbitals $= 2 + 1 + 3 = 6$.

(B) Ethane $(CH_3-CH_3)$ contains a single bond between two carbon atoms which is one $\sigma$-bond. Thus,$A-III$.
Ethene $(CH_2=CH_2)$ contains a double bond between two carbon atoms which consists of one $\sigma$-bond and one $\pi$-bond. Thus,$B-IV$.
Carbon molecule $(C_2)$ in its ground state,according to molecular orbital theory,has two $\pi$-bonds and no $\sigma$-bond between the two carbon atoms. Thus,$C-II$.
Ethyne $(HC \equiv CH)$ contains a triple bond between two carbon atoms which consists of one $\sigma$-bond and two $\pi$-bonds. Thus,$D-I$.
The correct matching is $A-III, B-IV, C-II, D-I$.

(D) Paramagnetism is due to the presence of unpaired electrons.
$1$. $Na_2O_2$ contains the peroxide ion $O_2^{2-}$. Its molecular orbital configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. It has $0$ unpaired electrons,so it is diamagnetic.
$2$. $O_3$ (Ozone) is diamagnetic as all electrons are paired.
$3$. $N_2O$ is diamagnetic as all electrons are paired.
$4$. $KO_2$ contains the superoxide ion $O_2^-$. Its molecular orbital configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. It has $1$ unpaired electron,making it paramagnetic.
Therefore,$KO_2$ is the paramagnetic compound. The correct option is $(D)$.

(A) The bond order of $CO$ ($14$ electrons) is $3$.
$NO^{+}$ ($14$ electrons),$CN^{-}$ ($14$ electrons),and $N_2$ ($14$ electrons) are isoelectronic with $CO$,so they all have a bond order of $3$.
$NO^{-}$ has $16$ electrons. Using the molecular orbital configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$.
Bond order $= \frac{1}{2} (N_b - N_a) = \frac{1}{2} (10 - 6) = 2$.
Thus,$NO^{-}$ has a bond order of $2$,which is different from $CO$.

(A) $B_2$: $\sigma 1s^2 \sigma^{\star} 1s^2 \sigma 2s^2 \sigma^{\star} 2s^2 \pi 2p_y^1 \pi 2p_z^1$. It is paramagnetic. Bond order $= (6-4)/2 = 1$. Mixing of $s$ and $p$ orbitals occurs. $B_2$ undergoes oxidation and reduction.
$(B)$ $N_2$: $\sigma 1s^2 \sigma^{\star} 1s^2 \sigma 2s^2 \sigma^{\star} 2s^2 \sigma 2p_x^2 \pi 2p_y^2 \pi 2p_z^2$. It is diamagnetic. Bond order $= (10-4)/2 = 3$. Mixing of $s$ and $p$ orbitals occurs. $N_2$ undergoes oxidation and reduction.
$(C)$ $O_2^{-}$: Paramagnetic with bond order $= 1.5$. $O_2^{-}$ undergoes oxidation and reduction.
$(D)$ $O_2$: $\sigma 1s^2 \sigma^{\star} 1s^2 \sigma 2s^2 \sigma^{\star} 2s^2 \sigma 2p_x^2 \pi 2p_y^2 \pi 2p_z^2 \pi^{\star} 2p_y^1 \pi^{\star} 2p_z^1$. Paramagnetic with bond order $= 2$. $O_2$ undergoes oxidation and reduction.
Correct mapping: $(A)$ $\rightarrow p, q, r, t; (B)$ $\rightarrow q, r, s, t; (C)$ $\rightarrow p, q, r; (D)$ $\rightarrow p, q, r, s$.

(B) The correct statements are $(A)$ and $(C)$.
$(A)$ $C_2^{2-}$ has $14$ electrons. Its configuration is $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2$. Since all electrons are paired,it is diamagnetic.
$(B)$ Bond order of $O_2^{2+}$ is $3$ and $O_2$ is $2$. Higher bond order implies shorter bond length. Thus,$O_2$ has a longer bond length than $O_2^{2+}$.
$(C)$ Bond order of $N_2^{+}$ ($13$ electrons) is $(9-4)/2 = 2.5$. Bond order of $N_2^{-}$ ($15$ electrons) is $(10-5)/2 = 2.5$. They have the same bond order.
$(D)$ $He_2^{+}$ has a bond order of $0.5$. Since the bond order is non-zero,it is more stable than two isolated $He$ atoms,meaning it has lower energy.

(C) The $F_2$ molecule has $18$ electrons. The molecular orbital configuration is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$.
In the $F_2$ molecule,the energy of the $\sigma 2p_z$ orbital is lower than the $\pi 2p_x$ and $\pi 2p_y$ orbitals due to the lack of $s-p$ mixing for elements with higher atomic numbers $(Z > 7)$.
Comparing the given options with the standard molecular orbital diagram for $F_2$,option $C$ correctly represents the energy levels and electron filling.

(D) To determine the magnetic nature,we use Molecular Orbital Theory $(MOT)$ configurations:
$H_2 (2e^-): \sigma 1s^2$ (Diamagnetic)
$He_2^+ (3e^-): \sigma 1s^2, \sigma^* 1s^1$ (Paramagnetic)
$Li_2 (6e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2$ (Diamagnetic)
$Be_2 (8e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2$ (Diamagnetic,though unstable)
$B_2 (10e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1, \pi 2p_y^1$ (Paramagnetic)
$C_2 (12e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2, \pi 2p_y^2$ (Diamagnetic)
$N_2 (14e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2, \pi 2p_y^2, \sigma 2p_z^2$ (Diamagnetic)
$O_2^- (17e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^2, \pi^* 2p_y^1$ (Paramagnetic)
$F_2 (18e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^2, \pi^* 2p_y^2$ (Diamagnetic)
The diamagnetic species are $H_2, Li_2, Be_2, C_2, N_2$,and $F_2$. Excluding $Be_2$ due to its non-existence,the count is $5$.

(A) The electronic configuration of $B_2$ ($10$ electrons) is $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p_x}^2$.
Bond order is calculated as $\frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2}$.
Bond order = $\frac{6 - 4}{2} = 1$.
Since all electrons are paired in the $\pi_{2p_x}$ orbital under the assumption that Hund's rule is violated,the molecule is diamagnetic.

(C) If $2s-2p$ mixing is not operative,the energy order of molecular orbitals is: $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < \pi 2p_x = \pi 2p_y < \pi^* 2p_x = \pi^* 2p_y < \sigma^* 2p_z$.
Electronic configurations:
$Be_2$ ($8$ electrons): $(\sigma 1s)^2, (\sigma^* 1s)^2, (\sigma 2s)^2, (\sigma^* 2s)^2$ (Diamagnetic).
$B_2$ ($10$ electrons): $(\sigma 1s)^2, (\sigma^* 1s)^2, (\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2$ (Diamagnetic).
$C_2$ ($12$ electrons): $(\sigma 1s)^2, (\sigma^* 1s)^2, (\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^1, (\pi 2p_y)^1$ (Paramagnetic due to two unpaired electrons).
$N_2$ ($14$ electrons): $(\sigma 1s)^2, (\sigma^* 1s)^2, (\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2, (\pi 2p_y)^2$ (Diamagnetic).

(D) The overlap of two $2p_z$ orbitals along the $z$-axis (internuclear axis) leads to the formation of $\sigma$ and $\sigma^*$ molecular orbitals.
$(A)$ The $\sigma$ molecular orbital formed by $2p_z-2p_z$ overlap has two nodal planes perpendicular to the internuclear axis,one for each $p_z$ orbital. This is correct.
$(B)$ The $\sigma^*$ antibonding molecular orbital has a nodal plane perpendicular to the internuclear axis between the two nuclei. It does not have a node in the $xz$-plane containing the molecular axis. This is incorrect.
$(C)$ The $\pi$ molecular orbital is formed by the lateral overlap of $p_x$ or $p_y$ orbitals. The question specifically asks about $2p_z$ overlap,which forms $\sigma$ bonds. However,if we consider the nodal properties of $\pi$ orbitals in general,they have one nodal plane containing the molecular axis. The statement provided is incorrect regarding the description of the $\pi$ orbital node.
$(D)$ The $\pi^*$ antibonding molecular orbital has two nodal planes: one containing the molecular axis (the $xz$-plane) and one perpendicular to the molecular axis. The statement describes a node in the $xy$-plane,which is correct for the $\pi^*$ orbital formed from $p_x$ orbitals. Given the options,$(A)$ and $(D)$ are the correct statements.

(D) The thermal decomposition of $AgNO_3$ is given by the reaction: $2AgNO_3(s) \rightarrow 2Ag(s) + 2NO_2(g) + O_2(g)$.
The two paramagnetic gases produced are $NO_2$ and $O_2$.
$NO_2$ is an odd-electron molecule with $1$ unpaired electron.
$O_2$ has $2$ unpaired electrons in its antibonding $\pi^*2p_y$ and $\pi^*2p_z$ orbitals.
Since $O_2$ has a higher number of unpaired electrons $(2 > 1)$,we consider the antibonding molecular orbitals of $O_2$.
The molecular orbital configuration of $O_2$ ($16$ electrons) is: $\sigma1s^2, \sigma^*1s^2, \sigma2s^2, \sigma^*2s^2, \sigma2p_x^2, \pi2p_y^2 = \pi2p_z^2, \pi^*2p_y^1 = \pi^*2p_z^1$.
The antibonding molecular orbitals are $\sigma^*1s$,$\sigma^*2s$,and $\pi^*2p_y, \pi^*2p_z$.
The number of electrons in these orbitals are $2 + 2 + 1 + 1 = 6$.

(D) For the formation of molecular orbitals,the atomic orbitals must have the same symmetry with respect to the internuclear axis ($z$-axis in this case).
$A$: $2p_z$ and $2p_x$ have different symmetries ($p_z$ is $\sigma$-type,$p_x$ is $\pi$-type),so they do not combine.
$B$: $2s$ and $2p_x$ have different symmetries ($s$ is $\sigma$-type,$p_x$ is $\pi$-type),so they do not combine.
$C$: $3d_{xy}$ and $3d_{x^2-y^2}$ have different symmetries,so they do not combine.
$D$: $2s$ and $2p_z$ both have cylindrical symmetry about the $z$-axis ($\sigma$-type),so they can combine to form a $\sigma$ molecular orbital.
$E$: $2p_z$ and $3d_{x^2-y^2}$ have different symmetries,so they do not combine.
Therefore,only the combination in $D$ leads to the formation of a molecular orbital.

(D) According to Molecular Orbital Theory $(MOT)$,a molecule is paramagnetic if it contains one or more unpaired electrons.

Molecule	No. of unpaired $e^-$
$A. O_2$	$2$
$B. N_2$	$0$
$C. F_2$	$0$
$D. S_2$	$2$
$E. Cl_2$	$0$

Since $O_2$ and $S_2$ contain unpaired electrons,they exhibit paramagnetic behavior. Therefore,the correct option is $A \& D$.

(B) Statement $I$: $A$ molecule with a bond order of $0$ implies that the number of bonding electrons equals the number of antibonding electrons,making the molecule unstable or non-existent (e.g.,$He_2$,$Be_2$,$Ne_2$). Thus,Statement $I$ is false.
Statement $II$: According to Molecular Orbital Theory,bond order is inversely proportional to bond length $(\text{Bond order} \propto \frac{1}{\text{Bond length}})$. Therefore,as bond order increases,bond length decreases. Thus,Statement $II$ is false.

(A) To solve this,we analyze the molecular orbital configuration and properties of each species:
$A$. $B_2$ ($10$ electrons) is paramagnetic with bond order $1$. $O_2$ ($16$ electrons) is paramagnetic with bond order $2$. Both are paramagnetic $(Q)$.
$B$. $Be_2$ ($8$ electrons) has bond order $0$ (does not exist). $H_2^{2-}$ ($4$ electrons) has bond order $0$ (does not exist). Both do not exist $(S)$.
$C$. $N_2^{+}$ ($13$ electrons) has bond order $2.5$ and is paramagnetic. $N_2^{-}$ ($15$ electrons) has bond order $2.5$ and is paramagnetic. Both have bond order $2.5$ $(P)$ and are paramagnetic $(Q)$.
$D$. $O_2^{+}$ ($15$ electrons) has bond order $2.5$ and is paramagnetic. $O_2^{-}$ ($17$ electrons) has bond order $1.5$ and is paramagnetic. Both are paramagnetic $(Q)$.
Correct matching: $A-Q, B-S, C-P, Q, D-Q$.

(D) According to Molecular Orbital Theory $(MOT)$,the electronic configuration of $C_2$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2$.
Since all electrons are paired,$C_2$ is diamagnetic in nature.
In $C_2$,the bond order is $\frac{8-4}{2} = 2$. Both bonds are $\pi$ bonds because the $\sigma 2p_z$ orbital is higher in energy than the $\pi 2p$ orbitals in $C_2$.
Thus,both Assertion and Reason are true,and the Reason correctly explains the nature of the bonding in $C_2$.

(B) The stability of a molecular species is directly proportional to its bond order.
The electronic configuration of $Li_2$ ($6$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2$. Bond order $= (4-2)/2 = 1$.
The electronic configuration of $Li_2^+$ ($5$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^1$. Bond order $= (3-2)/2 = 0.5$.
The electronic configuration of $Li_2^-$ ($7$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^1$. Bond order $= (4-3)/2 = 0.5$.
Comparing $Li_2^+$ and $Li_2^-$,$Li_2^+$ is more stable than $Li_2^-$ because $Li_2^-$ has an electron in the antibonding $\sigma^* 2s$ orbital,which decreases stability.
Thus,the order of stability is $Li_2^- < Li_2^+ < Li_2$.

(C) The molecular orbital configuration of $C_2$ ($12$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2$. It has a bond order of $2$ (consisting of $2 \pi$ bonds).
When $C_2$ changes to $C_2^{2-}$ ($14$ electrons),the configuration becomes $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. The bond order increases to $3$ (consisting of $2 \pi$ bonds and $1 \sigma$ bond).
Therefore,the number of $\sigma$ bonds increases from $0$ to $1$,and the bond order increases from $2$ to $3$. The statement that 'Bond order increases' is correct,and '$\sigma$ bond increases' is correct. The statement '$\sigma$ bond decreases' is incorrect.

(D) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as $\text{Bond Order} = \frac{N_b - N_a}{2}$.
For $O_2^{2-}$ ($18$ electrons): Bond order $= 1$,unpaired electrons $= 0$.
For $O_2^{-}$ ($17$ electrons): Bond order $= 1.5$,unpaired electrons $= 1$.
For $O_2$ ($16$ electrons): Bond order $= 2$,unpaired electrons $= 2$.
For $O_2^{+}$ ($15$ electrons): Bond order $= 2.5$,unpaired electrons $= 1$.
$1$. Bond length is inversely proportional to bond order: $O_2^{2-} (1) > O_2^{-} (1.5) > O_2 (2) > O_2^{+} (2.5)$. This is correct.
$2$. Bond strength is directly proportional to bond order: $O_2^{+} (2.5) > O_2 (2) > O_2^{-} (1.5) > O_2^{2-} (1)$. This is correct.
$3$. Number of unpaired electrons: $O_2 (2) > O_2^{+} (1) = O_2^{-} (1) > O_2^{2-} (0)$. This is correct.
Since all statements are correct,the answer is $D$.

(D) According to molecular orbital theory:
$1$. $B_2$ has $10$ electrons,configuration is $\sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \pi_{2p_x}^1 = \pi_{2p_y}^1$. It is paramagnetic due to unpaired electrons.
$2$. Bond order of $O_2^{+}$ is $2.5$ and $O_2^{-}$ is $1.5$. Higher bond order implies higher stability,so $O_2^{+}$ is more stable than $O_2^{-}$.
$3$. In $N_2$ ($14$ electrons),the energy of $\pi_{2p_x} = \pi_{2p_y}$ is lower than $\sigma_{2p_z}$ due to $s-p$ mixing. Thus,the statement that $\sigma_{2p_z}$ is higher than $\pi_{2p_x}$ is correct.
$4$. In $C_2$ ($12$ electrons),the configuration is $\sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \pi_{2p_x}^2 = \pi_{2p_y}^2$. Both bonds are pi bonds; there is no sigma bond. Thus,the statement in option $D$ is incorrect.

(A) $1$. $B_2$ ($10$ electrons) is paramagnetic ($HOMO$ $\pi_{2p_y}^1 \pi_{2p_z}^1$) and $O_2$ ($16$ electrons) is paramagnetic ($HOMO$ $\pi_{2p_y}^{*1} \pi_{2p_z}^{*1}$). Thus,$A-Q$.
$2$. $Be_2$ ($8$ electrons) and $H_2^{2-}$ ($4$ electrons) both have bond order $0$ and do not exist. Thus,$B-S$.
$3$. $N_2^{+}$ ($13$ electrons) and $N_2^{-}$ ($15$ electrons) both have a bond order of $2.5$. Thus,$C-P$.
$4$. $O_2^{+}$ ($15$ electrons) and $O_2^{-}$ ($17$ electrons) are both paramagnetic. Thus,$D-Q$.
Therefore,the correct match is $A-Q, B-S, C-P, D-Q$.

(D) The electronic configuration of $H_{2}$ is $(\sigma_{1s})^2$,giving a bond order of $\frac{2-0}{2} = 1$.
The electronic configuration of $Li_{2}$ is $(\sigma_{1s})^2, (\sigma_{1s}^*)^2, (\sigma_{2s})^2$,giving a bond order of $\frac{4-2}{2} = 1$.
However,in $Li_{2}$,the presence of inner $(\sigma_{1s})^2$ and $(\sigma_{1s}^*)^2$ electrons effectively shields the valence $\sigma_{2s}$ electrons from the nucleus,reducing the effective nuclear charge experienced by the bonding electrons and making the bond weaker compared to $H_{2}$.

(D) 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 = $\frac{1}{2}(N_b - N_a) = \frac{1}{2}(10 - 4) = 3$.
Since all electrons are paired,$N_2$ is diamagnetic.
$O_2$ ($16$ electrons) has a bond order of $2$ and is paramagnetic due to two unpaired electrons in $\pi^* 2p$ orbitals.
$N_2$ has $10$ bonding electrons,while $O_2$ has $8$ bonding electrons.
Therefore,the statement '$N_2$ is diamagnetic' is correct,and the statement '$N_2$ is paramagnetic' would be incorrect.
Given the options,the statement '$N_2$ is diamagnetic' is correct,but if the option was written as 'It is dimagnetic' (a typo for diamagnetic),it is still the intended correct property.
However,all statements $A$,$B$,$C$,and $D$ are actually correct descriptions of $N_2$.
If we assume the question asks for the incorrect statement,there might be a typo in the provided options.
Based on standard chemistry,$N_2$ is diamagnetic,so option $D$ is a correct statement.

(D) To determine the number of electrons in antibonding molecular orbitals,we write the molecular orbital configuration for each molecule:
$1$. $Li_2$ ($6$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2$. Antibonding electrons = $2$ (in $\sigma^* 1s$).
$2$. $N_2$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. Antibonding electrons = $2+2 = 4$.
$3$. $O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Antibonding electrons = $2+2+2 = 6$.
$4$. $F_2$ ($18$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. Antibonding electrons = $2+2+4 = 8$.
Thus,$F_2$ contains the maximum number of electrons in antibonding molecular orbitals.

(B) The total number of electrons in an $F_2$ molecule is $18$ $(9 + 9)$.
The molecular orbital configuration for $F_2$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$.
Counting the electrons in bonding molecular orbitals (those without an asterisk): $2 (\sigma 1s) + 2 (\sigma 2s) + 2 (\sigma 2p_z) + 4 (\pi 2p_x, \pi 2p_y) = 10$.
Counting the electrons in antibonding molecular orbitals (those with an asterisk): $2 (\sigma^* 1s) + 2 (\sigma^* 2s) + 4 (\pi^* 2p_x, \pi^* 2p_y) = 8$.
Therefore,the number of electrons in bonding and antibonding molecular orbitals are $10$ and $8$ respectively.

(C) The total number of electrons in $NO^{+}$ is $7 (N) + 8 (O) - 1 = 14$ electrons.
According to Molecular Orbital Theory,the electronic configuration of $NO^{+}$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$.
The bond order is calculated as: $\text{Bond Order} = \frac{N_b - N_a}{2} = \frac{10 - 4}{2} = \frac{6}{2} = 3$.

(B) According to Molecular Orbital Theory,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 $CO$ molecule,the total number of electrons is $6 + 8 = 14$.
The molecular orbital configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$.
Here,$N_b = 10$ and $N_a = 4$.
Bond order = $\frac{10 - 4}{2} = \frac{6}{2} = 3$.

(A) The electronic configuration of $Li_2$ molecule is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2$.
Bond order of $Li_2$ molecule is calculated as:
$\text{Bond Order} = \frac{N_b - N_a}{2} = \frac{4 - 2}{2} = 1$.
Since all electrons are paired in the $Li_2$ molecule,it is diamagnetic.

(C) The electronic configuration of the $O_2$ molecule 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 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 $O_2$,$N_b = 10$ and $N_a = 6$.
Bond order $= \frac{10 - 6}{2} = \frac{4}{2} = 2$.

(C) The electronic configuration of $O_2$ molecule 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 $(*)$.
These are $(\sigma 1s)^2$,$(\sigma 2s)^2$,$(\sigma 2p_z)^2$,$(\pi 2p_x)^2$,and $(\pi 2p_y)^2$.
Total number of electrons in bonding orbitals = $2 + 2 + 2 + 2 + 2 = 10$.

(B) The electronic configuration of $N_2$ molecule ($14$ electrons) is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2$.
Antibonding orbitals are those denoted with an asterisk $(*)$.
These are $(\sigma^* 1s)^2$ and $(\sigma^* 2s)^2$.
Total number of electrons in antibonding orbitals $= 2 + 2 = 4$.

(B) The total number of electrons in the $NO$ molecule is $7 + 8 = 15$.
According to Molecular Orbital Theory,the electronic configuration of $NO$ is:
$\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$.
There is $1$ unpaired electron in the antibonding $\pi^* 2p_x$ orbital.
Therefore,the number of unpaired electrons is $1$.

(B) The total number of electrons in an $F_2$ molecule is $18$.
The molecular orbital configuration is: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^2$.
Number of bonding electrons $(N_b)$ $= 10$.
Number of antibonding electrons $(N_a)$ $= 8$.
Bond order $= \frac{N_b - N_a}{2} = \frac{10 - 8}{2} = 1$.

(D) The total number of electrons in $N_2^{+}$ is $13$.
The molecular orbital configuration is: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^1$.
Number of bonding electrons $(N_b)$ $= 9$.
Number of antibonding electrons $(N_a)$ $= 4$.
Bond order $= \frac{N_b - N_a}{2} = \frac{9 - 4}{2} = \frac{5}{2} = 2.5$.

(A) According to Molecular Orbital Theory,the energy of molecular orbitals increases as follows:
$\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$.
For the $F_2$ molecule,the order of the lower energy orbitals is $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s$.

(C) According to Molecular Orbital Theory,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 = $\frac{N_b - N_a}{2} = \frac{10 - 6}{2} = \frac{4}{2} = 2.0$.

(D) The bond order is calculated using the formula: $\text{Bond Order} = \frac{1}{2} (N_b - N_a)$.
For $F_2$ ($18$ electrons): $\text{Bond Order} = 1$.
For $H_2$ ($2$ electrons): $\text{Bond Order} = 1$.
For $Li_2$ ($6$ electrons): $\text{Bond Order} = 1$.
For $N_2$ ($14$ electrons): $\text{Bond Order} = 3$.
Since $3 > 1$,the molecule $N_2$ has a bond order greater than $1$.

(C) According to Molecular Orbital Theory $(MOT)$,the electronic configuration of the $O_2$ molecule ($16$ electrons) is:
$\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$.
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.
$N_b = 10$
$N_a = 6$
$\text{Bond Order} = \frac{10 - 6}{2} = \frac{4}{2} = 2$.
Thus,the bond order of $O_2$ is $2$.

(D) The $CO$ molecule has a triple bond between the carbon and oxygen atoms,represented as $:C \equiv O:$.
Using the Molecular Orbital Theory,the bond order is calculated as:
$B.O. = \frac{1}{2} (N_b - N_a) = \frac{1}{2} (10 - 4) = 3$.
Therefore,the bond order of $CO$ is $3$.

(D) The concept of the combination of atomic orbitals to form molecular orbitals belongs to the Molecular Orbital Theory $(MOT)$,not the Valence Bond Theory $(VBT)$.

(B) According to Molecular Orbital Theory,the electronic configuration of the $O_2$ molecule ($16$ electrons) is:
$(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 = (\pi 2p_y)^2 (\pi^* 2p_x)^1 = (\pi^* 2p_y)^1$
The antibonding molecular orbitals (ABMOs) are $\sigma^* 1s$,$\sigma^* 2s$,$\pi^* 2p_x$,and $\pi^* 2p_y$.
The electrons in these antibonding orbitals are:
$\sigma^* 1s$: $2$ electrons
$\sigma^* 2s$: $2$ electrons
$\pi^* 2p_x$: $1$ electron
$\pi^* 2p_y$: $1$ electron
Total antibonding electrons = $2 + 2 + 1 + 1 = 6$ electrons.

(B) The electronic configuration of $B_{2}$ molecule ($10$ 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 $(N_b)$ = $2+2+1+1 = 6$.
Number of antibonding electrons $(N_a)$ = $2+2 = 4$.
Bond order = $\frac{N_b - N_a}{2} = \frac{6 - 4}{2} = \frac{2}{2} = 1$.

(D) 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: $\text{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
Here,$N_b = 10$ and $N_a = 4$.
$\text{Bond Order} = \frac{10 - 4}{2} = \frac{6}{2} = 3$.
Thus,the bond order in $N_{2}$ is $3$.

(C) The electronic configuration of the $Be_{2}$ molecule ($8$ electrons) is: $\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2}$.
Here,the number of bonding electrons $(N_{b})$ is $4$ and the number of antibonding electrons $(N_{a})$ is $4$.
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 $Be_{2}$ molecule does not exist.

(B) The bond energy is directly proportional to the bond order. The bond order 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.
$(A)$ $O_{2}^{2-}$: Total electrons = $18$. Configuration: $KK(\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\pi^* 2p_x)^2, (\pi^* 2p_y)^2$. Bond order = $\frac{10-8}{2} = 1$.
$(B)$ $O_{2}^{+}$: Total electrons = $15$. Configuration: $KK(\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\pi^* 2p_x)^1$. Bond order = $\frac{10-5}{2} = 2.5$.
$(C)$ $O_{2}^{-}$: Total electrons = $17$. Configuration: $KK(\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\pi^* 2p_x)^2, (\pi^* 2p_y)^1$. Bond order = $\frac{10-7}{2} = 1.5$.
$(D)$ $O_{2}$: Total electrons = $16$. Configuration: $KK(\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\pi^* 2p_x)^1, (\pi^* 2p_y)^1$. Bond order = $\frac{10-6}{2} = 2$.
Comparing the bond orders: $O_{2}^{+} (2.5) > O_{2} (2) > O_{2}^{-} (1.5) > O_{2}^{2-} (1)$.
Since $O_{2}^{+}$ has the highest bond order,it has the highest bond energy.

(A) The bond energy is directly proportional to the bond order. Let us calculate the bond order for each species using Molecular Orbital Theory:
$(A)$ $N_{2}^{2-}$: Total electrons = $14 + 2 = 16$. Configuration: $KK(\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$. Bond order = $(8 - 6) / 2 = 1$.
$(B)$ $N_{2}^{-}$: Total electrons = $14 + 1 = 15$. Configuration: $KK(\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1$. Bond order = $(8 - 5) / 2 = 1.5$.
$(C)$ $N_{2}^{+}$: Total electrons = $14 - 1 = 13$. Configuration: $KK(\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^1 (\pi 2p_x)^2 (\pi 2p_y)^2$. Bond order = $(7 - 2) / 2 = 2.5$.
$(D)$ $N_{2}$: Total electrons = $14$. Configuration: $KK(\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2$. Bond order = $(8 - 2) / 2 = 3$.
Since $N_{2}^{2-}$ has the lowest bond order $(1)$,it has the least bond energy.

(A) $\delta$ bond is formed by the lateral overlap of two $d$-orbitals where all four lobes of one orbital overlap with the four lobes of another orbital.
This occurs when two $d_{xy}$ orbitals or two $d_{x^2-y^2}$ orbitals approach each other along the $z$-axis.
Specifically,the $d_{xy}$ orbitals (lying in the $xy$-plane) and $d_{x^2-y^2}$ orbitals (lying along the $x$ and $y$ axes) are capable of forming $\delta$ bonds due to their specific spatial orientation.
Therefore,the correct pair is $d_{xy}$ and $d_{x^2-y^2}$.

(B) The electronic configuration of $N_2$ molecule ($14$ electrons) is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2$.
The antibonding orbitals are those marked with an asterisk $(*)$.
These are $(\sigma^* 1s)^2$ and $(\sigma^* 2s)^2$.
Total number of electrons in antibonding orbitals $= 2 + 2 = 4$.

(A) The bond order is calculated using the formula: $\text{Bond Order} = \frac{1}{2} [\text{Number of bonding electrons} - \text{Number of antibonding electrons}]$.
$1$. For $NO$ ($15$ electrons): $\text{Bond Order} = \frac{1}{2} [10 - 5] = 2.5$ (Matches $iii$).
$2$. For $CO$ ($14$ electrons): $\text{Bond Order} = \frac{1}{2} [10 - 4] = 3.0$ (Matches $iv$).
$3$. For $O_2^{-}$ ($17$ electrons): $\text{Bond Order} = \frac{1}{2} [10 - 7] = 1.5$ (Matches $i$).
$4$. For $O_2$ ($16$ electrons): $\text{Bond Order} = \frac{1}{2} [10 - 6] = 2.0$ (Matches $ii$).
Thus,the correct matching is $1-iii, 2-iv, 3-i, 4-ii$.