Molecular orbital theory Questions in English

(C) According to Molecular Orbital Theory $(MOT)$,the electronic configuration of $O_2$ molecule ($16$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$.
There are $2$ unpaired electrons in the antibonding $\pi^*$ molecular orbitals.

(C) According to Molecular Orbital Theory $(MOT)$,the electronic configuration of $O_2$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$.
Since it contains two unpaired electrons in the antibonding $\pi^*$ orbitals,it is paramagnetic.

(A) The formula for Bond Order $(B.O.)$ is given by:
$B.O. = \frac{N_b - N_a}{2}$
where $N_b$ is the number of electrons in bonding molecular orbitals and $N_a$ is the number of electrons in antibonding molecular orbitals.
Bond order can be a positive value,zero,or a fraction. For example,the bond order of $He_2$ is $0$,and the bond order of $O_2^+$ is $2.5$.

(C) The total number of electrons in $NO$ is $7 + 8 = 15$.
According to Molecular Orbital Theory,the electronic configuration of $NO$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$.
Bond order is calculated as: $B.O. = \frac{N_b - N_a}{2}$.
Here,$N_b$ (bonding electrons) $= 10$ and $N_a$ (antibonding electrons) $= 5$.
$B.O. = \frac{10 - 5}{2} = \frac{5}{2} = 2.5$.

(B) According to the $Molecular \ Orbital \ Theory$ $(MOT)$,the number of molecular orbitals formed is equal to the number of atomic orbitals that combine.
When two atomic orbitals combine,they form two molecular orbitals:
$1$. One bonding molecular orbital $(BMO)$ which is lower in energy.
$2$. One anti-bonding molecular orbital $(ABMO)$ which is higher in energy.
Therefore,the correct option is $B$.

(B) The stability of a species is directly proportional to its bond order.
Using Molecular Orbital Theory $(MOT)$,the bond orders are calculated as follows:
$O_2^{+}$: Bond order = $(10 - 5) / 2 = 2.5$
$O_2$: Bond order = $(10 - 6) / 2 = 2.0$
$O_2^{-}$: Bond order = $(10 - 7) / 2 = 1.5$
$O_2^{2-}$: Bond order = $(10 - 8) / 2 = 1.0$
Since $O_2^{2-}$ has the lowest bond order of $1.0$,it is the least stable species among the given options.

(C) The bond order $(B.O.)$ is calculated using the formula: $B.O. = \frac{1}{2} (N_b - N_a)$.
For $O_2$ ($16$ electrons): $B.O. = \frac{1}{2} (10 - 6) = 2.0$.
For $O_2^{-}$ ($17$ electrons): $B.O. = \frac{1}{2} (10 - 7) = 1.5$.
For $O_2^{+}$ ($15$ electrons): $B.O. = \frac{1}{2} (10 - 5) = 2.5$.
For $O_2^{2-}$ ($18$ electrons): $B.O. = \frac{1}{2} (10 - 8) = 1.0$.
Comparing these values,the bond order is maximum in $O_2^{+}$.

(C) Molecular orbital theory ($MO$ theory) was developed in the years after valence bond theory had been established $(1927)$,primarily through the efforts of Friedrich Hund,Robert Mulliken,John $C$. Slater,and John Lennard-Jones.
$MO$ theory was originally called the Hund-Mulliken theory,and Robert Mulliken is credited as a primary developer.

(D) According to the Molecular Orbital Theory $(MOT)$,the bond order of a molecule is defined as half the difference between the number of electrons in bonding molecular orbitals $(N_b)$ and the number of electrons in antibonding molecular orbitals $(N_a)$.
Mathematically,it is expressed as: $\text{Bond Order} = \frac{1}{2} (N_b - N_a)$.

(B) According to $MOT$,the electronic configuration of $O_2$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$.
Since it contains $2$ unpaired electrons in the $\pi^*$ antibonding molecular orbitals,it is paramagnetic.

(C) According to Molecular Orbital Theory $(MOT)$,a species is paramagnetic if it contains one or more unpaired electrons.
$1$. $O_2^-$ has $17$ electrons. Its electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. Since it has an unpaired electron,it is paramagnetic.
$2$. $NO$ has $15$ electrons. Its electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Since it has an unpaired electron,it is paramagnetic.
$3$. $CN^-$ has $14$ electrons. All electrons are paired,so it is diamagnetic.
Therefore,both $(a)$ and $(b)$ are paramagnetic.

(C) The total number of electrons in $N_2^+$ is $7 + 7 - 1 = 13$.
According to Molecular Orbital Theory,the electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^1$.
Number of bonding electrons $(N_b)$ = $2 + 2 + 2 + 2 + 1 = 9$.
Number of antibonding electrons $(N_a)$ = $2 + 2 = 4$.
Bond Order $(B.O.)$ = $\frac{N_b - N_a}{2} = \frac{9 - 4}{2} = \frac{5}{2} = 2.5$.

(B) The bond order is inversely proportional to the bond length.
The bond orders for the given species are:
$O_2^+ = 2.5$
$O_2 = 2.0$
$O_2^- = 1.5$
$O_2^{2-} = 1.0$
Since $O_2^+$ has the highest bond order $(2.5)$,it has the smallest bond length.

(C) Oxygen is paramagnetic due to the presence of two unpaired electrons in its antibonding molecular orbitals.
The molecular orbital configuration of $O_2$ is:
$O_2 = \sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x)^2 \pi(2p_y)^2 \pi^*(2p_x)^1 \pi^*(2p_y)^1$

(A) Bond order is defined as the number of chemical bonds between a pair of atoms.
According to Molecular Orbital Theory $(MOT)$:
$N_2$ has a bond order of $3$.
$Li_2$ has a bond order of $1$.
$He_2$ has a bond order of $0$.
$O_2$ has a bond order of $2$.
Therefore,$N_2$ has the highest bond order.

(C) The $H_2^-$ ion contains a total of $3$ electrons.
According to Molecular Orbital Theory,the electrons fill the molecular orbitals in increasing order of energy.
The order of filling is $\sigma 1s < \sigma^* 1s$.
Filling $3$ electrons: $(\sigma 1s)^2 (\sigma^* 1s)^1$.

(C) According to the Molecular Orbital Theory $(MOT)$,the electronic configuration of the oxygen molecule $(O_2)$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$.
Due to the presence of two unpaired electrons in the antibonding $\pi^*$ molecular orbitals,the oxygen molecule exhibits paramagnetic behavior. This phenomenon cannot be explained by the Valence Bond Theory or other classical models.

(B) The correct answer is $(B)$.
According to Molecular Orbital Theory,the electronic configuration of $H_2$ is $(\sigma 1s)^2$.
Since all electrons are paired,$H_2$ is diamagnetic in nature.
In contrast,$He_2^+$ has $3$ electrons $(\sigma 1s)^2 (\sigma^* 1s)^1$,$H_2^+$ has $1$ electron $(\sigma 1s)^1$,and $H_2^-$ has $3$ electrons $(\sigma 1s)^2 (\sigma^* 1s)^1$. All these species have unpaired electrons and are therefore paramagnetic.

(C) The electronic configuration of $N_2$ ($14$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$.
Number of bonding electrons $(N_b)$ = $2 + 2 + 2 + 2 + 2 = 10$.
Number of antibonding electrons $(N_a)$ = $2 + 2 = 4$.
Bond order = $\frac{N_b - N_a}{2} = \frac{10 - 4}{2} = \frac{6}{2} = 3$.
Thus,the bond order in $N_2$ is $3$.

(D) The electronic configuration of $H_2^+$ is $\sigma 1s^1$.
Bond order is calculated as $\frac{N_b - N_a}{2} = \frac{1 - 0}{2} = 0.5$ or $1/2$.
Since it contains one unpaired electron,it is paramagnetic.
Therefore,the correct option is $(D)$.

(D) $(d)$ Molecules that contain unpaired electrons are attracted by a magnetic field and exhibit paramagnetism.

(B) $NO_2$ has an odd number of valence electrons ($5 + 6 \times 2 = 17$ electrons),which results in the presence of an unpaired electron in its molecular structure.
Therefore,$NO_2$ is paramagnetic.
In contrast,$H_2O$,$SO_2$,and $CO_2$ have all their electrons paired,making them diamagnetic.

(C) According to Molecular Orbital Theory,the bond order is calculated as: $Bond \ Order = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of electrons in bonding orbitals and $N_a$ is the number of electrons in antibonding orbitals.
For $He_2$,the electronic configuration is $\sigma 1s^2, \sigma^* 1s^2$.
Bond order = $\frac{1}{2} (2 - 2) = 0$.
Since the bond order is $0$,the molecule $He_2$ does not exist.

(A) According to Molecular Orbital Theory $(MOT)$:
$1$. The electronic configuration of $O_2$ ($16$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. It has two unpaired electrons in the antibonding $\pi^*$ orbitals,making it paramagnetic.
$2$. 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 in the antibonding $\pi^*$ orbital,making it paramagnetic.
Therefore,both $O_2$ and $NO$ are paramagnetic due to the presence of unpaired electrons.

(C) The electronic configuration of $C_2$ molecule ($12$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2$.
Number of bonding electrons $(N_b)$ = $8$.
Number of antibonding electrons $(N_a)$ = $4$.
Bond order $(B.O.)$ = $\frac{N_b - N_a}{2} = \frac{8 - 4}{2} = 2$.

(A) The bond order is calculated using the formula: $B.O. = \frac{N_b - N_a}{2}$
Where $N_b$ is the number of electrons in bonding molecular orbitals and $N_a$ is the number of electrons in antibonding molecular orbitals.
From the given configuration:
$N_b = 2 (\sigma 1s) + 2 (\sigma 2s) + 2 (\sigma 2p_x) + 2 (\pi 2p_y) + 2 (\pi 2p_z) = 10$
$N_a = 2 (\sigma^* 1s) + 2 (\sigma^* 2s) = 4$
$B.O. = \frac{10 - 4}{2} = \frac{6}{2} = 3$.

(C) In Molecular Orbital Theory,when two atomic orbitals combine,they form two molecular orbitals: a bonding molecular orbital and an antibonding molecular orbital.
The bonding molecular orbital is lower in energy than the combining atomic orbitals,and this lowering in energy is called the stabilization energy.
Conversely,the antibonding molecular orbital is higher in energy,and this increase is called the destabilization energy.

(D) The electronic configuration of the $O_2$ molecule ($16$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$.
According to Hund's rule,the last two electrons enter the degenerate antibonding $\pi^*$ molecular orbitals singly.
These two unpaired electrons in the $\pi^* 2p_x$ and $\pi^* 2p_y$ orbitals are responsible for the paramagnetic nature of the $O_2$ molecule.

(B) The total number of electrons in $O_2^+$ is $16 - 1 = 15$.
According to Molecular Orbital Theory,the electronic configuration of $O_2^+$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$.
Number of bonding electrons $(N_b)$ = $10$.
Number of antibonding electrons $(N_a)$ = $5$.
Bond Order = $\frac{N_b - N_a}{2} = \frac{10 - 5}{2} = \frac{5}{2} = 2.5$.

(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.
$A$ molecule is considered stable if the bond order is greater than zero,which implies that the number of electrons in bonding orbitals must be greater than the number of electrons in antibonding orbitals $(N_b > N_a)$.
In the context of the given variables,$N_x > N_y$ ensures a positive bond order,indicating stability.

(C) The $\pi^* \, 2p_y$ antibonding molecular orbital is formed by the out-of-phase overlap of $2p_y$ atomic orbitals.
It possesses two nodal planes: one nodal plane is the molecular plane (containing the internuclear axis),and the second nodal plane is perpendicular to the internuclear axis between the two nuclei.

(B) The bond order is calculated using the Molecular Orbital Theory $(MOT)$ formula: $B.O. = \frac{1}{2} (N_b - N_a)$.
For $O_2^+$ ($15$ electrons): $B.O. = \frac{1}{2} (10 - 5) = 2.5$.
For $O_2$ ($16$ electrons): $B.O. = \frac{1}{2} (10 - 6) = 2.0$.
For $O_2^-$ ($17$ electrons): $B.O. = \frac{1}{2} (10 - 7) = 1.5$.
Thus,the correct sequence is $O_2^+ > O_2 > O_2^-$.

(A) To determine paramagnetism,we look for the presence of unpaired electrons in the molecular orbitals.
$A$. $S^{2-}$: The sulfur atom has an atomic number of $16$. The $S^{2-}$ ion has $18$ electrons. Its electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6$. Since all electrons are paired,it is diamagnetic.
$B$. $N_2^{-}$: Total electrons = $7 + 7 + 1 = 15$. The molecular orbital configuration is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2 (\pi^* 2p_x)^1$. It has one unpaired electron,so it is paramagnetic.
$C$. $O_2^{-}$: Total electrons = $8 + 8 + 1 = 17$. It has one unpaired electron in the $\pi^* 2p$ orbital,so it is paramagnetic.
$D$. $NO$: Total electrons = $7 + 8 = 15$. It has one unpaired electron in the $\pi^* 2p$ orbital,so it is paramagnetic.
Therefore,$S^{2-}$ is not paramagnetic.

(C) According to Molecular Orbital Theory $(MOT)$,a molecule is paramagnetic if it contains one or more unpaired electrons.
$CO_2$ has $22$ electrons (diamagnetic).
$SO_2$ has $32$ electrons (diamagnetic).
$H_2O$ has $10$ electrons (diamagnetic).
$NO$ has $15$ electrons. Its electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Due to the presence of one unpaired electron in the $\pi^* 2p_x$ orbital,$NO$ is paramagnetic.

(B) 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 = $(10-4)/2 = 3$.
For $N_2^-$,the extra electron enters the $\pi^* 2p$ orbital. Bond order = $(10-5)/2 = 2.5$. Since bond order decreases,the $N-N$ bond weakens. $N_2^-$ has one unpaired electron,so it is paramagnetic.
The electronic configuration of $O_2$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Bond order = $(10-6)/2 = 2$.
For $O_2^-$,the extra electron enters the $\pi^* 2p$ orbital. Bond order = $(10-7)/2 = 1.5$. Since bond order decreases,the bond length increases.
Therefore,the statement that the $O-O$ bond order increases is wrong.

(C) Bond order is defined as the number of chemical bonds between a pair of atoms.
According to Molecular Orbital Theory,the stability of a molecule or ion is directly proportional to the bond order.
$A$ higher bond order implies a greater number of bonding electrons,which results in stronger attraction between the nuclei and the shared electron pair.
Therefore,as the bond order increases,the bond becomes stronger and more stable.

(C) The electronic configuration of $O_2^{2-}$ (peroxide ion) 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,$O_2^{2-}$ is diamagnetic.

(A) The total number of electrons in $O_2^{2-}$ is $8 + 8 + 2 = 18$ electrons.
The molecular orbital configuration is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$.
The antibonding orbitals are $\sigma^* 1s$,$\sigma^* 2s$,$\pi^* 2p_x$,and $\pi^* 2p_y$.
Each of these four antibonding orbitals contains $2$ electrons,forming $4$ pairs.
Thus,the number of antibonding electron pairs is $4$.

(C) The total number of electrons in $He_2^+$ is $3$ $(2 + 2 - 1 = 3)$.
The electronic configuration in molecular orbitals is $\sigma 1s^2, \sigma^* 1s^1$.
The bond order is calculated as: $B.O. = \frac{N_b - N_a}{2} = \frac{2 - 1}{2} = 0.5$.
Therefore,the correct option is $C$.

(B) Paramagnetism is exhibited by species that contain at least one unpaired electron.
$ClO_2$ has $19$ valence electrons $(7 + 6 \times 2 = 19)$,so it has an unpaired electron.
$ClO_2^-$ has $20$ valence electrons $(7 + 6 \times 2 + 1 = 20)$. The Lewis structure shows all electrons are paired.
$NO_2$ has $17$ valence electrons $(5 + 6 \times 2 = 17)$,so it has an unpaired electron.
$NO$ has $11$ valence electrons $(5 + 6 = 11)$,so it has an unpaired electron.
Therefore,$ClO_2^-$ is diamagnetic and does not exhibit paramagnetism.

(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.
For $N_2$ ($14$ electrons): $B.O. = \frac{1}{2}[10 - 4] = 3$.
For $O_2^{2+}$ ($14$ electrons): $B.O. = \frac{1}{2}[10 - 4] = 3$.
Since both $N_2$ and $O_2^{2+}$ have a bond order of $3$,they have identical bond orders.

(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.
$(A)$ For $N_2^+$ ($13$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^1$. $B.O. = \frac{1}{2}[9 - 4] = 2.5$.
$(B)$ For $O_2^{2+}$ ($14$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. $B.O. = \frac{1}{2}[10 - 4] = 3$.
$(C)$ For $N_2$ ($14$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. $B.O. = \frac{1}{2}[10 - 4] = 3$.
$(D)$ For $NO^+$ ($14$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. $B.O. = \frac{1}{2}[10 - 4] = 3$.
Thus,the bond order is not $3$ for $N_2^+$.

(C) The bond energy depends on the bond order.
$N_2$ has a triple bond $(N \equiv N)$ with a bond order of $3$.
$O_2$ has a double bond $(O=O)$ with a bond order of $2$.
$F_2$ has a single bond $(F-F)$ with a bond order of $1$.
$C_2$ has a bond order of $2$.
Since bond energy is directly proportional to bond order,$N_2$ has the highest bond energy.

(A) For a molecule to be isoelectronic,it must have the same number of electrons.
For $CN^{-}$: Total electrons = $6 (C) + 7 (N) + 1 = 14$.
For $CO$: Total electrons = $6 (C) + 8 (O) = 14$.
Both $CN^{-}$ and $CO$ are isoelectronic with $14$ electrons.
Using Molecular Orbital Theory,the bond order is calculated as: $B.O. = \frac{1}{2} [N_b - N_a] = \frac{1}{2} [10 - 4] = 3$.
Thus,both molecules have a bond order of $3$ and are isoelectronic.

(A) To determine the magnetic nature,we use Molecular Orbital Theory $(MOT)$.
$1$. $O_2^+$: Total electrons = $8 + 8 - 1 = 15$. The configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Since it has one unpaired electron,it is paramagnetic.
$2$. $CN^-$: Total electrons = $6 + 7 + 1 = 14$. All electrons are paired,so it is diamagnetic.
$3$. $CO$: Total electrons = $6 + 8 = 14$. All electrons are paired,so it is diamagnetic.
$4$. $N_2$: Total electrons = $7 + 7 = 14$. All electrons are paired,so it is diamagnetic.
Therefore,the correct option is $A$.

(A) The total number of electrons in a neutral $N_2$ molecule is $14$ $(7 + 7)$.
For the $N_2^+$ ion,the total number of electrons is $14 - 1 = 13$.
According to the molecular orbital theory for $N_2$ (where the $\pi(2p_x) = \pi(2p_y)$ orbitals have lower energy than the $\sigma(2p_z)$ orbital),the filling order is: $\sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p_x)^2 \pi(2p_y)^2 \sigma(2p_z)^2$.
For $N_2^+$ ($13$ electrons),we remove one electron from the highest energy orbital,which is $\sigma(2p_z)$.
Thus,the configuration becomes: $\sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p)^4 \sigma(2p)^1$.

(C) According to the Molecular Orbital Theory $(MOT)$,the electronic configuration of the $O_2$ molecule is:
$KK(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_x)^2(\pi 2p_y)^2(\pi 2p_z)^2(\pi^* 2p_y)^1(\pi^* 2p_z)^1$.
The two unpaired electrons are present in the antibonding molecular orbitals $(\pi^* 2p_y)$ and $(\pi^* 2p_z)$.
These unpaired electrons are responsible for the paramagnetic nature of the oxygen molecule.