NCERT Exemplar MCQs

Hint

\(
\text { (b) } \mathrm{NF}_3 \text { is pyramidal whereas } \mathrm{BF}_3 \text { is planar triangular. }
\)
\(
\mathrm{BF}_4^{-} \text {and } \mathrm{NH}_4^{+} \text { ions both are tetrahedral. }
\)
\(\mathrm{BCl}_3\) is triangular planar whereas \(\mathrm{BrCl}_3\) is pyramidal.
\(\mathrm{NH}_3\) is pyramidal whereas \(\mathrm{NO}_3^{-}\) is triangular planar.

Hint

(c) \(\mathrm{H}_2 \mathrm{O}\) will have the highest dipole moment due to the maximum difference in electronegativity of \(\mathrm{H}\) and \(\mathrm{O}\) atoms.

Explanation: \(\mathrm{H}_2 \mathrm{O}\) has the highest dipole moment because it is highly electronegative and has two lone pair on it.

Hint

(b) The number of orbitals involved in hybridization can be determined by the application of formula:
\(
H=\frac{1}{2}[V+M-C+A]
\)
where \(\mathrm{H}=\) number of orbitals involved in hybridization
\(V=\) valence electrons of central atom
\(\mathrm{M}=\) number of monovalent atoms linked with central atom
\(\mathrm{C}=\) charge on the cation
\(A=\) charge on the anion
\(
\begin{aligned}
& \mathrm{NO}_2^{+}: H=\frac{1}{2}[5+0-1+0]=2 \text { or } s p \\
& \mathrm{NO}_3^{-}: H=\frac{1}{2}[5+0-0+1]=3 \text { or } s p^2 \\
& \mathrm{NH}_4^{+}: H=\frac{1}{2}[5+4-1+0]=4 \text { or } s p^3
\end{aligned}
\)

Hint

HINT: Boiling point depends on the extent of hydrogen bonding.
Explanation:
(b) Strength of \(\mathrm{H}\)-bonding depends on the electronegativity of the atom which follows the order: \(\mathrm{F}>O>N\).
Strength of \(\mathrm{H}\)-bond is in the order:
\(
\mathrm{H} {\ldots} \ldots . \mathrm{F}>\mathrm{H} \ldots \ldots \mathrm{O}>\mathrm{H} {\ldots} \ldots \ldots \mathrm{N}
\)
But each \(\mathrm{H}_2 \mathrm{O}\) molecule is linked to 4 other \(\mathrm{H}_2 \mathrm{O}\) molecules through \(\mathrm{H}\)-bonds whereas each \(\mathrm{HF}\) molecule is linked only to two other \(\mathrm{HF}\) molecules.
Hence, the correct decreasing order of the boiling points is \(\mathrm{H}_2 \mathrm{O}>\mathrm{HF}>\mathrm{NH}_3\).

Hint

(b) Formal charge of the atom in the molecule or ion = (Number of valence electrons in free atom) (Number of lone pair electrons \(+1 / 2\) Number of bonding electrons)

Hint

(d) In \(\mathrm{N}\)-atom, number of valence electrons \(=5\)
Due to the presence of one negative charge, number of valence electrons \(=5+1=6\). One O -atom forms two bonds ( \(=\) bond) and two O -atom are shared with two electrons of \(\mathrm{N}\)-atom.
Thus, 3 O-atoms are shared with 8 electrons of \(\mathrm{N}\)-atom.
Number of bond pairs (or shared pairs) \(=4\)
Number of lone pairs \(=0\)

Hint

(a) \(\mathrm{BH}_4^{-}=\)tetrahedral, \(\mathrm{NH}_2^{-}=\mathrm{V}\)-shape, \(\mathrm{CO}_3^{2-}=\) triangular planar, \(\mathrm{H}_3 \mathrm{O}^{+}=\) pyramidal.

Hint

(c) HINT: \(\pi\) bond is double bond and \(\sigma\) bond is single bond. Explanation:
The given compound will have the correct structure as

\(
\text { There are } 5 \mathrm{C}=\mathrm{C} \pi \text { bonds }
\)
\(
8 \mathrm{C}-\mathrm{H}+11 \mathrm{C}-\mathrm{C} \sigma \text {-bonds, i.e., } 19 \sigma \text {-bonds are present in the above molecule. }
\)

Hint

(c)
\(
\begin{aligned}
& \mathrm{N}_2^{+}=\mathrm{KK} \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^2=\pi 2 p_y^2 \sigma 2 p_z^1 \\
& \mathrm{O}_2=\mathrm{KK} \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi^* 2 p_x^1=\pi^* 2 p_y^1 \\
& \mathrm{O}_2^{2-}=\mathrm{KK} \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi^* 2 p_x^2=\pi^* 2 p_y^2 \\
& \mathrm{~B}_2=\mathrm{KK} \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^1=\pi 2 p_y^1
\end{aligned}
\)
\(
\text { Thus, } \mathrm{O}_2^{2-} \text { has no unpaired electrons. }
\)

Hint

(c) \(\mathrm{C}_2 \mathrm{H}_4\) has one double bond and four single bonds. Bond length of double bond \((\mathrm{C}=\mathrm{C})\) is smaller than single bond \((\mathrm{C}-\mathrm{H})\).

Hint

(b) \(\mathrm{HCl} \mathrm{HI}\) and \(\mathrm{H}_2 \mathrm{~S}\) do not from \(\mathrm{H}\)-bonds. Only \(\mathrm{H}_2 \mathrm{O}\) forms hydrogen bonds. One \(\mathrm{H}_2 \mathrm{O}\) molecule forms four \(\mathrm{H}\)-bonding.

Hint

(d) In transition elements ( \(n-1)\) d and \(ns\) orbitals take part in bond formation.

Explanation:
STEP 1:
The given electronic configuration shows that an element is a vanadium \((Z=22)\). It belongs to the d-block of the periodic table. In transition elements i.e., \(d\) block elements, electrons of \(n s\) and ( \(n-1) d\) subshell take part in bond formation.
STEP 2:
Hence, \(3 \mathrm{~d}\) and 4 s electrons participate in bond formation. Therefore, option 4 is the correct answer.

Hint

(b) \(\mathrm{sp}^2\) hybridisation gives three \(\mathrm{sp}^2\) hybrid orbitals which are planar triangular forming an angle of \(120^{\circ}\) with each other. The electronic configurations of three elements \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are given below. Answer the questions from 14 to 17 on the basis of these configurations.
A \(1s^2 2 \mathrm{~s}^2 2 \mathrm{p}^6\)
B \(1s^2 2 s^2 2 p^6 3 s^2 3 p^3\)
C \(1s^2 2 s^2 2 p^6 3 s^2 3 p^5\)

Hint

(a) The given electronic configuration shows that A represents noble gas because the octet is complete. \(\mathrm{A}\) is neon which has 10 atomic number.

Hint

\(
\text { (b) The electronic configuration of } \mathrm{C} \text { represent chlorine. Its stable form is dichlorine }\left(\mathrm{Cl}_2\right) \text {, i.e., } \mathrm{C}_2 \text {. }
\)

Hint

\(
\text { (d) } B \text { represent } P \text { and } C \text { represents } \mathrm{C l} \text {. The stable compound is } P C l_3 \text { i.e., } \mathrm{BC}_3 \text {. }
\)

Hint

\(
\text { (b) Both } \mathrm{B} \text { and } \mathrm{C} \text { are non-metals and therefore, bond formed between them will be covalent. }
\)

Hint

(a) Hint: Energy of \(\sigma 2 \mathrm{p}_{\mathrm{z}}\) molecular orbital is more than \(\pi^* 2 \mathrm{p}_{\mathrm{x}} \approx \pi^* 2 \mathrm{p}_{\mathrm{y}}\) molecular orbital.
Explanation: Molecules like \(\mathrm{B}_2, \mathrm{C}_2 \& \mathrm{~N}_2\) have 1 to 3 electrons in p orbital energy of \(\sigma 2 \mathrm{p}\) molecular orbital are greater than that of \(\pi 2 p_x\) and \(\pi 2 p_y\) molecular orbitals.
Step 1: The MOT diagram for \(\mathrm{N}_2\) is as follows:

Step-2:
So According to the above molecular orbital diagram The order of energies of molecular orbitals of \(\mathrm{N}_2\) is as follows: \(\left(\pi 2 \mathrm{p}_{\mathrm{y}}\right)<\left(\sigma 2 \mathrm{p}_{\mathrm{z}}\right)<\left(\pi * 2 \mathrm{p}_{\mathrm{x}}\right) \approx\left(\pi * 2 \mathrm{p}_{\mathrm{y}}\right)\)

Hint

(a) \(\mathrm{Be}_2(4+4=8)=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 1 s^2, \sigma^* 2 s^2\)
Bond order \((\mathrm{BO})=\frac{1}{2}\) [Number of bonding electron \(\left(\mathrm{N}_{\mathrm{b}}\right)-\) Number of antibonding electrons \(\mathrm{N}_{\mathrm{a}}\) ]
\(
=\frac{4-4}{2}=0
\)
Here, bond order of \(\mathrm{He}_2\) is zero. Thus, it does not exist.
(b) \(\mathrm{He}_2(2+2=4)=\sigma 1 s^2, \sigma^* 1 s^2\)
\(
\mathrm{BO}=\frac{2-2}{2}=0
\)
Here, bond order of \(\mathrm{He}_2\) is zero. Thus, it does not exist.
\(
\begin{aligned}
& \mathrm{He}_2^{+}(2+2-1=3)=\sigma 1 s^2, \sigma^* 1 s^1 \\
& \mathrm{BO}=\frac{2-1}{2}=0.5
\end{aligned}
\)
Since, the bond order is not zero, this molecule is expected to exist.
(c)
\(
\begin{aligned}
\mathrm{N}_2(7+7 & =14)=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \sigma 2 p_z^2 \\
\mathrm{BO} & =\frac{10-4}{2}=3
\end{aligned}
\)
Thus, dinitrogen \(\left(\mathrm{N}_2\right)\) molecule contains triple bond and no other molecule of second period have more than double bond. Hence, bond strength of \(\mathrm{N}_2\) is maximum amongst the homonuclear diatomic molecules belonging to the second period.
(d) It is incorrect. The correct order of energies of molecular orbitals in \(\mathrm{N}_2\) molecule is
\(
\sigma 2 s<\sigma^* 2 s<\left(\pi 2 p_x \approx \pi 2 p_y\right)<\sigma 2 p_z<\pi^* 2 p_x \approx \pi^* 2 p_y<\sigma^* 2 p_z
\)

Hint

(b) 

\(
\begin{aligned}
& \mathrm{O}_2: \mathrm{KK} \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z{ }^2 \pi 2 p_x{ }^2=\pi 2 p_y{ }^2 \pi^* 2 p_x{ }^1=\pi^* 2 p_y{ }^1 \\
& \mathrm{O}_2^{-}: \mathrm{KK} \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z{ }^2 \pi 2 p_x{ }^2=\pi 2 p_y{ }^2 \pi^* 2 p_x{ }^2=\pi^* 2 p_y{ }^1 \\
& \mathrm{O}_2^{+}: \mathrm{KK} \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z{ }^2 \pi 2 p_x{ }^2=\pi 2 p_y{ }^2 \pi^* 2 p_x{ }^1
\end{aligned}
\)
Bond order, \(\mathrm{O}_2 \quad \frac{8-4}{2}=2.0\)
Bond order, \(\mathrm{O}_2^{-} \quad \frac{8-5}{2}=1.5\)
Bond order, \(\mathrm{O}_2^{+} \quad \frac{8-3}{2}=2.5\)
\(
\therefore \quad \mathrm{O}_2^{+}>\mathrm{O}_2>\mathrm{O}_2^{-} \quad \text { or } \quad \mathrm{O}_2^{-}<\mathrm{O}_2<\mathrm{O}_2^{+}
\)

Hint

(a) The electronic configuration represents
\(2 s^2 2 p^5=\) fluorine \(=\) most electronegative element
\(3 s^2 3 p^5=\) chlorine
\(4 s^2 4 p^5=\) bromine
\(5 s^2 5 p^5=\) iodine

Hint

(b)
(b) and (d) have exactly half-filled p-orbitals but (b) is smaller in size than Hence, (b) has highest ionization enthalpy.

Alternate:

Hint: Elements having half-filled and fully-filled electronic configurations has high ionization enthalpy.
The electronic configuration of options (b) and (d) have exactly half-filled 3p orbitals (b) represents phosphorus and (c) represents arsenic but (b) is smaller in size than (d).
Hence, (b) has the highest ionization enthalpy increases left to right in the periodic table as the size decreases.

Hint

(a, b)

\(
\begin{aligned}
& \mathrm{CN}^{-} \text {(number of electrons }=6+7+1=14 \text { ) } \\
& \mathrm{NO}^{+} \text {(number of electrons }=7+8-1=14 \text { ) } \\
& \mathrm{O}_2^{-} \text {(number of electrons }=8+8+1=17 \text { ) } \\
& \mathrm{O}_2^{2-} \text { (number of electrons }=8+8+2=18 \text { ) }
\end{aligned}
\)
Thus, \(\mathrm{CN}^{-}\)and \(\mathrm{NO}^{+}\)because of the presence of same number of electrons, have the same bond order.

Hint

\((\mathrm{a}, \mathrm{d}) \mathrm{BeCl}_2(\mathrm{Cl}-\mathrm{Be}-\mathrm{Cl})\) and \(\mathrm{CS}_2(\mathrm{~S}=\mathrm{C}=\mathrm{S})\) both are linear, \(\mathrm{NCO}^{+}\)is non-linear. However, remember that \(\mathrm{NCO}(\mathrm{N}=\mathrm{C}=0)\) is linear because it is isoelectronic with \(\mathrm{CO}_2\). \(\mathrm{NO}_2\) is angular with bond angled \(132^{\circ}\) and each \(\mathrm{O}-\mathrm{N}\) bond length of \(1.20 \mathrm{~A}^{\circ}\) (intermediate between single and double bond).

Hint

\((a, b)\) Number of electrons in \(\mathrm{CO}=14\)
Number of electrons in \(\mathrm{NO}^{+}=14\)
Number of electrons in \(\mathrm{N}_2=14\)
Number of electrons in \(\mathrm{SnCl}_2=84\)
Number of electrons in \(\mathrm{NO}_2^{-}=24\)

Hint

\(\text { (c, d) } \mathrm{CO}_2 \rightarrow \text { Linear, } \mathrm{CCl}_4 \rightarrow \text { Tetrahedral, } \mathrm{O}_3 \rightarrow \text { Angular }\left(\mathrm{V} \text {-shaped), } \mathrm{NO}_2^{-} \rightarrow\right. \text { Angular (V-shaped) }\)

Hint

(c,d)
Formal charge on \(\mathrm{O}\) atom (1): \(6-4-\frac{1}{2} \times 4=0\)
Formal charge on \(\mathrm{O}\) atom (2): \(6-6-\frac{1}{2} \times 2=-1\)
Formal charge on \(\mathrm{O}\) atom (3): \(6-6-\frac{1}{2} \times 2=-1\)
Average formal charge on each oxygen atom: \(\frac{-1+(-1)+0}{3}=-0.67\) All \(\mathrm{C}-\mathrm{O}\) bonds are equal due to resonance.

Hint

(a,d)

(a) Electronic configuration of \(\mathrm{N}_2=\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^2 \approx \pi 2 p_y^2 \sigma 2 p_z^2\). It has no unpaired electron. This indicates it is a diamagnetic species.
(b) Electronic configuration of \(\mathrm{N}_2^{2-}\) ion \(=\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2\) \(\pi 2 p_x^2 \approx \pi 2 p_y^2 \sigma 2 p_z^2 \pi 2 p_x^1 \approx \pi * 2 p_y^1\).
It has two unpaired electrons, so it is paramagnetic in nature.
(c) Electronic configuration of \(\mathrm{O}_2=\dot{\sigma} 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2\)
\(
\sigma 2 p_z^2 \pi 2 p_x^2 \approx \pi 2 p_y^2 \pi^* 2 p_x^1 \approx \pi^* 2 p_y^1 \text {. }
\)
The presence of two unpaired electrons shows its paramagnetic nature.
(d) Electronic configuration of \(\mathrm{O}_2^{2-}\) ion \(=\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2\)
\(
\sigma 2 p_z^2 \pi 2 p_{\mathrm{x}}^2 \approx \sigma 2 p_y^2 \pi^* 2 p_x^2 \approx \pi^* 2 p_y^2 .
\)
It contains no unpaired electron, therefore, it is diamagnetic in nature.

Hint

\(
\text { (c, d) The total number of electrons in } \mathrm{F}_2^{+} \text {and } \mathrm{O}_2^{-} \text {are same, i.e. } 17 \text { electrons. }
\)
\(
\mathrm{F}_2^{+}(17): \text { B.O. }=\frac{10-7}{2}=1.5
\)
\(
\mathrm{O}_2^{-}(17): \text { B.O. }=\frac{10-7}{2}=1.5
\)

Hint

(a) Ionic compounds are good conductors only in a molten state or aqueous solution since ions are not furnished in the solid state.
(b) In canonical structures there is a difference in the arrangement of electrons.

Hint

(a)

\(
\text { (i) } \rightarrow \text { (e); (ii) } \rightarrow \text { (a); (iii) } \rightarrow \text { (b); (iv) } \rightarrow \text { (c) }
\)

Hint

(b)

(i) \(\rightarrow\) c ; (ii) \(\rightarrow\) d; (iii) \(\rightarrow\) a ; (iv) \(\rightarrow\) b
\(
\begin{aligned}
& N O \rightarrow \text { B.O. }=\frac{1}{2}(10-5)=2.5 \\
& C O \rightarrow \text { B.O. }=\frac{1}{2}(10-4)=3 \\
& O_2^{-} \rightarrow \text { B.O. }=\frac{1}{2}(10-7)=1.5 \\
& O_2 \rightarrow \text { B.O. }=\frac{1}{2}(10-6)=2
\end{aligned}
\)

Hint

(c) Definition of hybridisation.
Hybridization is the process of intermixing of the orbitals of slightly different energies to redistribute their energies, which results in the formation of new set of orbitals of equivalent energies and shape.
Hybridization helps in deciding the structure of the molecules.
Total number of hybrid orbitals \(=\) No. of sigma bonds made by central atom + No. of lone pairs of electrons on central atom.
Molecules Bond pairs(bp) lone pairs(lp) Total no.of hybrid orbitals Hybridisation
(i) \(\mathrm{SF}_4=\) number of \(\mathrm{bp}(4)+\) number of \(\mathrm{lp}(1)\) \(=\mathrm{sp}^3 \mathrm{~d}\) hybridisation

(ii) \(I F_5=\) number of \(b p(5)+\) number of \(l p(1)\) \(=\mathrm{sp}^3 \mathrm{~d}^2\) bybridisation

(iii) \(\mathrm{NO}_2^{+}=\)number of \(b p(2)+\) number of \(\mathrm{lp}(0)\) \(=\mathrm{sp}\) hybridisation

(iv) \(\mathrm{NH}_4^{+}=\)number of \(b p(4)+\) number of \(\mathrm{lp}(0)\) \(=\mathrm{sp}^3\) hybridisation.

Hint

(a) Hydrogen bond – HF
Resonance \(-\mathrm{O}_3\)
Ionic solid – LiF
Covalent solid – C

Hint

(d)

(i) Tetrahedral shape-sp \({ }^3\) hybridisation
(ii) Trigonal shape- \(\mathrm{sp}^2\) hybridisation
(iii) Linear shape-sp hybridisation

Hint

(a) Sodium chloride \(\left(\mathrm{Na}^{+} \mathrm{CL}^{-}\right)\)is stable ionic compound because both \(\mathrm{Na}^{+}\)and \(\mathrm{CL}^{-}\)ions have complete octet in outermost shell.

Explanation:
Assertion and reason both are correct and reason is the correct explanation of assertion.
\(
\underset{(2,8,1)}{\mathrm{Na}}+\underset{(2,8,7)}{\mathrm{Cl}} \rightarrow \underset{(2,8, 8)}{\mathrm{NaCl}}
\)
Here both \(\mathrm{Na}^{+}\)and \(\mathrm{Cl}^{-}\)have complete octet hence \(\mathrm{NaCl}\) is stable.

Hint

(a)

\(\mathrm{H}_2 \mathrm{O}\) has two lone pairs while \(\mathrm{NH}_3\) has single lone pair, hence, \(\mathrm{H}_2 \mathrm{O}\) involves greater lone pair-bond pair repulsion.

Hint

\(
\text { (d) Bond energy of two }(-\mathrm{O}-\mathrm{H}) \text { bonds in } \mathrm{H}_2 \mathrm{O} \text { will be different. }
\)
Correct assertion The bond enthalpies of the two \(\mathrm{O}-\mathrm{H}\) bonds in \(\mathrm{H}-\mathrm{O}-\mathrm{H}\) are not equal.
Correct reason This is because the electronic environment around \(\mathrm{O}\) is not the same after the breakage of one \(\mathrm{O}-\mathrm{H}\) bond.

Hint

(a) In the formation of dioxygen from oxygen atoms, ten molecular orbitals will be formed.
\(
\mathrm{O}_2=\frac{\sigma 1 \mathrm{~s}^2}{1} \frac{\sigma^* 1 \mathrm{~s}^2}{2} \frac{\sigma 2 \mathrm{~s}^2}{3} \frac{\sigma^* 2 \mathrm{~s}^2}{4} \frac{\sigma_2 \mathrm{p}_{\mathrm{z}}^2}{5} \frac{\pi 2 \mathrm{p}_{\mathrm{x}}^2}{6} \frac{\pi 2 \mathrm{p}_{\mathrm{y}}^2}{7} \frac{\pi^* 2 \mathrm{p}_{\mathrm{x}}^1}{8} \frac{\pi^* 2 \mathrm{p}_{\mathrm{y}}^1}{9} \frac{\sigma^* 2 \mathrm{p}_{\mathrm{z}}^0}{10}
\)

Hint

(b) The number of nodal planes are as follows
\(
\begin{aligned}
\sigma * 1 \mathrm{~s} & \rightarrow 1 . \\
\sigma * 2 \mathrm{p}_z & \rightarrow 3 \\
\pi 2 \mathrm{p}_{\mathrm{x}} & \rightarrow 1 \\
\pi * 2 \mathrm{p}_{\mathrm{y}} & \rightarrow 2
\end{aligned}
\)
Hence option (b) is the correct answer

Hint

(b)
Bond order \(=\left(\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right) / 2\)
In \(\mathrm{O}_2^{+}\), number of electrons in bonding orbitals are 10 and in anti bonding are 5.
\(\sigma 1 \mathrm{~s}^2<\sigma * 1 \mathrm{~s}^2<\sigma 2 \mathrm{~s}^2<\sigma * 2 \mathrm{~s}^2<\sigma 2 \mathrm{p}_{\mathrm{z}}^2\left(\pi 2 \mathrm{p}_{\mathrm{x}}^2=\pi 2 \mathrm{p}_{\mathrm{x}}^2\right)<\left(\pi * 2 \mathrm{p}_{\mathrm{x}}^1=\pi * 2 \mathrm{p}_{\mathrm{x}}\right)<\sigma * 2 \mathrm{p}_{\mathrm{z}}\)
Bond order \(=(10-5) / 2=2.5\)
In \(\mathrm{N}_2{ }^{-}\), number of electrons in bonding orbitals are 10 and in anti- bonding are 5.
\(\sigma 1 \mathrm{~s}^2<\sigma * 1 \mathrm{~s}^2<\sigma 2 \mathrm{~s}^2<\sigma * 2 \mathrm{~s}^2<\left(\pi 2 \mathrm{p}_{\mathrm{x}}^2=\pi 2 \mathrm{p}_{\mathrm{x}}^2\right)<\sigma 2 \mathrm{p}_{\mathrm{z}}^2<\left(\pi * 2 \mathrm{p}_{\mathrm{x}}^1=\pi * 2 \mathrm{p}_{\mathrm{x}}\right)<\sigma *\)
\(2 \mathrm{p}_z\)
Bond order \(=(10-5) / 2=2.5\).
Bond order for
\(\mathrm{O}_2 \rightarrow 2\)
\(\mathrm{N}_2 \rightarrow 3\)
\(\mathrm{O}_2{ }^{-} \rightarrow 1.5\)
\(\mathrm{N}_2^{+} \rightarrow 2.5\)

Hint

(c) The first atomic orbitals on two atoms form two molecular orbitals designated as
\(\sigma 1 s\) and \(\sigma^* 1 s\)
Similarly, the \(2 \mathrm{~s}\) and \(2 \mathrm{p}\) atomic orbitals give rise to the eight molecular orbitals given below
Antibonding \(\operatorname{MOs} \sigma^* 2 s \sigma^* 2 p_z \pi^* 2 p_x \pi^* 2 p_y\)
Bonding \(\operatorname{Mos} \sigma 2 s \sigma 2 p_z \pi 2 p_x \pi 2 p_y\)
For molecules such as \(\mathrm{B}_2, \mathrm{C}_2, \mathrm{~N}_2\), etc. it is observed that the increasing order of energies of various molecular orbitals is –
Here, the important characteristic feature is that the energy of \(\sigma 2 p_z\) molecular orbital is higher than that of \(\pi 2 p_x\) and \(\pi 2 p_y\) molecular orbitals