NEET PYQs

Hint

(b) The compounds \(\mathrm{IBr}_{2}^{-}\)and \(\mathrm{XeF}_{2}\) are isostructural. Both have the same number of valence electrons. Both have linear geometry as they have 3 lone pairs and 2 bond pairs of electrons. Electron pair geometry is trigonal bipyramidal and molecular geometry is linear.

Note: Isoelectronic compounds have the same number of electrons. This includes core electrons and valence electrons.

Hint

(c) \(\mathrm{BCl}_{3}\)-Trigonal planar, \(s p^{2}\)-hybridised, \(120^{\circ}\) angle.

Hint

(b) Molecular orbital electronic configurations and bond order values are
\(\mathrm{O}_{2}: \sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{z}^{2}, \pi 2 p_{x}^{2}=\pi 2 p_{y}^{2}\)
\(\pi^{*} 2 p_{x}^{1}=\pi^{*} 2 p_{y}^{1}\)
B.O. \(=\frac{1}{2}\left(N_{b}-N_{a}\right)=\frac{1}{2}(10-6)=2\)
\(\mathrm{NO}^{+}: \sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{z}^{2}, \pi 2 p_{x}^{2}=\pi 2 p_{y}^{2}\)
B. O. \(=\frac{1}{2}(10-4)=3\)
\(\mathrm{CN}^{-}: \sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \pi 2 p_{x}^{2}=\pi 2 p_{y}^{2}, \sigma 2 p_{z}^{2}\)
B.O. \(=\frac{1}{2}(10-4)=3\)
\(\mathrm{CO}: \sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \pi 2 p_{x}{ }^{2}=\pi 2 p_{y}{ }^{2}, \sigma 2 p_{z}{ }^{2}\)
B. O. \(=\frac{1}{2}(10-4)=3\)
\(N_{2}: \sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \pi 2 p_{x}^{2}=\pi 2 p_{y}^{2}, \sigma 2 p_{z}^{2}\)
B. O. \(=\frac{1}{2}(10-4)=3\)
\(\mathrm{O}_{2}^{-}=\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{z}^{2}, \pi 2 p_{x}^{2}\)
\(=\pi 2 p_{y}{ }^{2}, \pi^{*} 2 p_{x}{ }^{2}=\pi^{*} 2 p_{y}{ }^{1}\)
B.O. \(=\frac{1}{2}(10-7)=1.5\)
\(\mathrm{NO}: \sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{z}^{2}, \pi 2 p_{x}^{2}=\pi 2 p_{y}^{2}\),
\(\pi^{*} 2 p_{x}^{1}\)
B.O. \(=\frac{1}{2}(10-5)=2.5\)

Hint

(c) \(\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{HCN}\) and conc. \(\mathrm{CH}_{3} \mathrm{COOH}\) form intermolecular hydrogen bonding while cellulose has intramolecular hydrogen bonding.

Hint

\(\text { (c) } X=\frac{1}{2}(V E+M A-c+a)\)
For \(\mathrm{NO}_{2}^{+}, X=\frac{1}{2}(5+0-1)=2\) i.e., \(s p\) hybridisation
For \(\mathrm{NO}_{3}^{-}, X=\frac{1}{2}(5+0+1)=3\) i.e., \(s p^{2}\) hybridisation
For \(\mathrm{NH}_{4}^{+}, X=\frac{1}{2}(5+4-1)=4\) i.e., \(s p^{3}\) hybridisation

Hint

(d) Number of electrons
\(
\begin{aligned}
&\mathrm{CO}_{3}^{2-}=6+2+24=32 \\
&\mathrm{SO}_{3}^{2-}=16+2+24=42 \\
&\mathrm{CIO}_{3}^{-}=17+24+1=42 \\
&\mathrm{CO}_{3}^{2-}=6+24+2=32 \\
&\mathrm{NO}_{3}^{2-}=7+2+24=33
\end{aligned}
\)
Hence, \(\mathrm{CIO}_{3}^{-}\)and \(\mathrm{SO}_{3}^{2-}\) are isoelectronic and are pyramidal in shape.

Hint

(a) The correct option is ‘a’, Octahedral, \(\mathrm{sp}^{3} \mathrm{~d}^{2}\)
\(\mathrm{XeF}_{4}\) is \(\mathrm{AB}_{4} \mathrm{~L}_{2}\) type in which the xenon is bonded with four fluorine atoms with 4 sigma bonds and xenon has two lone pairs (4 sigma bonds and 2 lone pairs).
So, the hybridisation is \(\mathrm{sp}^{3} \mathrm{~d}^{2}\).
Geometry is octahedral and shape is square planar.

Hint

(c) 

Hint

(d)

Hint

(c) According to VSEPR theory, the repulsive forces between lone pair and lone pair are greater than between lone pair and bond pair which are further greater than bond pair and bond pair

Hint

(c) In diamond and silicon carbide, central atom is \(s p^{3}\) hybridised, and hence, both are isostructural. \(\mathrm{NH}_{3}\) and \(\mathrm{PH}_{3}\), both are pyramidal and central atom in both cases is
\(s p^{3}\) hybridised.
\(\mathrm{SiCl}_{4}\) and \(\mathrm{PCl}_{4}^{+}\), both are tetrahedral and central atom in both cases is \(s p^{3}\) hybridised.

In \(\mathrm{XeF}_{4}\), Xe is \(s p^{3} d^{2}\) hybridised and structure is square planar while in \(\mathrm{XeO}_{4}\), Xe is \(s p^{3}\) hybridised and structure is tetrahedral.

Hint

(d) \(
\begin{array}{r}
\mathrm{O}_2(16): K K \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_{\mathrm{z}}^2 \pi 2 p_x^2=\pi 2 p_y^2 \\
\pi^* 2 p_x{ }^1=\pi^* 2 p_y{ }^1
\end{array}
\)
\(
\text { Bond order }=\frac{1}{2}(8-4)=2
\)
\(
\begin{array}{l}
\mathrm{O}_2^{-}(17): K K \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_{\mathrm{z}}^2 \pi 2 p_x^2=\pi 2 p_y^2 \\
\pi^* 2 p_x^2=\pi^* 2 p_y{ }^1
\end{array}
\)
\(
\text { Bond order }=\frac{1}{2}(8-5)=1.5
\)
\(
\mathrm{O}_2^{+}(15): K K \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi^* 2 p_x{ }^1
\)
\(
\text { Bond order }=\frac{1}{2}(8-3)=2.5
\)

Bond order \(\propto\) stability. The decreasing order of stability is
\(
\mathrm{O}_{2}^{+}>\mathrm{O}_{2}>\mathrm{O}_{2}^{-}>\mathrm{O}_{2}{ }^{2-}
\)

Hint

(b) \(
\begin{array}{|l|c|c|c|}
\hline \text { Species } & \text { Hybridisation } & \text { Shape } & \begin{array}{c}
\text { No. of } \\
\boldsymbol{e}^{-} \mathbf{s}
\end{array} \\
\hline \mathrm{SO}_{3}{ }^{2-} & s p^{3} & \text { Pyramidal } & 42 \\
\hline \mathrm{ClO}_{3}{ }^{-} & s p^{3} & \text { Pyramidal } & 42 \\
\hline \mathrm{CO}_{3}{ }^{2-} & s p^{2} & \begin{array}{c}
\text { Triangular } \\
\text { planar }
\end{array} & 32 \\
\hline \mathrm{NO}_{3}^{-} & s p^{2} & \begin{array}{c}
\text { Triangular } \\
\text { planar }
\end{array} & 32 \\
\hline
\end{array}
\)

Hint

(b) \(\quad \mathrm{O}_{2}^{-} \quad \mathrm{O}_{2} \quad \mathrm{O}_{2}^{+} \quad \mathrm{O}_{2}{ }^{2+}\)
B.O.\(\quad 1.5 \quad 2.0 \quad 2.5 \quad 3.0\)

Hint

(d) \(\quad \mathrm{O}_{2}^{-} \lt \quad \mathrm{O}_{2} \lt \quad \mathrm{O}_{2}^{+}\)
B.O.\(\quad 1.5 \quad \quad 2.0 \quad \quad 2.5\)

Hint

(a)
\(
\begin{array}{|l|c|l|c|c|}
\hline \text { Species } & \mathrm{NO}_{3}^{-} & \mathrm{NO}_{2} & \mathrm{NO}_{2}^{-} & \mathrm{NO}_{2}^{+} \\
\hline \text { Hybridisation } & s p^{2} & s p^{2} & s p^{2} & s p \text { (linear) } \\
\hline \text { Bond angle } & 120^{\circ} & 134^{\circ} & 115^{\circ} & 180^{\circ} \\
\hline
\end{array}
\)
So, \(\mathrm{NO}_{2}^{+}\)has maximum bond angle.

Hint

(c) In \(\mathrm{NH}_{3}, \mathrm{H}\) is less electronegative than \(\mathrm{N}\) and hence dipole moment of each \(\mathrm{N}-\mathrm{H}\) bond is towards \(\mathrm{N}\) and create high net dipole moment whereas in \(\mathrm{NF}_{3}, \mathrm{~F}\) is more electronegative than \(\mathrm{N}\), the dipole moment of each \(\mathrm{N}-\mathrm{F}\) bond is opposite to that of lone pair, hence reducing the net dipole moment.

Hint

(b) 

Hint

(a) Boron hydrides are electron deficient compound

Hint

(d) 

Hint

(d) \(\mathrm{SF}_{4}\) has \(s p^{3} d\)-hybridisation and see-saw shape with \((4 b p+1 l p)\)

Hint

(d) \(\mathrm{O}_{2}^{-}\)(17) superoxide has one unpaired electron.

\(
\begin{aligned}
\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}^{2} \pi 2 p_{x}^{2}=& \pi 2 p_{y}^{2} \\
& \pi^{*} 2 p_{x}^{2}=\pi^{*} 2 p_{y}^{1}
\end{aligned}
\)

Hint

(a) \(\mathrm{HCl}\) is polar \((\mu \neq 0)\) and \(\mathrm{He}\) is non-polar \((\mu=0)\) gives dipole-induced dipole interaction

Hint

(a) \(\mathrm{CO}=6+8=14\) electrons, \(\mathrm{NO}^{+}=7+8-1=14\) electrons
Both have electronic configuration : \(\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}^{2} \pi 2 p_{x}^{2} \pi 2 p_{y}^{2}\)
So both have bond order \(=\frac{10-4}{2}=3\)

Hint

(c) 

Hint

\({ (b) } \mathrm{NF}_{3} \text { and } \mathrm{H}_{2} \mathrm{O} \text { are } s p^{3} \text {-hybridisation. }\)

Hint

(c) In \(C_{2}-C_{2}^{+}\)electron is removed from bonding molecular orbital so bond order decreases. In \(\mathrm{NO} \longrightarrow \mathrm{NO}^{+}\), electron is removed from anti bonding molecular orbital so bond order increases and nature changes from paramagnetic to diamagnetic.

Molecular orbital configuration of \(\mathrm{O}_{2} \Rightarrow \sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}{ }^{2} \pi 2 p_{x}{ }^{2} \pi 2 p_{y}{ }^{2} \pi^{*} 2 p_{x}{ }^{1}\) \(\pi^{*} 2 p_{y}{ }^{1}\)
\(\Rightarrow\) Paramagnetic
Bond order \(=\frac{10-6}{2}=2\)
\(\mathrm{O}_{2}^{+} \Rightarrow \sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}{ }^{2} \pi 2 p_{x}{ }^{2} \pi 2 p_{y}{ }^{2} \pi^{*} 2 p_{x}{ }^{1}\)
\(\Rightarrow\) Paramagnetic
Bond order \(=\frac{10-5}{2}=2.5\)
\(\mathrm{C}_{2} \Rightarrow \sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \pi 2 p_{x}^{2} \pi 2 p_{y}^{2}\)
\(\Rightarrow\) Diamagnetic
Bond order \(=\frac{8-4}{2}=2\)
\(\mathrm{C}_{2}^{+} \Rightarrow \sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \pi 2 p_{x}{ }^{2} \pi 2 p_{y}{ }^{1}\)
\(\Rightarrow\) Paramagnetic
Bond order \(=\frac{7-4}{2}=1.5\)
\(\mathrm{NO} \Rightarrow \sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}{ }^{2} \pi 2 p_{x}{ }^{2} \pi 2 p_{y}{ }^{2} \pi^{*} 2 p_{x}{ }^{1}\)
\(\Rightarrow\) Paramagnetic
Bond order \(=\frac{10-5}{2}=2.5\)
\(\mathrm{NO}^{+} \Rightarrow \sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}{ }^{2} \pi 2 p_{x}{ }^{2} \pi 2 p_{y}{ }^{2}\)
\(\Rightarrow\) Diamagnetic.
Bond order \(=\frac{10-4}{2}=3\)
\(\mathrm{N}_{2} \Rightarrow \sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \pi 2 p_{x}^{2} \pi 2 p_{y}^{2} \sigma 2 p_{z}^{2}\)
\(\Rightarrow\) Paramagnetic
Bond order \(=\frac{10-4}{2}=3\)
\(\mathrm{N}_{2}^{+} \Rightarrow \sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \pi 2 p_{x}{ }^{2} \pi 2 p_{y}{ }^{2} \sigma 2 p_{z}{ }^{1}\)
\(\Rightarrow\) Paramagnetic
Bond order \(=\frac{9-4}{2}=2.5\).

Hint

(d)

\(\begin{aligned}
\mathrm{BCl}_{3} \Rightarrow s p^{2}, & \text { trigonal planar } \\
\mathrm{BrCl}_{3} \Rightarrow s p^{3} d, & \text { T-shaped } \\
\mathrm{NH}_{3} \Rightarrow s p^{3}, & \text { pyramidal } \\
\mathrm{NO}_{3}^{-} \Rightarrow s p^{2}, & \text { trigonal planar } \\
\mathrm{NF}_{3} \Rightarrow s p^{3}, & \text { pyramidal } \\
\mathrm{BF}_{3} \Rightarrow s p^{2}, & \text { trigonal planar } \\
\mathrm{BF}_{4}^{-} \Rightarrow s p^{3}, & \text { tetrahedral } \\
\mathrm{NH}_{4}^{+} \Rightarrow s p^{3}, & \text { tetrahedral }
\end{aligned}\)

Hint

(b) Configuration of \(\mathrm{O}_{2}\)
\(\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}^{2} \pi 2 p_{x}{ }^{2} \pi 2 p_{y}{ }^{2} \pi^{*} 2 p_{x}{ }^{1} \pi^{*} 2 p_{y}{ }^{1}\)
                       No. of \(e^{-}\)in                   No. of \(e^{-}\)in
Bond order \(=\frac{\text { bonding M.O. }  – \quad \text { antibonding M.O. }}{2}\)
Bond order of \(\mathrm{O}_{2}^{+}=\frac{10-5}{2}=2.5\)
Bond order of \(\mathrm{O}_{2}^{-}=\frac{10-7}{2}=1.5\)
Bond order of \(\mathrm{O}_{2}^{2-}=\frac{10-8}{2}=1.0\)
Bond order of \(\mathrm{O}_{2}=\frac{10-6}{2}=2\)

Hint

(d) 

Hint

\(
\begin{aligned}
&\text { (a) } \mathrm{O}_{2}^{2-} \rightarrow 1 \quad \mathrm{~B}_{2} \rightarrow 1\\
&\mathrm{O}_{2}^{+} \rightarrow 2.5 \quad \mathrm{NO}^{+} \rightarrow 3\\
&\mathrm{NO} \rightarrow 2.5 \quad \mathrm{CO} \rightarrow 3\\
&\mathrm{N}_{2} \rightarrow 3 \quad \mathrm{O}_{2} \rightarrow 2.0
\end{aligned}
\)

Hint

(a) Electronic configuration of \(\mathrm{O}_{2}\)
\(\sigma(1 s)^{2} \sigma^{*}(1 s)^{2} \sigma(2 s)^{2} \sigma^{*}(2 s)^{2} \sigma\left(2 p_{z}\right)^{2} \pi\left(2 p_{x}\right)^{2} \pi\left(2 p_{y}\right)^{2}\) \(\pi^{*}\left(2 p_{x}\right)^{1} \pi^{*}\left(2 p_{y}\right)^{1}\)

Thus the incoming electron will enter in \(\pi^{*} 2 p_{x}\) to form \(\mathrm{O}_{2}^{-}\).

Hint

(d)

\(
\begin{array}{|c|c|}
\hline \text { Diatomic species } & \text { Bond order } \\
\hline \mathrm{NO} & 2.5 \\
\mathrm{O}_{2}{ }^{-} & 1.5 \\
\mathrm{C}_{2}{ }^{2-} & 3.0 \\
\mathrm{He}_{2}{ }^{+} & 0.5 \\
\hline
\end{array}
\)

\(
\text { Thus increasing order : } \mathrm{He}_{2}^{+}<\mathrm{O}_{2}^{-}<\mathrm{NO}<\mathrm{C}_{2}^{2-}
\)

Hint

(a) Electronic configuration \(\mathrm{O}_{2}: K K(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\sigma 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}\left(\pi 2 p_{z}\right)^{2}\left(\pi^{*} 2 p_{y}\right)^{1}\)
\(\left(\pi^{*} 2 p_{z}\right)^{1}\)
Bond order \(=\frac{1}{2}(8-4)=2\)
\(\mathrm{O}_{2}^{+}:\)Bond order \(=\frac{1}{2}(8-3)=2 \frac{1}{2}\)
\(\mathrm{O}_{2}^{-}:\)Bond order \(=\frac{1}{2}(8-5)=1 \frac{1}{2}\)
\(\mathrm{O}_{2}^{2-}:\) Bond order \(=\frac{1}{2}(8-6)=1\)
As bond order increases, bond length decreases.

Hint

(a)

\(
\begin{array}{lc}
\text { Ions } & \text { Hybridisation } \\
\mathrm{NO}_{2}^{-} & s p^{2} \\
\mathrm{NO}_{3}^{-} & s p^{2} \\
\mathrm{NH}_{2}^{-} & s p^{3} \\
\mathrm{NH}_{4}^{+} & s p^{3} \\
\mathrm{SCN}^{-} & s p
\end{array}
\)

Hint

(a) Increasing order of bond length is
\(
\underset{107 \mathrm{pm}}{\mathrm{C}-\mathrm{H}}<\underset{134 \mathrm{pm}}{\mathrm{C}} {\mathrm{=}}{\mathrm{C}}<\underset{141 \mathrm{pm}}{\mathrm{C}-\mathrm{O}}<\underset{154 \mathrm{pm}}{\mathrm{C}-\mathrm{C}}
\)

Hint

Hint

(c) \(\mathrm{O}_{2}^{+}\)and \(\mathrm{O}_{2}\) are paramagnetic in nature as they contain one and two unpaired electrons resnectively

Hint

(b) The hybridisation of the central atom can be calculated as
\(
\mathrm{H}=\frac{1}{2}\left[\begin{array}{c}
\left.\begin{array}{l}
\text { No. of electrons } \\
\text { in valence shell } \\
\text { of atom }
\end{array}\right)+\left(\begin{array}{l}
\text { No. of monovalent } \\
\text { atoms around } \\
\text { central atom }
\end{array}\right) \\
-\left(\begin{array}{l}
\text { Charge on } \\
\text { cation }
\end{array}\right)+\left(\begin{array}{l}
\text { Charge on } \\
\text { anion }
\end{array}\right)
\end{array}\right]
\)
\(\begin{aligned}
\therefore \quad \text { For } \mathrm{BF}_{3}, \mathrm{H} &=\frac{1}{2}[(3)+(3)-(0)+(0)] \\
&=3 \Rightarrow s p^{2} \text { hybridisation. } \\
\text { For } \mathrm{NO}_{2}^{-}, \mathrm{H} &=\frac{1}{2}[(5)+(0)-(0)+(1)] \\
&=3 \Rightarrow s p^{2} \text { hybridisation. }
\end{aligned}
\)

Hint

(b) \(\mathrm{Be}_{2}\) does not exist.
\(\mathrm{Be}_{2}\) has an electronic configuration of :
\(
\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2}
\)
\(\therefore \quad\) Bond order \(=\frac{4-4}{2}=0\)
Thus, \(\mathrm{Be}_{2}\) does not exist.

Hint

(c) Hybridisation of the central atom can be calculated as:
\(
\mathrm{H}=\frac{1}{2}\left[\begin{array}{c}
\left(\begin{array}{l}
\text { No. of valence } \\
\text { electrons in the } \\
\text { central atom }
\end{array}\right)+\left(\begin{array}{l}
\text { No. of monovalent } \\
\text { atoms around } \\
\text { central atom }
\end{array}\right) \\
-\left(\begin{array}{l}
\text { Charge on } \\
\text { cation }
\end{array}\right)+\left(\begin{array}{l}
\text { Charge on } \\
\text { anion }
\end{array}\right)
\end{array}\right]
\)
Applying this formula we find that all the given species except \(\left[\mathrm{SbCl}_{5}\right]^{2-}\) have central atom with \(s p^{3} d\) (corresponding to \(\mathrm{H}=5\) ) hybridization. In \(\left[\mathrm{SbCl}_{5}\right]^{2-}\), Sb is \(s p^{3} d^{2}\) hybridized.

Hint

(b) For neutral molecules,
No. of electron pairs \(=\) No. of atoms bonded to it \(+\) \(1 / 2\) [Gp. no. of central atom – Valency of central atom]
\(\therefore \quad\) For \(\mathrm{CH}_{4}\), no. of \(e^{-}\)pairs \(=4+\frac{1}{2}[4-4]\) \(=4\left(s p^{3}\right.\) hybridisation)

For \(\mathrm{SF}_{4}\), no. of \(e^{-}\)pairs \(=4+\frac{1}{2}[6-4]\) \(=5\left(s p^{3} d\right.\) hybridisation \()\)
For ions,
No. of electron pairs \(=\) No. of atoms bonded to it \(+\) \(1 / 2\) [Gp. no. of central atom – Valency of central atom \(\pm\) No. of electrons]
\(
\begin{aligned}
&\therefore \quad \text { For } \mathrm{BF}_{4}^{-}, \text {no. of } e^{-} \text {pairs }=4+\frac{1}{2}[3-4+1] \\
&=4\left(s p^{3} \text { hybridisation }\right) \\
&\text { For } \mathrm{NH}_{4}^{+}, \text {no. of } e^{-} \text {pairs }=4+\frac{1}{2}[5-4-1] \\
&=4\left(s p^{3} \text { hybridisation }\right)
\end{aligned}
\)

Hint

(a) No. of electron pairs at the central atom \(=\) No. of atoms bonded to it \(+1 / 2\) [Group number of central atom – Valency of the central atom \(\pm\) no. of electrons]
No. of electron pairs at the central atom in \(\mathrm{NO}_{3}^{-}\) \(=3+\frac{1}{2}[5-6+1]=3 \quad\left(s p^{2}\right.\) hybridisation \()\).
No. of electron pairs at the central atom in \(\mathrm{H}_{3} \mathrm{O}^{+}=3+\frac{1}{2}[6-3-1]=4 \quad\left(s p^{3}\right.\) hybridisation \()\).

Hint

(d) Methanol can undergo intermolecular association through \(\mathrm{H}\)-bonding as the \(-\mathrm{OH}\) group in alcohols is highly polarised.

As a result, in order to convert liquid \(\mathrm{CH}_{3} \mathrm{OH}\) to gaseous state, the strong hydrogen bonds must be broken.

Hint

(a) According to MOT, the molecular orbital electronic configuration of
\(\mathrm{N}_{2}:(\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}\left(\sigma 2 p_{z}\right)^{2}\)
\(
\begin{aligned}
\therefore \text { B.O }=\frac{10-4}{2}=3 \\
\mathrm{~N}_{2}^{-}:(\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi p_{y}\right)^{2} \\
\quad\left(\sigma 2 p_{z}\right)^{2}\left(\pi^{*} 2 p_{x}\right)^{1}
\end{aligned}
\)
\(
\begin{aligned}
&\therefore \quad \text { B.O }=\frac{10-5}{2}=2.5 \\
&\mathrm{~N}_{2}^{2-}:(\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2} \\
&\therefore \quad\left(\sigma 2 p_{z}\right)^{2}\left(\pi^{*} 2 p_{x}\right)^{1}\left(\pi^{*} 2 p_{y}\right)^{1} \\
&\therefore \quad \text { B.O. }=\frac{10-6}{2}=2 .
\end{aligned}
\)
Hence the order : \(\mathrm{N}_{2}^{2-}<\mathrm{N}_{2}^{-}<\mathrm{N}_{2}\)

Hint

(c) \(\mathrm{BF}_{3} \rightarrow s p^{2}, \mathrm{NO}_{2}^{-} \rightarrow s p^{2}, \mathrm{NH}_{2}^{-} \rightarrow s p^{3}\),
\(
\mathrm{H}_{2} \mathrm{O} \rightarrow s p^{3}
\)

Hint

The correct order of increasing bond angles in the following triatomic species is
\(
\mathrm{NO}_{2}^{-}<\mathrm{NO}_{2}>\mathrm{NO}_{2}^{+}
\)

Hint

(c) Hybridisation of \(\mathrm{Br}\) in \(\mathrm{BrO}_{3}^{-}\): \(H=1 / 2(7+0-0+1)=4\) i.e.sp \(p^{3}\) hybridisation Hybridisation of \(\mathrm{Xe}\) in \(\mathrm{XeO}_{3}\) : \(H=\frac{1}{2}(8+0-0+0)=4\) i.e. \(s p^{3}\) hybridisation In both \(\mathrm{BrO}_{3}^{-}\)and \(\mathrm{XeO}_{3}\), the central atom is \(s p^{3}\) hybridised and contains one lone pair of electrons, hence in both the cases, the structure is trigonal pyramidal.

Hint

More single bond character in resonance hybrid, more is the bond length. Hence the increasing bond length is
\(
\mathrm{CO}<\mathrm{CO}_{2}<\mathrm{CO}_{3}{ }^{2-}
\)

Hint

(d) For \(A B_{5}\) molecules, there are three possible geometries i.e. planar pentagonal, square pyramidal, and trigonal bipyramidal

Out of these three geometries, it is only trigonal pyramidal shape in which bond pair-bond pair repulsions are minimum and hence this geometry is the most probable geometry of \(A B_{5}\) molecule.

Hint

Hint

(b) \(\mathrm{SiCl}_{4}, \mathrm{NH}_{4}^{+}, \mathrm{SO}_{4}^{2-}\) and \(\mathrm{PO}_{4}^{3-}\) ions are the examples of molecules/ions which are of \(A B_{4}\) type and have tetrahedral structure. \(\mathrm{SCl}_{4}\) is \(\mathrm{AB}_{4}\) (lone pair) types species. Although the arrangement of five \(s p^{3} d\) hybrid orbitals in space is trigonal bipyramidal, due to the presence of one lone pair of electron in the basal hybrid orbital, the shape of \(A B_{4}\) (lone pair) species gets distorted and becomes distorted tetrahedral or see-saw.

Hint

Hint

(d) Bond lengths of \(\mathrm{O}-\mathrm{O}\) in \(\mathrm{O}_{2}\) is 1.21 Å, in \(\mathrm{H}_{2} \mathrm{O}_{2}\) is 1.48 Å and in \(\mathrm{O}_{3}\) is 1.28 Å. Therefore, correct order of the \(\mathrm{O}-\mathrm{O}\) bond length is \(\mathrm{H}_{2} \mathrm{O}_{2}>\mathrm{O}_{3}>\mathrm{O}_{2}\).

Hint

(c) Hydrogen bonding in \(\mathrm{H}_{2} \mathrm{O}>\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)
\(>\mathrm{CH}_{3} \mathrm{OH}\)
Hence, \(\mathrm{H}_{2} \mathrm{O}\) has maximum surface tension.

Hint

(a) \(\mathrm{SiF}_{4}\) has symmetrical tetrahedral shape which is due to \(s p^{3}\) hybridisation of the central silicon atom in its excited state configuration. \(\mathrm{SF}_{4}\) has distorted tetrahedral or sea-saw geometry which arises due to \(s p^{3} d\) hybridisation of central sulphur atom and due to the presence of one lone pair of electrons in one of the equatorial hybrid orbital.

Hint

(a) In octahedral molecule six hybrid orbitals directed towards the corners of a regular octahedron with a bond angle of \(90^{\circ}\).

According to this geometry, the number of \(X-M-X\) bonds at \(180^{\circ}\) must be three.

Hint

(d) The overall value of the dipole moment of a polar molecule depends on its geometry and shape, i.e. vectorial addition of dipole moment of the constituent bonds. Water has angular structure with bond angle \(105^{\circ}\) as it has dipole moment. However \(\mathrm{BeF}_{2}\) is a linear molecule since dipole moment summation of all the bonds present in the molecule cancel each other.

Hint

Bent T-shaped geometry in which both lone pairs occupy the equatorial positions of the trigonal bipyramid. Here \((l p-l p)\) repulsion \(=0,(l p-b p)\) repulsion \(=4\) and \((b p-b p)\) repulsion \(=2\).

Hint

(b) We know C – C \(=347 \mathrm{~kJ} / \mathrm{mol}\) \(\mathrm{C}=\mathrm{C}=619 \mathrm{~kJ} / \mathrm{mol}\)

Hint

Hint

Hint

(c) In \(\mathrm{CO}\), the number of electrons \(=6+8=14[\mathrm{Z}\) of \(\mathrm{C}=6\) and \(\mathrm{O}=8]\)
Electronic configuration of molecular orbital of \(\mathrm{CO}\) : \((\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}\left(\sigma 2 p_{z}\right)^{2}\) \(\mathrm{CN}^{-}\)have also get \((6+7+1) 14\) electrons and the configuration is similar to that of \(\mathrm{CO}\). \(\mathrm{CN}^{-}\)and \(\mathrm{CO}\) are isoelectronic.

Hint

(a) Compounds having same shape with same hybridisation are known as isostructural.
\(\mathrm{XeF}_{2}, \mathrm{IF}_{2}^{-} \rightarrow\) both are \(s p^{3} d\) hybridised linear molecules.

Hint

(b) Bond angle is maximum in \(\mathrm{NH}_{4}^{+}\)tetrahedral molecule with bond angle \(109^{\circ}\).

Hint

(b) For strong \(\pi\)-bonding, \(p \pi-p \pi\) bonding should be strong. In case of \(\mathrm{P}\), due to larger size as compared to \(\mathrm{N}\)-atom, \(p \pi-p \pi\) bonding is not so strong.

Hint

(a) In X-H–Y, ‘ \(\mathrm{H}\) ‘ is directly bonded to ‘ \(\mathrm{X}\) ‘ and there is a hydrogen bond between ‘ \(\mathrm{H}\) ‘ and ‘ \(\mathrm{Y}\) ‘. As ‘ \(\mathrm{X}\) ‘ and ” \(\mathrm{Y}\) ‘ are electronegative elements that pull electron density towards themselves. As ‘ \(\mathrm{X}\) ‘ is directly bonded, it pulls more electron density. So electron density on \(\mathrm{X}\) increases on ‘ \(\mathrm{H}\) ‘ electron density decreases.

Hint

(b) In \(\mathrm{PO}_{4}^{3-}, \mathrm{P}\) atom has vacant \(d\)-orbitals, thus it can form \(p \pi-d \pi\) bond. ‘ \(\mathrm{N}\) ‘ and ‘ \(\mathrm{C}\) ‘ have no vacant ‘ \(d\) ‘ orbital in their valence shell, so they cannot form such bond.

Hint

(a) \(\mathrm{N}_{2}(14) \rightarrow(\sigma 1 s)^{2},\left(\sigma^{*} 1 s\right)^{2},(\sigma 2 s)^{2},\left(\sigma^{*} 2 s\right)^{2}\)
\(
\left(\sigma^{*} 2 s\right)^{2},\left(\pi 2 p_{x}\right)^{2},\left(\pi 2 p_{y}\right)^{2},\left(\sigma 2 p_{z}\right)^{2}
\)

In \(\mathrm{N}_{2}\), bond order \(=\frac{N_{b}-N_{a}}{2}=\frac{10-4}{2}=3\)
In \(\mathrm{N}_{2}^{+}\), bond order \(=\frac{9-4}{2}=2 \cdot 5\)
As the bond order in \(\mathrm{N}_{2}\) is more than \(\mathrm{N}_{2}^{+}\)so the dissociation energy of \(\mathrm{N}_{2}\) is higher than \(\mathrm{N}_{2}^{+}\).

Hint

(c) In ‘ \(\mathrm{CO}\) ‘ (14 electrons), there is no unpaired electron in its molecular orbital. Therefore this does not exhibit paramagnetism.

Hint

(d) \(\mathrm{O}_{2}^{2-}(18) \rightarrow(\sigma 1 s)^{2},(\sigma * 1 s)^{2}(\sigma 2 s)^{2},\left(\sigma^{*} 2 s\right)^{2}\) \(\left(\sigma 2 p_{x}\right)^{2},\left(\pi 2 p_{y}\right)^{2},\left(\pi 2 p_{z}\right)^{2},\left(\pi * 2 p_{y}\right)^{2},\left(\pi^{*} 2 p_{z}\right)^{2}\) \(\rightarrow\) represents antibonding molecular orbitals.

Thus the no. of antibonding electrons in \(\mathrm{O}_{2}^{2-}\) ion is \(=8(4\) pairs \()\)

Hint

(a) The total charge \(=-3\)
So the average formal charge on each ‘ \(\mathrm{O}\) ‘ atom is \(-3 / 4=-0.75\)

Again total no. of electrons in the valence shell of \(\mathrm{PO}_{4}^{3-}\) ion \(=5+8=13\)

No. of electrons involved in bond formation in \(\mathrm{PO}_{4}^{3-}\) ion \(=13-3=10\)
No. of bonds in \(\mathrm{PO}_{4}^{3-}=\frac{10}{2}=5\)
\(\Rightarrow\) Average \(\mathrm{P}-\mathrm{O}\) bond order \(=\frac{5}{4}=1.25\)

Hint

(b) Diamagnetism is caused due to the absence of unpaired electrons. But in \(\mathrm{N}^{2+}\), there is unpaired electron. So, it is paramagnetic.

Hint

(d) In \(\mathrm{O}_{2}\) bond, the order is 2 and in \(\mathrm{O}_{2}^{-}\)bond, the order is \(1.5\).

Hint

(b) The \(\mathrm{C}-\mathrm{C}\) bond length \(=1.54 Å, \mathrm{C}=\mathrm{C}\) bond length \(=1.34 Å\) and \(\mathrm{C} \equiv \mathrm{C}\) bond length \(=1.20 Å\). Since propyne has a triple bond, therefore it has minimum bond length.

Hint

(d) \(\mathrm{BeCl}_{2}\) and \(\mathrm{C}_{2} \mathrm{H}_{2}\) have \(s p\)-hybridisation and \(\mathrm{C}_{2} \mathrm{H}_{6}\) has \(s p^{3}\)-hybridisation.

Hint

(d) Paramagnetism is caused by the presence of atoms, ions or molecules with unpaired electrons, 

\(
\text { i.e }: \dot{\mathrm{N}}=\ddot{\mathrm{O}} \text { : }
\)

Hint

(b) Bond length of \(\mathrm{O}-\mathrm{O}\) in \(\mathrm{O}_{2}\) is \(1.21 Å\) \((\mathrm{O}=\mathrm{O})\); in \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(1.48 Å(\mathrm{HO}-\mathrm{HO})\) and in \(\mathrm{O}_{3}\) is
\(
1.28 Å\left(\begin{array}{c}
\mathrm{O}=\mathrm{O} \\
\downarrow \\
\mathrm{O}
\end{array}\right) \text {. }
\)

Hint

(b) Number of electrons in bonding orbitals \(N_{b}=10\) and number of electrons in antibonding orbitals \(N_{a}=4\).

Therefore bond order \(=1 / 2\left(N_{b}-N_{a}\right)=1 / 2(10-4)=3\)

Hint

(c) The bond order of \(\mathrm{O}_{2}^{+}=2.5, \mathrm{O}_{2}^{2-}=1\), \(\mathrm{O}_{2}{ }^{-}=1.5\) and that of \(\mathrm{O}_{2}=2\).

Hint

(a) The structure of \(\mathrm{CS}_{2}\) is linear and therefore it does not have permanent dipole moment. It is represented as \(\mathrm{S}=\mathrm{C}=\mathrm{S}\)

Hint

(b) Smaller the atom, stronger is the bond, and greater the bond dissociation energy. Therefore the bond \(C-D\) has the greatest energy or smallest atoms.

Hint

(b) For compounds containing ions of same charge, lattice energy increases as the size of ions decreases. Thus, NaF has highest lattice energy.

Hint

(b) The strength of interaction follow the order van der Waals’ < hydrogen-bond \(<\) dipole-dipole \(<\) covalent. It is so because bond length of H-bond is larger than that of a covalent bond.

And also covalent bond is strongest because, the greater the extent of overlapping, the stronger is the bond formed.

Hint

(c) There is a triple bond in ethyne molecule \((\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H})\) and due to this triple bond, carboncarbon bond distance is shortest in ethyne.

Hint

(b) \(\mathrm{CO}_{2}\) molecule is \(s p\)-hybridised and thus it is linear, while \(\mathrm{CO}_{3}{ }^{2-}\) is planar ( \(s p^{2}\)-hybridised), \(\mathrm{SO}_{2}\) is an angular molecule with \(s p^{2}\) hybridisation \(\mathrm{SO}_{4}^{2-}\) is tetrahedral ( \(s p^{3}\)-hybridised).

Hint

(c) HF shows strongest H-bonds because fluorine is most electronegative.

Hint

(b) As all \(\mathrm{C}-\mathrm{Cl}\) bonds are directed towards the corner of a regular tetrahedron.

Hint

(c) Along the period, electronegativity \((E N)\) increases and hence as we move from \(\mathrm{Li} \rightarrow \mathrm{Be} \rightarrow \mathrm{B} \rightarrow \mathrm{C}\), the electronegativity increases.

\(\text { Thus } \mathrm{LiCl}<\mathrm{BeCl}_{2}<\mathrm{BCl}_{3}<\mathrm{CCl}_{4} \text { is correct. }\)

Hint

(a) According to VSEPR theory, a molecule with 6 bond pairs must be octahedral.

Hint

(a) A sigma bond is formed by the end-to-end overlapping, so the extent of overlapping is more in case of sigma bond and hence sigma bond is stronger than pi-bond

Hint

(a) \(\mathrm{H}_{2} \mathrm{O}\) shows maximum H-bonding because each \(\mathrm{H}_{2} \mathrm{O}\) molecule is linked to four \(\mathrm{H}_{2} \mathrm{O}\) molecules through H-bonds.

Hint

(a) When two hybridised orbitals of two atoms undergoes linear combination, they form sigma bond.

Hint

(d)

The asterick \((*)\) marked carbon has a valency of 5 and hence this formula is not correct because carbon has a maximum valency of 4 .

Hint

(a) For linear arrangement of atoms the hybridisation is \(s p\). (bond angle \(=180^{\circ}\) ).

Only \(\mathrm{H}_{2} \mathrm{~S}\) has \(s p^{3}\)-hybridisation and hence it has angular shape while \(\mathrm{C}_{2} \mathrm{H}_{2}, \mathrm{BeH}_{2}\) and \(\mathrm{CO}_{2}\) all involve \(s p\)-hybridisation and hence has linear arrangement of atoms.

Hint

(d) Metallic bonds have electrostatic attraction on all sides and hence do not have directional characteristics.

Hint

(b) \(\mathrm{BF}_{3} \text { involves } s p^{2} \text {-hybridisation }\)

Hint

Hint

Hint

(b) Equilateral or triangular planar shape involves \(s p^{2}\) hybridisation. e.g. \(-\mathrm{BCl}_{3}\).

Hint

(b) On moving down the group from \(\mathrm{F}\) to \(\mathrm{I}\), the size of atom increases. Order of the size of halogen atoms is \(\mathrm{I}>\mathrm{Br}>\mathrm{Cl}>\mathrm{F}\).
So, the bond length of \(C-X\) bond also increases from \(\mathrm{F}\) to \(\mathrm{I}\) and hence, the bond enthalpy decreases from \(\mathrm{F}\) to \(\mathrm{I}\).
Correct order of bond length of \(C-X\) bond is
\(
\mathrm{H}_{3} \mathrm{C}-\mathrm{l}>\mathrm{H}_{3} \mathrm{C}-\mathrm{Br}>\mathrm{H}_{3} \mathrm{C}-\mathrm{Cl}>\mathrm{H}_{3} \mathrm{C}-\mathrm{F} \text {. }
\)
Correct order of bond enthalpy is
\(
\mathrm{H}_{3} \mathrm{C}-\mathrm{F}>\mathrm{H}_{3} \mathrm{C}-\mathrm{Cl}>\mathrm{CH}_{3}-\mathrm{Br}>\mathrm{H}_{3} \mathrm{C}-\mathrm{I} \text {. }
\)

Hint

(c) \(\mathrm{SbCl}_{5}\) Hybridisation \(=\frac{1}{2} \times 10=5\left(s p^{3} d\right)\)
Shape \(=\) Trigonal bipyramidal

Dipole moment, \(\mu=0\)
\(\mathrm{SbCl}_{5}\) is non-polar in nature.

Hint

(c) \(P C l_{5}\) – Trigonal bipyramidal ( 5 bond pairs)
\(S F_{6}\) – Octahedral ( 6 bond pairs)
\(\mathrm{BrF}_{5}\) – Square pyramidal ( 5 bond pairs \(+1\) lone pair)
\(B F_{3}\) – Trigonal planar ( 3 bond pairs)

Hint

(c)
Hybridisation of a central atom can be calculate by using the formula:
Hybridisation \(=\frac{1}{2}\) [ number of valence
electrons + Number of side atoms –
Positive charge \(+\) Negative charge \(]\)
Electronic configuration of \(B\)
\(
=1 s^{2}, 2 s^{2}, 2 p^{1}
\)
Number of valence electrons in \(B=3\)
electrons in last shell, \(n=2\)
Number of side atoms in
\(\mathrm{BF}_{3}=3 \mathrm{~F}\)-atoms.
So, hybridisation \(=\frac{1}{2}(3+3)=\frac{1}{2} \times 6=3\).
Hybridisation of \(B\) in \(B F_{3}\) is \(s p^{2}\).
Number of electrons around central
atom, \(\mathrm{B}\) in \(\mathrm{BF}_{3}\) is equal to the number of
electrons in three sigma bonds \((B-F)\) i.e.
\(=3 \mathrm{~B}-\mathrm{F}\) bonds \(\times 2\) electrons in one
\(\sigma\)-bond.
\(=6\) electrons

Hint

(c) \(\mathrm{BF}_{3}, \mathrm{BeF}_{2}, \mathrm{CO}_{2}\) and 1,4 -dichlorobenzene all are symmetrical molecules.

Hint

(c) Dipole moment of a molecule is the vector sum of dipoles of bonds. So based on molecular geometry of following molecules,

Hint

(b) The structure of \(\mathrm{ClF}_{3}\) is

The number of lone pair of electrons on central \(\mathrm{Cl}\) is 2

Hint

Hint

(d) For \(\mathrm{He}_{2}\) molecule, Electronic configuration is \(\sigma 1 s^{2}, \sigma * 1 s^{2}\)
Bond order \(=\frac{1}{2}\left(N_{b}-N_{a}\right)=\frac{1}{2}(2-2)=0\)
Since, bond order of \(\mathrm{He}_{2}\) is zero, so it does not exist.

Hint

(a) Molecular orbital configuration of \(\mathrm{O}_{2}\) is given as :
\(
\begin{aligned}
&\mathrm{O}_{2}\left(16 \mathrm{e}^{-}\right): \sigma 1 \mathrm{~s}^{2} \sigma * 1 \mathrm{~s}^{2} \sigma 2 \mathrm{~s}^{2} \sigma * 2 \mathrm{~s}^{2} \sigma 2 p_{z}^{2} \\
&\pi 2 p_{x}^{2}=\pi 2 p_{y}^{2} \pi^{*} 2 p_{x}^{1}=\pi * 2 p_{y}^{1}
\end{aligned}
\)
So, in \(\mathrm{O}_{2}\) molecule, there are two (2) unpaired electrons, so, it is a “paramagnetic” substance in nature.

Hint

(c) Only \(\pi\) bond is present in \(\mathrm{C}_{2}\) molecule.
\(
\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \pi 2 p_{x}^{2}=\pi 2 p_{y}^{2}
\)

Hint

(b) 

\(\mathrm{CN}^{+}\): Valence electrons \(=4+5-1=8\)
\(
\begin{array}{l}
{\left[(\sigma 2 \mathrm{~s})^2\left(\sigma^* 2 \mathrm{~s}\right)^2(\sigma 2 \mathrm{p})^2(\pi 2 \mathrm{p})^2\right.} \\
\mathrm{BO}=\frac{1}{2} \times(4-0)=2
\end{array}
\)

\(\mathrm{NO}\): Valence electrons \(=5+6=11\)
\(
\begin{array}{l}
(\sigma 2 \mathrm{~s})^2\left(\sigma^* 2 \mathrm{~s}\right)^2(\sigma 2 \mathrm{p})^2(\pi 2 \mathrm{p})^4\left(\pi^* 2 \mathrm{p}\right)^1 \\
\mathrm{BO}=\frac{1}{2} \times(6-1)=2.5
\end{array}
\)
\(\mathrm{CN}\) : Valence electrons \(=4+5=9\)
\(
\begin{array}{l}
(\sigma 2 \mathrm{~s})^2\left(\sigma^* 2 \mathrm{~s}\right)^2(\sigma 2 \mathrm{p})^2(\pi 2 \mathrm{p})^3 \\
\mathrm{BO}=\frac{1}{2} \times(5-0)=2.5
\end{array}
\)
\(\mathrm{CN}^{-}\): Valence electrons \(=4+5+1=10\)
\(
(\sigma 2 \mathrm{~s})^2\left(\sigma^* 2 \mathrm{~s}\right)^2(\sigma 2 \mathrm{p})^2(\pi 2 \mathrm{p})^4
\)
\(
\mathrm{BO}=\frac{1}{2} \times(6-\mathrm{o})=3
\)

Hence, highest bond order is for \(\mathrm{CN}^{-}\)

Hint

(d)

\(
\mathrm{HF}>\mathrm{NH}_3>\mathrm{H}_2 \mathrm{S}>\mathrm{CH}_4 \text { (Non-polar) }
\)

Hint

(d) For Isoelectronic species total numbers electrons are same \(\mathrm{Ca}^{+2}, \mathrm{Ar}, \mathrm{K}^{+}, \mathrm{Cl}^{-} \rightarrow 20\) electrons

Hint

(c) In the formation of BMO, the two electron waves of the bonding atoms reinforce each other due to constructive interference. Molecular orbitals obtained from \(2 \mathrm{P}_{\mathrm{x}}\) and \(2 \mathrm{P}_{\mathrm{y}}\) orbitals are ‘unsymmetrical’ around bond axis.

Hint

(d) Hydrated chlorides and Bromides of \(\mathrm{Ca}, \mathrm{Sr}\) and \(\mathrm{Ba}\) are lonic so undergo dehydration after heating. Hydrated chlorides and Bromides of \(\mathrm{Be}\) and \(\mathrm{Mg}\) are covalent so undergo hydrolysis on Heating.

Hint

(d) Total number of species =3

Hint

(d) Molecular orbital (energy) diagram / sequence of \(\mathrm{N}_2\)

Hint

(c) The stability of a compound is often related to the charge on the central atom and the nature of its bonding with the ligands. In this case, \(\mathrm{TlI}\left(\mathrm{Tl}^{+1}\right)\) refers to thallium(I) ions, and \(\mathrm{TlI}_3\left(\mathrm{Tl}^{3+}\right)\) refers to thallium(III) ions.

Generally, lower oxidation states (e.g., \(\mathrm{Tl}^{+}\)) are more stable than higher oxidation states (e.g., \(\mathrm{Tl}^{3+}\) ) for heavy elements like thallium. This is due to the presence of additional electron shells that can provide increased shielding and stability for lower oxidation states.

Option [C] \(\mathrm{TlI}>\mathrm{TlI}_3\) reflects this principle and correctly represents the relationship based on stability.

Note: due to inert pair effect \(\mathrm{T} l^{+}\)is more stable than \(\mathrm{T} l^{+3}\).

Hint

(b) Intermolecular forces means force of attraction between two or more molecules
dipole-dipole (attraction between two or more polar molecules).
Dipole induced dipole (attraction between polar and non-polar molecules)
Hydrogen bonding (it is a special type of dipole-dipole and ion-dipole attraction)
Dispersion forces (mainly acts between non polar molecules).
Covalent bonding (acts between atom not between molecules)

Hint

(c)

Hint

(a)

\(\mathrm{CO}_2 \Rightarrow \mathrm{sp}^2\) hybridisation, bond angle \(=180^{\circ}\)
\(\mathrm{NH}_4^{+} \Rightarrow \mathrm{sp}^3\) hybridisation, bond angle \(=109^{\circ} 28^{\prime}\)
\(\mathrm{NH}_3 \Rightarrow \mathrm{sp}^3\) hybridisation with one lone pair on central atom, bond angle \(\simeq 107^{\circ}\)
\(\mathrm{H}_2 \mathrm{O} \Rightarrow \mathrm{sp}^3\) hybridisation with two lone pairs on central atom, bond angle \(\simeq 104.5^{\circ}\)

Hint

(a) In general, interhalogen compounds are more reactive than halogens (except fluorine). This is because \(X-X^{\prime}\) bond in interhalogens is weaker than \(\mathrm{X}-\mathrm{X}\) bond in halogens excepts \(\mathrm{F}-\mathrm{F}\) bond. Therefore \(\mathrm{I}-\mathrm{Cl}\) is more reactive than \(\mathrm{I}_2\) because of weaker \(\mathrm{I}-\mathrm{Cl}\) bond then \(\mathrm{I}\) – \(\mathrm{I}\) bond.

Hint

(d)

\(\mathrm{XeF}_2\) having maximum lone pairs, so, it has maximum ‘Ione pair-lone pair’ electron repulsions.

Hint

(d)

\(\mathrm{O}_2^{+}\)ion is having 15 electrons, so it contain one unpaired electron. Hence it is paramagnetic in nature.

\(
\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2
\)
\(
\begin{array}{cc}
\pi 2 p_x^2 & \pi^* 2 p_x^1 \\
\mid & \mid \\
\pi 2 p_y^2 & \pi 2 p_y
\end{array}
\)
Due to one unpaired electron in \(\pi * 2 \mathrm{p}\) molecular orbital, \(\mathrm{O}_2^{+}\)is a paramagnetic ion.

Hint

(b)

Note: Dipole moment of a molecule depends both on shape and bond dipole.

Hint

(a) Since \(CHCl _3\) is an organic solvent so, covalent (non-polar) compounds will be more soluble in it. As the dipole moment of solute increases, solubility in chloroform decreases. Hence increasing order of solubility.
\(
NaCl < CH _3 CN < CH _3 OH <\text { Cyclohexane }
\)

Hint

(b) It is \(s p^3 d\) hybridised with axial to equatorial angle of \(90^{\circ}\) and equatorial bond angles of \(120^{\circ}\). It has five \(P – Cl\) sigma bonds. Axial bonds are longer than equatorial bonds.

Hint

(d)

\(
\begin{aligned}
&\begin{array}{|c|c|}
\hline \text { Molecule } & \text { Bond enthalpy ( } kJ mol ^{-1} \text { ) } \\
\hline HCl & 431.0 \\
\hline N _2 & 946.0 \\
\hline H _2 & 435.8 \\
\hline O _2 & 498 \\
\hline
\end{array}\\
&\text { Molecule }\\
&\text { Bond enthalpy ( } kJ mol ^{-1} \text { ) }\\
&HCl\\
&431.0\\
&N _2\\
&946.0\\
&H _2\\
&435.8\\
&O _2\\
&498
\end{aligned}
\)

Hint

(c)

Hint

(a) The given question involves understanding the boiling points of Group 16 hydrides and the effect of molecular mass and hydrogen bonding on these boiling points.

Statement I claims that the boiling points of Group 16 hydrides follow this order:
\(
H _2 O > H _2 Te > H _2 Se > H _2 S
\)

This statement is indeed true. Water \(\left( H _2 O \right)\) has a much higher boiling point compared to the other hydrides in Group 16. This anomaly primarily arises due to the extensive hydrogen bonding present in water, which greatly increases its boiling point. In the absence of such strong hydrogen bonding, heavier molecules like \(H _2 Te , H _2 Se\), and \(H _2 S\) would typically have higher boiling points due to increased van der Waals forces (attributable to larger molecular mass and size). But in this case, \(H _2 O\) surpasses them because hydrogen bonds are much stronger than van der Waals forces.
Statement II states that based on molecular mass alone, \(H _2 O\) should have a lower boiling point than the other Group 16 hydrides but has a higher boiling point due to extensive hydrogen bonding. This statement is also true. Water, having a lower molecular mass, would typically have a lower boiling point if we consider only molecular mass. However, the strong hydrogen bonds between water molecules contribute to a much higher boiling point. Hydrogen bonding significantly impacts physical properties like boiling point because it requires more energy to break these bonds during the phase change from liquid to gas.
In summary, based on the analysis of the two statements:
Option A: Both Statement I and Statement II are true is the correct answer. Each statement is correct, and both relate accurately to the anomalous behavior of water among the hydrides of Group 16 elements.

Hint

(d) In o-nitrophenol intramolecular H-bonding is present.

Hint

(a)

\(NH _3 \Rightarrow s p^3\) hybridised with 1 lone pair.
Structure will be Trigonal Pyramidal.
\(BrF _5 \Rightarrow s p^3 d^2\) hybridised with 1 lone pair.
Structure will be Square Pyramidal.
\(XeF _4 \Rightarrow s p^3 d^2\) with two lone pairs.
Structure will be Square Planar.
\(SF _6 \Rightarrow s p^3 d^2\) with no lone pair.
Structure will be Octahedral.
A-I, B-IV, C-II, D-III

Hint

(d)

(1) In ozone; there are two resonating structures.