Additional Questions - Chapter 10 Chemical Bonding 11th Chemistry Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Questions Solved
I. Choose the correct answer.
Question 1.
Which is the correct Lewis structure of Helium?
(a) $\dot{\mathrm{He}}$.
(b) $\mathrm{He}$.
(c) $\mathrm{H} \mathrm{e}$
(d) $\ddot{\mathrm{H}} \epsilon$
Answer:
(d) $\mathrm{He}$
Solution:
Helium has only two electrons in its valence shell which is represented as a pair of dots (duplet).
Question 2.
Which one of the following form only covalent bonds?
(a) Alkali metals
(b) Metals
(c) Non metals
(d) Metalloids
Answer:
(c) Non metals
Question 3.
In which one of the following molecule triple bond is present?
(a) $\mathrm{O}_2$
(b) $\mathrm{H}_2$

(c) $\mathrm{CO}_2$
(d) $\mathrm{N}_2$
Answer:
(d) $\mathrm{N}_2$
Solution:
$
\mathrm{N} \equiv \mathrm{N}
$
Question 4.
Which of the following is the lewis structure of water?
(a) $\mathrm{H}-\ddot{\mathrm{O}}-\mathrm{H}$

Answer:

Question 5.
Which one of the following element forms only one bond?
(a) Carbon
(b) Oxygen
(c) Fluorine
(d) Nitrogen
Answer:
(c) Fluorine
Question 6 .
Which one is the preferred structure of $\mathrm{CO}_2$ ?

Answer:

Question 7.
Which is the correct lewis structure of $\mathrm{BF}_3$ ?

Answer:

Question 8 .
Statement I: In sulphur hexafluoride, the central atom has more than eight valence electrons.
Statement II: The central atom can accommodate additional electron pairs by using outer vacant d orbitals.
(a) Statements I and II are correct and statement II is the correct explanation of statement I.
(b) Statements I and II are correct but statement II is not the correct explanation of statement I.
(c) Statement I is correct but statement II is wrong.
(d) Statement $I$ is wrong but statement II is correct.
Answer:
(a) Statements I and II are correct and statement II is the correct explanation of statement I.
Question 9.
Which one of the following molecule has complete octet?
(a) $\mathrm{BF}_3$
(b) $\mathrm{BeCl}_2$
(c) $\mathrm{BCl}_3$
(d) $\mathrm{CCI}_4$
Answer:
(d) $\mathrm{CCI}_4$
Question 10.
Which one of the following does not have electrovalent bond?
(a) $\mathrm{KCI}$
(b) $\mathrm{NaI}$
(c) $\mathrm{MgO}$
(d) $\mathrm{CCI}_4$
Answer:
(d) $\mathrm{CCI}_4$
Question 11.
Which one of the following has an ionic bond?
(a) $\mathrm{CO}_2$
(b) $\mathrm{CH}_4$

(c) $\mathrm{CaF}_2$
(d) $\mathrm{BeCI}_2$
Answer:
(c) $\mathrm{CaF}_2$
Question 12 .
During the formation of 1 mole of $\mathrm{KCI}$ crystal, the amount of energy released is
(a) $418.81 \mathrm{~kJ}$
(b) $348.56 \mathrm{~kJ}$
(c) $718 \mathrm{~kJ}$
(d) $70.25 \mathrm{~kJ}$
Answer:
(c) $718 \mathrm{~kJ}$
Question 13.
Which one of the following has coordinate covalent bond?
(a) $\mathrm{CaF}_2$
(b) $\mathrm{MgO}$
(c) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$
(d) $\mathrm{KCI}$
Answer:
(c) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$
Question 14.
The distance between the nuclei of the two covalently bonded atoms is called
(a) bond order
(b) bond length
(c) bond angle
(d) bond enthalpy
Answer:
(b) bond length
Question 15.
The length of a bond can be determined by
(a) spectroscopic method
(b) $\mathrm{x}$ - ray diffraction method
(c) electron-diffraction method
(d) all the above
Answer:
(d) all the above
Question 16.
The value of carbon-carbon single bond length is
(a) $1.43 \mathrm{Å}$
(b) $1.54 Å$
(c) $1.33 \mathrm{Å}$
(d) $1.20 \mathrm{Å}$
Answer:
(b) $1.54 Å$

Question 17.
The value of carbon - carbon double bond length is
(a) $1.43 Å$
(b) $1.20 Å$
(c) $1.54 \mathrm{Å}$
(d) $1.33 \mathrm{Å}$
Answer:
(d) $1.33 \mathrm{Å}$
Question 18.
The value of carbon - carbon triple bond length is
(a) $1.33 \mathrm{Å}$
(b) $1.20 Å$
(C) $1.54 \mathrm{Å}$
(d) $1.43 \mathrm{Å}$
Answer:
(b) $1.20 Å$
Question 19.
Among the following which one has bond order as 3 ?
(a) $\mathrm{N}_2$
(b) $\mathrm{O}_2$
(c) $\mathrm{HCHO}$
(d) $\mathrm{CH}_4$
Answer:
(a) $\mathrm{N}_2$
Question 20.
Which one of the flowing has bond order as 2 ?
(a) $\mathrm{N}_2$
(b) $\mathrm{C}_2-\mathrm{H}_4$
(c) $\mathrm{CH}_4$
(d) $\mathrm{HCN}$
Answer:
(b) $\mathrm{C}_2 \mathrm{H}_4$

Question 21.
Identify the molecule with bond order 1 .
(a) $\mathrm{N}_2$
(b) $\mathrm{O}_2$
(c) $\mathrm{H}_2$
(d) $\mathrm{C}_2 \mathrm{H}_4$
Answer:
(c) $\mathrm{H}_2$
Question 22.
Which one of the following has zero dipole moment?
(a) $\mathrm{HF}$
(b) $\mathrm{H}_2$
(c) $\mathrm{CO}$
(d) NO
Answer:
(b) $\mathrm{H}_2$
Question 23.
Which one of the following is called polar molecule?
(a) $\mathrm{H}_3$
(b) $\mathrm{O}_2$
(c) $\mathrm{F}_2$
(d) $\mathrm{NO}$
Answer:
(d) $\mathrm{NO}$
Question 24.
Statement I: $\mathrm{CuCl}$ is more covalent than $\mathrm{NaCI}$.
Statement II: As compared to $\mathrm{Na}^{+} \cdot \mathrm{Cu}^{+}$is small and have $3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^{10}$ configuration and show greater polarisation.
(a) Statement I \& II are correct and II is the correct explanation of I
(b) Statement I \& II are correct but II is not the correct explanation of I
(c) Statement I \& II are correct but II is wrong
(d) Statement I \& II are wrong and II is the correct.
Answer:

(a) Statement I \& II are correct and II is the correct explanation of I
Question 25 .
Which of the following has see saw shape?
(a) $\mathrm{PCl}_5$
(b) $\mathrm{IO}_2 \mathrm{~F}_2^{-}$
(c) $\mathrm{SOF}_4$
(d) $\mathrm{ClO}_3^3$
Answer:
(b) $\mathrm{IO}_2 \mathrm{~F}_2^{-}$
Question 26.
Which one of the following has trigonal bipyramidal shape?
(a) $\mathrm{SF}_6$
(b) $\mathrm{IF}_4^{+}$
(c) $\mathrm{AsF}_5$
(d) $\mathrm{SF}_4$
Answer:
(c) $\mathrm{AsF}_5$
Question 27.
Which one of the following does not have tetrahedral shape?
(a) $\mathrm{NH}_4^{+}$
(b) $\mathrm{ClO}_4^{-}$
(c) $\mathrm{HCHO}$
(d) $\mathrm{CH}_4$
Answer:
(c) $\mathrm{HCHO}$
Question 28.
Which one of the following has linear shape?

(a) $\mathrm{O}_3$
(b) $\mathrm{CO}_3^{2-}$
(c) $\mathrm{NO}_3$
(d) $\mathrm{BCl}_3$
Answer:
(a) $\mathrm{O}_3$
Question 29.
Which of the following has linear shape?
(a) $\mathrm{PCI}_5$
(b) $\mathrm{SnBr}_2$
(c) $\mathrm{BeCl}_2$
(d) $\mathrm{CCl}_2 \mathrm{~F}_2$
Answer:
(c) $\mathrm{BeCl}_2$
Question 30.
Which one of the following has tetrahedral shape?
(a) $\mathrm{HCHO}$
(b) $\mathrm{BeCl}_2$
(c) $\mathrm{PbCl}_2$
(d) $\mathrm{CF}_2 \mathrm{Cl}_2$
Answer:
(d) $\mathrm{CF}_2 \mathrm{CI}_2$
Question 31.
Which one of the following pair has $T$ - shapcd structure?
(a) $\mathrm{BrF}_3, \mathrm{CIF}_3$
(b) $\mathrm{SF}_4, \mathrm{IF}_4^{+}$
(c) $\mathrm{PCl}_5, \mathrm{AsF}_5$
(d) $\mathrm{NH}_3, \mathrm{PF}_3$
Answer:
(a) $\mathrm{BrF}_3, \mathrm{CIF}_3$
Question 32.
Which one of the following has pentagonal bipyramidal shape?
(a) $\mathrm{XeF}_4$
(b) $\mathrm{XeOF}_4$
(c) $\mathrm{IF}_7$
(d) $\mathrm{IOF}_5$
Answer:
(c) $\mathrm{IF}_7$
Question 33.
Which one of the following has linear shape?

(a) $\mathrm{I}_3^{-}$
(b) $\mathrm{ICI}_4^{-}$
(c) $\mathrm{BrF}_5$
(d) $\mathrm{IOF}_5$
Answer:
(a) $\mathrm{I}_3^{-}$
Question 34 .
Which one of the following is the correct increasing order of bond angle?
(a) $\mathrm{H}_2 \mathrm{O}<\mathrm{CH}_4<\mathrm{BF}_3<\mathrm{BeCI}_2$
(b) $\mathrm{BeCI}_2<\mathrm{BF}_3<\mathrm{CH}_4<\mathrm{H}_2 \mathrm{O}$
(c) $\mathrm{BF}_3<\mathrm{CH}_4<\mathrm{BeCI}_2<\mathrm{H}_2 \mathrm{O}$
(d) $\mathrm{CH}_4<\mathrm{BeCI}_2<\mathrm{H}_2 \mathrm{O}<\mathrm{BF}_3$
Answer:
(a) $\mathrm{H}_2 \mathrm{O}<\mathrm{CH}_4<\mathrm{BF}_3<\mathrm{BeCI}_2$
Question 35 .
Which one of the following hybridisation takes place in the formation of $\mathrm{BeCI}_2$ ?
(a) $\mathrm{sp}^2$
(b) $\mathrm{sp}$
(c) $\mathrm{sp}^3$
(d) $\mathrm{dsp}^2$
Answer:
(b) $\mathrm{sp}$
Question 36.
Which hybridisation is possible in $\mathrm{BF}_3$ ?
(a) $\mathrm{sp}^2$
(b) $\mathrm{sp}$
(c) $\mathrm{sp}^3$
(d) $\mathrm{sp}^3 \mathrm{~d}$
Answer:
(a) $\mathrm{sp}^2$

Question 37.
Which one of the following has bond order as 2.5 ?
(a) $\mathrm{O}_2$
(b) $\mathrm{NO}$
(c) $\mathrm{CO}$
(d) $\mathrm{H}_2$
Answer:
(b) $\mathrm{NO}$
Question 38.
Which one of the following is an electron deficient compound?
(a) $\mathrm{Al}_2 \mathrm{Cl}_6$
(b) $\mathrm{AlBr}_3$
(c) $\mathrm{SF}_6$
(d) $\mathrm{BF}_3$
Answer:
(d) $\mathrm{BF}_3$
Question 39.
Apply the VSEPR model to $\mathrm{XeF}_4$, which of the following molecular shape is consistent with the model?
(a) Square planar
(b) Tetrahedral
(c) Square pyramidal
(d) Octahedral
Answer:
(a) Square planar
Question 40.
On the basis of molecular orbital theory, select the most appropriate option.
(a) The bond order of $\mathrm{O}_2$ is 2.5 and it is paramagnetic
(b) The bond order of $\mathrm{O}_2$ is 1.5 and it is paramagnetic
(c) The bond order of $\mathrm{O}_2$ is 2 and it is diamagnetic
(d) The bond order of $\mathrm{O}_2$ is 2 and it is paramagnetic
Answer:
(d) The bond order of $\mathrm{O}_2$ is 2 and it is paramagnetic

Question 41 .
Which of the following molecule does not exist due to its zero bond order?
(a) $\mathrm{H}_2^{-}$
(b) $\mathrm{He}_2^{+}$
(c) $\mathrm{He}_2$
(d) $\mathrm{H}_2^{+}$
Answer:
(c) $\mathrm{He}_2$
Question 42
Which of the following molecules have bond order equal to 1 ?
(a) $\mathrm{NO}, \mathrm{HF}, \mathrm{HCl}, \mathrm{Li}_2, \mathrm{CO}$
(b) $\mathrm{H}_2, \mathrm{Li}_2, \mathrm{HF}, \mathrm{Br}_2, \mathrm{HCI}$
(c) $\mathrm{Li}_2, \mathrm{~B}_2, \mathrm{CO}, \mathrm{NO}, \mathrm{He}_2^{+}$
(d) $\mathrm{B}_2, \mathrm{CO}, \mathrm{He}_2^{+}, \mathrm{NO}, \mathrm{HF}$
Answer:
(b) $\mathrm{H}_2, \mathrm{Li}_2, \mathrm{HF}, \mathrm{Br}_2, \mathrm{HCI}$
Solution:
Bond order of $\mathrm{He}_2^{+}=0.5$
Bond order of $\mathrm{NO}=2.5$
Bond order of $\mathrm{CO}=3$
Question 43.
Arrange the following molecules in decreasing order of bond length.
(a) $\mathrm{O}_2>\mathrm{O}_2{ }^{-}>\mathrm{O}_2{ }^{+}>\mathrm{O}_2{ }^{2-}$
(b) $\mathrm{O}_2^{2-}>\mathrm{O}_2^{-}>\mathrm{O}_2>\mathrm{O}_2^{-}$
(c) $\mathrm{O}_2^{2-}>\mathrm{O}_2{ }^{-}>\mathrm{O}_2{ }^{+}>\mathrm{O}_2$
(d) $\mathrm{O}_2{ }^{+}>\mathrm{O}_2^{+}>\mathrm{O}_2{ }^{2-}>\mathrm{O}_2$
Answer:
(b) $\mathrm{O}_2^{2-}>\mathrm{O}_2^{-}>\mathrm{O}_2>\mathrm{O}_2^{-}$
Solution:

Since the bond length is inversely proportional to the bond order so option ' $\mathrm{b}$ ' is correct.

Question 44.
Among the following which shows the maximum covalent character?
(a) $\mathrm{MgCI}_2$
(b) $\mathrm{FeCl}_2$
(c) $\mathrm{SnCI}_2$
(d) $\mathrm{AICI}_3$
Answer:
(d) $\mathrm{AICI}_3$
Question 45.
Which of the following has maximum number of lone pairs associated with $\mathrm{Xe}$ ?
(a) $\mathrm{XeF}_2$
(b) $\mathrm{XeO}_3$
(c) $\mathrm{XeF}_4$
(d) $\mathrm{XeF}_6$
Answer:
(a) $\mathrm{XeF}_2$
Question 46.
During the formation of a chemical bond
(a) energy decreases
(b) energy increases
(c) energy remains zero
(d) energy remains constant
Answer:
(a) energy decreases
Question 47.
Using MO theory, predict which of the following species has the shortest bond length?
(a) $\mathrm{O}_2^{+}$
(b) $\mathrm{O}_2^{-}$
(c) $\mathrm{C}_2^{2-}$
(d) $\mathrm{O}_2^{2+}$
Answer:
(d) $\mathrm{O}_2^{2+}$
Question 48.
Identify the incorrect statement regarding the molecule $\mathrm{XeO}_4$.
(a) $\mathrm{XeO}_4$ molecule is tetrahedral
(b) $\mathrm{XeO}_4$ molecule is square planar
(c) There are four $\mathrm{p} \pi-\mathrm{d} \pi$ bonds
(d) There are four $\mathrm{sp}^3-\mathrm{p}, \mathrm{s}$ bonds
Answer:
(b) $\mathrm{XeO}_4$ molecule is square planar

Question 49.
Which of the following contains maximum number of lone pairs on the central atom?
(a) $\mathrm{ClO}_3^{-}$
(b) $\mathrm{XeF}_4$
(c) $\mathrm{SF}_4$
(d) $\mathrm{I}_3{ }^{-}$
Answer:
(d) $\mathrm{I}_3^{-}$
Question 50.
Which one of the following is a correct set?
(a) $\mathrm{H}_2 \mathrm{O}, \mathrm{sp}^3$, bent
(b) $\mathrm{H}_2 \mathrm{O}, \mathrm{sp}^2$, linear
(c) $\mathrm{NH}_4^{+}, \mathrm{dsp}$, square planar
(d) $\mathrm{CH}_4^{+}, \mathrm{dsp}^2$ tetrahedal
Answwer:
(a) $\mathrm{H}_2 \mathrm{O}, \mathrm{sp}^3$, bent
II. Match the following
Question 1.

Answer:
(b) 4312
Question 2.

Answer:
(a) 3412
Question 3.

Answer:
(a) 3421
Question 4.

Answer:
(b) 3412
Question 5.

Answer:
(c) 2341
Question 6

Answer:
(b) 2413
Question 7.

Answer:
(c) 4312
Question 8.

Answer:
(d) 2341
Question 9.

Answer:
(a) 2413
Question 10

Answer:
(a) 4321
Question 11.

Answer:
(a) 4123
III. Fill in the blanks.
Question 1.
The electrovalent bond is present in ...................
Answer:
$\mathrm{NaCI}$
Solution:
$\mathrm{Na}^{+}$cation and $\mathrm{Cl}^{-}$anion are held together by electrostatic attractive forces and this is called electrovalent bond.
Question 2.
The structure (or) shape of water molecule is .....................
Answer:
inverted ' $\mathrm{V}$ ' shape
Solution:

Question 3.
The structure of $\mathrm{CO}_2$ is ....................
Answer:
linear
Solution:

Question 4.
In the formation of a chemical bond between $\mathrm{Na}$ and $\mathrm{C}[$, they attain the stable configuration of  ...............
Answer:
$\mathrm{Ne}, \mathrm{Ar}$
Solution:
$
\begin{aligned}
& \mathrm{Na}^{+}: 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6=[\mathrm{Ne}] \\
& \mathrm{Cl}^{-}: 1 \mathrm{~s}^2 2 \mathrm{~s}^2 3 \mathrm{~s}^2 3 \mathrm{p}^6=[\mathrm{Ar}]
\end{aligned}
$
Question 5.
The mutual sharing of one or more pair of electrons between the two combining atoms results in the formation of ..................
Answer:
Covalent bond
Question 6.
Formal charge of an atom can be calculated by the formula ..............
Answer:
$
\mathrm{N}_v-\left[\mathrm{N}_l+\frac{\mathrm{N}_b}{2}\right]
$
Question 7. 
The formal charge on the carbon atom in the following structure 

 is ......................

Answer:
zero
Solution:
Formal charge on carbon atom
$
=\mathrm{N}_{\mathrm{V}}-\left[\mathrm{N}_l+\frac{\mathrm{N}_b}{2}\right]=4-\left[0+\frac{8}{2}\right]=4-4=0
$
Question 8.
The formal charge on both oxygen atoms in the structure

 is .......................

Answer:
0
Solution:
Formal charge on oxygen
$
=\mathrm{N}_v-\left[\mathrm{N}_l+\frac{\mathrm{N}_b}{2}\right]=6-\left[4+\frac{4}{2}\right]=6-6=0
$
Question 9.
The formal charge on singly bonded oxygen atom in the structure

 is ...................

Answer:
$-1$
Solution:
Formal charge on singly bonded oxygen atom
$
=\mathrm{N}_{\mathrm{V}}-\left[\mathrm{N}_l+\frac{\mathrm{N}_b}{2}\right]=6-\left[6+\frac{2}{2}\right]=6-7=-1
$
Question 10.
The formal charge on the triply bonded oxygen atom in the structure

 is ....................... 

Answer:
$
+1
$
Solution:
Formal charge on triply bonded oxygen atom
$
=\mathrm{N}_v-\left[\mathrm{N}_l+\frac{\mathrm{N}_b}{2}\right]=6-\left[2+\frac{6}{2}\right]=6-5=+1
$
Question 11.
The complete transfer of one or more valence electron from one atom to another leads to the formation of ,.....................
Answer:
Ionic bond
Question 12.
The shape of the molecule is determined approximately by ........................
Answer:
bond angle
Question 13.
The unit of bond enthalpy is .............................
Answer:
$\mathrm{kJ} \mathrm{mol}^{-1}$
Question 14.
The electronegativity of hydrogen and fluorine on Pauling's scale are ...............
Answer:
2.1 and 4

Question 15.
The unit of dipole moment is ..................
Answer:
Coulomb $\mathrm{m}^{-1} \mathrm{~m}^2$
Question 16.
The dipole moment of $\mathrm{CO}_2$ is .................
Answer:
0
Solution:
$
\mu=\mu_1+\mu_2=\mu_1+\left(-\mu_1\right)=0
$
Question 17.
The shape of sulphur hexafluoride is ...................
Answer:
Octahedral
Question 18.
The type of hybridisation takes place in methane is ..................
Answer:
$\mathrm{sp}^3$
Question 19.
The type of hybridisation takes place in $\mathrm{SF}_6$ is
Answer:
$s p^3 \mathrm{~d}^2$
Question 20.
The number of lone pair of electrons on $\mathrm{C}$ - atom present in $\mathrm{CO}_2$ are  ......................
Answer:
4
Qustion 21.
In $\mathrm{SF}_6$, the bond angle is
Answer:
$90^0$
Question 22.
The ions have noble gas electronic configuration was suggested by .....................

Answer:
Kossel
Question 23.
Tetrachlorornethane is a molecule ......................
Answer:
non polar

Question 24.
In $\mathrm{C}_2 \mathrm{H}_4$, type of bonds present are ................
Answer:
Covalent bonds only
Question 25.
Molecule with bond of shape trigonal pyramid is .......................
Answer:
$\mathrm{BF}_3$
Question 26.
When magnesium reacts with oxygen, nature of bond formed is ....................
Answer:
ionic
Question 27.
The number of lone pair of electrons in water molecule is ....................
Answer:
2
Question 28.
Double bonds as compared to single bonds are .....................
Answer:
Shorter
Question 29.
Number of chlorine atoms which form equatorial bonds in $\mathrm{PCI}_5$ molecule are/is ....................
Answer:
3
Question 30 .
The hybridisation of $\mathrm{B}$ in $\mathrm{BF}_3$ is ............
Answer:
$\mathrm{sp}^2$
Question 31.
Bond order of $\mathrm{O}_2, \mathrm{~F}_3, \mathrm{~N}_2$ respectively are.... .................
Answer:
$2,1,3$

Question 32.
Hybridisation which takes place in acetylene is ...................
Answer:
$\mathrm{sp}$
Question 33.
Bond order of $\mathrm{O}_2, \mathrm{~F}_2, \mathrm{~N}_2$ respectively are ..............
Answer:
$2,1,3$
Question 34.
Hybridisation which takes place in acetylene is ...............
Answer:
$\mathrm{sp}$
Question 35.
The hybndisation of orbitais of $\mathrm{N}$ atom in $\mathrm{NO}_3{ }^{-}, \mathrm{NO}_3{ }^{+}$and $\mathrm{NH}_4{ }^{+}$are respectively .......................
Answer:
$\mathrm{sp}^2, \mathrm{sp}, \mathrm{sp}^3$
Question 36.
Malleability and ductility of metals can be accounted due to the capacity of layers of ......................... to slide over one another.
Answer:
metal ions
Question 37.
For a stable molecule, the value of bond order must be ..................
Answer:
positive
Question 38.
In acetylene molecule between the carbon atoms there are ............................ $\sigma$ and ........................ bonds.
Answer:
one, two

IV. Choose the odd one out.
Question 1.
(a) Hydrogen
(b) Chlorine
(c) Neon
(d) Argon
Answer:
(c) Neon. It is mono atomic whereas others are diatomic.
Question 2.
(a) $\mathrm{NaCl}$
(b) $\mathrm{CO}_2$
(c) $\mathrm{LiF}$
(d) $\mathrm{MgO}$
Answer:
(b) $\mathrm{CO}_2$. It contains covalent bond whereas others have ionic bond.
Question 3.
(a) Methane
(b) Ceasium chloride
(c) Ammonia
(d) Boron trifluoride
Answer:
(b) Ceasium chloride. It is an ionic compound whereas others are covalent compounds.
Question 4.
(a) $\mathrm{H}_2$
(b) $\mathrm{O}_2$
(c) $\mathrm{Cl}_2$
(d) $\mathrm{F}_2$
Answer:
(b) $\mathrm{O}_2$. It's bond order is 2 whereas in others bond order is 1 .

Question 5.
(a) $\mathrm{BeCI}_2$
(b) $\mathrm{CS}_2$
(c) $\mathrm{BF}_3$
(d) $\mathrm{HCN}$
Answer:
(c) $\mathrm{BF}_3$. It is $\mathrm{AB}_3$ type whereas others are $\mathrm{AB}_2$ type.
Question 6.
(a) $\mathrm{XeO}_2 \mathrm{~F}_2$
(b) $\mathrm{PCI}_5$
(c) $\mathrm{AsF}_5$
(d) $\mathrm{SOF}_4$
Answer:
(a) $\mathrm{XeO}_2 \mathrm{~F}_2$. It is $\mathrm{AB}_4 \mathrm{~L}$ type whereas others are $\mathrm{AB}_5$ type.
V. Choose the correct pair.
Question 1.
(a) $\mathrm{NaCI}$ - ionic compound
(b) $\mathrm{NH}_3$ - coordinate compound
(c) $\mathrm{BF}_3$ - ionic compound
(d) $\mathrm{H}_2$ - ionic compound
Answer:
(a) $\mathrm{NaCl}$ - ionic compound
Question 2.
(a) $\mathrm{O}_2-\mathrm{Bond}$ order 3
(b) $\mathrm{H}_2$ - Bond order 2
(c) $\mathrm{N}_2$ - Bond order 3
(d) $\mathrm{Cl}_2-\mathrm{Bond}$ order 2
Answer:
(c) $\mathrm{N}_2$-Bond order 3
$[\mathrm{N}=\mathrm{N}]$ Bond order is 3 .
Question 3.
(a) $\mathrm{CH}_4$ - ionic bond
(b) $\mathrm{BF}_3$ - dative bond
(c) $\mathrm{NH}_3-$ metallic bond
(d) $\mathrm{CCI}_4$ - covalent bond
Answer:
(d) $\mathrm{CCI}_4$ - covalent bond

Question 4.
(a) $\mathrm{CH}_4-107^{\circ} 18^{\prime}$
(b) $\mathrm{H}_2 \mathrm{O}-109^{\circ} 28^{\prime}$
(c) $\mathrm{NH}_3-104^{\circ} 35^{\prime}$
(d) $\mathrm{BF}_3-120^{\circ}$
Answer:
(d) $\mathrm{BF}_3-120^{\circ}$
Question 5.
(a) $\mathrm{AB}_3$ - Linear
(b) $\mathrm{AB}_3-\mathrm{V}$-shape(or)bent
(c) $\mathrm{AB}_4$ - Trigonal planar
(d) $\mathrm{AB}_5-\mathrm{T}$-shape
Answer:
(a) $\mathrm{AB}_3$ - Linear
VI. Choose the incorrect pair.
Question 1.
(a) $\mathrm{CS}_2$ - Linear
(b) $\mathrm{BF}_1$ - Trigonal planar
(c) $\mathrm{CH}_4-\mathrm{T}$-shape
(d) $\mathrm{NH}_3$ - Pyramidal
Answer:
(c) $\mathrm{CH}_4-\mathrm{T}$-shape
Question 2.
(a) $\mathrm{AB}_3$ - Trigonal planar
(b) $\mathrm{AB}_3 \mathrm{~L}_2-\mathrm{T}$-shape
(c) $\mathrm{AB}_5$ - Trigonal bipyramidal
(d) $\mathrm{AB}_3 \mathrm{~L}-$ Bent
Answer:
(a) $\mathrm{AB}_3 \mathrm{~L}$ : Bent.

Actually $\mathrm{AB}_3 \mathrm{~L}$ is pyramidal.
Question 3.
(a) $\mathrm{AB}_7-\mathrm{IF}_7$
(b) $\mathrm{AB}_4 \mathrm{~L}_2-\mathrm{ICI}_4^{-}$
(c) $\mathrm{AB}_6-\mathrm{XeOF}_4$
(d) $\mathrm{AB}_5 \mathrm{~L}-\mathrm{IF}_5$
Ans.
(c) $\mathrm{AB}_6-\mathrm{XeOF}_4$
Actually $\mathrm{XeOF}_4$ is $\mathrm{AB}_5 \mathrm{~L}$ type.
Question 4.
(a) Fluorine-Bond order 1
(b) Oxygen - Bond order 2
(c) Nitrogen - Bond order 2
(d) Cyanide - Bond order 3
Answer:
(c) Nitrogen - Bond order 2 .
Actually $\mathrm{N}=\mathrm{N}$ bond order is 3 .
Question 5 .
(a) $\mathrm{CH}_4-\mathrm{sp}^3$
(b) $\mathrm{PCI}_5-\mathrm{sp}^3 \mathrm{~d}$
(c) $\mathrm{BeCl}_2-\mathrm{sp}$
(d) $\mathrm{BF}_3-\mathrm{sp}^3 \mathrm{~d}^2$
Answer:
(d) $\mathrm{BF}_3-\mathrm{sp}^3 \mathrm{~d}^2$
Actually $\mathrm{BF}^4$ is $\mathrm{sp}^2$ hybridised.

VII. Assertion and Reason.
Question 1.
Assertion (A): Diatomic molecules such as $\mathrm{H}_2, \mathrm{O}_2, \mathrm{~F}_2$ are non-polar molecules.
Reason (R): $\mathrm{H}_2, \mathrm{O}_2, \mathrm{~F}_2$ have zero dipole moment.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) $(A)$ is correct but $(R)$ is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
Question 2.
Assertion (A): $\mathrm{HF}, \mathrm{HCl}, \mathrm{CO}$ and No are polar molecules.
Reason (R): They have non zero dipole moments and so they are polar molecules.
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of $(A)$.
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).
Question 3.
Assertion (A): $\mathrm{H}_2, \mathrm{Li}_2, \mathrm{C}_2, \mathrm{~N}_2$ are diamagnetic.
Reason (R): All have no unpaired electrons and so they are diamagnetic.
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of $(A)$.
(c) $(A)$ is correct but $(R)$ is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
Question 4.
Assertion (A): $\mathrm{B}_2, \mathrm{O}_2, \mathrm{NO}$ are paramagnetic in nature.
Reason (R): They have unpaired electrons and are paramagnetic.
(a) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(b) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(c) (A) is correct but $(R)$ is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 5.
Assertion (A): Metals have high thermal conductivity.
Reason (R): Absence of bond gap is the main reason for high thermal conductivity.
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
Question 6.
Assertion (A): Metals have high thermal conductivity.
Reason (R): Due to thermal excitation of many electrons from the valence band to the conductance band, metals have high thermal conductivity.
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(c) (A) is wrong but (R) is correct.
(d) (A) is correct but (R) is wrong.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).
VIII. Choose the correct statement.
Question 1.
(a) The metallic luster is due to reflection of light by the electron cloud.
(b) Metals have low inciting point and low boiling point.
(c) Metals have low thermal conductivity.
(d) Electrical conductivity of metals is low.
Answer:
(a) The metallic luster is due to reflection of light by the electron cloud.
Question 2.
(a) NO molecules is diamagnetic
(b) $\mathrm{O}_2$ molecules is paramagnetic
(c) $\mathrm{N}_2$ molecules is paramagnetic
(d) $\mathrm{Li}_2$ molecules is paramagnetic
Answer:
(b) $\mathrm{O}_2$ molecules is paramagnetic
Question 3.
(a) $\mathrm{BeCl}_2$ undergoes $\mathrm{sp}^3$ hybridisation
(b) $8 \mathrm{~F}_3$ undergoes $\mathrm{sp}^3 \mathrm{~d}$ hybridisation
(c) $\mathrm{CH}_4$ undergoes $\mathrm{sp}^3 \mathrm{~d}^2$ hybridisation
(d) $\mathrm{PCl}_5$ undergoes $\mathrm{sp}^3 \mathrm{~d}$ hybridisation
Answer:
(d) $\mathrm{PCl}_5$ undergoes $\mathrm{sp}^3 \mathrm{~d}$ hybridisation

Question 1.
What are chemical bonds?
Answer:
The interatomic attractive forces which holds the constituent atoms/ions together in a molecule are called chemical bonds.
Question 2.
State octet rule.
Answer:
The atoms transfer or share electrons so that all the atoms involved in chemical bonding obtain eight electrons in their outer shell (valence shell). It is called octet rule.
Question 3.
What is meant by covalent bond?
Answer:
The mutual sharing of one or more pair of electrons between two combining atoms results in the formation of a chemical bond called a covalent bond.
Question 4.
Draw the lewis structure of
1. $\mathrm{H}_2 \mathrm{O}$
2. $\mathrm{SO}_3$.
Answer:

(i)

(ii)

Question 5 .
Calculate the formal charge on the carbon atom and oxygen atom in the structure

Answer:
Formal charge on carbon
$
=\mathrm{N}_v-\left[\mathrm{N}_l+\frac{\mathrm{N}_b}{2}\right]=4-\left[0+\frac{8}{2}\right]=4-4=0
$
Formal charge on oxygen
$
=\mathrm{N}_v-\left[\mathrm{N}_l+\frac{\mathrm{N}_b}{2}\right]=6-\left[4+\frac{4}{2}\right]=6-6=0
$
Question 6.
Calculate the formal charge on the carbon atom and oxygen atom in the structure

Answer:
Formal charge on carbon
$
=\mathrm{N}_v-\left[\mathrm{N}_l+\frac{\mathrm{N}_b}{2}\right]=4-\left[0+\frac{8}{2}\right]=4-4=0
$
Formal charge on singly bonded oxygen
$
=\mathrm{N}_v-\left[\mathrm{N}_l+\frac{\mathrm{N}_b}{2}\right]=6-\left[6+\frac{2}{2}\right]=6-7=-1
$
Formal charge on triply bonded oxygen
$
=\mathrm{N}_v-\left[\mathrm{N}_l+\frac{\mathrm{N}_b}{2}\right]=6-\left[2+\frac{6}{2}\right]=6-5=+1
$

Question 7.
Among  and  $: \mathrm{O} \equiv \mathrm{C}-\ddot{\mathrm{O}}$ which is a preferable structure for $\mathrm{CO}_2$ molecule why?
Answer:
Structure I of

Formal charge on carbon
$
=\mathrm{N}_v-\left[\mathrm{N}_{/}+\frac{\mathrm{N}_b}{2}\right]=4-\left[0+\frac{8}{2}\right]=4-4=0
$
Formal charge on oxygen
$
=\mathrm{N}_v-\left[\mathrm{N}_l+\frac{\mathrm{N}_b}{2}\right]=6-\left[4+\frac{4}{2}\right]=6-6=0
$
Structure II of
$
\mathrm{CO}_2-\mathrm{O} \equiv \mathrm{C}-\ddot{\mathrm{O}}
$
Formal charge on carbon
$
=\mathrm{N}_v-\left[\mathrm{N}_l+\frac{\mathrm{N}_b}{2}\right]=4-\left[0+\frac{8}{2}\right]=4-4=0
$
Formal charge on singly bonded oxygen
$
=\mathrm{N}_v-\left[\mathrm{N}_l+\frac{\mathrm{N}_b}{2}\right]=6-\left[6+\frac{2}{2}\right]=6-7=-1
$
Formal charge on triply bonded oxygen
$
=\mathrm{N}_v-\left[\mathrm{N}_l+\frac{\mathrm{N}_b}{2}\right]=6-\left[2+\frac{6}{2}\right]=6-5=+1
$
A structure in which all formal charges are zero is preferred over the one with non - zero charges. In case of $\mathrm{CO}_2$ structure , structure $\mathrm{I}$ is preferred over the structure II as it has zero formal charge for all the atoms.
Question 8.
Draw the lewis structures of a few molecules containing odd electrons.
Answer:
Few molecules have a central atom with an odd number of valence electrons. For example, in nitrogen dioxide  and nitric oxide all the atoms does not have octet configuration. 

Question 9.
Draw the lewis structure of $\mathrm{PCl}_5$ and $\mathrm{SF}_6$
Answer:

Question 10.
Define bond length.
Answer:
The distance between the nuclei of two covalently bonded atoms is called bond length. For e.g., in a covalent molecule $\mathrm{A}-\mathrm{B}$. the bond length is equal to the sum of the radii of bonded atoms. i.e., $\mathrm{r}_{\mathrm{A}}+\mathrm{r}_{\mathrm{B}}=$ bond length.
Question 11.
Prove that bond order is inversely proportional to bond length.
Answer:
1. Bond order $\propto \frac{1}{\text { Bond length }}$
2. An example for illustrating the above equation is Carbon - carbon single bond length (I .54\&) is longer than the carbon-carbon double bond length $(1.34 \AA)$ and the carbon- carbon triple bond length $(1.20 \AA)$.
Question 12.
Define Bond angle.
Answer:
Covalent bonds are directional in nature and are oriented in specific direction in space. This directional nature creates a fixed angle between two covalent bonds in a molecule and this angle is termed as bond angle.
Question 13.
Define Resonance.
Answer:
The similar structures in which the relative position of the atoms are same but they differ in the position of bonding and lone pair of electrons are called resonance structures and this phenomenon is called resonance.
Question 14.
What are polar and non-polar molecules?
Answer:
1. Diatomic molecules such as $\mathrm{H}_2, \mathrm{O}_2, \mathrm{~F}_2$ have zero dipole moment and are called non polar molecules.
2. Molecules such as $\mathrm{HF}, \mathrm{HCl}, \mathrm{CO}$, NO have non zero dipole moment values and are called polar molecules.

Question 15.
What is meant by polarisaion?
Answer:
The ability of a cation to polarise an anion is called its polarising ability and the tendency of the anion to get polarised is called its polarisability. This phenomenon is known as polarisation.
Question 16.
Among $\mathrm{NaCI}, \mathrm{MgCI}_2$ and $\mathrm{AICI}_3$ which shows more covalent character? Why?
Answer:
Among, the ionic compounds $\mathrm{NaCI}, \mathrm{MgCl}_2$ and $\mathrm{AICI}_3$ the charge of the cation increases in the order $\mathrm{Na}^{+}<$ $\mathrm{Mg}^{2+}<\mathrm{Al}^{3+}$, thus the covalent character also follows the same order $\mathrm{NaCl}<\mathrm{MgCI}_2<\mathrm{AlCI}_3$. So $\mathrm{AICI}_3$ shows more covalent character.
Question 17.
Lithium chloride is more covalent than sodium chloride. Justify this statement.
Answer:
1. The smaller cation and larger anion shows greater covalent character due to greater extent of polarisation.
2. The size of $\mathrm{Li}^{+}$ion is smaller than $\mathrm{Na}^{+}$ion and hence the polarising power of $\mathrm{Li}^{+}$ion is more. So lithium chloride is more covalent than sodium chloride.
Question 18 .
Lithium iodide is more covalent than Lithium chloride. Give reason.
Answer:
Lithium iodide is more covalent than Lithium chloride as the size of $\mathrm{I}^{-}$ion is larger than $\mathrm{Cl}^{-}$ion. Hence $\mathrm{I}^{-}$ion will be more polarised than $\mathrm{Cl}^{-}$ion by the cation $\mathrm{Li}^{+}$. So $\mathrm{LiI}$ is more covalent than $\mathrm{LiCl}$.
Question 19.
Draw the structure of $\mathrm{AB}_4 \mathrm{~L}$ and $\mathrm{AB}_3 \mathrm{~L}_2$ type of molecules with example.
Answer:
$\mathrm{AB}_4 \mathrm{~L}$ :
Bond pairs $=4$
Lone pair $=1$
Shape $=$ see saw
e.g., $\mathrm{SF}_4$ Strucutre

$\mathrm{AB}_3 \mathrm{~L}_2$
Bond pairs $=3$
Lone pair $=2$
Shape $=T$ shaped
e.g., $\mathrm{CIF}_3$ Strucutre

Question 20.
Draw the structure of $\mathrm{AB}_4 \mathrm{~L}_2$ and $\mathrm{AB}_7$ type of molecules with example
Answer:
1. $\mathrm{AB}_4 \mathrm{~L}_2$
Bond pairs $=4$
Lone pairs $=2$
Shape $=$ Square planare.g., $\mathrm{XeF}_4$

2. $\mathrm{AB}_7$
Bond pairs $=7$
Lone pairs $=\mathrm{Nil}$
Shape $=$ pentagonal bipyramidal e.g., IF $_7$

Question 21.
Explain the bond formation of hydrogen molecule.
Answer:
1. Electronic configuration of hydrogen atom is $1 \mathrm{~s}^1$.
2. During the formation of $\mathrm{H}_2$ molecule, the $1 \mathrm{~s}$ orbitais of two hydrogen atoms containing one unpaired electron with opposite spin overlaps with each other along the internuclear axis.
This overlap is called $s-$ s overlap. Such axial overlap results in the formation of a sigma $(\sigma)$ covalent bond.

Question 22.
Explain the bond formation of fluorine molecule.
Answer:
1. Valence shell electronic confIguration of fluorine atom is $2 \mathrm{~s}^2 2 \mathrm{p}_{\mathrm{x}}^2 2 \mathrm{p}_{\mathrm{y}}^2 2 \mathrm{p}_z^1$
2. When the half filled $p_z$ orbitais of two fluorine atoms overlap along the $z$ - axis, a $\sigma$ covalent bond is formed between them.

Question 23.
How is HF molecule formed?
Answer:
1. Electronic configuration of hydrogen atom is $1 \mathrm{~s}^1$.
2. Valence shell electronic configuration of fluorine atom is $2 \mathrm{~s}^2 2 \mathrm{p}_{\mathrm{x}}^2 2 \mathrm{p}_{\mathrm{y}}^2 2 \mathrm{p}_{\mathrm{z}}^1$.
3. When half filled is orbital of hydrogen linearly overlaps with a halt filled $2 p_z$ orbital of fluorine, a $\sigma$ covalent bond is formed between hydrogen and fluorine.

Question 24.
What is meant by metallic bond?
Answer:
The forces that keep the atoms of the metal so closely in a metallic crystal constitute what is generally known as the metallic bond.
Question 25.
Why metallic bonding is referred as electronic bonding?
Answer:
1. Metallic crystals are an assemblage of positive ions immersed in a gas of free electrons. The free electrons are due to the ionisation of the valence electrons of the atom of the metal.
2. As the valence electrons of the atoms are freely shared by all the ions in the crystal, the metallic bonding is also referred to as electronic bonding.
Question 26.
Metals have high density. Give reason.
Answer:
The electrostatic attraction between the metal ions and the free electrons yields a three dimensional close packed crystal with a large number of nearest metal ions. So metals have high density.
Question 27.
Metals are ductile in nature. why?
Answer:
In the close packed structure of metallic crystal. it contains many slip planes along which movement can occur during mechanical loading, so the metal acquires ductility.
Question 28.
Give reason behind the lustrous nature, high melting point and boiling point of metals?
Answer:
1. The metallic lustre is due to the reflection of light by the electron cloud.
2. As the metallic bond is strong enough. the metal atoms are reluctant to break apart into a liquid or gas, so the metals have high melting and boiling points.

Question 29.
Metals are very good electrical conductors. Why?
Answer:
1. The bonding in metal is explained by molecular orbital theory. As per this theory. the atomic orbitals of large number of atoms in a crystal overlap to form numerous bonding and anti-bonding molecular orbitals without any band gap.
2. The bonding molecular orbitals are completely filled with an electron pair in each, and the anti-bonding molecular orbitals are empty.
3. Absence of band gap accounts for high electrical conductivity of metals.
Question 30 .
Metals have high thermal conductivity. Give reason.
Answer:
High thermal conductivity of metals is due to thermal excitation of many electrons from the valence band to the conduction band.
Question 31.
Except $\mathrm{Cu}, \mathrm{Ag}$ and $\mathrm{Au}$, most metals are black. Why?
Answer:
Most metals are black except copper, silver and gold. It is due to the absorption of light of all wavelengths. Absorption of light of all wavelengths is due to the absence of band gap in metals.
Question 32 .
Write the favourable factors for the formation of ionic bond.
Answer:
1. Low ionisation enthalpy of metal atoms.
2. High electron gain enthalpy of non-metal atoms.
3. High lattice enthalpy of compound formed.
Question 33.
Although geometries of $\mathrm{NH}_3$ and $\mathrm{H}_2 \mathrm{O}$ molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.

Answer:
$
\begin{aligned}
& \mathrm{NH}_3 \rightarrow \mathrm{NH}_3 \\
& \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{O}:
\end{aligned}
$
Because of two lone pair of electrons on $\mathrm{O}$ - atom, repulsion on bond pairs is greater in $\mathrm{H}_2 \mathrm{O}$ in comparison to $\mathrm{NH}_3$. Thus, the bond angle is less in $\mathrm{H}_2 \mathrm{O}$ molecules.
Question 34.
Write the significance/applications of dipole moment.
Answer:
1. In predicting the nature of the molecules: Molecules with specific dipole moments are polar in nature and those with zero dipole moments are non-polar in nature.
2. In the determination of shapes of molecules.
3. In calculating the percentage of ionic character.
Question 35.
Although both $\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$ are triatomic molecules, the shape of $\mathrm{H}_2 \mathrm{O}$ molecule is bent while that of $\mathrm{CO}_2$ is linear. Explain this on the basis of dipole moment.
Answer:
In $\mathrm{CO}_2$, there are two $\mathrm{C}=\mathrm{O}$ bonds. Each $\mathrm{C}=\mathrm{O}$ bond is a polar bond. The net dipole moment of $\mathrm{CO}_2 \mathrm{moleculc}$ is zero. This is possible only if $\mathrm{CO}_2$ is a linear molecule. $(\mathrm{O}=\mathrm{C}=\mathrm{O})$. The bond dipoles of two $\mathrm{C}=\mathrm{O}$ bonds cancels the dipole moment of each other.
Whereas, $\mathrm{H}_2 \mathrm{O}$ molecule has a net dipole moment (1.84 D). $\mathrm{H}_2 \mathrm{O}$ molecule has a bent structure because here the $\mathrm{O}-\mathrm{H}$ bonds are oriented at an angle of $104.5^{\circ}$ and do not cancel the bond dipole moments of each other.
Question 36.
What is the total number of sigma and pi bonds in the following molecules?
1. $\mathrm{C}_2 \mathrm{H}_2$
2. $\mathrm{C}_2 \mathrm{H}_4$
Answer:
1. $\mathrm{H}-\mathrm{C}=\mathrm{C}-\mathrm{H}$
Sigma bond $=3$
$\pi$ bond $=2$

2. 

Sigma bond $=5$
$\pi$ bond $=1$
Question 37.
Use molecular orbital theory to explain why the $\mathrm{Be}_2$ molecule does not exist
Answer:
E.C of $\mathrm{Be}=1 \mathrm{~s}^2 2 \mathrm{~s}^2$
M.O.E.C of $\mathrm{Be}_2=\sigma 1 \mathrm{~s}^2 \sigma 2 \mathrm{~s}^2 \sigma^* 2 s^2$
Bond order $=\frac{1}{2}(4-4)=0$
Hence, $\mathrm{Be}_2$
Question 38.
Compare the relative stabililty of the following species and indicate their magnetic properties $\mathrm{O}_2, \mathrm{O}_2{ }^{+}, \mathrm{O}_2^{-}$ (superoxide), $\mathrm{O}_2^{2-}$ (peroxide)
Answer:
$\mathrm{O}_2-\mathrm{Bond}$ order $=2$ paramagnetic
$\mathrm{O}_2^{+}-\mathrm{Bond}$ order $=2.5$, paramagnetic
$\mathrm{O}_2{ }^{-}$- Bond order $=1.5$, paramagnetic
$\mathrm{O}_2^{2-}-$ Bond order $=1$, diagmagnetic
Order of relative stability is:

Question 39.
Account for the following:
1. Water is a liquid while $\mathrm{H}_2 \mathrm{~S}$ is a gas
2. $\mathrm{NH}_3$ has higher boiling point than $\mathrm{PH}_3$.
Answer:
1. In case of water, hydrogen bonding causes association of the $\mathrm{H}_2 \mathrm{O}$ molecules. There is no such hydrogen bonding in $\mathrm{H}_2 \mathrm{~S}$, that is why it is a gas.
2. There is hydrogen bonding in $\mathrm{NH}_3$ but not in $\mathrm{PH}_3$.
Question 40 .
Why $B_2$ is paramagnetic in nature while $\mathrm{C}_2$ is not?
Answer:
The molecular orbital electronic configuration of both $\mathrm{B}_2$ and $\mathrm{C}_2$ are:
$\mathrm{B}_2$ :
$[\sigma 1 s]^2\left[\sigma^* 1 s\right]^2[\sigma 2 s]^2[\sigma * 2 s]^2\left[\pi 2 p_x\right]^1\left[\pi 2 p_y\right]^1$
$\mathrm{C}_2$
$[\sigma 1 s]^2[\sigma * 1 s]^2[\sigma 2 s]^2[\sigma * 2 s]^2\left[\pi 2 p_x\right]^2\left[\pi 2 p_y\right]^2$
Since, $B_2$ has two unpaired electrons, therefore, $B_2$ is paramagnetic $C_2$ has no unpaired electron, therefore, $C_2$ is diamagnetic.
Question 1.
Draw the lewis structure of
1. Nitrogen
2. Carbon
3. Oxygen
Answer:

Lewis structure of nitrogen atom 

Lewis structure of Oxygen atom 

Lewis structure of Carbon atom 
Question 2.
Draw the lewis structure of
1. Ammonia
2. Methane
3. Dinitrogen pentoxide.
Answer:
Lewis dot structures of Molecules

Question 3.
Calculate the bond enthalpy of $\mathrm{OH}$ bond in water.
Answer:
1. In the case of polyatomic molecules with two or more same bond types, the arithmetic mean of the bond energy value of the same type of bonds is considered as average bond enthalpy.
2. For e.g., in water, there are two $\mathrm{OH}$ bonds present and the energy needed to break them are not same.
3. $\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{OH}(\mathrm{g})$
$\Delta \mathrm{H}_1=502 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{OH}(\mathrm{g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{O}(\mathrm{g})$
$\Delta \mathrm{H}=427 \mathrm{~kJ} \mathrm{~mol}$
The average bond enthalpy of $\mathrm{OH}$ bond in water $=\frac{502+427}{2}=464.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Question 4.
Explain how the ionic character in a covalent bond is related to electronegativity?
Answer:
1. The extent of ionic character in a covalent bond can be related to the electronegativity difference of the bonded atoms.
2. In a typical polar molecule $\mathrm{A}^{\tilde{\delta}-}-\mathrm{B}^{\hat{}+}$ the electronegativity difference $\left(\mathrm{X}_{\mathrm{A}}-\mathrm{X}_{\mathrm{B}}\right)$ can be used to predict the percentage of the ionic character as follows
3. If the electronegativity difference $\mathrm{X}_{\mathrm{A}}-\mathrm{X}_{\mathrm{B}}$ is equal to 1.7 , then the bond $\mathrm{A}-\mathrm{B}$ has $50 \%$ ionic character.
4. If it is greater than 1.7 , then the bond $\mathrm{X}_{\mathrm{A}}-\mathrm{X}_{\mathrm{B}}$ has more than $50 \%$ ionic character.
5. If it is greater than 1.7 , then the bond $\mathrm{A}-\mathrm{B}$ has more than $50 \%$ ionic character.
6. If it is lesser than 1.7 , then the bond $\mathrm{A}-\mathrm{B}$ has less than $50 \%$ ionic character.
Question 5.
$\mathrm{CuCI}$ is more covalent than $\mathrm{NaCl}$. Give reason.
Answer:
1. Cations having $n s^2 n p^6 n d^{10}$ configuration show greater polansing power than the cations with $n s^2 n p^6$ configuration. Hence they show greater covalent character.

2. $\mathrm{CuCI}$ is more covalent than $\mathrm{NaCI}$. As compared to $\mathrm{Na}^{+}(1.13 \AA), \mathrm{Cu}^{+}(0.6 \AA)$ is small and has $3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^{10}$ configuration.
3. Electronic configuration of $\mathrm{Cu}^{+}:[\mathrm{Ar}] 3 \mathrm{~d}^{10}$
Electronic configuration of $\mathrm{Na}^{+}:[\mathrm{He}] 2 \mathrm{~s}^2 2 \mathrm{p}^6$
So $\mathrm{CuCI}$ is more covalent than $\mathrm{NaCI}$
Question 6.
Draw the structure of $\mathrm{AB}_2, \mathrm{AB}_3, \mathrm{AB}_3 \mathrm{~L}$ type of molecules with example.
Answer:
1. $\mathrm{AB}_2$
Number of bond pairs $=2$
Shape $=$ Linear
Example- $\mathrm{BeCl}_2$

2. $\mathrm{AB}_3$
Number of bond pairs $=3$
Shape $=$ Tirgonal planar
Example $-\mathrm{BF}_3$

3. $\mathrm{AB}_2 \mathrm{~L}$
Number of bond pairs $=2$
Number of lone pairs $=1$
Shape = Bent (or) inverted $\mathrm{V}$ shape
Example $-\mathrm{O}_3$

Question 7.
Give example and structure of
1. $\mathrm{AB}_3 \mathrm{~L}$
2. $\mathrm{AB}_5$
3. $\mathrm{AB}_2 \mathrm{~L}_2$
type of molecules with example.
Answer:
1. $\mathrm{AB}_3 \mathrm{~L}$
Number of bond pairs $=3$
Number of lone pairs $=1$
Shape $=$ Pyramidal
Example $-\mathrm{NH}_3$

2. $\mathrm{AB}_5$
Number of bond pairs $=5$
Number of lone pairs $=0$
Shape $=$ Trigonal bipyramidal
Example $=\mathrm{PCIC}_5$

3. $\mathrm{AB}_2 \mathrm{~L}_2$
Number of bond pairs $=2$
Number of lone pairs $=2$
Shape $=$ Bent
Example $-\mathrm{H}_2 \mathrm{O}$

Question 8 .
Draw the shape of
1. $\mathrm{XeF}_2$
2. $\mathrm{IOF}_5$
3. $\mathrm{XeOF}_4$
Answer:
1. $\mathrm{XeF}_2$

$\mathrm{XeF}_2$ : Linear
2. $\mathrm{IOF}_5$

$\text { 3. } \mathrm{XeOF}_4$

Question 9.
Explain the bonding in oxygen molecule.
Answer:
1. Valence shell electronic configuration of oxygen atom is $2 \mathrm{~s}^2 2 \mathrm{p}_{\mathrm{x}}^2 2 \mathrm{p}_{\mathrm{y}}^1 2 \mathrm{p}_{\mathrm{z}}^1$

2. When the half filled $p_z$ orbitaIs of two oxygen atoms overlap along the $z$ - axis a $\sigma$ covalent bond is formed between them. Other two hail filled $p_y$ orbitais of two oxygen atoms overlap laterally to form a $\pi$-covalent bond between the oxygen atoms.
3. Thus in oxygen molecule, two oxygen atoms are connected by two covalent bonds (double bond). The other two pair of electrons present in the $2 \mathrm{~s}$ and $2 \mathrm{p}_{\mathrm{x}}$ orbital do not involve in bonding and remains as lone pair on the respective oxygen atoms.

Question 10.
Explain about the molecular orbital diagram of hydrogen molecule.
Answer:

1. Electronic configuration of $\mathrm{H}$ atom $1 \mathrm{~s}^1$
2. Electronic configuration of $\mathrm{H}$, molecule $-\sigma 1 \mathrm{~s}^1$
Bond order
Bond order $=\frac{\mathrm{N}_b-\mathrm{N}_a}{2}=\frac{2-0}{2}=1$
3. Molecule $\left(\mathrm{H}_2\right)$ has no unpaired electrons, hence it is diamagnetic.
Question 11.
Draw and explain the M.O. diagram of lithium molecule.
Answer:

1. Electronic configuration of $\mathrm{Li}$ atom $-1 \mathrm{~s}^1$
2. Electronic configuration of $\operatorname{Li}_2$ molecule is ais $2 \sigma^* 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2 \sigma \mathrm{s}^2$
3. Bondorder $=\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{b}} / 2=4-2 / 2$
4. $\mathrm{Li}_2$ molecule has no unpaired electrons, hence it is diamagnetic.
Question 12.
Draw and explain the M.O. diagram of Boron molecule.

Answer:

Molecular orbitals of $\mathrm{B}_2$
1. Electronic configuration of $B=1 s^2 2 s^2 2 p^3$
2. Electronic configuration of $B=\sigma 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2 \sigma 2 \mathrm{~s}^2 \sigma^* 2 \mathrm{~s}^2 \pi 2 \mathrm{p}_{\mathrm{x}}{ }^1 \pi 2 \mathrm{p}_z^1$
$
=\frac{\mathrm{N}_b-\mathrm{N}_a}{2}=\frac{6-4}{2}=1
$
3. Bond order
4. $B_2$ molecule has two unpaired electrons hence it is paramagnetic.
Question 13.
Draw and explain the molecular orbital diagram of carbon molecule.
Answer:

1. Electronic configuration of $\mathrm{C}$ atom $-1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^2$
2. Electronic configuration of $\mathrm{C}_2$ molecule is $\sigma 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2 \sigma^* 2 \mathrm{~s}^2 \sigma^* 2 \mathrm{~s}^2 \pi 2 \mathrm{p}_{\mathrm{x}}^2 \pi 2 \mathrm{p}_{\mathrm{y}}^2$
$
=\frac{\mathrm{N}_b-\mathrm{N}_a}{2}=\frac{8-4}{2}=2
$

Question 14.
write Lewis symbols for the following atoms and ions: $\mathrm{S}$ and $\mathrm{S}^{2-} ; \mathrm{Al}$ and $\mathrm{Al}^{3+} ; \mathrm{H}_{\text {and }} \mathrm{OH}^{-}$ Answer:

Answer:

Question 15.

Write Lewis dot symbols for atoms of the following elements: Mgq Naq B O, N, Br.

Question 16.
Draw the Lewis structures for the following molecules and ions $\mathrm{H}_2 \mathrm{~S}, \mathrm{SiCl}_4, \mathrm{BeF}_2, \mathrm{CO}_3{ }^{2-}, \mathrm{HCOOH}$

Answer:

Question 17.
Define Octet rule. Write its significance and limitations.
Answer:
Octet rule:
Atoms of elements combine with each other in order to complete their respective octet so as to acquire the stable nearest noble gas configuration.
Significance:
It helps to explain why dilfferent atoms combine with each other to form ionic compounds or covalent compounds.
Limitations of Octet rule:
1. According to octet rule, atoms take part in chemical combination to achieve the configuration of nearest noble gas elements.

However, some of noble gas elements like Xenon have formed compounds with fluorine and oxygen. For example: $\mathrm{XeF}_2, \mathrm{XeF}_4, \mathrm{XeO}_3$ etc. Therefore, validity of the octet rule has been challenged.
2. This theory does not account for the shapes of molecules.
Question 18.
Write the resonance structure for $\mathrm{SO}_3, \mathrm{NO}_2$ and $\mathrm{NO}_3$
Answer:

Question 19.
What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.
Answer:
The electron pair involved in sharing between two atoms during covalent bonding is called shared pair or bond pair. At the same time, the electron pair which is not involved in sharing is called lone pair of electrons.

For example, in   there are only 4 bond pairs, but in  there are two bond pairs and two lone pairs. 

Question 20.
Distinguish between a sigma bond and a pi bond
Answer:
Sigma $(\sigma)$ Bond
1. $\sigma-$ bond is formed by the axial overlap of the atomic orbitais.
2. The bond is quite strong.
3. Only one lobe ofthep-orbitals is involved in the overlap.
4. Electron cloud of the molecular orbital is symmetrkal around the internuclear axis.
$\operatorname{Pi}(\pi)$ Bond
1. $\pi$-bonnd is formed by the sidewise overlap of atomic orbitais.
2. It is comparatively a weaker bond.
3. Both lobes of the p-orbitais are involved in the overlap.
4. The electron cloud is not symmetrical.
Question 21.
Write the Important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
Answer:
1. The combining atomic orbitals should have comparable energies. For example, is orbital of one atom can combine with $1 \mathrm{~s}$ atomic orbital of another atom, $2 \mathrm{~s}$ orbitai can combine with $2 \mathrm{~s}$ orbital and so on.
2. The combining atomic orbitals must have proper orientations so that they are able to overlap to a considerable extent.
3. The extent of overlapping should be large.
Question 22.
What are Lewis structures? Write the Lewis structure of $\mathrm{H}_2, \mathrm{BeF}_2$ and $\mathrm{H}_2 \mathrm{O}$.
Answer:
The outer shell electrons are shown as dots surrounding the symbol of the atom. These symbols are known as Lewis symbols or Lewis structures. The Lewis structure of 

Question 23.
What are the main postulates of Valence Shell Electron Pair Repulsion (VSEPR) theory?
Answer:
1. The shape of a molecule depends upon the no. of electron pairs around the central atom.
2. There is a repulsive force between the electron pairs, which tend to repel one another.
3. The electron pairs in space tend to occupy such positions that they arc at maximum distance, so that the repulsive force will be minimum.
4. A multiple bond is treated as lilt is a single bond and the remaining electron pairs which constitute the bond may be regarded as single super pair.
Question 24.
Apart from tetrahedral geometry, another possible geometry for $\mathrm{CH}_4$ is square planar with four $\mathrm{H}$ atoms at the corners of the square and $\mathrm{C}$ atom at its centre. Explain why $\mathrm{CH}_4$ is not square planar?
Answer:
Electronic configuration of carbon atom: $\mathrm{C}: \sigma 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^2$. in the excited state, the orbital picture of carbon can be represented as:

$\text { Hence, carbon atom undergoes } \mathrm{sp}^3 \text { hybridisation in } \mathrm{CH}_4 \text { molecule and takes tetrahedral shape. }$

For a square planar shape, the hybridisation of the central atom has to be $\mathrm{dsp}^3$. However, an atom of carbon does not have $\mathrm{d}$ - orbitals to undergo $\mathrm{dsp}^3$ hybridisation. Hence, the structure of $\mathrm{CH}_4$ is tetrahedral.
Question 25.
Explain why $\mathrm{BeH}_2$ molecule has a zero dipole moment although the $\mathrm{Be}-\mathrm{H}$ bonds are polar.
Answer:

The Lewis structure for $\mathrm{BeH}_2$ molecule is as follows: $\mathrm{H}: \mathrm{Be}: \mathrm{H}$
There is no lone pair at the central atom $(\mathrm{Be})$ and there are two bond pairs. Hence, $\mathrm{BeH}_2$, is of the type $\mathrm{AB}_2$. It has a linear structure.
Dipole moments of each $\mathrm{Be}-\mathrm{H}$ bond are equal and opposite in direction. Therefore, they nullify each other. Hence, $\mathrm{BeH}_2$ has a net zero dipole moment.
Question 1.
Explain about Kossel-Lewis approach to chemical bonding.
Answer:
1. Kossel and Lewis approach to chemical bonding is based on the inertness of the noble gases which have little or no tendency to combine with other atoms.
2. They proposed that noble gases are stable due to their completely filled outer electronic configuration.
3. Elements other than noble gases try to attain the completely filled outer electronic configuration by losing, gaining or sharing one or more electrons from their outer shell.
4. For e.g., sodium loses one electron to form $\mathrm{Na}$ ion and chlorine accepts that electron to give chloride ion, $\mathrm{Cl}^{-}$. These two ions are held together by electrostatic attractive forces, a bond known as an electrovalent bond.

5. In diatomic molecules such as nitrogen and oxygen, they achieve the stable noble gas electronic configuration by mutual sharing of electrons.
6. Lewis introduced a scheme to represent the chemical bond and the electrons present in the outer shell of the atom called Lewis dot structure.

7. For example, the electronic configuration of nitrogen is $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^3$. It has 5 electrons in its outer shell. The lewis structure of nitrogen is 
8. In $\mathrm{N}$, molecule, equal sharing of 3 electrons from each nitrogen atom takes place as fol lows

Question 2.
What is meant by covalent bond?
Explain the covalent bonding in $\mathrm{H}_2, \mathrm{O}_2, \mathrm{~N}_2$.
Answer:
1. Mutual sharing of one or more pair of electrons between two combining atoms results in the formation of a chemical bond called a covalent bond.
2. If two atoms share just one pair of electron, a single covalent bond is formed as in the çase of hydrogen molecule $\left(\mathrm{H}_2\right)$
3. If two or three electron pairs are shared benveen the two combining atoms, then the covalent bond is called double bond and triple bond respectively, as in the case of $\mathrm{O}_2$ and $\mathrm{N}_2$ molecules respectively.

Question 3.
What is an ionic bond?
Explain about the formation of ionic bond with a suitable example.
Answer:
1. The complete transfer of electrons leads to the fomiation of a cation and an anion. Both these ions are held together by electrostatic attractive forces which is known as ionic bond.
2. $\mathrm{KCl}$ P Potassium chloride
Electronic configuration of $\mathrm{K}[\mathrm{Ar}] 4 \mathrm{~s}$
Eleçtronic configuration of $\mathrm{Cl}=[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^5$
3. Potassium has 1 electron in its valence shell and chlorine has 7 electrons in its valence shell.
4. By losing one electron potassium attains the nearest inert gas configuration of Argon and becomes a unipositive cation $(\mathrm{K})$ and chlorine accepts this electron to become uninegative chloride ion (CI) to attain the stable configuration of nearest noble gas, Argon.
5. These two ions combine to form an ionic crystal in which they are held together by electrostatic attractive forces.
6. During the formation of one mole of potassium chloride crystal from its constituent ions, $718 \mathrm{~kJ}$ of energy is released. This favours the formation of $\mathrm{KCl}$ and its stabilisation.
Question 4.
Define coordinate covalent bond. Illustrate the formation of coordinate covalent bond with a suitable example.

Answer:
1. In the bond formation, one of the combining atoms donates a pair of electrons i.e., two electrons which are necessary for the covalent bond formation and these electrons are shared by both the combining atoms, and the bond formed is called coordinate covalent bond.
2. The combining atom which donates the pair of electron is called the donor atom and the other atom is called the acceptor atom. This bond is denoted by an arrow starting from the donor atom pointing towards the acceptor atom.
3. For example, in ferricyanide ion $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ each cyanide ion $\left(\mathrm{CN}^{-}\right)$donates a pair of electrons to form a coordinate bond with iron $\left(\mathrm{Fe}^{2+}\right)$ and these electrons are shared by $\mathrm{Fe}^{2+}$ and $\mathrm{CN}^{-}$ions.

4.

5. Ammonia having a lone pair of electrons donates its pair to an electron deficient molecule such as $\mathrm{BF}_3$ to form a coordinate covalent bond

Question 5.
Explain about valence bound theory for the formation of $\mathrm{H}_2$ molecule.
Answer:
1. Two hydrogen atoms $\mathrm{H}_{\mathrm{a}}$ and $\mathrm{H}_{\mathrm{b}}$ are separated by infinite distance. At this stage, there is no interaction between these two atoms and the potential energy of this system is arbitrarly taken as zero.
2. As these two atoms approach each other, in addition to electrostatic attractive forces between the nucleus and its own electrons, the following new forces begins to operate

3. The new attractive forces arise between:
- nucleus of $\mathrm{H}_{\mathrm{a}}$ and valence electron of $\mathrm{H}_{\mathrm{b}}$
- nucleus of $\mathrm{H}_{\mathrm{b}}$ and the valence electron of $\mathrm{H}_{\mathrm{a}}$
4. The new repulsive forces  arise between:
- the nucleus Of $\mathrm{H}_{\mathrm{a}}$ and $\mathrm{H}_{\mathrm{b}}$
- the valence electrons of $\mathrm{H}_{\mathrm{a}}$ and $\mathrm{H}_{\mathrm{b}}$
5. The attractive forces tend to bring $\mathrm{H}_{\mathrm{a}}$ and $\mathrm{H}_{\mathrm{b}}$ together whereas the repulsive forces tends to push them apart.
6. At the initial stage, as the two hydrogen atoms approach each other, the attractive forces are stronger than repulsive forces and the potential energy decreases.
7. A stage is reached where the net attractive forces are exactly balanced by repulsive forces and the potential energy of the system acquires a minimum energy.
8. At this stage, there is a maximum overlap between the atomic orbitals of $\mathrm{H}_{\mathrm{a}}$ and $\mathrm{H}_{\mathrm{b}}$ and atoms $\mathrm{H}_{\mathrm{a}}$ and $\mathrm{H}_{\mathrm{b}}$ are now said to be bonded together by a covalent bond.
Question 6.
What arc the salient features of Valence Bond (VB) theory?
Answer:
1. When half filled orbitals of two atoms overlap, a covalent bond will be formed between them.
2. The resultant overlapping orbitals are occupied by the two electrons with opposite spins. For example when $\mathrm{H}_2$ is formed, the two is electron of two hydrogen atoms get paired up and occupy the overlapped orbitals.
3. The strength of a covalent bond depends upon the extent of overlap of atomic orbitals. Greater the overlap, larger is the energy released and stronger will be the bond formed.
4. Each atomic orbital has a specific direction (excepts-orbital which is spherical) and hence orbital overlap takes place in the direction that maximises overlap.
5. Depending upon the nature of overlap, the bonds are classified as $\sigma$ covalent bond and $\pi$ it covalent bond.
6. When two atomic orbitals overlap linearly along the axis, the resultant bend is called a sigma $(\sigma)$ bond. This overlap is also called or "axial overlap".

7. When two atomic orbitals overlap sideways the resultant covalent bond is called a pi $(\pi)$ bond.
Question 7.
Explain about $\mathrm{sp}$ hybridisation with suitable example.
Answer:
1. bond rormation in Beryllium chloride takes place by sp hybridisation.
2. The valence shell of Beryllium has the electronic configuration as follows:

3. In $\mathrm{BeCl}_2$, both the $\mathrm{Be}-\mathrm{Cl}$ bonds are equivalent and it was observed that the molecule is linear. VB theory explains this observed behaviour by $\mathrm{sp}$ hybridisation. One of the paired electrons in the $2 \mathrm{~s}$ orbital gets excited to $2 \mathrm{p}$ orbital.
4. Now the $2 \mathrm{~s}$ and $2 \mathrm{p}$ orbitals hybridise and produce two equivalent $\mathrm{sp}$ hybridised orbitals which have $50 \% \mathrm{~s}-$ character and $50 \% \mathrm{p}$-character. These $\mathrm{sp}$ hybridised orbitals are oriented in opposite direction.
5. Each of the $\mathrm{sp}$ hybridised orbitals linearly overlap with $\mathrm{p}$ orbital of the chlorine to form a covalent bond between $\mathrm{Be}$ and $\mathrm{Cl}$ atoms as follow:

Question 8.
Explain the formation of methane using VB theory?
Answer:
1. Methane is formed by $\mathrm{sp}^3$ hybridisation. In $\mathrm{CH}_4$ molecule, the central carbon atom is bounded to four hydrogen atoms.
2. The ground state valence shell electronic configuration of carbon is $[\mathrm{He}] 2 \mathrm{~s}^2 2 \mathrm{p}_{\mathrm{x}}^2 2 \mathrm{p}_{\mathrm{y}}^1 2 \mathrm{p}_{\mathrm{x}}^0$
3. En order to form four covalent bonds with the four hydrogen atoms, one of the paired electrons in $2 \mathrm{~s}$ orbital of carbon is promoted to its $2 \mathrm{P}_z$ orbital in the excited state.
4. The one $2 s$ orbital and three $2 \mathrm{p}$ orbitals of carbon atom mixes to give four equivalent $\mathrm{sp}^3$ hybridised orbitals. The angle between any of the two $\mathrm{sp}^3$ hybridised orbitals is $109^{\circ} \cdot 28^{\prime}$,

5. The Is orbital of the four hydrogen atoms overlap linearly with the four $\mathrm{sp}^3$ hybridised orbitais of carbon to form four $\mathrm{C}-\mathrm{H} \sigma$ bonds in the methane molecule as follows

Question 9.
Explain $\mathrm{sp}^3 \mathrm{~d}$ hybridisation with a suitable example.
Answer:
1. In the $\mathrm{PCl}_5$ molecule, the central atom phosphorous is covalently bonded to five chlorine atoms. Here the atomic orbitals of phosphorous undergoes $\mathrm{sp}^3 \mathrm{~d}^2$ hybridisation which involves its one $3 \mathrm{~s}$ orbital, three $3 \mathrm{p}$ orbitals and one vacant $3 \mathrm{~d}$ orbital $\left(\mathrm{d}_{\mathrm{z}}^2\right)$
2. The ground state electronic configuration of phosphorous is $[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}_{\mathrm{x}}{ }^2 3 \mathrm{p}_{\mathrm{y}}{ }^1 3 \mathrm{p}_z{ }^1$

3. One of the paired electrons in the $3 \mathrm{~s}$ orbital of phosphorous is promoted to one of its vacant $3 \mathrm{~d}$ orbital (dz ${ }^2$ ) in the excited state.

4. The $3 p_z$ orbitals of the five chlorine atoms linearly overlap along the axis with the five $\mathrm{sp}^3 \mathrm{~d}$ hybridised orbitals of phosphorous to form the five $\mathrm{P}-\mathrm{CI}$ bonds as follows.

Question 10.
Explain about. $\mathrm{sp}^3 \mathrm{~d}^2$ hybridisation with an example.
Answer:
1. In sulphur hexafluoride $\mathrm{SF}_6$, the central atom sulphur extend its octet to undergo $\mathrm{sp}^3 \mathrm{~d}$ hybridisation to generate $\operatorname{six} \mathrm{sp}^3 \mathrm{~d}^2$ hybridised orbitals which accounts for $\operatorname{six}$ equivalent $\mathrm{S}-\mathrm{F}$ bonds.
2. The ground state electronic configuration of sulphur is $[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}_{\mathrm{x}}^1 3 \mathrm{p}_{\mathrm{y}}^1 3 \mathrm{p}_{\mathrm{z}}^1$

$\text { 3. One electron each form } 3 \mathrm{~s} \text { orbital and } 3 \mathrm{p} \text { orbital of sulphur is promoted to its two vacant } 3 \mathrm{~d} \text { orbitals } \mathrm{d}_z \text { and }$

4. A total of six valence orbitals from sulphur (one $3 \mathrm{~s}$ orbital, three $3 \mathrm{p}$ orbitals and two $3 \mathrm{~d}$ orbitals) (d $\mathrm{d}^2$ and $\mathrm{d}_{\mathrm{x}} \mathrm{x}^2 \mathrm{y}^2$ ) which mixes to give six equivalent $\mathrm{sp}^3 \mathrm{~d}^2$ hybridised orbitals. The orbital geometry is octahedral.
5. The $\operatorname{six} \mathrm{sp}^3 \mathrm{~d}^2$ hybridised orbitals of sulphur overlaps linearly with $2 \mathrm{p}_{\mathrm{z}}$ orbital of six fluorine atoms to form the $\operatorname{six} S-F$ bonds in sulphur hexa fluoride.

Question 11.
Explain about the salient features of molecular orbital theory.
Answer:
1. When atoms combine to form molecules, their individual atomic orbitals lose their identity and form new orbitals called molecular orbitals.
2. The shape of molecular orbitals depend upon the shapes of combining atomic orbitals.
3. The number of molecular orbitals formed is the same as the number of combining atomic orbitals. half the number of molecular orbitals formed will have lower energy and are called bonding orbitals, while the remaining half molecular orbitals will have higher energy and are called anti-bonding molecular orbitals.
4. The bonding molecular orbitals are represented as $\sigma$ (sigma), $\pi$ (pi), $\delta$ (delta) and the corresponding antibonding orbitals are called $\sigma^*, \pi^*$ and $\delta^*$.
5. The electrons in the molecule are accommodated in the newly formed molecular orbitals. The filling of electrons in these orbitals follow Autbau's Principle, Pauli's exclusion principle and Hund's rule as in the case of filling of electrons in the atomic orbitals.
6. Bond order gives the number of covalent bonds between the two combining atoms. The bond order of a molecule can be calculated using the foLlowing equation:
$
=\frac{\mathrm{N}_b-\mathrm{N}_a}{2}
$
$\mathrm{N}_{\mathrm{b}}=$ Number of electrons in bonding molecular orbitals.
$\mathrm{N}_{\mathrm{a}}=$ Number of electrons in anti-bonding molecular orbitals.
7. A bond order of zero value indicates that the molecule does not exist.
Question 12 .
Explain the MO diagram for NO molecule.

Answer:
1. Electronic configuration of $N$ atom is $1 s^2 2 s^2 2 p^3$
2. Electronic configuration of $\mathrm{O}$ atom is $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^4$
3. Electronic configuration of NO molecule is
$
\begin{gathered}
\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 \pi^2 p_x^2 \pi 2 p_y{ }^1 \\
=\frac{\mathrm{N}_b-\mathrm{N}_a}{2}=\frac{10-5}{2}=2.5
\end{gathered}
$
4. Bond order
5. NO molecule has one unpaired electron, hence it is paramagnetic.
Question 13.
Explain about metallic bonding.
Answer:
1. The forces that keep the atoms of the metal so closely in a metallic crystal constitute what is known as metallic bond.
2. According to Drude and Lorentz, metallic crystal is an assemblage of positive ions immersed in a gas of free electrons. The free electrons are due to ionisation of the valence electrons of the atoms of the metal.
3. As the valence electrons of the atoms are freely shared by all the ions in the crystal, the metallic bonding is referred to as electronic bonding.
4. The electrostatic attraction between the metal ions and the free electrons yield a three dimensional close packed crystal with a large number of nearest metal ions. So metals have high density.
5. As the close packed structure contains many slip planes along which movement can occur during mechanical loading, metal acquires ductility.
6. As metal ion is surrounded by electron cloud in all directions, the metallic bonding has no and thermal conductivity. The metallic lustre is due to the reflection of light by the electron cloud.As the metallic bond is strong enough, the metal atoms are reluctant to break apart into a liquid or gas, so the metals have high melting and boiling points.
7. High thermal conductivity of metals is due. to thermal excitation of many electrons from the valence bond to the conduction band.

Question 14.
Explain about the bonding in metals by molecular orbital theory.
Answer:
1. According to molecular orbital theory the atomic orbitals of large number of atoms in a crystal overlap to form numerous bonding and anti-bonding molecular orbitals without any band gap.
2. The bonding molecular orbitals are completely filled with an electron pair in each and the anti-bonding molecular orbitals are empty.
3. Absence of band gap accounts for high electrical conductivity of metals.
4. High thermal conductivity is due to thermal excitation of many electrons from the valence band to the conduction band.
5. With an increase in temperature, the electrical conductivity decreases due to vigorous thermal motion of lattice ions that disrupts the uniform lattice structure. that is required for free motion of electrons within the crystal.
Common Errors
1. The number of bonds formed by elements may go wrong.
2. When writing Lewis structure, electrons may be written in an irregular way.
3. Coordinate covalent bond should not be written as a line
Rectifications
1. Always hydrogen and fluorine form 1 bond Oxygen 2 bonds, Nitrogen 3 bonds, Carbon 4 bonds
2. When writing lewis structure, each atom should be surrounded by eight electrons in such a way as 4 pairs of electrons.
3. Coordinate covalent bond should be written from donor atom to acceptor atom as an arrow mark Donor $\rightarrow$ Acceptor