CBSE 11th Standard Chemistry Subject Chemical Bonding and Molecular Structure Ncert Exemplar 3 Mark Questions With Solution 2021
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CBSE 11th Standard Chemistry Subject Chemical Bonding and Molecular Structure Ncert Exemplar 3 Mark Questions With Solution 2021
11th Standard CBSE
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Reg.No. :
Chemistry
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Write lewis structure of the following compounds and show formal charge each atom.
(a) -
Write the resonance structures for NO2
(a) -
What is meant by the term average bond enthalpy? Why is there difference in bond enthalpy of O - H bond in ethanol (C2H5OH) and water?
(a) -
Give reasons for the following.
Water molecule has bent structure whereas carbon dioxide molecule is linear.(a) -
Predict the hybridisation of each carbon in the molecule of organic compound given below. Also indicate the total number of \(\sigma \ and \ \pi \)-bonds in this molecule.
(a)
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CBSE 11th Standard Chemistry Subject Chemical Bonding and Molecular Structure Ncert Exemplar 3 Mark Questions With Solution 2021 Answer Keys
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Formal charge on an atom in a Lewis structure
= [total number of valence electrons in free atom]
- [total number of non-bonding (lone pairs) electrons]
- \(\frac { 1 }{ 2 } \)[total number of bonding or shared electron]
Formal charge on H = 1 - 0 x 2 = 0
Formal charge on N = 5 - 0 x 8 = 1
Formal charge on O (1) = 6 - 4 - x 4 = 0
Formal charge on O (2) = 6 - 4 - x 4 = 0
Formal charge on O (3) = 6 - 6 - x 2 = -1
On solving we get, formal charges on O(1), N and O(2) as 0, +1 and -1 respectively.
Similarly, on solving we get formal charge on H(1), H(2), O(2), O(3), O(4) ans S, as 0, 0, 0, -1 , 0, -1 and +2 respectively. -
Nitrogen dioxide, NO2
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All the similar bonds in a polyatomic molecule do not have the same bond enthalpies. e.g. in H2O (H-O-H) molecule bond undergoes some change because of changed chemical environment. Therefore, in polyatomic molecules the term mean or average bond enthalpy is used. It is obtained by dividing total bond dissociation enthalpy by the number of bonds broken.
e.g H2O(g) -> H(g) + OH(g);
\({ \triangle }_{ a }{ H }_{ 1 }^{ o }=502 \ KJ{ mol }^{ -1 }\)
Oh (g) -> H = O(g); \({ \triangle }_{ a }{ H }_{ 2 }^{ o }=427 \ KJ{ mol }^{ -1 }\)
Average O-H bond enthalpy
= \(\frac { 502+427 }{ 2 } \) = 464.5 Kj mol-1
The bond enthalpies of O-H bond in C2H5OH and H2O are different because of the different chemical (electronic) environment around oxygen atom.
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In H2O, oxygen atom is sp3 hybridised with two lone pairs. The four sp3 hybridised orbitals acquire a tetrahedral geometry with two corners occupied by hydrogen atoms while other two by the lone pairs.
The bond angle is reduced to 104.5o due to greater repulsive forces between lp - lp and the molecule thus acquires a V-shape or bent structure (angular structure).
In CO2 molecule, carbon atom is sp-hybridised. The two sp hybrid are oriented in opposite direction forming an angle of 180o \(O\overset { \pi }{ = } C\overset { \pi }{ = } O\\ \quad \quad \sigma \quad \quad \sigma \)
That's why H2O molecule has bent structure whereas CO2 molecule is linear. -