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7 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter Four

Which of the following molecular orbitals has maximum number of nodal planes?

(i) σ∗1

(ii) σ*2p_x

(iii) π2p_x

(iv) π*2p_y

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Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Ans: Option (ii)

Nodal plane is a plane which passes through the nucleus and the probability of finding an electron on a nodal plane is zero. Amongst the above given molecular orbital, only π*2p_x contains two nodal planes. Rest of the molecular orbitals, σ∗1, π2p_x, π*2p_y contains one nodal plane.

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Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter Four

Comprehension given below is followed by some multiple choice questions. Each question has one correct option. Choose the correct option.

Molecular orbitals are formed by the overlap of atomic orbitals. Two atomic orbitals combine to form two molecular orbitals called bonding molecular orbital (BMO) and anti bonding molecular orbital (ABMO). Energy of anti bonding orbital is raised above the parent atomic orbitals that have combined and the energy of the bonding orbital is lowered than the parent atomic orbitals. Energies of various molecular orbitals for elements hydrogen to nitrogen increase in the order :

σ1s $<$ σ*1s< 2s < *2s< [ 2p_x= π2p_y] < 2p_z< [ *2p_x=

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Comprehension given below is followed by some multiple choice questions. Each question has one correct option. Choose the correct option.

σ1s $<$ σ*1s< σ2s < σ*2s< [ π2p_x= π2p_y] < σ2p_z< [ π*2p_x= π*2p_y] σ*2p_z

for oxygen and fluorine order of energy of molecular orbitals is given below :

σ1s $<$ σ*1s< σ2s < σ*2s< σ2p_z< [ π2p_x= π2p_y] < [ π*2p_x= π*2p_y] σ*2p_z

Different atomic orbitals of one atom combine with those atomic orbitals of the second atom which have comparable energies and proper orientation. Further, if the overlapping is head on, the molecular orbital is called 'Sigma', (σ) and if the overlap is lateral, the molecular orbital is called 'pi', (π). The molecular orbitals are filled with electrons according to the same rules as followed for filling of atomic orbitals. However, the order for filling is not the same for all molecules or their ions. Bond order is one of the most important parameters to compare the strength of bonds.

Which of the following statements is correct?

(i) In the formation of dioxygen from oxygen atoms 10 molecular orbitals will be formed.

(ii) All the molecular orbitals in the dioxygen will be completely filled.

(iii) Total number of bonding molecular orbitals will not be same as total number of anti bonding orbitals in dioxygen. (iv) Number of filled bonding orbitals will be same as number of filled anti bonding orbitals.

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Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Ans: (i) The electronic configuration of dioxygen is:

σ1s² σ∗1s² σ2s² σ∗2s²s2p_z²π2p²_y π2p²_x π*2p¹_y π*2p¹_x

(ii) As it can be seen from the electronic configuration of oxygen atom π*2p¹_X π*2p¹_Y the are partially. So, the statement given is incorrect.

(iii) The statement given is correct because there are five bonding molecular orbitals and four antibonding molecular orbital in oxygen molecules. Hence, the bonding and antibonding molecular orbitals are no equal.

(iv) The filled bonding orbitals are not the same as the number of filled antibonding orbi

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7 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter Four

(i) Discuss the concept of hybridisation. What are its different types in a carbon atom.

(ii) What is the type of hybridisation of carbon atoms marked with star.

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Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Ans: (i) Different types of hybridization in a carbon atom are:

(a) sp hybridization: The carbon atoms forming triple bonds with each other determines hybridized carbon. ( Cº C ).

(b) sp² hybridization: The carbon atoms forming double bonds with each other determine sp² hybridized carbon. (C=C). (c) sp³hybridization: The carbon atoms forming single bonds with each other determine sp³ hybridized carbon. ( C-C).

(ii)The starred C atom is sp² hybridised.

B] The starred C atom is sp³ hybridised.

C] The starred C atom is sp² hybridised.

D] The starred C atom is sp³ hybridis

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7 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter Four

Describe hybridisation in the case of PCl₅ and _SF6. The axial bonds are longer as compared to equatorial bonds in PCl₅whereas in SF₆ both axial bonds and equatorial bonds have the same bond length. Explain.

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Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Ans: Hybridisation in the case of PCl₅ :

The hybridisation of phosphorus PCl₅ is sp₃d and it has a trigonal bipyramidal geometry. The axial bonds in PCl₅ are slightly longer as compared to the equatorial bonds because axial bonds experience greater repulsive forces from other bonds as compared to the equatorial bonds.

Hybridisation in the case of SF₆:

The hybridisation of sulphur in SF₆ is sp³d² and the molecule has an octahedral geometry. The bond length of axial as well as the equatorial bonds are similar because all the bonds in octahedral geometry experience similar r

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7 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter Four

Use the molecular orbital energy level diagram to show that N₂ would be expected to have a triple bond, F₂, a single bond and Ne₂, no bond.

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Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Ans: Formation of N₂ molecule: Electronic Configuration

σ1s^{2
$<$}σ* < 1s² < 2s² < *2s² < [ 2p²_x= π2p²_y] < 2p²_z

The bond order will be:

$\frac{1}{2}$ ( 10-4)= 3.0

Bond order indicates the number of bonds in a diatomic molecule. Triple bond N=N.

Formation of F₂ molecule: Electronic Configuration

σ1s² < *1s²< 2s² < *2s²< 2p²_z < [ 2p²_x= π2p²_y] < [ *2p²_x= π*2p²_y] σ*2p²_z

The bond order will be:

$\frac{1}{2}$ ( 10-10)= 0

Hence there will be no bond formation in between the neon atoms.

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7 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter One

Assertion (A) : Combustion of 16 g of methane gives 18 g of water.

Reason (R) : In the combustion of methane, water is one of the products.

(i) Both A and R are true but R is not the correct explanation of A.

(ii) A is true but R is false.

(iii) A is false but R is true.

(iv) Both A and R are false.

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alok kumar singh

Contributor-Level 10

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Option (C) The combustion reaction of methane is given below

CH₄ + O₂ -> CO₂ + H₂O

In this reaction, water and carbon dioxide are formed.

According to the reaction, 1 mole of methane gives 2 moles of water. So, 16 g of methane gives 36 g of water

Thus, X is false but R is true

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7 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter One

Assertion (A) : Significant figures for 0.200 is 3 where as for 200 it is 1.

Reason (R) : Zero at the end or right of a number are significant provided they are not on the right side of the decimal point.

(i) Both A and R are true and R is correct explanation of A.

(ii) Both A and R are true but R is not a correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

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alok kumar singh

Contributor-Level 10

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Option (C)

In the decimal portion, only trailing or final zeros are significant. This is in accordance with the rules of significant figures.

The given number, 0.200 has 3 significant figures whereas the number, 200 has only one significant figure because zeros present on the right side or at the end of the decimal number are significant and it is provided that they are present on the right side of the decimal point.

Also, zeros are not significant in numbers without decimals. Thus, X is true but R is false.

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7 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter One

Assertion (A) : One atomic mass unit is defined as one twelfth of the mass of one carbon-12 atom.

Reason (R) : Carbon-12 isotope is the most abundunt isotope of carbon and has been chosen as standard.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

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alok kumar singh

Contributor-Level 10

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Option (B)

Atomic mass unit is defined as a mass that is equal to accurately the mass of carbon-12 atom.

When scientists compare the relative atomic masses of the elements to the mass of carbon, they find that they are near to a whole number value.

So, both X and B are true but R is the correct explanation of A.

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7 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter One

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion (A): The empirical mass of ethene is half of its molecular mass.

Reason (R): The empirical formula represents the simplest whole number ratio of various atoms present in a compound.

(i) Both A and R are true and R is the correct explanation of A.

(ii) A is true but R is false.

(iii) A is false but R is true.

(iv) Both A and R are false

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alok kumar singh

Contributor-Level 10

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

(i) Both A and R are true and R is the correct explanation of A.

Explanation: The molecular formula gives the actual number of different atoms present in a molecule and the empirical formula gives a simple whole number ratio of different atoms present in a molecule.

Since ethene C₂H₄ can be further simplified, the empirical formula becomes CH₂. The molecular mass of C₂H₄ is 28 g/mol and the empirical mass of CH2 is 14 g/mol.

Hence, the empirical mass of ethene is half of its molecular mass.

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7 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter One

Match the following physical quantities with units.

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alok kumar singh

Contributor-Level 10

This is a Matching Type Questions as classified in NCERT Exemplar

(i) → (e) (ii) → (d) (iii) → (b) (iv) → (g) (v) → (c), (h) (vi) → (f) (vii) → (a) (viii) → (i)

Explanation:

(i) It is defined as the number of moles of solute dissolved in 1 litre of solution. So, the unit of molarity is mol L-1
(ii) Mole fraction is defined as the number of moles of a constituent divided by the total number of moles. So, it is unitless.

(iii) Mole is the unit to measure the amount of an atom or molecule in SI system. it is symbolized as "mol".

(iv) It is defined as the number of moles of solute dissolved in 1 kg of solvent, so, the unit o

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