Chemistry NCERT Exemplar Solutions Class 11th Chapter Four Chemical Bonding and Molecular Structure

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Ans: In BrF₅ the central bromine atom has 7 valence electrons. It makes five bonds with the fluorine atom and one lone pair of electron is left. Due to the lone pair-bond pair repulsions, BrF₅ makes a structure of square pyramidal geometry. Due to the distortion of fluorine ions, each fluorine ion makes an angle of 90° . The square pyramidal shape of BrF₅ is

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Ans: (a) Which of the two compounds will have intermolecular hydrogen bonding and which compound is expected to show intramolecular hydrogen bonding.

Ans: The intramolecular hydrogen bonding is shown by compound (I) and the intermolecular hydrogen bonding is shown by compound (II) . In compound (I) the NO₂ and OH group are close together in comparison to that in compound (II) . So, that is why compound (I) shows intramolecular hydrogen bonding. The intermolecular hydrogen bonding in compound (II) is shown as:

(b) The melting point of a compound depends on, among other things, the extent of hydrogen bonding. On this basis explain which of the above two compounds will show higher melting point.

Ans: Due to the formation of intermolecular hydrogen bonds, compound (II) will have a higher melting point. Thus, due to hydrogen bond formation more molecules are joined together.

(c) Solubility of compounds in water depends on power to form hydrogen bonds with water. Which of the above compounds will form hydrogen bond with water easily and be more soluble in it?

Ans: Compound (I) will be less soluble in water due to intramolecular hydrogen bonding; it will not be able to form hydrogen bonds with water. Whereas, compound (II) can form hydrogen bonds with water more easily and is thus more soluble in water.

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Ans: In the first figure given above, the area which is under + overlapping is equal to the area under +- overlap. Both the overlaps cancel out with each other as they are oppositely charged. Due to cancelling out of the overlaps the net overlap will be zero. In the second figure given above, both the p-orbitals are perpendicular to each other. Due to the,   p_x p_y orbitals being perpendicular with each other, no overlap will be possible.

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Ans: In PCl₅, P has 5 valence electrons in orbitals and make 5 bonds with 5 Cl atoms, it will share one of its electrons from 3s to 3d orbital, therefore the hybridization will be sp3d and the geometry will be trigonal bipyramidal. IF₅, the Iodine atom has 7 valence electrons in molecular orbitals it will form 5 bonds with 5 Cl atoms using 5 electrons from its molecular orbital, two electrons will form one lone pair on Iodine atom, which gives the square pyramidal geometry.

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Ans: Although the hybridisation of the central atom oxygen in both the molecules is 3 sp but dimethyl ether will have a higher bond angle than water molecule.

Due to the presence of two bulky methyl groups in dimethyl ether, the repulsive forces will be greater in them than the two hydrogens in water molecules. In dimethyl ether the -CH₃ is a group attached to three hydrogen atom through s bonds. Thus, the C- H bond pairs increase the electron density on the carbon atom which results in lone pair-bond pair repulsions. Due to this lone pair-bond pair repulsions, the bond angle of dimethyl ether is greater than that of water molecules.

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Ans: The Lewis structures of the compounds are:

Lewis structure for nitric acid:

Lewis structure of nitrogen dioxide:

Lewis structure of sulphuric acid:

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Ans: The general sequence of the energy level of the molecular orbital is σ1s < σ*1s < σ2s< σ*2s < π2px = π2py < σ2pz

N₂    =   σ1s² σ*1s² σ2s² σ*2s² π2p²_x = π2p²_y σ2p²_z

N₂⁺  =   σ1s² σ*1s² σ2s² σ*2s² π2p²_x = π2p²_y σ2p¹_z

N₂^-   =   σ1s² σ*1s² σ2s² σ*2s² π2p²_x= π2p²_y σ2p²_z σ2p²_x

N₂²⁺ =   σ1s² σ*1s² σ2s² σ*2s² π2p²_x = π2p²_y

The formula for finding the bond order (B. O) for any molecular species is:

Bond order = $\frac{1}{2}$ ( N_b- N_a)

Hence, the bond orders of the given molecular species are:

N₂= $\frac{1}{2}$ ( 10-4)= 3

N₂⁺= $\frac{1}{2}$ ( 9-4)= 2.5

N₂^-= $\frac{1}{2}$ ( 10-5)= 2.5

N₂²⁺= $\frac{1}{2}$ (8-4)= 2

The order for the stability of the given molecular species thus will be:

N_2.> N₂^-> N₂⁺> N₂²⁺

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Ans: (i) N₂ → N₂⁺ + e^-

The electronic configuration of N₂ is:

σ1s² σ*1s² σ2s² σ*2s² π2p²_x = π2p²_y σ2p²_z

So, its bond order will be 3

The electronic configuration of N₂⁺ is:

σ1s² σ*1s² σ2s² σ*2s² π2p²_x = π2p²_y σ2p¹_z

Its bond order will be 2.5

Hence the bond order decreases in this reaction N₂ → N₂⁺ + e^-

(ii) O₂→ O₂⁺ + e-

The electronic configuration of O₂ is:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z² π2p²_y π2p²_x π*2p¹_y π*2p¹_x

Its bond order will be 2

The electronic configuration of O₂⁺ will be:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z², π2p²_y , π2p²_x, π*2p¹_x

Its bond order will be 2.5 .

Hence, the bond order in this reaction increases O₂→ O₂⁺ + e-

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Ans: (i) Due to the electrostatic forces between the two opposite charges, ionic bonds are non-directional. The bonding direction does not matter as the electrostatic field of an ion is non-directional. Whereas, the formation of covalent bonds happens with the overlap of the atomic orbitals. The direction of the bonds is given by the direction of overlapping.

(ii) The hybridization of the oxygen atom in water molecules is 3 sp due to the presence of two lone pairs of electrons on the oxygen atom. Tetrahedral geometry is acquired by these four 3 sp hybridized orbitals with two bonds of hydrogen atoms and two lone pairs of electrons. Due to the presence of lone pairs of electron the bond angle gets reduced as there is a repulsion between the LP-LP and thus the molecule acquires a bent shape. The hybridization of carbon atoms in the carbon dioxide molecule is sp. The two sp are arranged in such a way that it makes an angle of 180° . That is why carbon dioxide has a linear shape.

 

 

(iii) Both the carbon atoms in ethyne molecules are sp hybridized. Ethyne has two unhybridized 2p_x 2p_y orbitals. The two sp hybrid orbitals of both the carbon atoms are arranged such that they form an angle of 180° with each other. That is why ethyne molecules are linear in shape.

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Ans: Ionic bonds are such bonds in which complete transfer of electrons from one atom to another atom. Due to complete transfer positive and negative ions are formed in this bond. The ions in this bond are held together by electrostatic force of attraction. The formation of calcium fluoride CaF₂ leads to the formation of an ionic bond.

Ca→  Ca²⁺ + e ^-   The electronic configuration of  Ca= [Ar] 4s²   and Ca²⁺= [Ar]

F +e ^-→F^- The electronic configuration of F= [He] 2p⁵  and F = [He ]2p⁶

Thus Ca²⁺ +2F^- → CaF₂

A covalent bond is formed due to mutual sharing of electrons between two atoms of a non-metal. This mutual sharing of electrons between the two non-metal species leads to the formation of a covalent bond. The making of Cl₂ leads to a formation of a covalent bond between two chloride ions.

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N—H, F—H, C—H and O—H

Ans: The ionic character in a molecular species is decided by the electronegativity difference between the two bonded atoms. Greater the electronegativity difference between two bonded pairs, the greater will be the ionic character. The electronegativity difference of the given species is

C - F = (2.5 - 2.1) = 0.4

N -H = (3.0 - 2.1) = 0.9

O - H = (3.5 - 2.1) = 1.4

F - H = (4.0 - 2.1) = 1.9

According to the above given difference in the electronegativities, the order of increasing ionic character is:

C – H< N - H

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Ans: The carbonate ion CO₃^2- can be best represented by its resonating structures. The carbonate ion cannot be represented by a single Lewis structure because the three carbon oxygen bond lengths are the same. This cannot be shown by a single Lewis structure. For showing the similar lengths of all the carbon to oxygen bonds three hybrid structures are constructed which are in resonance with each other.

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Ans: The hybridization of Carbon 1 is sp, carbon 2 is sp, carbon 3 sp2, carbon 4 is sp3 and carbon 5 is sp². The triple bond has 2 pie bonds and one sigma bond. Each double bond has one sigma and one pie bond. Every single bond is a sigma bond. Thus, the total number of sigma bonds is 11 and pie bonds are 4.

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Ans: BeCl₂ has a linear structure

HOCl is also non-linear in structure.

H₂O has a V-shaped structure.

Cl₂O has a V-shaped structure.

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Ans: (i) According to the electronic configuration, a total of 8 electrons must be present in the valence shell of an element. Element X has 4 valence electrons, so it will share the remaining 4 electrons for the formation of the bond, the molecular formula will be XH₄ . The element Y has 5 valence shell electrons, so it will form 3 bonds and the formula will be YH₃ . The element Z has 7 valence shell electrons, so it will form one bond with hydrogen and has the molecular formula H-Z .

(ii) Elements X, Y and Z having, 5 and 4 7 valence electrons respectively belongs to the second period and group 14^th ,15^th and 17^th . The electronegativity of the elements increases on moving across the period and the dipole moment is calculated by the difference in the electronegativity of the atoms in a molecular structure. Higher the difference in the electronegativity will be the dipole moment. So, the dipole moment of H - Z will be the highest.

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Ans: (i)

(ii)

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Ans: In compound BCl₃, Boron has sp²-hybridisation and the shape is Triangular Planar.

In methane CH₄, Carbon has sp³ -hybridization and shape are Tetrahedral.

In carbon dioxide CO₂, carbon has sp-hybridisation and shape is Linear.

In ammonia NH₃, nitrogen has sp³-hybridisation and shape is Pyramidal.

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Ans: All the C-Obonds in carbonate ion (CO₃^2-) are equal in length due to the equivalent resonance of all the three carbon-oxygen bonds gets a double bond character atleast once. This type of double bond character happened throughout 3C- O skeleton, and hence all the bonds acquired equal bond length.

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Ans: The average bond enthalpy is defined as the ratio of total bond dissociation enthalpy to the number of bonds broken in the structure.

The identical O- H bonds in water molecule does not have the same bond enthalpies. According to the structure of a water molecule, there are two O- H bonds, but there is a change in the breaking of the first O- H bond than the second because of the different charge.

Hence in water molecule the average bond enthalpy will be:

The average O-H bond enthalpy = $\frac{502+427}{2}$ = 464 mol^-1

In ethanol C₂H₅OH  the bond enthalpy of O-H is different because the electronic environment around oxygen atom is different. In ethanol the O-H is attached to the carbon atom and in water molecule the O-H is attached to the hydrogen atom.

$\mathcal{l}=2\pi r$

$\Rightarrow 314cm=2\times 3.14r$

$\Rightarrow r=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.m=0.5m$

$\mu =iA$

$=14\times \pi r^2$

$=14^2\times \frac{22}{7}\times \frac{1}{4}=11$

Reaction is

Mass of product, 1, 2, 2- tetrobromopropane obtained

= $\frac{1\times 360}{160\times 2}\times \frac{27}{100}$

= 3.30375

= 3.0375 × 10^-1

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Ans: Formation of N₂ molecule: Electronic Configuration

σ1s^{2 $<$} σ* < 1s² < σ2s² < σ*2s² < [ π2p²_x = π2p²_y] < σ2p²_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 < σ*1s²< σ2s² < σ*2s²  < σ2p²_z < [ π2p²_x = π2p²_y] < [ π*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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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 repulsive forces from other bonds.

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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) sp³ hybridization: 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 hybridised.

E] The starred C atom is sp² hybridised.

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

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z² π2p²_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 orbitals. The filled bonding orbitals are five and only two filled antibonding orbitals are present in dioxygen molecules.

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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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Ans: Option (ii)

(i) The electronic configuration of O₂ is:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z² π2p²_y π2p²_x π*2p¹_y π*2p¹_x

Bond order = $\frac{1}{2}$ ( N_b- N_a)

Bond order = $\frac{1}{2}$ (10-6)=2.0

The electronic configuration of N₂ :

σ1s² σ*1s² σ2s² σ*2s² π2p²_x, π2p²_y σ2p²_z

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

So, the bond order of O₂ , N₂ will not be equal.

(ii) The electronic configuration of O₂⁺ is:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z²,π2p²_y , π2p²_x, π*2p¹_x

Bond order = $\frac{1}{2}$ (10-5)=2.5

The electronic configuration of N₂^-

σ1s² σ∗1s² σ2s² σ∗2s² ,π2p²_x , π2p²_y, s2p_z² π*2p¹_x

Bond order = $\frac{1}{2}$ (10-5)=2.5

Hence, the bond order of O₂⁺,N₂^- will be equal.

(iii) The electronic configuration of O₂ ^- is:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z²,π2p²_y , π2p²_x, π*2p²_y π*2p¹_x

Bond order = $\frac{1}{2}$ (10-7)=1.5

The electronic configuration of N₂⁺ is:

σ1s² σ*1s² σ2s² σ*2s² π2p²_x , π2p²_y σ2p¹_z

Bond order = $\frac{1}{2}$ (9-4)=2.5

Hence, the bond order of O₂^-, N₂⁺ will not be equal.

(iv) The electronic configuration of O₂ ^- is:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z²,π2p²_y , π2p²_x, π*2p²_y π*2p¹_x

Bond order = $\frac{1}{2}$ (10-7)=1.5

The electronic configuration of N₂^-

σ1s² σ∗1s² σ2s² σ∗2s² ,π2p²_x , π2p²_y, s2p_z² π*2p¹_x

Bond order = $\frac{1}{2}$ (10-5)=2.5

Hence, the bond order of O₂ ^- , N₂^- will not be equal.

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Ans: Option (iii)

Molecules from hydrogen to nitrogen shows this type of electronic configuration:

σ1s $<$ σ*1s< σ2s < σ*2s< [ π2p_x = π2p_y] < σ2p_z< [ π*2p_x = π*2p_y] σ*2p_z  in which σ2p_z is filled after π2p_x and π2p_y

So, amongst the given elements only nitrogen will show this type of electronic configuration in which σ2p_z molecular orbital is filled after π2p_y  and π2p_x

The electronic configuration of N2 is:

σ1s² σ∗1s² σ2s² σ∗2s², π2p²_x , π2p²_y, s2p_z² π*2p¹_x

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Ans:

(i) NO                                                    (c) 2.5

(ii) CO    (d) 3.0

(iii) O₂^2-                                                (a) 1.5

(iv) O₂                                                   (b) 2.0

Explanation :-

(i) The electronic configuration of NO is:

σ1s² σ∗1s² σ2s² σ∗2s², π2p²_x , π2p²_y, s2p_z² π*2p¹_x

The bond order will be:

$\frac{1}{2}$ ( 10-5)= 2.5

(ii) The electronic configuration of CO is:

σ1s² σ∗1s² σ2s² σ∗2s², π2p²_x , π2p²_y, s2p_z²

The bond order will be:

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

(iii) The electronic configuration of O₂ ^- is:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z², π2p²_y , π2p²_x, π*2p²_y π*2p¹_x

The bond order will be:

$\frac{1}{2}$ ( 10-7)= 1.5

(iv) The electronic configuration of O₂ is:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z² π2p²_y π2p²_x π*2p¹_y π*2p¹_x

The bond order will be:

$\frac{1}{2}$ ( 10-6)= 2.0

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Ans:

(i) Hydrogen bond                              (d) HF

(ii) Resonance    (e) O₃

(iii) Ionic bond                                    (b) LiF

(iv) Covalent solid    (a) C

Explanation :-

(i) In hydrogen fluoride HF, hydrogen forms strong hydrogen bonds with fluorine because of the electronegativity of fluorine. Fluorine is more electronegative than hydrogen and thus the electrons in fluorine are more stabilized and they cannot be easily donated to an acceptor.

(ii) Ozone is a neutral molecule which is stabilized through resonance. The negative charge created by an extra electron is delocalized by resonance through terminal oxygen atoms.

(iii) In lithium fluoride LiF, the lithium metal has a positive charge ( 1) + and is a cation. The fluoride ion has a negative charge ( 1) - and is an anion. These two oppositely charged ions are held together by an ionic bond.

(iv) Carbon is a known covalent solid. The two allotropes of carbon, diamond and graphite are covalent network solids with different geometries. Due to the presence of covalent bonds in graphite and diamond, they have very high melting points and are poor conductors of heat and electricity.

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Ans:

(i) Tetrahedral    (c) sp³

(ii) Trigonal    (a) sp²

(iii) Linear    (b)sp

Explanation :-

(i) A tetrahedral molecule has four electron pairs and these make sigma bonds with each other. The s and p orbitals overlap with each other thus forming a sp³ hybridized molecule.

(ii) There are two possibilities for the central atom to have sp² Either all the bonds are in place i.e, the bond pairs are all sigma bonds or pi bonds or there are only two bonds and one lone pair of electrons.

(iii) The sp hybridization involves the mixing of the valence electrons in s orbital with another valence electron in the p orbital which yields two sp orbitals which are oriented in a linear geometry.

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Ans: Option (iv)

Both the statements given in the assertion and reason are false.

The energy required to break the first O-H bond is not same than that of the second O-H bond in water molecule, that means the bond enthalpies of the two O-Hare different. This is because the electronic environment around the oxygen atom in water molecules is not the same even after the O-H bond breaks.

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Ans: Option (iv)

The number of bond pairs on the nitrogen atom of NO₃^-  are 4 and the structure of - NO₃ does not have any lone pairs of electrons. The structure of  NO₃^- molecule is:

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Ans: Option (iii)

The molecular orbital diagram of O₂^2-,   π^∗2Px and π^∗2Py molecular orbitals are completely filled, which are partially filled in O2. These are no unpaired electrons. So,  O₂^2-does not contain unpaired electrons.

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Ans: Option (iv)

The electrons involved in the formation of any chemical bond are the valence shell electrons. In the given electronic configuration,1s² 2s² 2p⁶ 3s² 3d² 4s² the valence shell electrons are in d-orbital and s-orbital. So, the four electrons involved in the chemical bond formation will be 3d² 4s².

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Ans: Option (i)

The electronic configuration of A 1S² 2S²  2P⁶  is having no unpaired electrons in the outer most shells. So, an atom having no unpaired electron in its valence shell is said to be stable.

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Ans: Option (iv)

The electronic configuration of B given as 1S² 2S²  2P⁶ 3S² 3P³ which represents phosphorous atom. The electronic configuration of C is given as 1S² 2S²  2P⁶ 3S² 3P⁵   which represents chlorine molecule.

So, the molecular formula representing B and C will be phosphorous trichloride ( PCl₃ ) which corresponds to BC₃ .

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Ans: Option (iv)

In the molecules of boron ( B₂ ), carbon ( C₂ ) and nitrogen ( N₂ ) the energy of s2p_z  molecular orbital is greater than the energy of P2p_x and P2p_y molecular orbital.

So, the correct order of energies of molecular orbital of N2 is

  (π2p_y ) > (σ2p_z ) < (π* 2p_x ) ≈ (π*2p_y )

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Ans: Option (ii)

The electronic configurations and the bond orders of the given species are given below:

Bond order of O₂ ^- : σ1s² σ∗1s² σ2s² σ∗2s² σ2p²_z (π2p²_x = π2p²_y) (π∗2p¹_x= π∗2p¹_y)

Bond order = $\frac{1}{2}$ ( N_b- N_a)

BO= $\frac{1}{2}$ (10-7)= $\frac{3}{2}$ = 1.5

Bond order of O₂ ⁺ : σ1s² σ∗1s² σ2s² σ∗2s² σ2p²_z (π2p²_x = π2p²_y) (π∗2p¹_x= π∗2p⁰_y)

Bond order = $\frac{1}{2}$ (10-5)= $\frac{5}{2}$ =2.5

Bond order of O₂ : σ1s² σ∗1s² σ2s² σ∗2s² σ2p²_z (π2p²_x = π2p²_y) (π∗2p¹_x= π∗2p¹_y)

BO= $\frac{1}{2}$ (10-6) = $\frac{4}{2}$ =2

So, the correct order of bond order is (ii) O₂ ^– < O₂ < O₂ ⁺

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Ans: Option (ii)

Aluminium has [Ne]3s² 3p¹  electronic configuration. Phosphorus has [Ne]3s² 3p³. The electronic configuration of silicon is [Ne]3s² 3p² and the electronic configuration of arsenic is   [Ar]3d¹⁰4s²4p³.

Ionisation energy increases as the atomic number in a period increases and decreases on moving down the group. It is also known that the ionisation enthalpy of group 15 elements is more than group 16 elements as group 15 has half-filled p-subshells which gives extra stability.Thus, the electronic configuration having the highest ionization enthalpy is [Ne]3s² 3p³

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Ans: Option (iii) and (iv)

(i) The hybridization of the central carbon atom in CO₃ ^2- is 2 sp as the carbon is bonded with three oxygen atoms. So, the given statement is incorrect.

(ii) The resonance structure of CO₃^2- has one C= O double bonds and two C- O single bonds. So, the statement is incorrect.

(iii) The net charge on three oxygen atoms is -2, hence the average formal charge on each oxygen atom is 0.67 units. So, the given statement is correct.

(iv) As it can be seen from the resonating structures that the structures of the bonds are not fixed. Hence, all the C- O bond lengths are equal. So, the given statement is correct.

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Ans: Option (i) and option (ii)

(i) NaCl being an ionic compound is a good conductor of electricity in the molten state. It is not a bad conductor of electricity in solid state. So, the statement mentioned in the question is not correct. Hence, this is the correct answer.

(ii) There is no difference in the arrangement of atoms in the canonical structures. The difference is in the arrangement of electron pairs in the canonical structures. So, the statement mentioned in the question is not correct. Hence, this is the correct answer.

(iii) The hybrid orbitals have the same energy and shape and hence they are more effective in the formation of stable bonds than the pure atomic orbitals. So, the statement given is correct. Hence, this is not the correct answer.

(iv) XeF₄ acquires a square planar geometry, this can very well be explained by the VSEPR theory. The hybridisation of the central xenon atom will be sp³d² due to the presence of two lone pairs of electrons on the central xenon atom. So, the statement given is correct. Hence, this is not the correct answer.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (ii)

The hybrid orbitals of nitrogen in the given species will be confirmed by knowing the hybridization. In  NO₂⁺ the central nitrogen atom is sp hybridized as it has a linear shape. The molecule NO₃^- is sp²  due to the presence of one lone pair of electrons on a nitrogen atom and hence having a bent geometry. The molecule NH₄⁺ is sp³  hybridized having a tetrahedral geometry.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (ii)

The size and the electronegativity are the main factors on which the strength of a compound depends on. As the size of the atom decreases, the electronegativity increases and thus the hydrogen-bonding becomes stronger. Thus, the strength of hydrogen-bonding in the given compounds is: H₂O > HF > NH₃

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (ii)

In a polyatomic molecule or an ion, the formal charge of an atom is defined as the difference between the valence electrons present in that atom and the number of electrons assigned to that atom in the Lewis structure.

Formal charge = No. of valence electrons - (No. of lone pair $+\frac{1}{2}Nbelectrons)$

Formal Charge = 6 - $\left(6+\frac{1}{2}*2\right)$

Formal Charge = -1

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (i)

(i) The molecule of BH₄ ^- has four bond pairs and zero lone pair of electrons, so it will be a tetrahedral molecule.

(ii) NH₂ ^- has two bond pairs and two lone pairs of electrons on the nitrogen atom. So, it will have a bent geometry.

(iii) CO₃ ^2- has three bond pairs and no lone pairs of electrons on carbon atoms. So, it will have a trigonal planar geometry.

(iv) H₃O⁺ has three bond pairs and one lone pair of electrons on oxygen atoms. So, it has a pyramidal geometry.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (iii)

A double bond has one s bond and one π bond. In the above given structure 5π bonds are present. The s bonds are the additions of all the single bonds in the structure. There are 19s bonds in the above given structure.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (iii)

(i) The molecule XeF₄ has a square planar geometry and the bond lengths of all the fluoride atoms attached to xenon atoms are equal.

(ii) The molecule BF₄ ^- has a tetrahedral geometry and the bond lengths of all the fluoride bonds attached to boron are equal.

(iii) In the molecule C₂H₄ all the bonds are not equal. The bond length of the double bond between the carbon atoms is 134 pm and the bond length of CH- is 110 pm.

(iv) The molecule SiF₄ has a tetrahedral geometry and the bond lengths of all the fluoride atoms attached to silicon atoms are equal.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (ii)

Hydrogen bond is the strongest in that molecules have a higher difference in electronegativity. Due to the small size of the oxygen atom, it has the highest electronegative character. So, H₂O molecules will have the strongest hydrogen bonding

HCl, HI and H₂Sdo not have hydrogen bonding between their atoms. So, H₂O will have the strongest hydrogen bonding.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (ii)

For a molecule having a hybridization of sp², its geometry in plane is trigonal planar. For a molecule having a trigonal planar geometry, its bond angle will be 120°

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (ii)

The electronic configuration of C is given 1S² 2S²  2P⁶ 3S² 3P⁵  which the electronic configuration of chlorine atom ( Cl ). It is unstable due to one unpaired electron in the p-subshell. Hence, the C₂ will be the stable form of C as it will form a dichloride molecule ( Cl₂ ).

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (ii)

The element B represents the electronic configuration of phosphorus atom and element C represents the electronic configuration of chlorine atom. Both phosphorus and chlorine are non-metals hence the bonding forming between them will be a covalent bond.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (iv)

(a) A molecule which has a zero bond order never exists and a molecule having a non-zero bond order either exists or is expected to exist.

(b) If the value of bond order is high then the bond strength will also be higher. Electrons present in the bonding molecular orbital are known as the bonding molecular orbital ( N_b ) and the electrons present in the anti-bonding molecular orbital are known as anti-bonding electrons ( N_a ). The half of the difference of bonding and anti-bonding electrons is known as the bond order.

The bond order of Be₂ : Be₂=s 1s2, s *1s², s 2s², s*2s²

The bonding and anti-bonding electrons are both 4 . Then the bond order (BO) will be:

BO= (4-4)

BO= 0

Bond order of Be₂ is zero. So, it does not exist.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (i)

The most electronegative element in the periodic table is fluorine having an atomic number 9 . The electronic configuration of fluorine is 1s² 2s² 2p⁵. Hence, the electronic configuration of the outermost shell of fluorine is 2s² 2p⁵ .

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (i) and option (ii)

Bond order depends on the electronic configuration of the molecular species, and the electronic configuration depends on the number of electrons a particular atom is contributing in its molecular structure.

The CN^- species has a total of 14 electrons; 6 electrons from carbon atom, 7 electrons from nitrogen atom and 1 electron for the negative charge.

The NO⁺ species has a total of 14 electrons; 8 from oxygen atoms, 7 from nitrogen atom and 1 electron will be subtracted due to the positive charge on the species.

The O₂ ^- species has a total of 17 electrons; 8 from each oxygen atom and 1 for the negative charge on the species.

The O₂^2- species has a total of 18 electrons; 8 from each oxygen atom and 2 for the negative charge on the species. Hence, the molecular species / ion having the same bond order will be CN^- and NO⁺ .

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (i) and (iv)

Beryllium chloride BeCl₂ does have the presence of any lone pair of electrons in its structure. So, it will attain a linear structure.

     Carbon disulphide CS₂ does not have any lone pair of electron on carbon atom. Thus, it will attain a linear structure.

The structures of NO₂ and NCO⁺ has a lone pair of electron on nitrogen atom thus it will form a bent shape rather than a linear shape.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (i) and (ii)

The species having the same number of electrons are known as isoelectronic species.

The number of electrons in CO species is 14; 8 electrons from oxygen atom and 6 electrons from carbon atom.

The number of electrons in NO⁺ is 14; 8 electrons from oxygen atom, 7 electrons from nitrogen atom and 1 electron will be subtracted due to the positive charge on the species.

The number of electrons in N₂ is 14 ; 7 from each nitrogen atom.

The number of electrons in SnCl₂ is 84 ;50 from the tin atom and 17 from each chlorine atom.

The number of electrons in NO₂ ^- is 24 ; 7 from nitrogen atom, 8 from each oxygen atom and 1 from the negative charge on the species.

Hence, NO₊ and N₂ are the isoelectronic species of CO.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (iii) and (iv)

(i) The number of valence electrons in CO₂ is 14. The oxygen atom attached to the carbon atom forms a double bond. Therefore, the hybridization of the structure is sp². Thus, carbon dioxide is linear in shape.

(ii) The number of valence electrons in CCl₄ is 32. The structure of CCl4 is arranged in a tetrahedral manner. The chlorine atoms are arranged such that the hybridization is sp³.

(iii) The number of valence electrons in O₃ is 24. The hybridization of the molecule is sp² and due to the presence of lone pair of electrons on the central oxygen atom and due to the effect of resonance, ozone molecule has a bent shape.

(iv) The number of valence electrons in NO₂^- is 24. The hybridization of the molecule is sp² and due to the presence of lone pair on the nitrogen atom it attains a bent shape.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (i) and (iv)

(i) N₂ will have 14 electrons in its structure. So, its electronic configuration will be:

σ1s² σ∗1s² σ2s² σ∗2s², π2p²_x , π2p²_y, s2p_z²

According to the molecular electronic configuration of N₂ it does not contain any unpaired electrons, it will be diamagnetic in nature.

(ii) N₂^2– will have 16 electrons in its structure. So, its molecular electronic configuration will be:

σ1s² σ∗1s² σ2s² σ∗2s², π2p²_x , π2p²_y, s2p_z², , π*2p¹_y

According to the molecular electronic configuration of 2 N₂ ^-, it has two unpaired electrons in π*2p¹_x ,   π*2p¹_y  sub- shell. Hence, it will be paramagnetic in nature.

(iii) O₂ will have 16 electrons in its structure. So, its molecular electronic configuration will be:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z² π2p²_y π2p²_x π*2p¹_y π*2p¹_x

According to the molecular electronic configuration of O₂, it has two unpaired electrons in π2p¹_y π*2p¹_x  sub-shell. Hence, it will be paramagnetic in nature.

(iv) O₂^2- will have 18 electrons in its structure. So, its molecular electronic configuration will be:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z² π2p²_y π2p²_x π*2p²_y π*2p²_x

According to the molecular electronic configuration of O₂^2- it does not contain any unpaired electrons, it will be diamagnetic in nature.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (iii) and (iv)

The formula of bond order is:

Bond order = $\frac{1}{2}$ ( N_b- N_a)

Number of bonding and antibonding electrons can be found from the molecular electronic configuration of the species.

(i) Total number of electrons in N2 is 14. So, its molecular electronic configuration will be

σ1s² σ∗1s² σ2s² σ∗2s², π2p²_x , π2p²_y, s2p_z²

The bond order will be:

$\frac{1}{2}$ ( 10-4)= 3

(ii) The total number of electrons in N₂ ^- is 15. So, its molecular electronic configuration will be:

σ1s² σ∗1s² σ2s² σ∗2s², π2p²_x , π2p²_y, s2p_z² π*2p¹_x

The bond order will be:

$\frac{1}{2}$ ( 10-5)= 2.5

(iii) The total electrons in F₂⁺ is 17. So, its electronic configuration will be:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z², π2p²_y , π2p²_x, π*2p²_y π*2p¹_x

The bond order will be:

$\frac{1}{2}$ ( 10-7)= 1.5

(iv) The total electrons in O₂^- is 17 So, its electronic configuration will be

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z², π2p²_y , π2p²_x, π*2p²_y π*2p¹_x

$\frac{1}{2}$ ( 10-7)= 1.5

9.00

α - D - Glucose sucrose β – D - Fructose

Δ? > P
Pairing of electron will take place

Number of unpaired electron = 0
µ = 0.00

13.49
2 A (g) → A? (g)
Δn_g = 1 - 2
Δn_g = 1
ΔH = ΔU + Δn_gRT
ΔH = -20 × 10³ + (-1)8.31 × 298
ΔH = -22477.572 J mol? ¹
ΔG = ΔH - TΔS
ΔG = -22477.572 - (-30)298 = -22477.572 + 8940
ΔG = -13537.57 J mol? ¹

50.00
K? Cr? O? + FeC? O? → Fe³? + CO? + Cr³? + K? + H? O
n_f=6    n_f=3
Now apply equivalent concept
0.02 × 6 × V = (0.288 / (144/3) × 10³
Normality
V = (0.288 × 10³)/ (48 × 6 × 0.02) = 50.00 mL

16.00
X + Y? 2Z
Initial 1.0 1.5 0.5
At equ m 1-x 1.5-x 0.5+2x
equ m conc 0.75 1.25 1.0
0.5 + 2x = 1
x = 0.25
K = 1 / (0.75 × 1.25)
K = 16/15
x = 16.00

Molar mass of C₇H₅N₃O₆ = 12 × 7 + 1 × 5 + 14 × 3 + 16 × 6 = 84 + 42 + 96 = 227 g/mol

Number of moles =  $\frac{681}{227}$  = 3 mol

$\therefore$  number of N-atoms

$=3\times 6.02\times 10^{23}\times 3=9\times 6.02\times 10^{23}$  

$=5418\times 10^{21}$  

$\therefore x=5418$  

$d=\frac{ZM}{N_A{a}^3}=\frac{4\times 58.5}{6.02\times 10^{23}\times a^3}$

$43.1=\frac{4\times 58.5}{6.02\times 10^{23}\times a^3}$

$\therefore a=2.08\times 10^{-8}cm$

$\therefore r_{N{a}^{+}}+r_{C{l}^{-}}=1.04\times 10^{-10}$ m

$E_{1_H}=-2.2\times 10^{-18}J$

$Li\to L{i}^{+2}+2e^{-},n=2$

$E_{L{i}^{+2}}=E_{1H}\times \frac{Z^2}{n^2}=-2.2\times 10^{-18}\times \frac{3^2}{2^2}$

$\lambda =4\times 10^{-8}$ m

$\Delta H=-165kJ/mole$

T =?

$\Delta S=-550J{K}^{-1}$

At equilibrium $\Delta G=0$

$\therefore T=\frac{\Delta H}{\Delta S}=\frac{-165\times 1000}{-550}K$

$\begin{array}{l}=3\times 100K\\ =300K\end{array}$

1000 ml solution contains 0.02 milli mole (mm)

$\therefore$ 500 ml solution contains 0.02 m.m

$\therefore$ Solution made 1000 ml with H₂O

$\therefore$ m.m in final solution = 0.01 mm

Solution (A) + 0.01 m. m H₂SO₄

= 0.01 + 0.01

= 0.02 m.m

                             = 0.00002 × 10³ mm

$N_2{O}_4?2N{O}_2$

$\underset{=0.5mol}{\left(1-0.5\right)}$ $2\times 0.5=1mol$

$k_P=\frac{{\left(\frac{1}{1.5}\times 1\right)}^2}{\left(\frac{0.5}{1.5}\times 1\right)}$

$=\frac{4}{3}=1.33$

$=-710.15J/mol$

$F{e}^{+3}+l^{-}\to I_2+F{e}^{+2}$

Fe⁺² undergoes reduction & I₂ undergoes oxidation.

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

= (0.77 – 0.54) V = 0.23 V

 = 23 × 10^-2 V

$\therefore$  x = 23

 x = 23

$\frac{d\left[C\right]}{dt}=\left(\frac{20-10}{10}\right)=1mmold{m}^{-3}$

$+\frac{d\left[D\right]}{dt}=\left\{-\frac{d\left(B\right)}{dt}\right\}\times 1.5=\frac{3}{2}\left\{-\frac{d\left[B\right]}{dt}\right\}$

$\left\{-\frac{d\left[B\right]}{dt}\right\}=2\times \left\{\frac{-d\left[A\right]}{dt}\right\}$

$\therefore \frac{1}{6}\left\{\frac{-d\left[B\right]}{dt}\right\}=\frac{1}{3}\left\{\frac{-d\left[A\right]}{dt}\right\}=\frac{1}{9}\left\{\frac{+d\left[D\right]}{dt}\right\}=\frac{d\left[C\right]}{dt}$

$\therefore$ rate of reaction = $\frac{+d\left[C\right]}{dt}$  = 1m.m dm^-3 S^-1

${\Delta }_t=\frac{hc}{\lambda }=\frac{6.63\times 10^{-34}\times 3.08\times 10^8}{600\times 10^{-9}}$

$=\frac{6.63\times 3.08\times 10^{-17}}{600}$ J

$=765.765\times 10^{-21}J$

$?766\times 10^{-21}$ J