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CBSE Solutions for Class 11 English

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H3POcan be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H3PO3? If not, give reasons for the same.

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જવાબ : No, these cannot be taken as canonical forms because the positions of atoms have been changed.


Define the bond-length.

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જવાબ : Bond-length: It is the equilibrium distance between the nuclei of two bonded atoms in a molecule. Bond-lengths are measured by spectroscopic methods.


How do you express the bond strength in terms of bond order?

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જવાબ : Bond strength is directly proportional to the bond order. Greater the bond order, more is the bond strength.


Although geometries of NHand H20 molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.

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જવાબ :
Because of two lone pairs of electrons on O-atom, repulsion on bond pairs is greater in H20 in comparison to NH. Thus, the bond angle is less in H20 molecules.


Write the favourable factors for the formation of ionic bond.

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જવાબ :

  1. Low ionization enthalpy of metal atoms
  2. High electron gain enthalpy of non-metal atoms
  3. High lattice enthalpy of compound formed.


Draw the Lewis structures for the following molecules and ions:
H2S, SiCl4 , BeF2, C032-, HCOOH

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જવાબ :


Write Lewis symbols for the following atoms and ions: S and S2– ; Al and Al3+; H and H

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જવાબ :


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

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જવાબ :


Explain the formation of a chemical bond.

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જવાબ : According to Kossel and Lewis, atoms combine together in order to complete their respective octets so as to acquire the stable inert gas configuration. This can occur in two ways; by transfer of one or more electrons from one atom to other or by sharing of electrons between two or more atoms.


What is meant by the term average bond enthalpy? Why is there a difference in bond enthalpy of O—H bond in ethanol (C2H5OH) and water?

 

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જવાબ : Similar bonds in a molecule do not have the same bond enthalpies. Mainly the term average bond enthalpy is used in polyatomic molecules. It is obtained by dividing bond dissociation enthalpy by the number of bonds broken. The bond enthalpy of OH bond is different in ethanol and water because of the difference in electronegativity of hydrogen and carbon. As electronegativity differs in hydrogen and oxygen is higher than that in carbon and oxygen, so the O-H bond in water has more bond enthalpy than in ethanol.


All the C—O bonds in carbonate ion (CO2-3 ) are equal in length. Explain

 

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જવાબ : Carbonate ion is present in the form of a resonating structure. These structures are equivalenting to nature. Resonance all 3 C-O bonds get a double character in one of the resonating structures.


Predict the shapes of the following molecules based on hybridisation.

BCl3, CH4, CO2, NH3

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જવાબ : 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.


Draw the resonating structure of

(i) Ozone molecule

(ii) Nitrate ion

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જવાબ :


Elements X, Y and Z have 4, 5 and 7 valence electrons respectively.

(i) Write the molecular formula of the compounds formed by these elements individually with hydrogen.

(ii) Which of these compounds will have the highest dipole moment?

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જવાબ : (i); XH4, H3Y, and HZ Hydrogen has only one electron in its outermost shell it shares one electron to form a covalent bond or accepts or donates one electron to form an ionic bond. (ii) The compound HZ has a linear shape and the difference in the electronegativity of Hydrogen and element Z is maximum.


Group the following as linear and non-linear molecules :

H2O, HOCl, BeCl2, Cl2O

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જવાબ : BeCl2 has a linear structure HOCl is also non-linear in structure. H2O has a V-shaped structure. Cl2O has a V-shaped structure.


Predict the hybridisation of each carbon in the molecule of the organic compound given below. Also indicate the total number of sigma and pi bonds in this molecule.

 

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જવાબ : 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.


Explain why CO32- ion cannot be represented by a single Lewis structure. How can it be best represented?

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જવાબ : Carbonate ion is present in the form of a resonating hybrid structure. These structures are equivalent. Resonance all 3 C-O bonds get a double character in one of the resonating structures. Thus, all the bonds are equivalent and have equal length hence carbonate ion cannot be represented by a single Lewis structure.


Arrange the following bonds in order of increasing ionic character giving a reason

N—H, F—H, C—H and O—H

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જવાબ : The ionic character is greater in the molecules that are having the greatest electronegativity difference because the electron pair shifts toward a more electronegative atom increasing the ionic character. Thus, the ionic character order will be: C-H


What is an ionic bond? With two suitable examples explain the difference between an ionic and a covalent bond?

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જવાબ : when a positively charged ion forms a bond with a negatively charged ions and one atom transfers electrons to another. An example of an ionic bond is the chemical compound Sodium Chloride (NaCl). The difference between an ionic bond and covalent bond is that An ionic bond essentially donates an electron to the other atom participating in the bond, while electrons in a covalent bond are shared equally between the atoms.


Give reasons for the following :

(i) Covalent bonds are directional bonds while ionic bonds are nondirectional.

(ii) The water molecule has a bent structure whereas carbon dioxide molecule is linear.

(iii) Ethyne molecule is linear.

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જવાબ : (i) A covalent bond is formed by the overlapping of atomic orbitals. The direction of overlapping gives the direction of the bond. (ii) In a water molecule, the oxygen atom is sp3 hybridized and has two lone pairs of electrons. (iii) In the ethyne molecule, both the carbon atoms are sp hybridized. The two sp hybrid orbitals of both the carbon atoms are oriented in the opposite direction forming an angle of 180°


What is the effect of the following processes on the bond order in N2 and O2?

(i) N2→ N2+ + e-

(ii) O2→ O2+ + e-

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જવાબ : (i) N2 is having 14 electrons when it donates one electron, these electrons are removed from the Bonding molecular orbital. BO for N2 = 3 (ii) O2 has 16 electrons, 8 electrons in the molecular orbitals and 4 in the antibonding molecular orbitals. BO for O2 = 2


The energy of σ2pz molecular orbital is greater than π2px and π2py molecular orbitals in nitrogen molecule. Write the complete sequence of energy levels in the increasing order of energy in the molecule. Compare the relative stability and the magnetic behaviour of the following species :

N2, N2+ , N2- , N22+

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જવાબ : 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 π2p2x = π2p2y σ2p2z N2+ σ1s2 σ*1s2 σ2s2 σ*2s2 π2p2x = π2p2y σ2p1z N2- σ1s2 σ*1s2 σ2s2 σ*2s2 π2p2x = π2p2y σ2p2z σ2p2x N22+ σ1s2 σ*1s2 σ2s2 σ*2s2 π2p2x = π2p2y Bond order = ½[electrons in BMO – electrons in ABMO] For N2 = 10-4/2 = 3 Bond order for N2+= 9-4/2 = 2.5 Bond order for N2-= 10-5/2 = 2.5 Bond order for N22+= 8-4/2 = 2 Thus, the order of stability is: N2> N2- > N2+> N22+


Write Lewis structure of the following compounds and show a formal charge on each atom.

HNO3, NO2, H2SO4

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જવાબ : The formal charge is calculated by Formal charge = ½ [total no: of bonding or shared electrons] The formal charge on the oxygen with single bond =6-6-2/2 = -1 The formal charge on the oxygen with double bond 6-4-4/2 = 0 The formal charge on nitrogen=5-2-6/2 = 0 The formal charge on oxygen 1 and 4 = 6-4-4/2= 0 The formal charge on oxygen 2 and 3 = 6-4-4/2=0 The formal charge on hydrogen 1 and 2 = 1-0-2/2=0 The formal charge on sulfur =6-0-12/2 = 0


In both water and dimethyl ether (CH3 —Ο — CH3 ), the oxygen atom is the central atom and has the same hybridization, yet they have different bond angles. Which one has a greater bond angle? Give reason.

 

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જવાબ : Dimethyl ether will have a greater bond angle. There will be more repulsion between bond pairs of CH3 groups attached in ether than between bond pairs of hydrogen atoms attached to oxygen in the water.


Explain why PCl5 is trigonal bipyramidal whereas IF5 is square pyramidal.

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જવાબ : 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.


Why does the type of overlap given in the following figure not result in bond formation?

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જવાબ : Solution: In the figure (i) the area of the contact of ++ overlap is equal to the area of the +- overlap. The so-net overlap is zero. In figure (ii) there is no overlap of the orbitals due to the different symmetry.


Structures of molecules of two compounds are given below :

 

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

(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 a higher melting point.

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

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જવાબ : (a) Compound 1 will have intramolecular hydrogen bonding in o-nitrophenol Compound (II) will have intermolecular hydrogen bonding in p-nitrophenol. (b) The compound (II) has a higher melting point because of the intermolecular bonding, a large number of molecules that will get attached. (c) The compound (II) will be more soluble in water because it will form hydrogen bonding with the water molecules easily.


Explain the shape of BrF5.

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જવાબ : In BrF5 the central atom is bromine which has the hybridization sp3d2. Br atom has 7 valence electrons out of which 5 are used to make pair with the F atoms and two are used to make lone pair of electrons. The lone pair and bond pair repel each other. Thus, the shape is square pyramidal.


Using molecular orbital theory, compare the bond energy and magnetic character of O2+ and O2- species.

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જવાબ : The Molecular Orbital configuration of O2+ and O-2 is given below: O2+ (15): σ1s2 σ *1s2 σ 2s2 σ *2s2 σ 2pz2 π2px2 = π 2py2π *2px1 O2- (17): σ1s2 σ *1s2 σ 2s2 σ *2s2 σ 2pz2 π2px2 = π 2py2π *2px2= π*2py1 Bond order for O2+ = 10-5/2 = 2.5 Bond order for O-2 = 10-7/2 = 1.5 According to Molecular Orbital Theory, the greater the bond order greater is the bond energy. Thus, O2+ is more stable than O2-


Explain the non-linear shape of H2 S and non-planar shape of PCl3 using valence shell electron pair repulsion theory

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જવાબ : In H2S, the Sulphur atom is surrounded by the four electron pairs (two bond pairs and two lone pairs).These four electron pairs adopt tetrahedral geometry.The repulsion between the lone pair electrons brings distortion in the shape of the H2S.Thus, H2Sis not linear in shape


Define antibonding molecular orbital.

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જવાબ : The molecular orbital formed by the subtractive effect of the electron waves of the combining atomic orbitals, is called antibonding molecular orbital.


Out of bonding and antibonding molecular orbitals, which one has lower energy and which one has higher stability?

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જવાબ : Bonding molecular orbital has lower energy and higher stability.


How is bond order related to bond length of a molecule?

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જવાબ : Bond length is inversely proportional to bond order.


Why Nis more stable than  O2? Explain on the basis of molecular orbital theory.

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જવાબ : Bond order of N2 (= 3) is greater than that of O2 (= 2).


State the types of hybrid orbitals associated with (i) P in PCland (ii) S in SF6

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જવાબ : (i) sp3d of P in PCl5 (ii) sp3dof S in SF6


Which is more polar COor N2O? Give reason.

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જવાબ : N2O is more polar than CO2.
This is “because CO2 is linear and symmetrical. Its net dipole moment is zero.
N2O is linear but not symmetrical. It has a net dipole moment of sigma II6D.


Why ethyl alcohol is completely miscible with water?

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જવાબ : This is because ethyl alcohol forms H-bonds witfi water.


Why Bis paramagnetic in nature while C2 is not?

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જવાબ : The molecular orbital electronic configuration of both B2 and C2 are.

Since, B2has two impaired electrons, B2 is paramagnetic.
C2 has no unpaired electron. Thus, C2 is diamagnetic.


Define covalent bond according to orbital concept?

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જવાબ : Covalent bond can be formed by the overlap of the orbitals belonging to the two atoms having opposite spins of electrons.


Arrange the following, according to increasing covalent nature.
NaCl, MgCl2, AlCl3

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જવાબ : NaCl < MgCl2 < AlCl3


Which of the following has larger bond angle in each pair?
(i) CO2, BF3 (ii) NH3, CH4

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જવાબ : (i) CO2 (ii) CH4


What is meant by bond pairs of electrons?

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જવાબ : The electron pairs involved in the bond formation are known as bond pairs or shared pairs.


Arrange O2,O2,O22-, O2+in increasing order of bond energy. 

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જવાબ : O22-< O2 < O22-< O2+


Write the state of hybridisation of boron in BF3.

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જવાબ : SP2


Predict the shapes of the following molecules using VSEPR theory?
(i) BeCl2(ii) SiCl4

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જવાબ : (i) Linear
(ii) Tetrahedral


Write the Lewis dot symbols of the following elements and predict their valencies. (i) Cl (ii) P

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જવાબ :


Out of sigma and Π  bonds, which one is stronger and why?

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જવાબ : sigma-bond is stronger. This is because sigma-bond is formed by head-on overlapping of atomic orbitals and Π bond is formed by side wise overlapping.


Write the type of hybridisation involved in CH4,C2Hand C2H2.

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જવાબ : CH4= sp3
C2H4 = sp2
C2H2 = sp


How is bond order related to the stability of a molecule?

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જવાબ : Higher the bond order, greater is the stability.


Name the two conditions which must be satisfied for hydrogen bonding to take place in a molecule.

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જવાબ : (i) The molecule should contain highly electronegative atom like hydrogen atom. (ii) The size of electronegative atom should be small.


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