LOADING . . .
જવાબ : No, these cannot be taken as canonical forms because the positions of atoms have been changed.
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.
જવાબ : Bond strength is directly proportional to the bond order. Greater the bond order, more is the bond strength.
Because of two lone pairs of electrons on O-atom, repulsion on bond pairs is greater in H20 in comparison to NH3 . Thus, the bond angle is less in H20 molecules.
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.
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.
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.
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.
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.
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.
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.
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.
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:
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.
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°
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
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+
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
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.
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.
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.
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.
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.
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-
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
જવાબ : The molecular orbital formed by the subtractive effect of the electron waves of the combining atomic orbitals, is called antibonding molecular orbital.
જવાબ : Bonding molecular orbital has lower energy and higher stability.
જવાબ : Bond length is inversely proportional to bond order.
જવાબ : Bond order of N2 (= 3) is greater than that of O2 (= 2).
જવાબ : (i) sp3d of P in PCl5 (ii) sp3d2 of S in SF6
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.
જવાબ : This is because ethyl alcohol forms H-bonds witfi water.
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.
જવાબ : Covalent bond can be formed by the overlap of the orbitals belonging to the two atoms having opposite spins of electrons.
જવાબ : NaCl < MgCl2 < AlCl3
જવાબ : (i) CO2 (ii) CH4
જવાબ : The electron pairs involved in the bond formation are known as bond pairs or shared pairs.
જવાબ : O22-< O2– < O22-< O2+
જવાબ : SP2
જવાબ : (i) Linear
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.
જવાબ : CH4= sp3
C2H4 = sp2
C2H2 = sp
જવાબ : Higher the bond order, greater is the stability.
જવાબ : (i) The molecule should contain highly electronegative atom like hydrogen atom. (ii) The size of electronegative atom should be small.
આ પ્રકરણને લગતા વિવિધ એનિમેશન વિડીયો, હેતુલક્ષી પ્રશ્નો, ટૂંકા પ્રશ્નો, લાંબા પ્રશ્નો, પરિક્ષામાં પુછાઈ ગયેલા પ્રશ્નો તેમજ પરિક્ષામાં પુછાઈ શકે તેવા અનેક મુદ્દાસર પ્રશ્નો જોવા અમારી વેબસાઈટ પર રજીસ્ટર થાઓ અથવા અમારી App ફ્રી માં ડાઉનલોડ કરો.
The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.
The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.
For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.