{"id":35422,"date":"2022-09-14T09:42:42","date_gmt":"2022-09-14T04:12:42","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35422"},"modified":"2022-09-14T09:42:42","modified_gmt":"2022-09-14T04:12:42","slug":"ap-inter-1st-year-chemistry-study-material-chapter-3","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-inter-1st-year-chemistry-study-material-chapter-3\/","title":{"rendered":"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure"},"content":{"rendered":"<p>Andhra Pradesh BIEAP <a href=\"https:\/\/apboardsolutions.guru\/ap-inter-1st-year-chemistry-study-material\/\">AP Inter 1st Year Chemistry Study Material<\/a> 3rd Lesson Chemical Bonding and Molecular Structure Textbook Questions and Answers.<\/p>\n<h2>AP Inter 1st Year Chemistry Study Material 3rd Lesson Chemical Bonding and Molecular Structure<\/h2>\n<p><span style=\"color: #0000ff;\">Very Short Answer Questions<\/span><\/p>\n<p>Question 1.<br \/>\nWhat is Octet rule?<br \/>\nAnswer:<br \/>\nEvery atom must possess 8 electrons in its outermost energy level for its stability. Atoms combine in two ways to get octets either by transfer of electrons (or) by mutual sharing of electrons. They can attain ns<sup>2<\/sup> np<sup>6<\/sup> configuration.<br \/>\nThe tendency of an atom to achieve eight electrons in its outermost shell is known as the octet rule.<\/p>\n<p>Question 2.<br \/>\nWrite Lewis dot structures for S and S<sup>2-<\/sup>.<br \/>\nAnswer:<\/p>\n<ul>\n<li>Lewis dot structure for \u2018s\u2019 is<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25024\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-1.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 1\" width=\"134\" height=\"36\" \/><br \/>\nElectronic configuration \u2014 1s<sup>2<\/sup> 2s<sup>2<\/sup> 2p<sup>6<\/sup> 3s<sup>2<\/sup> 3p<sup>4<\/sup><\/li>\n<li>Lewis dot structure for s<sup>-2<\/sup> is<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25025\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-2.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 2\" width=\"148\" height=\"52\" \/><br \/>\nElectronic configuration of s<sup>-2<\/sup> is 1s<sup>2<\/sup> 2s<sup>2<\/sup> 2p<sup>6<\/sup> 3s<sup>2<\/sup> 3p<sup>6<\/sup><\/li>\n<\/ul>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 3.<br \/>\nWrite the possible resonance structures for SO<sub>3<\/sub>.<br \/>\nAnswer:<br \/>\nThe resonance structures of SO<sub>3<\/sub> as follows<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25026\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-3.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 3\" width=\"475\" height=\"163\" \/><\/p>\n<p>Question 4.<br \/>\nPredict the change, if any, in hybridization of Al atom in the following reaction<br \/>\nAlCl<sub>3<\/sub> + Cl<sup>&#8211;<\/sup> \u2192 \\(\\mathrm{AlCl}_4^{-}\\).<br \/>\nAnswer:<br \/>\nIn AlCl<sub>3<\/sub> Aluminium undergoes sp<sup>2<\/sup> hybridisation<br \/>\nIn \\(\\mathrm{AlCl}_4^{-}\\) Aluminium undergoes sp<sup>3<\/sup> hybridisation<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25027\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-4.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 4\" width=\"307\" height=\"73\" \/><\/p>\n<p>Question 5.<br \/>\nWhich of the two ions Ca<sup>2+<\/sup> or Zn<sup>2+<\/sup> is more stable and why ?<br \/>\nAnswer:<\/p>\n<ul>\n<li>Ca<sup>+2<\/sup> has electronic configuration 1s<sup>2<\/sup>2s<sup>2<\/sup>2p<sup>6<\/sup>3s<sup>2<\/sup>3p<sup>6<\/sup>. This configuration is noble gas (or) inert gas configuration.<\/li>\n<li>Zn<sup>+2<\/sup> has electonic configuration 1s<sup>2<\/sup>2s<sup>2<\/sup>2p<sup>6<\/sup>3s<sup>2<\/sup>3p<sup>6<\/sup>4s<sup>0<\/sup>3d<sup>10<\/sup>. This configuration is psuedo inert gas configuration.<br \/>\n\u2234 Ca<sup>+2<\/sup> is more stable than Zn<sup>+2<\/sup> ion.<\/li>\n<\/ul>\n<p>Question 6.<br \/>\nCl<sup>&#8211;<\/sup> ion is more stable than Cl atom\u2014Why ?<br \/>\nAnswer:<br \/>\nThe electronic configurations of Cl and Cl<sup>&#8211;<\/sup> is :<br \/>\nCl = 1s<sup>2<\/sup> 2s<sup>2<\/sup> 2p<sup>6<\/sup> 3s<sup>2<\/sup> 3p<sup>5<\/sup><br \/>\nCl<sup>&#8211;<\/sup> = 1s<sup>2<\/sup> 2s<sup>2<\/sup> 2p<sup>6<\/sup> 3s<sup>2<\/sup> 3p<sup>6<\/sup>.<br \/>\nThis electronic configuration clearly shows that chlorine atom has 7 electrons in the outermost orbit. Whereas chloride has a stable octet electronic configuration (3s<sup>2<\/sup> 3p<sup>6<\/sup>). After gaining one electron, chloride ion attains the electronic configuration of Argon. Hence chloride ion has greater stability than chlorine atom.<\/p>\n<p>Question 7.<br \/>\nWhy argon does not form Ar<sub>2<\/sub> molecule ?<br \/>\nAnswer:<br \/>\nAr<sub>2<\/sub> represents diatomic molecule. But Argon does not form diatomic molecule. So it cannot represented as &#8216;Ar<sub>2<\/sub>&#8216;.<br \/>\nReason : Since Ar&#8217; has only paired electrons with stable octet configuration. It cannot share its electrons with another Ar atom and does not form diatomic molecule.<\/p>\n<p>Question 8.<br \/>\nWhat is the best possible arrangement of four bond pairs in the valence shell of an atom to minimise repulsions ?<br \/>\nAnswer:<br \/>\nThe best possible arrangement of four bond pairs in the valency shell of an atom to minimise repulsions is Tetrahedral. (Bond angle 109\u00b0.28&#8242;)<br \/>\nEg. : Methane (CH<sub>4<\/sub>).<\/p>\n<p>Question 9.<br \/>\nIf A and B are two different atoms when does AB molecule become Covalent ?<br \/>\nAnswer:<\/p>\n<ol>\n<li>If the difference in electronegativity values between A and B is less than 1.7, then covalent compound formation is possible (according to Allred &#8211; Rochow scale).<\/li>\n<li>If A and B are sharing one or more electron pairs mutually then AB will be a covalent compound.<\/li>\n<\/ol>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 10.<br \/>\nWhat is meant by localized orbitals?<br \/>\nAnswer:<br \/>\nThe molecular orbital with bonded electron cloud localised between the two nuclei of bonded atoms is called localized orbital, (or) The orbitals which are involved in bond formation are called localized orbitals.<\/p>\n<p>Question 11.<br \/>\nHow many Sigma and Pi bonds are present in<br \/>\n(a) C<sub>2<\/sub>H<sub>2<\/sub> and<br \/>\n(b) C<sub>2<\/sub>H4<sub>4<\/sub>?<br \/>\nAnswer:<br \/>\na) C<sub>2<\/sub>H<sub>2<\/sub><br \/>\nH &#8211; C \u2261 C &#8211; H<br \/>\nC<sub>2<\/sub>H<sub>2<\/sub> contains 3 &#8211; sigma bonds and 2 &#8211; pi bonds.<\/p>\n<p>b) C<sub>2<\/sub>H<sub>4<\/sub><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25028\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-5.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 5\" width=\"277\" height=\"103\" \/><br \/>\nC<sub>2<\/sub>H<sub>4<\/sub> contains 5 &#8211; sigma bonds and 1-pi bond.<\/p>\n<p>Question 12.<br \/>\nIs there any change in the hybridization of Boron and Nitrogen atoms as a result of the following reaction? BF<sub>3<\/sub> + NH<sub>3<\/sub> \u2192 F<sub>3<\/sub>BNH<sub>3<\/sub><br \/>\nAnswer:<br \/>\na) Ammonia &#8211; Boron trifluoride formation (H<sub>3<\/sub>N \u2192 BF<sub>3<\/sub>):<br \/>\nAmmonia molecule contains Nitrogen atom with a lone pair of electrons (in sp<sup>3<\/sup> orbital). BF<sub>3<\/sub> has &#8216;B&#8217; atom with an incomplete octet (with a vacant P<sub>z<\/sub> orbital). Therefore, nitrogen of ammonia donates its lone pair to Boron and thus forms coordinate covalent bond. During this bond formation, the sp<sup>3<\/sup> orbital of nitrogen having a lone pair overlaps the vacant &#8216;p&#8217; orbital of Boron. The equation corresponding to the reaction is written as follows :<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25029\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-6.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 6\" width=\"539\" height=\"61\" \/><\/p>\n<p>b) Change in hybridised states N and B during [H<sub>3<\/sub>N \u2192 BF<sub>3<\/sub>] formation :<br \/>\nBoron in BF<sub>3<\/sub> undergoes (sp<sup>2<\/sup> hybridization with one vacant unhybrid &#8216;p&#8217; orbital. This orbital also undergoes) hybridization in presence of NH<sub>3<\/sub> so that the hybridised state of &#8216;B&#8217; changes from sp<sup>2<\/sup> to sp<sup>3<\/sup>. This vacant hybrid orbital is bonded to NH<sub>3<\/sub> through dative bond. During this process there is no change in the hybridized state of Nitrogen in NH<sub>3<\/sub>.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25030\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-7.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 7\" width=\"743\" height=\"91\" \/><\/p>\n<p>Question 13.<br \/>\nGive reasons for the following.<br \/>\na) Why H<sub>2<\/sub>O Boiling point is more than H<sub>2<\/sub>S<br \/>\nb) Why H<sub>2<\/sub>O Boiling point is more than HF<br \/>\nAnswer:<br \/>\na) H<sub>2<\/sub>O has high boiling point than H<sub>2<\/sub>S<br \/>\nReason:<br \/>\nIn H<sub>2<\/sub>O inter molecular hydrogen bonding is present where as in case of H<sub>2<\/sub>S such bonding is absent.<\/p>\n<p>b) H<sub>2<\/sub>O has high boiling point than HF<br \/>\nReason:<br \/>\nIn H<sub>2<\/sub>O and H<sub>2<\/sub>S inter molecular hydrogen bonding is present but in H<sub>2<\/sub>O the no. of hydrogen bonds are more than in HF.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure\" width=\"196\" height=\"17\" \/><\/p>\n<p><span style=\"color: #0000ff;\">Short Answer Questions<\/span><\/p>\n<p>Question 1.<br \/>\nExplain Kossel-Lewis approach to Chemical bonding.<br \/>\nAnswer:<br \/>\nKossel &#8211; Lewis Theory : This theory was also called as electronic theory of valency (or) chemical bond theory.<br \/>\nPostulates of Kossel &#8211; Lewis Theory : Kossel explained the formation of electrovalent bond while Lewis explained the formation of covalent bond. Their explanation of valency is mainly based on the inertness of noble gases.<\/p>\n<p>Postulates:<\/p>\n<ul>\n<li>The chemical inertness of noble gases is due to the presence of octet structure. Octet rule was stated as follows &#8220;An atom must possess eight electrons in the outermost energy level for its stability&#8221;.<\/li>\n<li>Even though &#8216;He&#8217; has only two electrons in the valency shell, it is highly stable and chemically inert.<\/li>\n<li>Elements other than zero group are chemically reactive because of having less than 8 electrons in their outer most shells.<\/li>\n<li>Every atom try to acquire the Eight electron configuration (octet) in its outer most shell. This can be possible by losing (or) sharing (or) gaining electrons.<\/li>\n<li>According to Lewis the valency electrons are represented by dots. These are called Lewis symbols.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25031\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-8.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 8\" width=\"197\" height=\"32\" \/><\/li>\n<li>Lewis dot structures can be used to calculate the group valency of the element.<br \/>\nEg:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25032\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-9.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 9\" width=\"29\" height=\"35\" \/> has four electrons.<br \/>\n\u2234 Valency of &#8216;c&#8217; is &#8216;4&#8217;.<\/li>\n<\/ul>\n<p>Question 2.<br \/>\nWrite the general properties of Ionic Compounds.<br \/>\nAnswer:<\/p>\n<ol>\n<li>Physical state : Due to close packing of ions, ionic compounds are crystalline solids.<\/li>\n<li>Melting and Boiling points : In ionic crystals the oppositely charged ions are bound by strong electrostatic force of attraction. To overcome these attractive force between ions, more thermal energy is required. Hence the melting and boiling points of ionic compounds are high.<\/li>\n<li>Solubility : Ionic compounds are soluble in polar solvents like water, liquid ammonia etc., but are insoluble in non &#8211; polar solvents like benzene, carbon disulphide etc.,<\/li>\n<li>Reactivity: Reactions between ionic compounds in aqueous solution are very fast due to strong attraction among ions.<br \/>\ne.g. : When AgNO<sub>3<\/sub> solution is added to NaCl solution, a white precipitate of AgCl is formed.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25033\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-10.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 10\" width=\"355\" height=\"54\" \/><\/li>\n<li>Isomerism : Ionic bond is non-directional.<br \/>\nSo ionic compounds cannot exhibit isomerism.<\/li>\n<li>Electrical conductivity : Ionic substances conduct electricity in molten state and in aqueous solution. The ionic compounds are, therefore, electrolytes.<\/li>\n<\/ol>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 3.<br \/>\nState Fajan&#8217;s rules and give suitable examples.<br \/>\nAnswer:<br \/>\nFajan&#8217;s rules:<\/p>\n<ol>\n<li>Ionic nature of the bond increases with increase in the size of cation, e.g.: The ionic nature increases in the order<br \/>\nLi<sup>+<\/sup> &lt; Na<sup>+<\/sup> &lt; K<sup>+<\/sup> &lt; Rb<sup>&#8211;<\/sup> &lt; Cs<sup>+<\/sup><\/li>\n<li>The formation of ionic bond is favoured with the decrease of the size of anion.<br \/>\ne.g.: CaF<sub>2<\/sub> is more ionic than CaI<sub>2<\/sub>.<\/li>\n<li>If the charge on cation (or) anion (or) both is less, then they can form ionic bonds, e.g.: The ionic nature increases in the order<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25034\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-11.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 11\" width=\"179\" height=\"47\" \/><\/li>\n<li>Cations with inert gas configurations form ionic compounds while those cations with pseudo inert gas configurations favour covalent bond formation.<br \/>\ne.g.: Na<sup>+<\/sup> in Na<sup>+<\/sup>Cl<sup>&#8211;<\/sup> has an inert gas configuration. So Na<sup>+<\/sup>Cl<sup>&#8211;<\/sup> is ionic. But CuCl is more covalent because Cu<sup>+<\/sup> has not acquired inert gas configuration in this compound, instead it has acquired pseudo inert gas configuration.<\/li>\n<li>The cation with inert gas configuration is more stable, e.g.: Ca<sup>2+<\/sup> is more stable than Zn<sup>2+<\/sup> ion.<\/li>\n<li><img loading=\"lazy\" class=\"alignnone size-full wp-image-25035\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-12.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 12\" width=\"767\" height=\"165\" \/><\/li>\n<\/ol>\n<p>Question 4.<br \/>\nWhat is Octet rule ? Briefly explain its significance and limitations.<br \/>\nAnswer:<br \/>\nOctet rule: Every atom must possess 8 electrons in its outermost energy level for its stability. Atoms combine in two ways to get octet either by transfer of electrons (or) by mutual sharing of electrons. They can attain ns<sup>2<\/sup> np<sup>6<\/sup> configuration.<br \/>\ne.g.:<\/p>\n<ol>\n<li>Na loose one electron to get Ne configuration by possessing 8 electrons.<br \/>\nNa : 1 s<sup>2<\/sup> 2s<sup>2<\/sup> 2p<sup>6<\/sup> 3s<sup>1<\/sup> and Na<sup>+<\/sup> : 1 s<sup>2<\/sup> 2s<sup>2<\/sup> 2p<sup>6<\/sup>.<\/li>\n<li>&#8216;Cl&#8217; atom take one electron to get &#8220;Ar&#8221; configuration. Cl<sup>&#8211;<\/sup> : 1s<sup>2<\/sup> 2s<sup>2<\/sup> 2p<sup>6<\/sup> 3s<sup>2<\/sup> 3p<sup>6<\/sup>.<br \/>\nHere Na<sup>+<\/sup> and Cl<sup>&#8211;<\/sup> ions obey octet rule.<\/li>\n<li>In H<sub>2<\/sub>O molecule oxygen obey octet rule.<\/li>\n<\/ol>\n<p>It is therefore, concluded that s2p6 configuration in the outer energy level constitutes a structure of maximum stability and therefore, of minimum energy.<\/p>\n<p>The atoms of all elements when enter into chemical combination try to attain noble gas configuration (i.e.,) they try to attain 8 electrons in their outermost energy level which is of maximum stability and hence of minimum energy.<br \/>\nThe tendency of an atoms to achieve eight electrons in their outermost shell is known as OCTET RULE.<\/p>\n<p>Octet rule was the basis of electronic theory of valency.<\/p>\n<p>Limitations : There are 3 types of exceptions to the octet rule. These are mentional below.<\/p>\n<ul>\n<li>Central atoms containing incomplete octet.<br \/>\nEg : BeH<sub>2<\/sub>, BCl<sub>3<\/sub> etc.,<\/li>\n<li>Molecules containing odd number of electrons.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25036\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-13.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 13\" width=\"146\" height=\"39\" \/><\/li>\n<li>Central atoms possessing more than 8 electrons which istermed as expanded octet.<br \/>\nEg : SF<sub>6<\/sub>, H<sub>2<\/sub>SO<sub>4<\/sub> etc.,<\/li>\n<li>This theory does not explained about shape of molecules.<\/li>\n<li>This theory does not explained about the formation of noble gas compounds like XeF<sub>2<\/sub>, XeOF<sub>2<\/sub> etc.,<\/li>\n<\/ul>\n<p>Question 5.<br \/>\nWrite the resonance structures for NO<sub>2<\/sub> and \\(\\mathrm{NO}_3^{-}\\)<br \/>\nAnswer:<br \/>\nResonance structure of NO<sub>2<\/sub><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25037\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-14.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 14\" width=\"575\" height=\"252\" \/><\/p>\n<p>Question 6.<br \/>\nUse Lewis symbols to show electron transfer between the following pairs of atoms to form cations and anions:<br \/>\n(a) K and S<br \/>\n(b) Ca and O<br \/>\n(c) Al and N.<br \/>\nAnswer:<br \/>\na) Between the atoms K and S<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25038\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-15.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 15\" width=\"400\" height=\"156\" \/><br \/>\nb) Between Ca and O<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25039\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-16.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 16\" width=\"449\" height=\"110\" \/><br \/>\nc) Between Al and N<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25040\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-17.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 17\" width=\"414\" height=\"109\" \/><\/p>\n<p>Question 7.<br \/>\nExplain why H<sub>2<\/sub>O has dipole moment while CO<sub>2<\/sub> does not have.<br \/>\nAnswer:<\/p>\n<ul>\n<li>H<sub>2<\/sub>O molecule is a polar molecule and it has un symmetrical structure i.e. Angular (or) V &#8211; shape<\/li>\n<li>CO<sub>2<\/sub> molecule is non polar and it is linear molecule.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25041\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-18.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 18\" width=\"320\" height=\"72\" \/><\/li>\n<li>SO H<sub>2<\/sub>O has dipolemoment (\u03bc = 1 ,835D) and CO<sub>2<\/sub> does not have dipolemoment (\u03bc = 0)<\/li>\n<\/ul>\n<p>Question 8.<br \/>\nDefine Dipole moment. Write its applications.<br \/>\nAnswer:<br \/>\nDipole moment : The product of magnitude of the charge and the distance between the two poles (bond length) is called dipole moment.<\/p>\n<ul>\n<li>Dipole moment \u03bc = q \u00d7 d<br \/>\nq = charge<br \/>\nd = bond length<\/li>\n<li>Units : Debye (D), 1 Debye = 3.34 \u00d7 10<sup>-30<\/sup> coulombs metres.<\/li>\n<\/ul>\n<p>Applications : &#8211;<\/p>\n<ul>\n<li>Dipole moment is used to calculate the percentage of ionic character in a molecule.<\/li>\n<li>It is used to know the shape of the molecule.<\/li>\n<li>Symmetry (symmetrical (or) non symmetrical) of the molecule can be known by dipole moment.<\/li>\n<\/ul>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 9.<br \/>\nExplain why BeF<sub>2<\/sub> molecule has zero dipole moment although the Be-F bonds are polar.<br \/>\nAnswer:<\/p>\n<ul>\n<li>Even though Be-F bonds in BeF<sub>2<\/sub> are polar, the dipole moment of BeF<sub>2<\/sub> molecule is zero. Because BeF<sub>2<\/sub> has linear shape.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25042\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-19.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 19\" width=\"328\" height=\"51\" \/><\/li>\n<li>Flere the vectrorial sum of the dipole moment of two Be-F bonds is zero. Hence dipole moment<br \/>\n(\u03bc) = 0.<\/li>\n<\/ul>\n<p>Question 10.<br \/>\nExplain the structure of CH<sub>4<\/sub> molecule.<br \/>\nAnswer:<br \/>\nFormation of Methane molecule :<\/p>\n<ol>\n<li>The central atom of methane is carbon.<\/li>\n<li>The electronic configuration of carbon in ground state is <img loading=\"lazy\" class=\"alignnone size-full wp-image-25043\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-20.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 20\" width=\"144\" height=\"30\" \/> and on excitation it is <img loading=\"lazy\" class=\"alignnone size-full wp-image-25044\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-21.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 21\" width=\"153\" height=\"29\" \/> During excitation the 2s pair splits and the electron jumps into the adjacent vacant 2p<sub>z<\/sub> orbital.<\/li>\n<li>The <img loading=\"lazy\" class=\"alignnone size-full wp-image-25045\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-22.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 22\" width=\"119\" height=\"29\" \/> undergo sp<sup>3<\/sup> hybridisation giving four equivalent sp<sup>3<\/sup> hybridised orbitals.<\/li>\n<li>Each sp<sup>3<\/sup> hybrid orbital overlaps with the 1s orbitals of hydrogen forming \\(\\sigma_{s p^3-s}\\) bond.<\/li>\n<li>In case of methane four \\(\\sigma_{s p^3-s}\\) bonds are formed. The bonds are directed towards the four corners of a regular tetrahedron. The shape of methane molecule is tetrahedral with a bond angle 109\u00b028\u2019.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25046\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-23.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 23\" width=\"734\" height=\"223\" \/><\/li>\n<\/ol>\n<p>Question 11.<br \/>\nExplain Polar Covalent bond with a suitable example.<br \/>\nAnswer:<br \/>\nThe covalent bond which is formed by the mutual sharing of electron pairs between two dissimilar atoms is called polar covalent bond.<\/p>\n<ul>\n<li>Dissimilar atoms means two atoms having different electronegativity values (or) atoms of different elements.<br \/>\nEg : HF, HCl, H<sub>2<\/sub>O, CO<sub>2<\/sub> etc.,<br \/>\nFormation of HCl :<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25047\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-24.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 24\" width=\"532\" height=\"45\" \/><\/li>\n<li>In the above example H and Cl are two atoms of different elements having different electro negativities.<\/li>\n<li>These two atoms (H, Cl) mutually share the electron pairs and form the polar covalent bond.<\/li>\n<\/ul>\n<p>Question 12.<br \/>\nExplain the shape and bond angle In BCl<sub>3<\/sub> molecule in terms of Valence Bond Theory.<br \/>\nAnswer:<br \/>\nBoron trichloride molecule formation :<\/p>\n<ol>\n<li>The electronic configuration of &#8216;B&#8217; in the ground state is <img loading=\"lazy\" class=\"alignnone size-full wp-image-25048\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-25.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 25\" width=\"148\" height=\"33\" \/><\/li>\n<li>On excitation the configuration is <img loading=\"lazy\" class=\"alignnone size-full wp-image-25049\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-26.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 26\" width=\"145\" height=\"28\" \/> Now there are three half filled orbitals are available for hybridisation.<\/li>\n<li>Now sp<sup>2<\/sup> hybridisation takes place at boron atom giving three sp<sup>2<\/sup> hybrid orbitals.<\/li>\n<li>Each of them with one unpaired electron forms a &#8216;\u03c3&#8217; bond with one chlorine atom. The overlapping is \u03c3<sub>sp<sup>2<\/sup> &#8211; p<\/sub> (Cl atom has the unpaired electron in 2p<sub>z<\/sub> orbital). In boron trichloride there are three &#8216;\u03c3&#8217; bonds.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25050\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-27.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 27\" width=\"499\" height=\"286\" \/><\/li>\n<\/ol>\n<p>Question 13.<br \/>\nWhat are \u03c3 and \u03c0 bonds ? Specify the differences between them.<br \/>\nAnswer:<br \/>\nDefinition of \u03c3 bond : &#8220;\u03c3 &#8211; bond is along the internuclear axis. It has a cylindrical symmetry&#8221;.<br \/>\nDefinition of \u03c0 bond : &#8220;A covalent bond formed by a sidewise overlap of &#8216;p&#8217; orbitals of atoms that are already bonded through a \u03c3 &#8211; bond, and in which the electron clouds are present above and below the internuclear axis is known as \u03c0 &#8211; bond&#8221;.<\/p>\n<p>Sigma bond (\u03c3)<\/p>\n<ol>\n<li>A \u2018\u03c3\u2019 bond is formed by the axial overlap of two half filled orbitals belonging to the valence shells of the two combining atoms.<\/li>\n<li>The \u2018\u03c3\u2019 bonding electron cloud is symmetric about the inter-nuclear axis.<\/li>\n<li>It is strong bond since the extent of overlap is much.<\/li>\n<li>It allows free rotation of atoms or groups about the bond.<\/li>\n<li>It can exist independently.<\/li>\n<li>It determines the shape of the molecule.<\/li>\n<li>There can be only one &#8216;\u03c3\u2019 bond between two atoms.<\/li>\n<li>Hybrid orbitaIs form only &#8216;\u03c3\u2019 bonds.<\/li>\n<\/ol>\n<p>Pi \u2014 bond (\u03c0)<\/p>\n<ol>\n<li>A \u03c0-bond is formed by the lateral overlap of orbitals.<\/li>\n<li>The \u03c0-bonding electron cloud lies above and below the flame of the internuclear axis.<\/li>\n<li>It is a weaker than &#8216;\u03c3\u2019 bond, since the extent of overlap is less.<\/li>\n<li>\u03c0-bond restricts such free rotation.<\/li>\n<li>It is formed only after a \u2018\u03c3\u2019 bond is formed.<\/li>\n<li>It does not determine the shape of the molecule.<\/li>\n<li>There can be one or two \u03c0-bonds between the atoms.<\/li>\n<li>Hybrid orbitals cannot form \u03c0-bonds.<\/li>\n<\/ol>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 14.<br \/>\nEven though nitrogen in ammonia is in sp<sup>3<\/sup> hybridization, the bond angle deviate from 109\u00b028. Explain.<br \/>\nAnswer:<br \/>\nIn NH<sub>3<\/sub> molecule the central nitrogen atom shares its &#8216;3&#8217; unpaired electrons with three hydrogen atoms to form 3\u03c3 bonds. Hence NH<sub>3<\/sub> molecule contains one lone pair, three bond pairs.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25051\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-28.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 28\" width=\"239\" height=\"197\" \/><br \/>\na) Because of the repulsion between the lone pair and the bond pairs the angle reduces to 107\u00b0.<br \/>\nb) According to VSEPR theory the geometry of the molecule is pyramidal with bond angle 107\u00b0.<\/p>\n<p>Question 15.<br \/>\nShow how a double and triple bond are formed between carbon atoms in<br \/>\n(a) C<sub>2<\/sub>H<sub>4<\/sub> and<br \/>\n(b) C<sub>2<\/sub>H<sub>2<\/sub> respectively.<br \/>\nAnswer:<br \/>\nFormation of double bond between the carbon atoms of C<sub>2<\/sub>H<sub>4<\/sub> : &#8211;<br \/>\nC &#8211; Ground state electronic configuration 1s<sup>2<\/sup> 2s<sup>2<\/sup> 2p<sup>2<\/sup><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25052\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-29.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 29\" width=\"417\" height=\"169\" \/><\/p>\n<ul>\n<li>\u00a0In ethylene two carbons undergo sp<sup>2<\/sup> hybridisation.<\/li>\n<li>One of sp<sup>2<\/sup> hybrid orbital of carbon overlaps with sp<sup>2<\/sup> hybrid orbital of another carbon atom to form C &#8211; C sigma bond.<\/li>\n<li>The two other sp<sup>2<\/sup> hybrid orbitals of each carbon overlap with &#8216;s&#8217; orbital of hydrogen atoms to form C &#8211; H bonds.<\/li>\n<\/ul>\n<p>The unhybridised orbital of one carbon atom overlap side wisely with the similar orbital to form weak \u03c0 bond.<\/p>\n<p><img loading=\"lazy\" class=\"alignnone size-full wp-image-25053\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-30.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 30\" width=\"675\" height=\"344\" \/><br \/>\nFormation of triple bond between the carbon atoms of C<sub>2<\/sub>H<sub>2<\/sub> : &#8211;<br \/>\nC &#8211; Ground state electronic configuration 1s<sup>2<\/sup> 2s<sup>2<\/sup> 2p<sup>2<\/sup><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25054\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-31.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 31\" width=\"275\" height=\"129\" \/><\/p>\n<ul>\n<li>In acetylene two carbons undergoes sp hybridisation.<\/li>\n<li>One of sp hybrid orbital of carbon overlaps with sp hybrid orbital of another carbon atom to form C &#8211; C sigma bond.<\/li>\n<li>another sp hybrid orbitals of each carbon overlap with s orbital of hydrogen atoms to form C &#8211; H bonds.<\/li>\n<li>The un hybridised orbitals of two carbon atom overlap side wisely to form two weak \u03c0 bonds.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25055\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-32.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 32\" width=\"695\" height=\"175\" \/><\/li>\n<\/ul>\n<p>Question 16.<br \/>\nExplain the hybridization involved in PCl<sub>5<\/sub> molecule. (T.S. Mar. &#8217;16, &#8217;15)<br \/>\nAnswer:<br \/>\n1) In PCl<sub>5<\/sub> the electron configuration of phosphorus is<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25056\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-33.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 33\" width=\"655\" height=\"110\" \/><\/p>\n<p>2) Phosphorus undergoes sp<sup>3<\/sup>d &#8211; hybridisation by intermixing of one s-orbital [3s], three p &#8211; orbitals [3p<sub>x<\/sub>, 3p<sub>y<\/sub>, 3p<sub>z<\/sub>] and one d &#8211; orbital.<br \/>\nThese five hybrid orbitals overlap. The p<sub>z<\/sub> orbitals of chlorine atoms forming five \\(\\sigma_{s p^3 d-s}\\) bonds. Out of these five p &#8211; Cl bonds three are coplanar and the remaining two are in the axial position. There by PCl<sub>5<\/sub> acquires the trigonal bipyramidal shape. The molecule contains two bond angles 90\u00b0 and 120\u00b0.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25057\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-34.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 34\" width=\"471\" height=\"207\" \/><\/p>\n<p>Question 17.<br \/>\nExplain the hybridization involved in SF<sub>6<\/sub> molecule.<br \/>\nAnswer:<br \/>\nIn this hybridisation one &#8216;s&#8217; orbital, three &#8216;p&#8217; orbitals and two &#8216;d&#8217; orbitals of the excited atom combine to form six equivalent sp<sup>3<\/sup>d<sup>2<\/sup> hybrid orbitals.<br \/>\ne.g. : SF<sub>6<\/sub><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25058\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-35.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 35\" width=\"448\" height=\"94\" \/><br \/>\nThese six sp<sup>3<\/sup> d<sup>2<\/sup> hybrid orbitals overlap six 2p<sub>z<\/sub> orbitals of fluorine atoms to form six \\(\\sigma_{s p^3 d^2}\\) bonds. The directions of the bonds give an octa-hedral shape to the molecule. The bond angle is 90\u00b0 or 180\u00b0 &amp; 90\u00b0.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25059\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-36.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 36\" width=\"459\" height=\"254\" \/><\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 18.<br \/>\nExplain the formation of Coordinate Covalent bond with one example.<br \/>\nAnswer:<br \/>\nCo-ordinate covalent bond (dative bond) is a special type of covalent bond. It is proposed by Sidgwick. It is formed by the sharing of electrons between two atoms in which both the electrons of the shared electron pair are contributed by one atom and the other atom nearly participates in sharing.<\/p>\n<p>The bond is represented as (&#8220;\u2192&#8221;) an arrow starting from the donar atom and directed towards the acceptor atom.<\/p>\n<p>Examples :<br \/>\n1) Ammonia &#8211; Boron trifluoride H<sub>3<\/sub>N : \u2192 BF<sub>3<\/sub><br \/>\nAmmonia combines with boron trifluoride to give ammonium boron trifluoride.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25060\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-37.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 37\" width=\"540\" height=\"127\" \/><br \/>\nIn ammonia nitrogen has a complete octet and also it has a lone pair of electrons. In BF<sub>3<\/sub> the boron atom has a total of six electrons after sharing with fluorine. Nitrogen donates the electron pair to boron to form a co-ordinate covalent bond between ammonia and boron trifluoride.<\/p>\n<p>2) Ammonium ion (\\(\\mathrm{NH}_4^{+}\\))<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25061\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-38.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 38\" width=\"366\" height=\"113\" \/><\/p>\n<p>3) Hydronium ion (\\(\\mathrm{H}_3 \\mathrm{O}^{+}\\))<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25062\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-39.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 39\" width=\"300\" height=\"92\" \/><\/p>\n<p>Properties of co-ordinate covalent bond :<\/p>\n<ol>\n<li>The bond do not ionise in water.<\/li>\n<li>The compounds are generally soluble in organic solvents and are sparingly soluble in water.<\/li>\n<li>These compounds exhibit space isomerism because the bond is rigid and directional.<\/li>\n<li>The bond is semipolar in nature &#8211; so their volatility lies in between covalent and ionic bonds.<\/li>\n<\/ol>\n<p>Question 19.<br \/>\nWhich hybrid orbitals are used by Carbon atoms in the following molecules?<br \/>\n(a) CH<sub>3<\/sub> -CH<sub>3<\/sub><br \/>\n(b) CH<sub>3<\/sub> &#8211; CH = CH<sub>2<\/sub><br \/>\n(c) CH<sub>3<\/sub> &#8211; CH<sub>2<\/sub> &#8211; OH<br \/>\n(d) CH<sub>3<\/sub> &#8211; CHO<br \/>\nAnswer:<br \/>\na) CH<sub>3<\/sub> -CH<sub>3<\/sub> (ethane)<br \/>\nThe two carbons of ethane undergo sp<sup>3<\/sup> hybridisation.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25063\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-40.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 40\" width=\"235\" height=\"53\" \/><br \/>\nCarbon &#8211; (1) &#8211; undergoes sp<sup>2<\/sup> hyrbidisation<br \/>\nCarbon &#8211; (2) &#8211; undergoes sp<sup>2<\/sup> hyrbidisation<br \/>\nCarbon &#8211; (3) &#8211; undergoes sp<sup>3<\/sup> hyrbidisation<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25064\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-41.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 41\" width=\"235\" height=\"47\" \/><br \/>\nCarbon (1) and (2) both undergo &#8216;sp<sup>3<\/sup>&#8216; hybridisation.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25065\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-42.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 42\" width=\"391\" height=\"120\" \/><\/p>\n<p>Question 20.<br \/>\nWhat is Hydrogen bond ? Explain the different types of Hydrogen bonds with examples. (A.P., T.S. Mar. &#8217;16)<br \/>\nAnswer:<br \/>\nHydrogen bond is a weak electrostatic bond formed between partially positive charged hydrogen atom and an highly electronegative atom of the same molecule or another molecule.<\/p>\n<p>Hydrogen bond is formed when the Hydrogen is bonded to small, highly electronegative atoms like F, O and N. A partial positive charge will be on hydrogen atom and partial negative charge on the electronegative atom.<\/p>\n<p>The bond dissociation energy of hydrogen bond is 40 KJ\/mole. Hydrogen bond is represented with dotted lines (&#8212;&#8211;). Hydrogen bond is stronger than Vander Waals&#8217; forces and weaker than covalent bond.<br \/>\nHydrogen bonding is of two types.<\/p>\n<p>(1) Intermolecular hydrogen bond and<br \/>\n(2) Intramolecular hydrogen bond.<\/p>\n<p>1) Intermolecular hydrogen bond :<br \/>\nIf the hydrogen bond is formed between two polar molecules it is called intermolecular hydrogen bond, i.e., the hydrogen bond is formed between hydrogen atom of one molecule and highly electronegative atom of another molecule is known as intermolecular hydrogen bond.<br \/>\nEx. : Water (H<sub>2<\/sub>O) ; HF : NH<sub>3<\/sub> ; p &#8211; nitrophenol, CH<sub>3<\/sub>COOH, ethyl alcohol etc.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25066\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-43.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 43\" width=\"601\" height=\"72\" \/><br \/>\nWater molecule forms oi. associated molecule through intermolecular hydrogen bond. Due to molecular association water possess high boiling point.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25067\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-44.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 44\" width=\"673\" height=\"117\" \/><\/p>\n<p>m or p &#8211; nitrophenol :<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25068\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-45.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 45\" width=\"361\" height=\"183\" \/><\/p>\n<p>2) Intramolecular hydrogen bond:<br \/>\nIf the hydrogen bond is formed within the molecule it is known as intramolecular hydrogen bond.<br \/>\nEx. : o &#8211; nitrophenol; o &#8211; hydroxy benzaldehyde.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25069\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-46.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 46\" width=\"489\" height=\"177\" \/><\/p>\n<p>Abnormal behaviour due to hydrogen bond:<\/p>\n<ol>\n<li>The physical state of substance may alter. They have high melting and boiling points.<\/li>\n<li>Ammonia has higher boiling point than HCl eventhough nitrogen and chlorine have same electronegativity values (3.0). Ammonia forms an associated molecule through intermolecular hydrogen bond.<\/li>\n<li>p &#8211; hydroxy benzaldehyde have higher boiling point than o- hydroxy benzaldehyde. This is due to intermolecular hydrogen bonding in para isomer.<\/li>\n<li>Ethyl alcohol is highly soluble in water due to association and co-association through intermolecular hydrogen bonding.<\/li>\n<\/ol>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 21.<br \/>\nExplain the formation of H<sub>2<\/sub> molecule on the basis of Valence Bond theory.<br \/>\nAnswer:<br \/>\nPostulates of valency bond theory:<\/p>\n<ol>\n<li>Covalent bond is formed by the overlap of an half filled atomic orbital of one atom with an half filled atomic orbital of the other atom involved in the bond formation.<\/li>\n<li>The electrons in these two orbitals involved in the overlap shall have opposite spins.<\/li>\n<li>Greater the overlap stronger the bond formed.<\/li>\n<li>The bonds are formed mostly in the direction in which the electron clouds are concentrated.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25070\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-47.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 47\" width=\"265\" height=\"78\" \/><\/li>\n<\/ol>\n<p>Formation of H<sub>2<\/sub> molecule:<br \/>\nHydrogen molecule is formed due to overlaping of s &#8211; s orbitals. When two Hydrogen atoms come together, is orbitals of the Hydrogen atoms overlap to form a strong \u2018\u03c3\u201d bond. This is \u03c3<sub>s-s&#8217;<\/sub><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25071\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-48.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 48\" width=\"601\" height=\"171\" \/><\/p>\n<p>Question 22.<br \/>\nUsing Molecular Orbital Theory explain why the B<sub>2<\/sub> molecule is paramagnetic ?<br \/>\nAnswer:<br \/>\nBoron electronic configuration is &#8211; 1s<sup>2<\/sup> 2s<sup>2<\/sup> 2p<sup>1<\/sup><br \/>\nThe molecular orbital energy level sequence for B<sub>2<\/sub> is<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25072\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-49.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 49\" width=\"273\" height=\"28\" \/><\/p>\n<ul>\n<li>Bond order = \\(\\frac{6-4}{2}\\) = \\(\\frac{2}{2}\\) = 1<\/li>\n<li>In the above sequence unpaired electrons are present.<\/li>\n<li>Presence of unpaired electrons leads to paramagnetic nature.<br \/>\n\u2234 B<sub>2<\/sub> molecule is paramagnetic.<\/li>\n<\/ul>\n<p>Question 23.<br \/>\nWrite the important conditions necessary for linear combination of atomic orbitals.<br \/>\nAnswer:<\/p>\n<ol>\n<li>The molecular orbitals are formed when the atomic orbitals combine linearly (i.e.,) when the atoms approach each other. The no. of molecular orbitals resulting are equal to the no.of atomic orbitals combining.<\/li>\n<li>Only such atomic orbitals which are of similar energies and symmetry with respect to the inter nuclear axis combine to form molecular orbitals.<\/li>\n<li>The total no. of molecular orbitals produced will be numerically equal to the no. of combining orbitals.<\/li>\n<li>The order of energies of bonding, anti bonding and non bonding orbitals can be written as bonding orbitals &lt; non bonding orbitals &lt; anti &#8211; bonding orbitals.<\/li>\n<\/ol>\n<p>Question 24.<br \/>\nWhat is meant by the term Bond order? Calculate the bond orders in the following<br \/>\n(a) N<sub>2<\/sub><br \/>\n(b) O<sub>2<\/sub><br \/>\n(c) \\(\\mathrm{O}_2^{+}\\) and<br \/>\n(d) \\(\\mathrm{O}_2^{-}\\)<br \/>\nAnswer:<br \/>\nBond order : The half of the difference between the no.of bonding electrons and anti bonding electrons is known as bond order. \u2019<br \/>\na) N<sub>2<\/sub> : Molecular orbital energy level sequence.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25073\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-50.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 50\" width=\"268\" height=\"45\" \/><br \/>\n\u2192 Bond Order = \\(\\frac{10-4}{2}\\) = \\(\\frac{6}{2}\\) = 3<\/p>\n<p>b) O<sub>2<\/sub> : Molecular orbital energy level sequence.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25074\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-51.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 51\" width=\"431\" height=\"313\" \/><\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 25.<br \/>\nOf BF<sub>3<\/sub> and NF<sub>3<\/sub>, dipole moment is observed for NF<sub>3<\/sub> and not for BF<sub>3<\/sub>. Why ?<br \/>\nAnswer:<br \/>\nOf BF<sub>3<\/sub> and NF<sub>3<\/sub> dipole moment is observed for NF<sub>3<\/sub> and not for BF<sub>3<\/sub>.<br \/>\nReasons:<\/p>\n<ul>\n<li>BF<sub>3<\/sub> molecule is non polar and is a symmetrical molecule. Symmetrical molecules have zero dipole moment.<\/li>\n<li>NF<sub>3<\/sub> molecule is polar and it is a unsymmetrical molecule so it has dipole moment.<\/li>\n<li>BF<sub>3<\/sub> molecule has trigonal planar structure.<br \/>\nNF<sub>3<\/sub> molecule has pyramidal shape.<\/li>\n<li>NF<sub>3<\/sub> has dipole moment \u00b5 = 0.8 \u00d7 10<sup>-30<\/sup> coloumb \u00d7 meter.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25075\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-52.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 52\" width=\"609\" height=\"179\" \/><\/li>\n<\/ul>\n<p>Question 26.<br \/>\nEventhough both NH<sub>3<\/sub> and NF<sub>3<\/sub> are Pyramidal, NH<sub>3<\/sub> has a higher dipole moment compared to NF<sub>3<\/sub>. Why? (A.P. Mar.&#8217;16)<br \/>\nAnswer:<\/p>\n<ul>\n<li>Both NH<sub>3<\/sub> and NF<sub>3<\/sub> molecules have pyramidal shape and in two molecules N atom has lone pair of electrons.<\/li>\n<li>Even though fluorine has more electronegativity than nitrogen the dipole moment of NH<sub>3<\/sub> is greater than that of NF<sub>3<\/sub><br \/>\n\u00b5 (NH<sub>3<\/sub>) = 4.9 \u00d7 10<sup>-30<\/sup> coloumbs \u00d7 meter.<br \/>\n\u00b5 (NF<sub>3<\/sub>) = 0.8 \u00d7 10<sup>-30<\/sup> coloumbs \u00d7 meter.<\/li>\n<li>In case of NH<sub>3<\/sub> the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of N &#8211; H bonds.<\/li>\n<\/ul>\n<p>Where as in case of NF<sub>3<\/sub> the orbital dipole is in the direction opposite to the resultant dipole<br \/>\nmoment of the three N &#8211; F bonds.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25076\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-53.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 53\" width=\"464\" height=\"158\" \/><\/p>\n<p>Question 27.<br \/>\nHow do you predict the shapes of the following molecules making use of VSEPR Theory ?<br \/>\n(a) XeF<sub>4<\/sub><br \/>\n(b) BrF<sub>5<\/sub><br \/>\n(c) ClF<sub>3<\/sub> and<br \/>\n(d) IC\\(l_4^{-}\\)<br \/>\nAnswer:<br \/>\nAccording to VSEPR theory the shape of the molecule can be predicted by counting no.of electron pairs (bond pairs, lone pairs) around the central atom.<br \/>\na) XeF<sub>4<\/sub>:<br \/>\nIn XeF<sub>4<\/sub> No.of bond pairs present are &#8216;4&#8217;.<br \/>\nNo.of lone pairs present are &#8216;2&#8217;.<br \/>\nAccording to VSEPR theory shape of molecule is square planar (Actual shape octahedral).<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25077\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-54.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 54\" width=\"134\" height=\"146\" \/><\/p>\n<p>b) BrF<sub>5<\/sub> :<br \/>\nIn BrF<sub>5<\/sub> No.of bond pairs present are &#8216;5&#8217;<br \/>\nNo.of lone pairs present are &#8216;1&#8217;<br \/>\nAccording to VSEPR theory shape of the molecule is square pyramid (actual shape octahedral)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25078\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-55.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 55\" width=\"200\" height=\"176\" \/><br \/>\nc) ClF<sub>3<\/sub>:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25079\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-56.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 56\" width=\"139\" height=\"133\" \/><br \/>\nIn ClF<sub>3<\/sub> No.of bond pairs present are &#8216;3&#8217;<br \/>\nNo.of lone pairs present are &#8216;2&#8217;<br \/>\nAccording to VSEPR Theory shape of ClF<sub>3<\/sub> Molecule is T &#8211; shape (Actual shape TBP)<\/p>\n<p>d) IC\\(l_4^{-}\\):<br \/>\nIn IC\\(l_4^{-}\\) No.of bond pairs present are &#8216;4\u2019<br \/>\nNo.of lone pairs present are \u20182\u2019<br \/>\nAccording to VSEPR Theory shape of ICl\\(l_4^{-}\\) is square planar<br \/>\n(Actual shape octahedral)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25080\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-57.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 57\" width=\"130\" height=\"155\" \/><\/p>\n<p><span style=\"color: #0000ff;\">Long Answer Questions<\/span><\/p>\n<p>Question 1.<br \/>\nExplain the formation of Ionic Bond with a suitable example.<br \/>\nAnswer:<br \/>\nThe electrostatic force that binds the oppositely charged ions which are formed by the transfer of electrons from one atom with low ionization potential to the other with high electron affinity is called ionic bond or electrovalent bond.<br \/>\nIt is formed when the electronegativity difference between the two atoms is more tha 1.7.<br \/>\nExample :<br \/>\nFormation of sodium chloride in terms of orbital concept:<br \/>\n1) Na (Z = 11). The electronic configuration is 1s<sup>2<\/sup> 2s<sup>2<\/sup> 2p<sup>6<\/sup> 3s<sup>1<\/sup>.<br \/>\nThis can be expressed as<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25081\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-58.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 58\" width=\"89\" height=\"55\" \/><\/p>\n<p>2) Cl (Z = 17). The electronic configuration is = 1s<sup>2<\/sup> 2s<sup>2<\/sup> 2p<sup>6<\/sup> 3s<sup>2<\/sup> 3p<sup>5<\/sup><br \/>\nThis can be expressed as<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25082\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-59.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 59\" width=\"196\" height=\"54\" \/><\/p>\n<p>3) The configurations after the transfer of electrons forming ions can be expressed as:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25083\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-60.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 60\" width=\"604\" height=\"140\" \/><br \/>\nIn the formation of sodium chloride the 3s electron of sodium atom is transferred to the 3p<sub>2<\/sub> orbital of chlorine atom. The Na<sup>+<\/sup> ion and Cl<sup>&#8211;<\/sup> ion so formed are now bound by strong coulombic electrostatic forces of attraction forming sodium chloride.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 2.<br \/>\nExplain the factors favourable for the formation of Ionic Compounds.<br \/>\nAnswer:<br \/>\nFactors favour the ionic bond formation:<br \/>\na) Cation formation:<\/p>\n<ol>\n<li>Lower ionization energy: Lower ionization energy of an atom greater is the ease of formation of cation<br \/>\ne.g. : The ionization energy of sodium is 117.9 kcal\/mole and that of potassium is 100 kcal\/mole So K<sup>+<\/sup> ion can readily form than Na<sup>+<\/sup> ion.<\/li>\n<li>Large size of the atom : Large atoms can easily lose the valence electrons. If the size large, the distance between the nucleus and the valence electrons is more and so the force of attraction is less. Therefore the electron can be removed easily from the atom forming cation.<\/li>\n<li>Ion with lower charge: Small magnitude of charge favours the formation of ions easily.<br \/>\ne.g. : The ease of ion formation increases in the order Na<sup>+<\/sup> &gt; Mg<sup>2+<\/sup> &gt; Al<sup>3+<\/sup>.<\/li>\n<li>Cations with inert gas configuration : Ion possessing electronic configuration similar to zero group elements are more stable than those ions which do not have such configuration.<br \/>\neg.: Ca<sup>2+<\/sup> (2, 8, 8) is more stable than Zn<sup>2+<\/sup> (2, 8, 18) because the former has inert gas<br \/>\nconfiguration.<\/li>\n<\/ol>\n<p>b) Anion formation:<\/p>\n<ol>\n<li>High electron affinity: If the electron affinity of an element is high its anion can be easily formed.<br \/>\ne.g.: Cl + e<sup>\u2014<\/sup> \u2192 Cl<sup>&#8211;<\/sup><\/li>\n<li>Smaller size of atom : Smaller the size of the atom lesser is the distance between the nucleus and the valence orbit. Hence the nuclear attraction on incoming electron is more. So the anion is readily formed.<\/li>\n<li>Lower charge : Ions with lower charge are more readily formed than those with higher charge.<br \/>\ne.g.: Cl<sup>&#8211;<\/sup> &gt; O<sup> 2-<\/sup> &gt; N<sup>3-<\/sup><\/li>\n<li>The ions with inert gas electronic configuration are more readily formed than others with the same charge. .<br \/>\nc) If the two bonded atoms differ by more than 1.70 in their EN values, the bond between them is ionic in nature.<\/li>\n<\/ol>\n<p>Question 3.<br \/>\nDraw Lewis Structures for the following molecules.<br \/>\n(a) H<sub>2<\/sub>S<br \/>\n(b) SiCl<sub>4<\/sub><br \/>\n(c) BeF<sub>2<\/sub><br \/>\n(d) HCOOH<br \/>\nAnswer:<br \/>\nLewis structures:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25084\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-61.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 61\" width=\"514\" height=\"162\" \/><\/p>\n<p>Question 4.<br \/>\nWrite notes on<br \/>\n(a) Bond Angle<br \/>\n(b) Bond Enthalpy<br \/>\n(c) Bond length and<br \/>\n(d) Bond order.<br \/>\nAnswer:<br \/>\na) Bond angle : The angle between the orbitals containing bonding electron pairs around the central atom in a molecule (or) complex ion is known as Bond angle.<\/p>\n<ul>\n<li>it is expressed in degrees.<\/li>\n<li>It is determined experimentally by spectroscopic methods.<br \/>\nEg: In H<sub>2<\/sub>O (H \u2014 0 \u2014 H) bond angle is 104.5\u00b0 .<\/li>\n<\/ul>\n<p>b) Bond Enthalpy : The amount of energy required to break one mole of bonds of a particular type between two atoms in a gaseous state is known as Bond Enthalpy.<br \/>\nUnits: KJ\/Mole.<br \/>\nEg: H &#8211; H bond enthalpy in hydrogen is 435.8 KJ\/mole<br \/>\nH<sub>2(g)<\/sub> \u2192 H<sub>(g)<\/sub> + H<sub>(g)<\/sub> \u2206H = 435.8 KJ\/mole<\/p>\n<ul>\n<li>In case of poly atomic molecules average bond enthalpy is used.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25085\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-62.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 62\" width=\"587\" height=\"52\" \/><\/li>\n<\/ul>\n<p>c) Bond length : The distance between the nuclei of the atoms in a molecule is known as bond length.<\/p>\n<ul>\n<li>Bond length is equal to the sum of the covalent radii of the two atoms that are bonded.<\/li>\n<li>Units of Bond length A\u00b0 (or) cm (or) m (or) pm<\/li>\n<li>As the number of bonds between two atoms increases the bond length decrease.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25086\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-63.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 63\" width=\"281\" height=\"106\" \/><\/li>\n<\/ul>\n<p>d) Bond Order:<br \/>\nAccording to Lewis the bond order is given by the number of bonds between the two atoms in a covalent molecule.<br \/>\nEg: Bond order of N<sub>2<\/sub> &#8211; 3<br \/>\nBond order of O<sub>2<\/sub> &#8211; 2<br \/>\nBond order of H<sub>2<\/sub> &#8211; 1<\/p>\n<ul>\n<li>In case of Iso electronic species and ions bond orders are same.<\/li>\n<li>Bond order is useful in predecting stabilities of molecules.<\/li>\n<li>Bond order increases bond enthalpy increases and Bond length decreases.<\/li>\n<\/ul>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 5.<br \/>\nGive an account of VSEPR Theory and its applications.<br \/>\nAnswer:<br \/>\nVSEPR theory was proposed by Sidgwick and Powell and later extended by Gillespie and Nyholm. It was developed by Ronald and Nyholm.<br \/>\nThis theory explains the shapes of simple molecules having electron pairs bonded or non-bonded. The repulsions among the electron pairs present in the valence shell of the central atom decides the shape of the molecules.<\/p>\n<p>According to this theory :<\/p>\n<p>a) The shape of the molecule is determined by repulsions between all of the electron pairs present in the valency shell of central atom.<br \/>\nb) The electron pairs orient in space so as to have minimum repulsions among them.<br \/>\nc) The magnitude of repulsions between bonding pairs of electrons depends on the electronegativity difference between the central atom and the other atoms.<br \/>\nd) The order of repulsions between various electron pairs is lone pair &#8211; lone pair &gt; lone pair &#8211; bond pair &gt; bond pair &#8211; bond pair.<br \/>\ne) The repulsive forces between different bonds is of the order triple bond &gt; double bond &gt; single bond.<br \/>\nf) The shapes of molecules can be predicted as.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25087\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-64.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 64\" width=\"703\" height=\"311\" \/><br \/>\nIn NH<sub>3<\/sub> molecule the central nitrogen atom shares its &#8216;3&#8217; unpaired electrons with three hydrogen atoms to form 3\u03c3 bonds. Hence NH<sub>3<\/sub> molecule contains one lone pair, three bond pairs.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25088\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-65.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 65\" width=\"191\" height=\"174\" \/><br \/>\na) Because of the repulsion between the lone pair and the bond pairs the angle reduces to 107\u00b0.<br \/>\nb) According to VSEPR theory the geometry of the molecule is pyramidal with bond angle 107\u00b0.<\/p>\n<p>Question 6.<br \/>\nHow do you explain the geometry of the molecules on the basis of Valence bond Theory ?<br \/>\nAnswer:<br \/>\nPostulates of valency bond theory :<\/p>\n<ol>\n<li>Covalent bond is formed by the overlap of an half filled atomic orbital of one atom with an half filled atomic orbital of the other atom involved in the bond formation.<\/li>\n<li>The electrons in these two orbitals involved in the overlap shall have opposite spins.<\/li>\n<li>Greater the overlap stronger the bond formed.<\/li>\n<li>The bonds are formed mostly in the direction in which the electron clouds are concentrated.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25089\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-66.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 66\" width=\"259\" height=\"78\" \/><\/li>\n<\/ol>\n<p>Formation of H<sub>2<\/sub> molecule :<br \/>\nHydrogen molecule is formed due to overlapping of s &#8211; s orbitals. When two Hydrogen atoms come together, 1s orbitals of the Hydrogen atoms overlap to form a strong &#8220;\u03c3&#8221; bond. This is \u03c3<sub>s-s<\/sub>.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25090\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-67.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 67\" width=\"605\" height=\"163\" \/><br \/>\nFormation of Cl<sub>2<\/sub> molecule :<br \/>\nThe electronic configuration of Chlorine atom is <img loading=\"lazy\" class=\"alignnone size-full wp-image-25091\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-68.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 68\" width=\"187\" height=\"34\" \/> It has one half filled 3p<sub>z<\/sub> orbital. The p<sub>z<\/sub> orbital of one chlorine atom overlaps the p<sub>z<\/sub> orbital of the other chlorine atom and the two electrons of opposite spins pair up to form covalent bond. As the overlap along the internuclear axis is maximum a strong bond is formed. The bond is formed due to \\(\\sigma_{p-p}\\) overlap.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25092\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-69.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 69\" width=\"681\" height=\"147\" \/><\/p>\n<p>Formation of O<sub>2<\/sub> molecule:<br \/>\nThe electronic configuration of oxygen atom is <img loading=\"lazy\" class=\"alignnone size-full wp-image-25093\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-70.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 70\" width=\"149\" height=\"33\" \/> It has two half filled 2p orbitals i.e., 2p<sub>y<\/sub> and 2p<sub>z<\/sub>.<br \/>\nThe p<sub>y<\/sub> orbital of one atom overlaps the p<sub>y<\/sub> orbital of the second atom to form a &#8216;\u03c3&#8217; bond \\(\\sigma_{p_y}-p_y\\).<br \/>\nThe p<sub>z<\/sub> orbital in the two atoms will be at right angles to the internuclear axis. These two can have lateral overlap. The electron density of the bonded pair is distributed in two banana like regions lying on either side of the internuclear axis. Thus the oxygen molecule has a double bond. The molecule has one \u03c3<sub>p &#8211; p<\/sub> and one \u03c0<sub>p &#8211; p<\/sub> between the two atoms.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25094\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-71.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 71\" width=\"611\" height=\"226\" \/><\/p>\n<p>Question 7.<br \/>\n&#8216;What do you understand by Hybridisation? Explain different types of hybridization involving s and p orbitals. (Mar. \u201913)<br \/>\nAnswer:<br \/>\nHybridisation is defined as the process of mixing of atomic orbitals of nearly equal energy of an atom to give the same number of new set of orbitals of equal energy and shapes.<br \/>\nDepending on the number and nature of orbitals involving hybridisation it is classified into different types. If &#8216;s&#8217; and &#8216;p&#8217; atomic orbitals are involved three types are possible namely sp<sup>3<\/sup>, sp<sup>2<\/sup> and sp.<\/p>\n<p>1. sp<sup>3<\/sup> hybridisation : In this hybridisation one&#8217;s and three &#8216;p&#8217; atomic orbitals of the excited atom combine to form four equivalent sp<sup>3<\/sup> hybridised orbitals.<br \/>\nThis hybridisation is known as tetrahedral or tetragonal hybridisation.<\/p>\n<p>Each sp<sup>3<\/sup> hybridised orbital possess 25% &#8216;s&#8217; nature and 75% of &#8216;p&#8217; nature. The shape of the molecule is tetrahedral with a bond angle 109\u00b028&#8242;, e.g. : Formation of Methane molecule :<\/p>\n<ol>\n<li>The central atom of methane is carbon.<\/li>\n<li>The electronic configuration of carbon in ground state is <img loading=\"lazy\" class=\"alignnone size-full wp-image-25095\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-72.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 72\" width=\"136\" height=\"30\" \/> and on excitation it is <img loading=\"lazy\" class=\"alignnone size-full wp-image-25096\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-73.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 73\" width=\"143\" height=\"31\" \/> During excitation the 2s pair splits and the electron jumps into the adjacent vacant 2p<sub>z<\/sub> orbital.<\/li>\n<li>The <img loading=\"lazy\" class=\"alignnone size-full wp-image-25097\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-74.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 74\" width=\"110\" height=\"30\" \/> undergo sp<sup>3<\/sup> hybridisation giving four equivalent sp<sup>3<\/sup> hybridised orbitals.<\/li>\n<li>Each sp<sup>3<\/sup> hybrid orbital overlaps with the 1s orbitals of hydrogen forming \\(\\sigma_{s p^3-s}\\) bond.<\/li>\n<li>In case of methane four \\(\\sigma_{s p^3}-s\\) bonds are formed. The bonds are directed towards the four corners of a regular tetrahedron. The shape of methane molecule is tetrahedral with a bond angle 109\u00b028&#8242;.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25098\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-75.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 75\" width=\"713\" height=\"220\" \/><\/li>\n<\/ol>\n<p>2. sp<sup>2<\/sup> hybridisation : In this hybridisation one \u2018s\u2019 and two p\u2019 atomic orbitals of the excited atom<br \/>\ncombine to form three equivalent sp<sup>2<\/sup> hybridised orbitals.<br \/>\nThis hybridisation is also known as trigorial hybridisation. In sp<sup>2<\/sup> hybridisation each sp<sup>2<\/sup> hybrid orbital has 33.33% \u2018s\u2019 nature and 66.66% \u2018p\u2019 nature. The shape of the molecule is trigonal with a bond angle 120\u00b0.<br \/>\nE.g.: Boron trichioride molecule formation:<\/p>\n<ol>\n<li>The electronic configuration of \u2018B\u2019 in the ground state is <img loading=\"lazy\" class=\"alignnone size-full wp-image-25099\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-76.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 76\" width=\"143\" height=\"31\" \/><\/li>\n<li>On excitation the configuration is <img loading=\"lazy\" class=\"alignnone size-full wp-image-25100\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-77.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 77\" width=\"132\" height=\"33\" \/> Now there are three half filled orbitals are available for hybridisation.<\/li>\n<li>Now sp<sup>2<\/sup> hybridisation takes place at boron atom giving three sp<sup>2<\/sup> hybrid orbitals.<\/li>\n<li>Each of them with one unpaired electron forms &#8216;\u03c3&#8217; bond with one chlorine atom. The overlapping is \\(\\sigma_{s p^2-p}\\) (Cl atom has the unpaired electron in 2P<sub>z<\/sub> orbital). In boron trichloride there are three \u2018\u03c3\u2019 bonds.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25101\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-78.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 78\" width=\"538\" height=\"282\" \/><\/li>\n<\/ol>\n<p>3. sp hybridisation : In this hybridisation one \u2018s and one \u2018p\u2019 atomic orbitals of the excited atom combine to form two equivalent sp hybridised orbitals.<br \/>\nThis hybridisation is also known as diagonal hybridisation. In sp hybridisation each sp hybrid orbital has 50% \u2018s\u2019 character and 50% \u2018p\u2019 character. The shape of the molecule is linear or diagonal with a bond angle 180\u00b0.<br \/>\nEx. : Beryllium chloride molecule formation:<\/p>\n<ol>\n<li>Be atom has <img loading=\"lazy\" class=\"alignnone size-full wp-image-25102\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-79.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 79\" width=\"136\" height=\"35\" \/> electronic configuration.<\/li>\n<li>In ground state it has no half filled orbitais. On excitation the configuration becomes \\(1 s^2 2 s^1 2 P_x^1\\)\\(2 p_y^0 2 p_z^0\\).<\/li>\n<li>Now sp hybridisation takes place at beryllium atom giving two sp hybrid orbitais. Each of them with one unpaired electron forms a \u2018\u03c3\u2019 bond with one chlorine atom.<\/li>\n<li>The overlaping is \u03c3<sub>sp-p<\/sub> (Cl atom has the unpaired electron in 2p<sub>z<\/sub> orbital). In beryllium chloride there are two \u2018\u03c3\u2019 bonds.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25103\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-80.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 80\" width=\"615\" height=\"233\" \/><\/li>\n<\/ol>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 8.<br \/>\nWrite the salient features of Molecular Orbital Theory.<br \/>\nAnswer:<br \/>\nMolecular orbital theory:<br \/>\nHund and Mulliken.<\/p>\n<ol>\n<li>Theory was proposed by<\/li>\n<li>Atomic orbitals (AO) of the bonded atoms combine loose their identity to form molecular orbitals (MO).<\/li>\n<li>The electrons in a molecule reside in molecular orbitals.<\/li>\n<li>Molecular orbital is the region around the nuclei where the probability of finding electon is maximum (or) the wave function of a molecule.<\/li>\n<li>The electrons of all the atoms in a molecule are revolving under the influence of all the nuclei in the molecule.<\/li>\n<li>The molecular orbitals are formed when the atomic orbitals combine linearly.<\/li>\n<li>The shape of the molecular orbitals depends on the shape of the atomic orbitals.<\/li>\n<li>Each molecular orbital can accommodate two electrons with opposite spins.<\/li>\n<li>The molecular orbitals are arranged in the increasing order of energy, and electrons are filled in the same order.<\/li>\n<li>Hund&#8217;s rule of maximum multiplicity is to be followed while filling molecular orbitals.<\/li>\n<li>Atomic orbitals with similar energy and symmetry can combine to give molecular orbitals.<\/li>\n<li>Molecular orbitals with energy lower than A.O are known as bonding molecular orbital; while those with higher energy are known as anti bonding molecular orbitals. Those which are not involved in combination are called non bonding orbitals.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25104\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-81.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 81\" width=\"428\" height=\"83\" \/><\/li>\n<li>The order of energies of molecular orbital is : bonding &lt; nonbonding &lt; antibonding molecular orbitals.<\/li>\n<li>The bonding orbitals are designated a \u03c3 and \u03c0.<\/li>\n<li>The antibonding orbitals are designated \u03c3* and \u03c0*.<\/li>\n<\/ol>\n<p>Filling of electrons into molecular orbitals :<br \/>\nThe sequence of energy levels of molecular orbitals is given by<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25105\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-82.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 82\" width=\"449\" height=\"46\" \/><br \/>\nsequence is valid for oxygen and other heavier elements.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25106\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-83.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 83\" width=\"446\" height=\"40\" \/><br \/>\nThis sequence is valid for lighter elements like B.CandN.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25107\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-84.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 84\" width=\"399\" height=\"44\" \/><\/p>\n<p>Question 9.<br \/>\nGive the Molecular Orbital Energy diagram of<br \/>\n(a) N<sub>2<\/sub> and<br \/>\n(b) O<sub>2<\/sub>. Calculate the respec- five bond order. Write the magnetic nature of N<sub>2<\/sub> and O<sub>2<\/sub> molecules.<br \/>\nAnswer:<br \/>\nMolecular orbital energy level diagram (MOED) of &#8216;N<sub>2<\/sub>\u2018 ; Electronic configuration of nitrogen (z = 7) is 1s<sup>2<\/sup> 2s<sup>2<\/sup> 2p<sup>3<\/sup>. Since nitrogen atom has 7 electrons, the molecular orbitals of nitrogen molecule (N<sub>2<\/sub>) has 14 electrons which are distributed as below :<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25108\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-85.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 85\" width=\"279\" height=\"47\" \/><\/p>\n<p><img loading=\"lazy\" class=\"alignnone size-full wp-image-25109\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-86.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 86\" width=\"551\" height=\"690\" \/><\/p>\n<ul>\n<li>Bond order = \\(\\frac{8-2}{2}\\) = 3 ( N \u2261 N)<\/li>\n<li>Absence of unpaired electrons showed that N<sub>2<\/sub> molecule is diamagnetic.<\/li>\n<\/ul>\n<p>MOED of O<sub>2<\/sub>:<br \/>\nElectronic configuration of Oxygen (Z = 8) is 1s<sup>2<\/sup> 2s<sup>2<\/sup> 2p<sup>4<\/sup>. Since Oxygen atom has 8 electrons, the molecular orbitais of Oxygen molecule (O<sup>2<\/sup>) has 16 electrons, which are distributed as below:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25110\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-87.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 87\" width=\"320\" height=\"50\" \/><\/p>\n<p><img loading=\"lazy\" class=\"alignnone size-full wp-image-25111\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-88.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 88\" width=\"488\" height=\"642\" \/><\/p>\n<ul>\n<li>Bond order = \\(\\frac{10-6}{2}\\) = 2 (O = O)<\/li>\n<li>Presence of two unpaired 6 electrons \\(\\left(\\pi_{2 p_y^1}^{\\star}, \\pi_{2 p_z^1}^{\\star}\\right)\\) showed that O<sub>2<\/sub> molecule is paramagnetic.<\/li>\n<\/ul>\n<p><span style=\"color: #0000ff;\">Solved Problems<\/span><\/p>\n<p>Question 1.<br \/>\nWrite the Lewis dot structure of CO molecule.<br \/>\nSolution:<br \/>\nStep 1. Count the total number of valence electrons of carbon and oxygen atoms. The outer (valence) shell configurations of carbon and oxygen atoms are: 2s<sup>2<\/sup> 2p<sup>2<\/sup> and 2s<sup>2<\/sup> 2p<sup>4<\/sup>, respectively. The valence electrons available are 4 + 6 = 10.<\/p>\n<p>Step 2. The skeletal structure of CO is written as: C O<\/p>\n<p>Step 3. Draw a single bond (one shared electron pair) between C and O and complete the octet on O, the remaining two electrons are the lone pair on C.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25112\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-89.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 89\" width=\"212\" height=\"37\" \/><br \/>\nThis does not complete the octet on carbon and hence we have to resort to multiple bonding (in this case a triple bond) between C and O atoms. This satisfies the octet rule condition for both atoms.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25113\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-90.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 90\" width=\"265\" height=\"50\" \/><\/p>\n<p>Question 2.<br \/>\nWrite the Lewis structure of the nitrite ion, \\(\\mathrm{NO}_2^{-}\\).<br \/>\nSolution:<br \/>\nStep 1. Count the total number of valence electrons of the nitrogen atom, the oxygen atoms and the additional one negative charge (equal to one electron).<br \/>\nN(2s<sup>2<\/sup> 2p<sup>3<\/sup>), O (2s<sup>2<\/sup> 2p<sup>4<\/sup>)<br \/>\n5 + (2 \u00d7 6) + 1 = 18 electrons<\/p>\n<p>Step 2. The skeletal structure of \\(\\mathrm{NO}_2^{-}\\) is written as: O N O<\/p>\n<p>Step 3. Draw a single bond (one shared electron pair) between the nitrogen and each of the oxygen atoms completing the octets on oxygen atoms. This, however, does not complete the octet on nitrogen if the remaining two electrons constitute lone pair on it.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25114\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-91.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 91\" width=\"265\" height=\"63\" \/><\/p>\n<p>Hence we have to resort to multiple bonding between nitrogen and one of the oxygen atoms (in this case a double bond). This leads to the following Lewis dot structures.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25115\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-92.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 92\" width=\"373\" height=\"64\" \/><\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 3.<br \/>\nExplain the structure of \\(\\mathrm{CO}_3^{2-}\\) ion interms of resonance.<br \/>\nSolution:<br \/>\nThe single Lewis structure based on the presence of two single bonds and one double bond between carbon and oxygen atoms is inadequate to represent the molecule accurately as it represents unequal bonds. According to the experimental findings, all carbon to oxygen bonds in CCO<sup>2<\/sup> are equivalent.<\/p>\n<p>Therefore the carbonate ion is best described as a resonance hybrid of the canonical forms I, II, and III shown below.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25116\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-93.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 93\" width=\"395\" height=\"284\" \/><\/p>\n<p>Question 4.<br \/>\nExplain the structure of CO<sub>2<\/sub> molecule.<br \/>\nSolution:<br \/>\nThe experimentally determined carbon to oxygen bond length in CO<sub>2<\/sub> is 115 pm. The lengths of a normal carbon to oxygen double bond (C = O) and carbon to oxygen triple bond(C \u2261 O) are 121 pm and 110 pm respectively. The carbon-oxygen bond lengths in CO<sub>2<\/sub> (115 pm) lie between the values for C = O and C \u2261 O. Obviously, a single Lewis structure cannot depict this position and it becomes necessary to write more than one Lewis structures and to consider that the structure of CO<sub>2<\/sub> is best described as a hybrid of the canonical or resonance forms I, II and III.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-25117\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/AP-Inter-1st-Year-Chemistry-Study-Material-Chapter-3-Chemical-Bonding-and-Molecular-Structure-94.png\" alt=\"AP Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 94\" width=\"373\" height=\"188\" \/><\/p>\n<p><span style=\"color: #0000ff;\">Additional Problems<\/span><\/p>\n<p>Question 1.<br \/>\nThe experimental dipole moment of HCl is 1.03D and its bond length (distance) is 1.27\u00c5. Calculate the % of ionic character of HCl.<br \/>\nAnswer:<br \/>\nCalculated dipole moment = q \u00d7 d<br \/>\n= 4.8 \u00d7 10<sup>-10<\/sup> \u00d7 1.27 \u00d7 10<sup>-8<\/sup> cm<br \/>\n= 6.09 Debye<br \/>\n% of ionic character = \\(\\frac{\\mu_{\\text {ods }}}{\\mu_{\\text {calc }}}\\) \u00d7 100<br \/>\n= \\(\\frac{1.03}{6.09}\\) \u00d7 100<br \/>\n= 16.9%<\/p>\n<p>Question 2.<br \/>\nThe dipole moment of H<sub>2<\/sub>S is 0.95D. Find the bond moment if the bond angle is 97\u00b0 (Cos 48.5\u00b0 = 0.662).<br \/>\nAnswer:<br \/>\n\\(\\mu_{\\text {obs }}\\) = 2 (bond moment) \\(\\left(\\cos \\frac{\\theta}{2}\\right)\\)<br \/>\n0.95 = 2 (bond moment) (Cos 48.5\u00b0)<br \/>\n0.95 = 2 \u00d7 bond moment \u00d7 0.662<br \/>\nBond moment = \\(\\frac{0.95}{2 \\times 0.662}\\) = 0.72D<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 3rd Lesson Chemical Bonding and Molecular Structure Textbook Questions and Answers. AP Inter 1st Year Chemistry Study Material 3rd Lesson Chemical Bonding and Molecular Structure Very Short Answer Questions Question 1. What is Octet rule? Answer: Every atom must possess 8 electrons in its outermost &hellip;<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35422"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35422"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35422\/revisions"}],"predecessor-version":[{"id":35423,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35422\/revisions\/35423"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35422"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35422"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35422"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}