{"id":22210,"date":"2024-01-11T17:08:45","date_gmt":"2024-01-11T11:38:45","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=22210"},"modified":"2024-01-12T16:24:31","modified_gmt":"2024-01-12T10:54:31","slug":"ap-inter-2nd-year-chemistry-study-material-chapter-6a","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-inter-2nd-year-chemistry-study-material-chapter-6a\/","title":{"rendered":"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements"},"content":{"rendered":"<p>Andhra Pradesh BIEAP <a href=\"https:\/\/apboardsolutions.in\/ap-inter-2nd-year-chemistry-study-material\/\">AP Inter 2nd Year Chemistry Study Material<\/a> Lesson 6(a) Group-15 Elements Textbook Questions and Answers.<\/p>\n<h2>AP Inter 2nd Year Chemistry Study Material Lesson 6(a) Group-15 Elements<\/h2>\n<p><span style=\"color: #0000ff;\">Very Short Answer Questions<\/span><\/p>\n<p>Question 1.<br \/>\nWhy does the reactivity of nitrogen differ from phosphorus?<br \/>\nAnswer:<br \/>\nNitrogen gas exists as diatomic molecule. Dut to the presence of triple bond between Base N &#8211; Atoms bond dissociation energy is high (941.4 KJ \/mol). Hence nitrogen is inert and unreactive.<br \/>\nPhosphorus is a tetra atomic molecule and P-P single bond is weaker than N \u2261 N. P &#8211; P bond dissociation energy is 213 KJ\/mole. Hence phosphorus is more reactive than Nitrogen.<\/p>\n<p>Question 2.<br \/>\nHow is nitrogen prepared in the laboratory ?<br \/>\nWrite the chemical equations of the reactions involved.<br \/>\nAnswer:<br \/>\nPreparation of di Nitrogen :<\/p>\n<ul>\n<li>Very pure nitrogen is obtained by the thermal decomposition of sodium (or) barium a-zide.<br \/>\nBa(N<sub>3<\/sub>)<sub>2<\/sub> \u2192 Ba + 3N<sub>2<\/sub><\/li>\n<li>In the laboratory dinitrogen is prepared by treating an aqueous solution of NH<sub>4<\/sub>Cl with NaNO<sub>2<\/sub><br \/>\nNH<sub>4<\/sub>Cl<sub>(aq)<\/sub> + NaNO<sub>2(aq) <\/sub> \u2192 N<sub>2(g)<\/sub> + 2H<sub>2<\/sub>O<sub>(I)<\/sub> + NaCl<sub>(aq)<\/sub><\/li>\n<li>Nitrogen can also be obtained by the thermal decompostion of ammonium dichromate.<br \/>\n(NH<sub>4<\/sub>)<sub>3<\/sub> Cr<sub>2<\/sub>O<sub>7<\/sub> <img loading=\"lazy\" class=\"alignnone size-full wp-image-23206\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-1.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 1\" width=\"51\" height=\"20\" \/> N<sub>2<\/sub> + 4H<sub>2<\/sub>O + Cr<sub>2<\/sub>O<sub>3<\/sub><\/li>\n<\/ul>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 3.<br \/>\nNitrogen exists as diatomic molecule and phosphorus as P<sub>4<\/sub> &#8211; Why ?<br \/>\nAnswer:<br \/>\nNitrogen exists as diatomic molecule :<\/p>\n<ul>\n<li>Nitrogen has small size and high electronegativity and nitrogen atom forms P\u03c0 &#8211; P\u03c0 multiple bonds with it self (triple bond). So it exists as a discrete diatomic molecule in elementary state.<\/li>\n<\/ul>\n<p>Phosphorus exists as tetra atomic molecule :<\/p>\n<ul>\n<li>Phosphorus has large size and less electronegative and it forms P-P single bonds. So it exists as tetra atomic i.e., P<sub>4<\/sub>.<\/li>\n<\/ul>\n<p>Question 4.<br \/>\nWhy does nitrogen show catenation properties less than phosphorus ?<br \/>\nAnswer:<br \/>\nExplanation :<br \/>\nThe single N-N bond is weaker than the single P-P bond due to high inter electronic repulsion of the non-boriding electrons in N<sub>2<\/sub> because of small bond length. Therefore the catenation property is weaker in nitrogen as compared to phosphorus.<\/p>\n<p>Question 5.<br \/>\nNitrogen molecule is highly stable &#8211; Why ?<br \/>\nAnswer:<br \/>\nNitrogen Molecule is more stable because in between two nitrogen atoms of N<sub>2<\/sub>, a triple bond is present. To break this triple bond high energy is required (941.4KJ\/mole).<\/p>\n<p>Question 6.<br \/>\nWhy are the compounds of bismuth more stable in +3 oxidation state ?<br \/>\nAnswer:<br \/>\nBismuth compounds are more stable in +3 oxidation state because &#8216;Bi&#8217; exhibits +3 stable oxidation state instead of +5 due to inert pair effect.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 7.<br \/>\nWhat is allotropy ? Explain the different allotropic forms of phosphorus.<br \/>\nAnswer:<br \/>\nAllotropy: The existance of an element in different physical forms having similar chemical properties is called allotropy.<br \/>\nAllotropes of &#8216;P&#8217;: \u2192 White &#8216;P&#8217; (or) Yellow &#8216;P&#8217;.<\/p>\n<ul>\n<li>Red &#8216;P&#8217;<\/li>\n<li>Scarlet &#8216;P&#8217;<\/li>\n<li>Violet &#8216;P&#8217;<\/li>\n<li>\u03b1 &#8211; black &#8216;P&#8217;<\/li>\n<li>\u03b2 &#8211; black &#8216;P&#8217;.<\/li>\n<\/ul>\n<p>White phosphorus :<\/p>\n<ul>\n<li>It is poisonous and insoluble in water and soluble in carbon disulphide and glows in dark.\u00a0It is a translucent white waxy solid. &#8216;<\/li>\n<li>\u00a0It dissolves in boiling NaOH solution and gives PH<sub>3<\/sub>.<br \/>\nP<sub>4<\/sub> + 3NaOH + 3H<sub>2<\/sub> \u2192 PH<sub>3<\/sub> + 3NaH<sub>2<\/sub>PO<sub>2<\/sub>. (sodium hypo phosphite)<\/li>\n<li>\u00a0It is more reactive than other solid phases.<\/li>\n<li>Bond angle is 60\u00b0 and it readily catches fire.<\/li>\n<\/ul>\n<p>Red phosphorus :<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23207\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-2.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 2\" width=\"324\" height=\"104\" \/><\/p>\n<ul>\n<li>Red &#8216;F possesses iron grey lustre.<\/li>\n<li>In is odour less, non poisonous and insoluble in water as well as CS<sub>2<\/sub>.<\/li>\n<li>Red F&#8217; is much less reactive than white &#8216;P&#8217;.<br \/>\nBlack &#8216;P&#8217;:<\/li>\n<li>\u03b1 &#8211; Black &#8216;P&#8217;: It is formed when red &#8216;P&#8217; is heated in a sealed tube 803K.<\/li>\n<li>\u03b2 &#8211; Black F : It is prepared by heating white P&#8217; at 473 K under high pressure.<\/li>\n<\/ul>\n<p>Question 8.<br \/>\nHow do you account for the inert character of dinitrogen ?<br \/>\nAnswer:<br \/>\nDi Nitrogen is chemically inert. .<br \/>\nExplanation : In nitrogen molecule there exists a triple bond between two nitrogen atoms. To break this triple bond high energy (bond dissociation energy) is required (941.4 KJ\/mole).<\/p>\n<p>Question 9.<br \/>\nExplain the difference in the structures of white and red phosphorus.<br \/>\nAnswer:<br \/>\nWhite &#8216;P&#8217; molecule has tetrahedral structure (discrete molecule). Discrete &#8216;P&#8217; molecules are held by vander waal&#8217;s forces<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23208\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-3.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 3\" width=\"134\" height=\"157\" \/><br \/>\nWhite phosphorus<br \/>\nRed &#8216;P&#8217; is polymeric consisting of chains of P<sub>4<\/sub> tetrahedron linked together through covalent bonds.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23209\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-4.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 4\" width=\"499\" height=\"178\" \/><\/p>\n<p>Question 10.<br \/>\nHow is \u03b1 &#8211; black phosphorus prepared from red phosphorus?<br \/>\nAnswer:<br \/>\n\u03b1 &#8211; Black \u2018P: It is formed when red? is heated in a sealed tube 803K.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 11.<br \/>\nWrite the difference between the properties of white phosphorus and red phosphorus.<br \/>\nAnswer:<br \/>\nWhite &#8216;P&#8217;<\/p>\n<ul>\n<li>It is white waxy solid (translucent).<\/li>\n<li>It is insolable in water and solable in CS<sub>2<\/sub>.<\/li>\n<li>It is more reactive.<\/li>\n<li>It is poisonous.<\/li>\n<\/ul>\n<p>Red &#8216;P&#8217;<\/p>\n<ul>\n<li>It possesses iron grey lustre.<\/li>\n<li>It is in soluble in CS<sub>2<\/sub> as cool water.<\/li>\n<li>It is more reactive than white \u2018P\u2019.<\/li>\n<li>\u00a0It is non &#8211; poisonous.<\/li>\n<\/ul>\n<p>Question 12.<br \/>\nWhat is inert pair effect ?<br \/>\nAnswer:<br \/>\nInert pair effect: The reluctance of ns pair of electrons to take part in bond formation is called inert pair effect.<br \/>\nBi exhibits +3 oxidation state instead of +5 due to inert pair effect.<\/p>\n<p>Question 13.<br \/>\nExplain why is NH<sub>3<\/sub> basic while BiH<sub>3<\/sub> is only feebly basic.<br \/>\nAnswer:<br \/>\nNH<sub>3<\/sub> is basic while BiH<sub>3<\/sub> is only feebly basic.<\/p>\n<p>Explanation :<br \/>\nDue to small atomic size of nitrogen, the electron density on nitrogen atom is greater than that on Bi atom. So electron releasing tendency is greater in NH<sub>3<\/sub>.<br \/>\nHence NH<sub>3<\/sub> is basic and BiH<sub>3<\/sub> is feebly basic.<\/p>\n<p>Question 14.<br \/>\nArrange the hydrides of group &#8211; 15 elements in the increasing order of basic strength and decreasing order of reducing character.<br \/>\nAnswer:<\/p>\n<ol>\n<li>Increasing order of basic strength of Group &#8211; 15 elements hydrides is<br \/>\nBiH<sub>3<\/sub> &lt; SbH<sub>3<\/sub> &lt; ASH<sub>3<\/sub> &lt; PH<sub>3<\/sub> &lt; NH<sub>3<\/sub>.<\/li>\n<li>Decreasing order of reducing character of Group &#8211; 15 elements hydrides is<br \/>\nBiH<sub>3<\/sub> &gt; SbH<sub>3<\/sub> &gt; ASH<sub>3<\/sub> &gt; PH<sub>3<\/sub> &gt; NH<sub>3<\/sub>.<\/li>\n<\/ol>\n<p>Question 15.<br \/>\nPH<sub>3<\/sub> is a weaker base than NH<sub>3<\/sub> &#8211; Explain.<br \/>\nAnswer:<br \/>\nPH<sub>3<\/sub> is a weaker base than NH<sub>3<\/sub>.<\/p>\n<ul>\n<li>In NH<sub>3<\/sub> nitrogen atom undergoes SP<sup>3<\/sup> hybridisation and due to small size it has high electron density than in &#8216;P&#8217; of PH<sub>3<\/sub>.<\/li>\n<li>Due to large size of &#8216;P&#8217; atom and availability of large surface area, lonepair of electron spread in PH<sub>3<\/sub>. Hence PH<sub>3<\/sub> is weaker base then NH<sub>3<\/sub>.<\/li>\n<\/ul>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 16.<br \/>\nA hydride of group -15 elements dissolves in water to form a basic solution. This solution dissolves the AgCl precipitate. Name the hydride. Write the chemical equations involved.<br \/>\nAnswer:<br \/>\nGiven a hydride of group &#8211; 15 elements dissolves in water to form a basic solution;This solution dissolves the Agcl precipitate.<\/p>\n<p>The given hydride is ammonia. It forms basic solution when dissolved in water due to formation of OH<sup>&#8211;<\/sup> ions.<br \/>\nNH<sub>3(g)<\/sub> + H<sub>x<\/sub>O<sub>l<\/sub> \u21cc \\(\\mathrm{NH}_4^{+}(\\mathrm{aq})\\) + OH<sup>&#8211;<\/sup><sub>(aq)<\/sub><br \/>\nThis solution dissolves AgCl. Ppt due to formation of complex compound.<br \/>\nAgCl<sub>(s)<\/sub> (white Ppt) + 2NH<sub>3(aq)<\/sub> \u2192 [Ag(NH<sup>3<\/sup>)<sup>2<\/sup>]Cl<sub>(aq)<\/sub> (colourless)<\/p>\n<p>Question 17.<br \/>\nWhat happens when white phosphorus is heated with cone. NaOH solution in an inert atmosphere of CO<sub>2<\/sub>? [A.P. Mar. 19, 15]<br \/>\nAnswer:<br \/>\nWhen white phosphorus heated with con. NaOH solution in an inert atomosphere of CO<sub>2<\/sub> forms PH<sub>3<\/sub>.<br \/>\nP<sub>4<\/sub> + 3NaOH + 3H<sub>2<\/sub>O \u2192 PH<sub>3<\/sub> + 3NaH<sub>2<\/sub>PO<sub>2<\/sub>.<\/p>\n<p>Question 18.<br \/>\nNH<sub>3<\/sub> forms hydrogen bonds but PH<sub>3<\/sub> does not &#8211; Why ?<br \/>\nAnswer:<br \/>\nNH<sub>3<\/sub> forms hydrogen bonds but PH<sub>3<\/sub> does not.<br \/>\nReason : Ammonia forms hydrogen bonds because it it a polar molecule and N-H bond is highly polar. Nitrogen has more electronegativity than phosphrous. In case of PH<sub>3<\/sub> P-H bond polarity decreases.<\/p>\n<p>Question 19.<br \/>\nThe HNH angle is higher than HPH, HAsH and HSbH angles &#8211; Why?<br \/>\nAnswer:<br \/>\nThe central atom \u03b5 (where, \u03b5 = N, P, As, Sb, Bi) in all given hydrides is Sp<sup>3<\/sup> hybridized. However, its electronegativity decreases and atomic size increase on moving down the group. Therefore is a gradual decrease in the force of repulsion between the shared electron pairs around the central atom. Thus, bond angle decreases as we move down the group.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23210\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-5.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 5\" width=\"637\" height=\"51\" \/><\/p>\n<p>Question 20.<br \/>\nHow do calcium phosphide and heavy water react?<br \/>\nAnswer:<br \/>\nCalcium phosphide reacts with heavy water to form Deutero phosphine.<br \/>\nCa<sub>3<\/sub>P<sub>2<\/sub> + 6D<sub>2<\/sub>O \u2192 3 Ca (OD)<sub>2<\/sub> + 2PD<sub>3<\/sub><\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 21.<br \/>\nAmmonia is a good complexing agent &#8211; Explain with an example. [Mar. 14]<br \/>\nAnswer:<br \/>\nNH<sub>3<\/sub> is a lewis base and it donates electron pair to form dative bond with metal ions. This results in the formation of complex compound.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23211\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-6.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 6\" width=\"336\" height=\"60\" \/><\/p>\n<p>Question 22.<br \/>\nA mixture of Ca<sub>3<\/sub>P<sub>2<\/sub> and CaC<sub>2<\/sub> is used in making Holme&#8217;s signal &#8211; Explain. [A.P. Mar. 16]<br \/>\nAnswer:<br \/>\nA Mixture of Ca<sub>3<\/sub>P<sub>2<\/sub> and CaC<sub>2<\/sub> is used in Holme&#8217;s signal.This Mixture containing containers are pierced and thrown in the sea, when the gas is evolved bum and serve as a signal.<br \/>\nThe spontaneous combustion of PH<sub>3<\/sub> is the technical use of Holme&#8217;s signal.<\/p>\n<p>Question 23.<br \/>\nWhich chemical compound is formed in the brown ring test of nitrate ions ?<br \/>\nAnswer:<br \/>\nIn the brown ring test of nitrate salts a brown ring is formed. It&#8217;s formula is [Fe(H<sub>2<\/sub>O)<sub>5<\/sub>N0]<sup>+2<\/sup>.<\/p>\n<p>Question 24.<br \/>\nGive the resonating structures of NO<sub>2<\/sub> and N<sub>2<\/sub>O<sub>5<\/sub>.<br \/>\nAnswer:<br \/>\nResonance structure of NO<sub>2<\/sub>:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23212\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-7.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 7\" width=\"397\" height=\"90\" \/><br \/>\nResonance structure of N<sub>2<\/sub>O<sub>5<\/sub><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23213\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-8.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 8\" width=\"443\" height=\"120\" \/><\/p>\n<p>Question 25.<br \/>\nWhy does R<sub>3<\/sub>P = O exist but R<sub>3<\/sub>N = O does not (R = alkyl group) ?<br \/>\nAnswer:<br \/>\nR<sub>3<\/sub>P = O exist but R<sub>3<\/sub>N = O does not.<br \/>\nExplanation :<br \/>\nNitrogen does not form d\u03c0 &#8211; P\u03c0 multiple bond with oxygen because of lack of d &#8211; orbitals in Nitrogen atom. But in case of R<sub>3<\/sub>N = O the value of nitrogen should be 5. So these compounds do not exist where as in case of &#8216;P&#8217; atom d-orbitals are available. So P-atom can able to form d\u03c0 &#8211; P\u03c0 multiple bonds hence R<sub>3<\/sub>P = O exist.<\/p>\n<p>Question 26.<br \/>\nHow is nitric oxide (NO) prepared ?<br \/>\nAnswer:<br \/>\nThe catalytic oxidation of NH<sub>3<\/sub> by atmospheric oxygen gives nitric oxide.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23214\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-9.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 9\" width=\"349\" height=\"41\" \/><\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 27.<br \/>\nGive one example each of normal oxide and mixed oxide of nitrogen.<br \/>\nAnswer:<\/p>\n<ul>\n<li>Nitric oxide (NO) is an example of normal oxide of Nitrogen.<\/li>\n<li>Dinitrogen trioxide (N<sub>2<\/sub>O<sub>3<\/sub>) is an example of mixed oxide of nitrogen.<\/li>\n<\/ul>\n<p>Question 28.<br \/>\nNO is paramagnetic in gaseous state but diamagnetic in liquid and solid states &#8211; Why ?<br \/>\nAnswer:<br \/>\nIn gaseous state NO<sub>2<\/sub> exists as a Monomer and contains one unpaired electron but in solid state it dimerises to N<sub>2<\/sub>O<sub>4<\/sub> so it doesnot contain unpaired electron.<br \/>\nHence NO<sub>2<\/sub> is parce magnetic is geseous state but diamagnetic in solid state.<\/p>\n<p>Question 29.<br \/>\nGive an example of<br \/>\na) acidic oxide of phosphorus<br \/>\nb) neutral oxide of nitrogen.<br \/>\nAnswer:<br \/>\na) P<sub>2<\/sub>O<sub>5<\/sub> (or) P<sub>4<\/sub>O<sub>10<\/sub>, phosphorus pentoxide is an example of acidic oxide of phosphorus<br \/>\nb) Nitrous oxide (N<sub>2<\/sub>O) and Nitric oxide (NO) are neutral oxides of nitrogen.<\/p>\n<p>Question 30.<br \/>\nExplain the following<br \/>\na) reaction of alkali with red phosphorus.<br \/>\nb) reaction between PCl<sub>3<\/sub> and H<sub>3<\/sub>PO<sub>3<\/sub>.<br \/>\nAnswer:<br \/>\na) Red phosphorus reacts with an alkali to form hypophosphoric acid (H<sub>4<\/sub>P<sub>2<\/sub>O<sub>6<\/sub>)<br \/>\nb) PCl<sub>3<\/sub> undergoes hydrolysis to form H<sub>3<\/sub>PO<sub>3<\/sub><br \/>\nPCl<sub>3<\/sub> + 3H<sub>2<\/sub>O \u2192 H<sub>3<\/sub>PO<sub>3<\/sub> + 3HCl<\/p>\n<p>Question 31.<br \/>\nHow does PCl<sub>3<\/sub> react with<br \/>\na) CH<sub>3<\/sub>COOH<br \/>\nb) C<sub>2<\/sub>H<sub>5<\/sub>OH and<br \/>\nc) water.<br \/>\nAnswer:<br \/>\na) PCl<sub>3<\/sub> reacts with CH<sub>3<\/sub>COOH and form phosphorous acid, acetyl chloride.<br \/>\n3CH<sub>3<\/sub>COOH + PCl<sub>3<\/sub> \u2192 3CH<sub>3<\/sub>COCl + H<sub>2<\/sub>PO<sub>3<\/sub><\/p>\n<p>b) PCl<sub>3<\/sub> reacts with phosphorous acid and forms phosphorus acid, Ethyl chloride<br \/>\n3C<sub>2<\/sub>H<sub>5<\/sub>OH + PCl<sub>3<\/sub> \u2192 2C<sub>2<\/sub>H<sub>5<\/sub>Cl<sub>3<\/sub> + H<sub>3<\/sub>PO<sub>3<\/sub><\/p>\n<p>c) PCl<sub>3<\/sub> reacts with water (hydrolysis) to form phosphorus acid. It undergo hydrolysis in presence of moisture.<br \/>\nPCl<sub>3<\/sub> + 3H<sub>2<\/sub>O \u2192 H<sub>3<\/sub>PO<sub>3<\/sub> + 3HCl<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 32.<br \/>\nPCl<sub>3<\/sub> can act as an oxidizing as well as a reducing agent &#8211; Justify.<br \/>\nAnswer:<br \/>\nPCl<sub>3<\/sub> acts as reducing agent. It is evidented by the following reaction.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23215\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-10.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 10\" width=\"170\" height=\"54\" \/><br \/>\nPCl<sub>3<\/sub> acts as oxidising agent. It is evidented by the following reaction.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23216\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-11.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 11\" width=\"386\" height=\"59\" \/><\/p>\n<p>Question 33.<br \/>\nWhich of the following are not known ?<br \/>\nPCl<sub>3<\/sub>, AsCl<sub>3<\/sub>, SbCl<sub>3<\/sub>, NCl<sub>5<\/sub>, BiCl<sub>5<\/sub>, PH<sub>5<\/sub><br \/>\nAnswer:<br \/>\nNCl<sub>5<\/sub>, BiCl<sub>5<\/sub>, PH<sub>5<\/sub> are not known in the given compounds.<\/p>\n<p>Question 34.<br \/>\nWhich of the following is more covalent &#8211; SbCl<sub>5<\/sub> or SbCl<sub>3<\/sub> ?<br \/>\nAnswer:<br \/>\nSbCl<sub>5<\/sub> (Penta halide) is more covalent than SbCl<sub>3<\/sub> (Tri halide). Because Sb in the higher oxidation state exert more polarising power.<\/p>\n<p>Question 35.<br \/>\nWrite the oxidation states of phosphorus in solid PCl<sub>5<\/sub>.<br \/>\nAnswer:<\/p>\n<ul>\n<li>In solid state PCl<sub>5<\/sub> exists as an ionic solid [PCl<sub>4<\/sub>)<sup>+<\/sup> [PCl<sub>6<\/sub>]<sup>&#8211;<\/sup><\/li>\n<li>&#8216;P&#8217; exhibits +5 oxidation state in [PCl<sub>4<\/sub>)<sup>+<\/sup> [PCl<sub>6<\/sub>]<sup>&#8211;<\/sup><\/li>\n<\/ul>\n<p>Question 36.<br \/>\nIllustrate how copper metal can give different products on reaction with HNO<sub>3<\/sub>.<br \/>\nAnswer:<br \/>\nReaction of Cu metal with dilute HNO<sub>3<\/sub>.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23217\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-12.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 12\" width=\"403\" height=\"56\" \/><br \/>\nReaction of Cu Metal with cone. HNO<sub>3<\/sub><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23218\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-13.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 13\" width=\"421\" height=\"68\" \/><\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 37.<br \/>\nWhich oxide of nitrogen has oxidation number of N same as that in nitric acid ?<br \/>\nAnswer:<\/p>\n<ul>\n<li>In HNO<sub>3<\/sub>, &#8216;N&#8217; has oxidation state +5.<\/li>\n<li>Among oxides of nitrogen N<sub>2<\/sub>O<sub>5<\/sub> exhibits &#8216;+5&#8217; oxidation state.<\/li>\n<\/ul>\n<p>Question 38.<br \/>\nWrite the chemical reactions that occur in the manufacture of nitric acid.<br \/>\nAnswer:<br \/>\nChemical reactions involved in the manufacturing of HNO<sub>3<\/sub>.<\/p>\n<ul>\n<li>Oxidation of NH<sub>3<\/sub><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23219\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-14.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 14\" width=\"296\" height=\"46\" \/><\/li>\n<li>NO<sub>2<\/sub> formation :<br \/>\n2NO + O<sub>2<\/sub> \u21cc 2NO<sub>2<\/sub><\/li>\n<li>Formation of HNO<sub>3<\/sub>:<br \/>\n3NO<sub>2<\/sub> + H<sub>2<\/sub>O \u2192 2HNO<sub>3<\/sub> + NO<\/li>\n<\/ul>\n<p>Question 39.<br \/>\nIron becomes passive in cone. HNO<sub>3<\/sub> &#8211; Why ?<br \/>\nAnswer:<br \/>\nIron becomes passive in cone. HNO<sub>3<\/sub> due to formation of a passive film of oxide on the surface of iron.<\/p>\n<p>Question 40.<br \/>\nGive the uses of<br \/>\na) nitric acid and<br \/>\nb) ammonia.<br \/>\nAnswer:<br \/>\nUse of HNO<sub>3<\/sub>:<\/p>\n<ul>\n<li>HNO<sub>3<\/sub> is used in the manufacture of ammonium nitrate for fertilisers and other nitrates for use in explosives and pyrotechnics.<\/li>\n<li>HNO<sub>3<\/sub> is used in the pickling of stainless steel.<\/li>\n<li>HNO<sub>3<\/sub> is used as oxidiser in rocket fuels.<\/li>\n<\/ul>\n<p>Question 41.<br \/>\nWhat are the oxidation states of phosphorus in the following ?<br \/>\ni) H<sub>3<\/sub>PO<sub>3<\/sub><br \/>\nii) PCl<sub>3<\/sub><br \/>\niii) Ca<sub>3<\/sub>P<sub>2<\/sub><br \/>\niv) Na<sub>3<\/sub>PO<sub>4<\/sub><br \/>\nv) POF<sub>3<\/sub><br \/>\nAnswer:<br \/>\ni) H<sub>3<\/sub>PO<sub>3<\/sub><br \/>\n3(1) + x + 3(-3) = 0<br \/>\nx = +3<\/p>\n<p>ii) PCl<sub>3<\/sub><br \/>\nx + 3(-1) = 0<br \/>\nx = 3<\/p>\n<p>iii) Ca<sub>3<\/sub>P<sub>2<\/sub><br \/>\n3(+2) + 2x = 0<br \/>\nx = -3<\/p>\n<p>iv) Na<sub>3<\/sub>PO<sub>4<\/sub><br \/>\n3(1) + x + 4(-2) = 0<br \/>\nx = +5<\/p>\n<p>v) POF<sub>3<\/sub><br \/>\nx + (-2) + 3(-1) = 0<br \/>\nx = +5<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 42.<br \/>\nH<sub>3<\/sub>PO<sub>3<\/sub> is diprotic while H<sub>3<\/sub>PO<sub>2<\/sub> is monoprotic &#8211; Why ?<br \/>\nAnswer:<br \/>\nH<sub>3<\/sub>PO<sub>3<\/sub> is diprotic :<br \/>\nStructure of H<sub>3<\/sub>PO<sub>3<\/sub>:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23220\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-15.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 15\" width=\"236\" height=\"143\" \/><\/p>\n<p>H<sub>3<\/sub>PO<sub>2<\/sub> is monoprotic :<br \/>\nStructure of H<sub>3<\/sub>PO<sub>2<\/sub> :<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23221\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-16.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 16\" width=\"110\" height=\"149\" \/><\/p>\n<p>Question 43.<br \/>\nGive the disproportionation reaction of H<sub>3<\/sub>PO<sub>3<\/sub>.<br \/>\nAnswer:<br \/>\nOrthophosphoric acid (H<sub>3<\/sub>PO<sub>3<\/sub>) on heating disproportionates to give orthophosphoric acid and phosphine.<br \/>\n4H<sub>3<\/sub>PO<sub>3<\/sub> \u2192 3H<sub>3<\/sub>PO<sub>4<\/sub> + PH<sub>3<\/sub>.<\/p>\n<p>Question 44.<br \/>\nH<sub>3<\/sub>PO<sub>2<\/sub> is a good reducing agent &#8211; Explain with an example.<br \/>\nAnswer:<br \/>\nIn H<sub>3<\/sub>PO<sub>2<\/sub>, two H-atoms are bonded directly to P-atom which imparts reducing character to the acid.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 45.<br \/>\nDraw the structures of<br \/>\na) Hypo phosphoric acid<br \/>\nb) Cyclic meta phosphoric acid.<br \/>\nAnswer:<br \/>\na) Structure of hypo phosphoric acid (H<sub>4<\/sub>P<sub>2<\/sub>O<sub>6<\/sub>) :<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23222\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-17.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 17\" width=\"444\" height=\"111\" \/><br \/>\nb) Structure of cyclic meta phosphoric acid (HPO<sub>3<\/sub>)<sub>3<\/sub><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23223\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-18.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 18\" width=\"155\" height=\"182\" \/><\/p>\n<p><span style=\"color: #0000ff;\">Short Answer Questions<\/span><\/p>\n<p>Question 1.<br \/>\nDiscuss the general characteristics of Group &#8211; 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionization enthalpy and electronegativity.<br \/>\nAnswer:<br \/>\n1) Electronic Configuration : All these elements of this group have s<sup>2<\/sup>p<sup>3<\/sup> configuration in their respective outmost orbits.<br \/>\nNitrogen (N) : [He] 2s<sup>2<\/sup>2p<sup>3<\/sup>; phosphorus (P) : [Ne] 3s<sup>2<\/sup>3p<sup>3<\/sup>, Arsenic (As) : [Ar] 3d<sup>10<\/sup>4s<sup>2<\/sup>4p<sup>3<\/sup><br \/>\nAtimony (Sb) : [Kr] 4d<sup>10<\/sup> 5s<sup>2<\/sup> 5p<sup>3<\/sup> and Bismuth (Bi) = [Xe] 4f<sup>14<\/sup> 5d<sup>10<\/sup> 6s<sup>2<\/sup>6p<sup>3<\/sup><\/p>\n<p>2) Oxidation Spates : As per their configurations, these elements may utilize either the three electrons iti the p-orbitals or the five electrons present in both s -and p -orbitals in the exhibition of oxidation states. Consequently, they exhibit common oxidation states +III and +V.<\/p>\n<p>3) Atomic size: In Group-15 elements atomic size increase down the group. A considerable increase in covalent radius observed from N to P and from As to Bi only a small increase in atomic size observed.<\/p>\n<p>4) Ionisation enthalpy : In group &#8211; 15 elements ionisation enthalpy decreases down the group due to gradual increase in atomic size.<\/p>\n<p>5) Electronegativity : In group-15 elements electronegativity values decrease down the group with increase in atomic size.<\/p>\n<p>Question 2.<br \/>\nDiscuss the trends in chemical reactivity of group &#8211; 15 elements.<br \/>\nAnswer:<br \/>\nNitrogen gas exists as diatomic molecule. Dut to the presence of triple bond between Base N &#8211; Atoms bond dissociation energy is high (941.4 KJ \/mol). Hence nitrogen is inert and unreactive.<\/p>\n<p>Phosphorus is a tetra atomic molecule and P-P single bond is weaker than N \u2261 N. P &#8211; P bond dissociation energy U 213 KJ\/mole. Hence phosphorus is more reactive than Nitrogen.<\/p>\n<p>Explanation :<br \/>\nThe single N-N bond is weaker than the single P-P bond due to high inter electronic repulsion of the non-bonding electrons in N<sub>2<\/sub> because of small bond length. There fore the catenation property is weaker in nitrogen as compoud to phosphorares.<br \/>\nR<sub>3<\/sub> P = O exist but R<sub>3<\/sub>N = O does not.<\/p>\n<p>Explanation :<br \/>\nNitrogen does not form d\u03c0 &#8211; P\u03c0 multiple bond with oxygen because of lack of d &#8211; orbitals in Nitrogen atom. But in case of R<sub>3<\/sub>N = 0 the value of nitrogen should be 5. So these compounds do not exist where as in case of P&#8217; atom d-orbitals are available. So P-atom can able to form d\u03c0 &#8211; P\u03c0 multiple bonds hence R<sub>3<\/sub>P = 0 exist.<\/p>\n<ol>\n<li>Reactivity towards hydrogen: Group -15 elements forms EH<sub>3<\/sub> type hydrides (E = Group &#8211; 15 elements) .<br \/>\nEg : PH<sub>3<\/sub>, NH<sub>3<\/sub>, AsH<sub>3<\/sub>, BiH<sub>3<\/sub>, SbH<sub>3<\/sub>.<\/p>\n<ul>\n<li>Among above hydrides NH<sub>3<\/sub> is mild reducing agent while BiH<sub>3<\/sub> is strong reducing agent.<\/li>\n<li>Stability of hydrides decreases from NH<sub>3<\/sub> to BiH<sub>3<\/sub>.<\/li>\n<li>Basicity of hydrides decreases as follows.<br \/>\nNH<sub>3<\/sub> &gt; PH<sub>3<\/sub> &gt; AsH<sub>3<\/sub> &gt; SbH<sub>3<\/sub> &gt; BiH<sub>3<\/sub>.<\/li>\n<\/ul>\n<\/li>\n<li>Reactivity towards Oxygen : These forms two types of oxides E<sub>2<\/sub>O<sub>3<\/sub> and E&lt;sub2O<sub>5<\/sub>.<br \/>\nEg : P<sub>2<\/sub>O<sub>3<\/sub> &gt; N<sub>2<\/sub>O<sub>5<\/sub>, N<sub>2<\/sub>O<sub>3<\/sub>, P<sub>2<\/sub>O<sub>5<\/sub>.<\/p>\n<ul>\n<li>Acidic character of oxides decrease down the group.<\/li>\n<li>E<sub>3<\/sub>O<sub>3<\/sub> of N and P are acidic, As and Sb are amphoteric while &#8216;Bi&#8217; is basic.<\/li>\n<\/ul>\n<\/li>\n<li>Reactivity towards halogens : These elements forms two types of halides EX<sub>3<\/sub> and EX<sub>5<\/sub>.<br \/>\n&#8216;N&#8217; &#8211; does not form penta halides due to lack of the d-orbitals.<br \/>\nPenta halides are more covalent than tri halides because the elements in higher oxidation state have more polarising power.<\/li>\n<li>Reactivity towards metals : All these elements react with metals to form their binary compounds containing &#8211; 3 oxidation state.<br \/>\nEg. : Ca<sub>3<\/sub>N<sub>2<\/sub>, Ca<sub>3<\/sub>P<sub>2<\/sub> etc.<\/li>\n<\/ol>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 3.<br \/>\nHow does P<sub>4<\/sub> react with the following ?<br \/>\na) SOCl<sub>2<\/sub><br \/>\nb) SO<sub>2<\/sub>Cl<sub>2<\/sub><br \/>\nAnswer:<br \/>\na) When P<sub>4<\/sub> reacts with SOCl<sub>2<\/sub> to form phosphorus trichloride.<br \/>\nP<sub>4<\/sub> + 8 SOCl<sub>2<\/sub> \u2192 4 PCl<sub>3<\/sub> + 4 SO<sub>2<\/sub> + 2 S<sub>2<\/sub>Cl<sub>2<\/sub><br \/>\nb) When P<sub>4<\/sub> reacts with SOCl<sub>2<\/sub> to form phosphorus pentachloride.<br \/>\nP<sub>4<\/sub> + 10 SOCl<sub>2<\/sub> \u2192 4 PCl<sub>5<\/sub> + 10 SO<sub>2<\/sub><\/p>\n<p>Question 4.<br \/>\nExplain the anomalous nature of nitrogen in group -15.<br \/>\nAnswer:<br \/>\nAnomalous properties of Nitrogen : Nitrogen differs from the remaining elements of this group due to its small size, high electronegativity, high ionisation enthalpy and non-availability of d-orbitals.<\/p>\n<p>Nitrogen gas exists as diatomic molecule. Due to the presence of triple bond between base N &#8211; Atoms bond dissociation energy is high (941.4 KJ \/mol). Hence nitrogen is inert and un-reactive.<\/p>\n<p>Explanation :<\/p>\n<ul>\n<li>The single N-N bond is weaker than the single P-P bond due to high inter electronic repulsion of the non-bonding electrons in N<sub>2<\/sub> because of small bond length. Therefore the catenation property is weaker in nitrogen as compound to phosphorus.<\/li>\n<li>Nitrogen does not form penta halides due to lack of d-orbitals.<br \/>\nR<sub>3<\/sub> P = O exist but R<sub>3<\/sub>N = O does not.<\/li>\n<\/ul>\n<p>Explanation :<br \/>\nNitrogen does not form d\u03c0 &#8211; P\u03c0 multiple bond with oxygen because of lack of d &#8211; orbitals in Nitrogen atom. But in case of R<sub>3<\/sub>N = O the value of nitrogen should be 5. So these compounds do not exist where as in case of &#8216;P atom d-orbitals are available. So P-atom can able to form d\u03c0 &#8211; P\u03c0 multiple bonds hence R<sub>3<\/sub>P = O exist.<\/p>\n<p>Question 5.<br \/>\nComplete the following reactions.<br \/>\na) Ca<sub>3<\/sub>P<sub>2<\/sub> + HaO \u2192<br \/>\nb) P<sub>4<\/sub> + KOH \u2192<br \/>\nc) CuSO<sub>4<\/sub> + NH<sub>3<\/sub> \u2192<br \/>\nd) Mg + N<sub>2<\/sub> \u2192<br \/>\nd) (NH<sub>4<\/sub>)<sub>2<\/sub> + Cr<sub>2<\/sub>O<sub>7<\/sub> <img loading=\"lazy\" class=\"alignnone size-full wp-image-23206\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-1.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 1\" width=\"51\" height=\"20\" \/><br \/>\nf) Decomposition of nitrous acid \u2192<br \/>\nAnswer:<br \/>\na) Ca<sub>3<\/sub>P<sub>2<\/sub> + 6 H<sub>2<\/sub>O \u2192 3 Ca(OH)<sub>2<\/sub> + 2PH<sub>3<\/sub><br \/>\nb) P<sub>4<\/sub> + 3 KOH + 3 H<sub>2<\/sub>O \u2192 PH<sub>3<\/sub> + 3KH<sub>2<\/sub> PO<sub>2<\/sub><br \/>\nc) CuSO<sub>4(aq)<\/sub> (blue) + 4NH<sub>3<\/sub> \u2192 [Cu (NH<sub>3<\/sub>)<sub>4<\/sub>] SO<sub>4<\/sub> (deep blue)<br \/>\nd) 3 Mg + N<sub>2<\/sub> \u2192 Mg<sub>3<\/sub>N<sub>2<\/sub><br \/>\ne) (NH<sub>4<\/sub>)<sub>2<\/sub> + Cr<sub>2<\/sub>O<sub>7<\/sub> <img loading=\"lazy\" class=\"alignnone size-full wp-image-23206\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-1.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 1\" width=\"51\" height=\"20\" \/> N<sub>2<\/sub> + 4 H<sub>2<\/sub>O + Cr<sub>2<\/sub>O<sub>3<\/sub><br \/>\nf) 3HNO<sub>2<\/sub> \u2192 HNO<sub>3<\/sub> + 2 NO + H<sub>2<\/sub>O<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 6.<br \/>\nHow does PCl<sub>5<\/sub> react with the following?<br \/>\na) Water<br \/>\nb) C<sub>2<\/sub>H<sub>5<\/sub>OH<br \/>\nc) CH<sub>3<\/sub>COOH<br \/>\nd) Ag<br \/>\nAnswer:<br \/>\na) PCl<sub>5<\/sub> undergo hydrolysis to form phosphoric acid.<br \/>\nPCl<sub>5<\/sub> + H<sub>2<\/sub>O \u2192 POCl<sub>3<\/sub> + 2HCl<br \/>\nPOCl<sub>3<\/sub> + 3H<sub>2<\/sub>O \u2192 H<sub>3<\/sub>PO<sub>4<\/sub> + 3 HCl<\/p>\n<p>b) PCl<sub>5<\/sub> reacts with C<sub>2<\/sub>H<sub>5<\/sub>OH to form Ethyl chloride.<br \/>\nC<sub>2<\/sub>H<sub>5<\/sub>OH + PCl<sub>5<\/sub> \u2192 C<sub>2<\/sub>H<sub>5<\/sub>Cl + POCl<sub>3<\/sub> + HCl<\/p>\n<p>c) PCl<sub>5<\/sub> reacts with CH<sub>3<\/sub>COOH to form acetyl chloride.<br \/>\nCH<sub>3<\/sub>COOH + PCl<sub>5<\/sub> \u2192 CH<sub>3<\/sub>COCl + POCl<sub>3<\/sub> + HCl<\/p>\n<p>d) PCl<sub>5<\/sub> reacts with Ag to form PCl<sub>3<\/sub> and AgCl<br \/>\nPCl<sub>5<\/sub> + 2 Ag \u2192 2 AgCl + PCl<sub>3<\/sub><\/p>\n<p>Question 7.<br \/>\nComplete the following.<br \/>\na) NH<sub>4<\/sub>NO<sub>3<\/sub> <img loading=\"lazy\" class=\"alignnone size-full wp-image-23206\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-1.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 1\" width=\"51\" height=\"20\" \/><br \/>\nb) HNO<sub>3<\/sub> + P<sub>4<\/sub>O<sub>10<\/sub> \u2192<br \/>\nc) Pb(NO<sub>3<\/sub>)<sub>2<\/sub> <img loading=\"lazy\" class=\"alignnone size-full wp-image-23224\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-19.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 19\" width=\"71\" height=\"24\" \/><br \/>\nd) Zn + dil.HNO<sub>3<\/sub> \u2192<br \/>\ne) P<sub>4<\/sub> + conc.HNO<sub>3<\/sub> \u2192<br \/>\nf) HgCl<sub>2<\/sub> + PH<sub>3<\/sub> \u2192<br \/>\nAnswer:<br \/>\na) NH<sub>4<\/sub>NO<sub>3<\/sub> <img loading=\"lazy\" class=\"alignnone size-full wp-image-23206\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-1.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 1\" width=\"51\" height=\"20\" \/> N<sub>2<\/sub>O (Nitrous oxide) + 2 H<sub>2<\/sub>O<br \/>\nb) 12 HNO<sub>3<\/sub> + P<sub>4<\/sub>O<sub>10<\/sub> \u2192 6 N<sub>2<\/sub>O<sub>5<\/sub> + 4 H<sub>3<\/sub>PO<sub>4<\/sub><br \/>\nc) 2 Pb(NO<sub>3<\/sub>)<sub>2<\/sub> <img loading=\"lazy\" class=\"alignnone size-full wp-image-23206\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-1.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 1\" width=\"51\" height=\"20\" \/> 2 PbO + 4 NO<sub>2<\/sub> + O<sub>2<\/sub><br \/>\nd) 4 Zn + 10 HNO<sub>3<\/sub> (dil) \u2192 4 Zn (NO<sub>3<\/sub>)<sub>2<\/sub> + 5 H<sub>2<\/sub>O + N<sub>2<\/sub>O<br \/>\ne) P<sub>4<\/sub> + 20 HNO<sub>3<\/sub> \u2192 4H<sub>3<\/sub>PO<sub>4<\/sub> + 20 NO<sub>2<\/sub> + 4 H<sub>2<\/sub>O<br \/>\nf) 3 HgCl<sub>2<\/sub> + 2 PH<sub>3<\/sub> \u2192 Hg<sub>3<\/sub>P<sub>2<\/sub> + 6 HCl<\/p>\n<p><span style=\"color: #0000ff;\">Long Answer Questions<\/span><\/p>\n<p>Question 1.<br \/>\nHow is ammonia manufactured by Habes process? Explain the reactions of ammonia with [A.P. Mar. 18]<br \/>\na) ZnSO<sub>4(aq)<\/sub><br \/>\nb) CuSO<sub>4(aq)<\/sub><br \/>\nc) AgCl<sub>(s)<\/sub><br \/>\nAnswer:<br \/>\nIn Haber process ammonia is directly synthesised from elements (nitrogen and hydrogen). The principle involved in this is<br \/>\nN<sub>2<\/sub> (g) + 3H<sub>2<\/sub>O (g) \u21cc 2NH<sub>3<\/sub> + 92.4kJ<br \/>\nThis is a reversible exothermic reaction.<br \/>\nAccording to Le Chatelier&#8217;s principle favourable conditions for the better yield of ammonia are low temperature and high pressure. But the optimum conditions are<br \/>\nTemperature : 720k<br \/>\nPressure : 200 atmospheres<br \/>\nCatalyst : Finely divided iron in the presence of molybdenum (Promoter).<br \/>\nProcedure : A mixture of nitrogen and hydrogen in the volume ratio 1 : 3 is heated to 725 &#8211; 775K at a pressure of 200 atmospheres is passed over hot finely divided iron mixed with small amount of molybdenum as promotor. The gases coming out of the catalyst chamber consists of 10 &#8211; 20% ammonia gas are cooled and compressed, so that ammonia gas is liquified, and the uncondensed gases are sent for recirculation.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23225\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-20.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 20\" width=\"722\" height=\"426\" \/><br \/>\nb) Aq.CuSO<sub>4<\/sub> reacts with ammonia to form a deep blue complex,<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23226\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-21.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 21\" width=\"314\" height=\"66\" \/><br \/>\nc) Solid AgCl reacts with ammonia to form a colourless complex.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23227\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-22.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 22\" width=\"322\" height=\"58\" \/><\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 2.<br \/>\nHow is nitric acid manufactured by Ostwald&#8217;s process ? How does it react with the following ? [T.S. Mar. 19]<br \/>\na) Copper<br \/>\nb) Zn<br \/>\nc) S8<br \/>\nd) P4<br \/>\nAnswer:<br \/>\nOstwald&#8217;s process : Ammonia, mixed with air in 1 : 7 or 1 : 8, when passed over a hot platinum gauze catalyst, is oxidised to NO mostly (about 95%).<br \/>\nThe reaction is<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23228\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-23.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 23\" width=\"441\" height=\"52\" \/><br \/>\nThe liberated heat keeps the catalyst hot. The &#8216;NO&#8217; is cooled and is mixed with oxygen to give the dioxide, in large empty towers (oxidation chamber). The product is then passed in to warm water, under pressure in the presence of excess of air, to give HNO<sub>3<\/sub>.<br \/>\n4NO<sub>2<\/sub> + O<sub>2<\/sub> + 2H<sub>2<\/sub>O \u2192 4HNO<sub>3<\/sub>.<br \/>\nThe acid formed is about 61% concentrated.<\/p>\n<p>a) Copper reacts with dil. HNO<sub>3<\/sub> and cone. HNO<sub>3<\/sub> and liberates Nitric Oxide and Nitrogen dioxide respectively<br \/>\n3 Cu + 5 HNO<sub>3<\/sub> (dil) \u2192 3 Cu (NO<sub>3<\/sub>)<sub>2<\/sub> + 2 NO + 4 H<sub>2<\/sub>O<br \/>\nCu + 4 HNO<sub>3<\/sub> (cone.) \u2192 Cu (NO<sub>3<\/sub>)<sub>2<\/sub> + 2 NO<sub>2<\/sub> + 2 H<sub>2<\/sub>O<\/p>\n<p>b) Zn reacts with dil. HNO<sub>3<\/sub> and Cone. HNO<sub>3<\/sub> and liberates Nitrous oxide and Nitrogen dioxide respectively.<br \/>\n4 Zn + 10 HNO<sub>3<\/sub> (dil) \u2192 4 Zn (NO<sub>3<\/sub>)<sub>2<\/sub> + 5 H<sub>2<\/sub>O + N<sub>2<\/sub>O<br \/>\nZn + 4 HNO<sub>3<\/sub> (cone.) \u2192 Zn (NO<sub>3<\/sub>)<sub>2<\/sub> + 2 H<sub>2<\/sub> + 2 NO<sub>2<\/sub><\/p>\n<p>c) S<sub>8<\/sub> reacts with cone, nitric acid to form Sulphuric acid, NO<sub>2<\/sub> gas.<br \/>\nS<sub>8<\/sub> + 48 HNO<sub>3<\/sub> \u2192 8 H<sub>2<\/sub>SO<sub>4<\/sub> + 48 NO<sub>2<\/sub> + 16 H<sub>2<\/sub>O<br \/>\nd) P<sub>4<\/sub> reacts with cone, nitric acid to form phosphoric acid and NO<sub>2<\/sub> gas.<br \/>\nP<sub>4<\/sub> + 20 HNO<sub>3<\/sub> \u2192 4 H<sub>3<\/sub>PO<sub>4<\/sub> + 20 NO<sub>2<\/sub> + 4 H<sub>2<\/sub>O<\/p>\n<p><span style=\"color: #0000ff;\">Textual Examples<\/span><\/p>\n<p>Question 1.<br \/>\nThough nitrogen exhibits +5 oxidation state, it does not form peniahalide. Give reason.<br \/>\nAnswer:<br \/>\nNitrogen with n = 2, has s and p orbitals only. It does not have d orbitals to expand its covalence beyond four. That is why it does not form pentahalide.<\/p>\n<p>Question 2.<br \/>\nPH<sub>3<\/sub> has lower boiling point than NH<sub>3<\/sub>. Why ? [T.S. Mar. 16]<br \/>\nAnswer:<br \/>\nUnlike NH<sub>3<\/sub>. PH<sub>3<\/sub> molecules are not associated through hydrogen bonding in liquid state. That is why the boiling point of PH<sub>3<\/sub> is lower than NH<sub>3<\/sub>.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 3.<br \/>\nWrite the reaction of thermal decomposition of sodium azide.<br \/>\nAnswer:<br \/>\nThermal decomposition of sodium azide gives dinitrogen gas.<br \/>\n2NaN<sub>3<\/sub> \u2192 2Na + 3N<sub>2<\/sub><\/p>\n<p>Question 4.<br \/>\nWhy does NH<sub>3<\/sub> act as a Lewis base ?<br \/>\nAnswer:<br \/>\nNitrogen atom in NH<sub>3<\/sub> has one lone pair of electrons with is available for donation. Therefore, it acts as a Lewis base.<\/p>\n<p>Question 5.<br \/>\nWhy does NO<sub>2<\/sub> dimerise ?<br \/>\nAnswer:<br \/>\nNO<sub>2<\/sub> contains odd number of valence electrons. It be haves as a typical odd molecule. On dimerisation. It is converted to stable N<sub>2<\/sub>O<sub>4<\/sub> molecule with even number of electrons.<\/p>\n<p>Question 6.<br \/>\nIn what way can it be proved that PH<sub>3<\/sub> is basic in nature ?<br \/>\nAnswer:<br \/>\nPH<sub>3<\/sub> reacts with acids like HI to form PH<sub>4<\/sub>I which shows that it is basic in nature.<br \/>\nPH<sub>3<\/sub> + HI \u2192PH<sub>4<\/sub>I<br \/>\nDue to lone pair on phosphorus atom, PH<sub>3<\/sub> is acting as a Lewis base in the above reaction.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 7.<br \/>\nWhy does PCl<sub>3<\/sub> fume in moisture ?<br \/>\nAnswer:<br \/>\nPCl<sub>3<\/sub> hydrolyses in the presence of moisture giving fumes of HCl.<br \/>\nPCl<sub>3<\/sub>+ 3H<sub>2<\/sub>O \u2192 H<sub>3<\/sub>PO<sub>3<\/sub> + 3HCl<\/p>\n<p>Question 8.<br \/>\nAre all the five bonds in PCl<sub>5<\/sub> molecule equivalent ? Justify your answer.<br \/>\nAnswer:<br \/>\nPCl<sub>5<\/sub> has a trigonal bipyramidal structure and the three equatorial P-Cl bonds are equivalent. While the two axial bonds are different and longer than equatorial bonds.<\/p>\n<p><span style=\"color: #0000ff;\">Intext Questions<\/span><\/p>\n<p>Question 1.<br \/>\nWhy are pentahalides more covalent than trihalides ?<br \/>\nAnswer:<br \/>\nHigher the positive oxidation state of central atom, more will be its polarising power, which in turn increases the covalent character of the bond formed between the central atom and the halogen atom. In pentahalides, the central atom is in +5 oxidation state while in trihalides, it is in +3 oxidation state. Therefore, pentahalides are more covalent than trihalides.<\/p>\n<p>Question 2.<br \/>\nWhy is BiH<sub>3<\/sub> the strongest reducing agent amongst all the hydrides of Group 15 elements ?<br \/>\nAnswer:<br \/>\nAmong the 15 group elements, the size of Bi atom is largest and hence, the Bi-H bond length is largest or Bi-H bond dissociation energy is lowest. That&#8217;s why Bi-H bond dissociates more readily than the other hydrides of the group and hence, BiH<sub>3<\/sub> is the strongest reducing agent.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 3.<br \/>\nWhy is N<sub>2<\/sub> less reactive at room temperature ?<br \/>\nAnswer:<br \/>\nDinitrogen is inert of less reactive because the bond enthalpy of N \u2261 N bond is very high.<\/p>\n<p>Question 4.<br \/>\nMention the conditions required to maximise the yield of ammonia.<br \/>\nAnswer:<br \/>\nAmmonia is produced by Haber\u2019s process as.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23229\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-24.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 24\" width=\"454\" height=\"75\" \/><br \/>\nYield of ammonia is favoured by high pressure according to Le-Chatelier\u2019s principle, Other conditions, that favour the production of ammonia are as follows :<\/p>\n<ol>\n<li>Temperature &#8211; approximately 700 K<\/li>\n<li>Pressure &#8211; 200 atm or 200 \u00d7 10<sup>5<\/sup> Pa<\/li>\n<li>Catalyst &#8211; Iron oxide<\/li>\n<li>Promotor &#8211; Molybdenum of K<sub>2<\/sub>O and Al<sub>2<\/sub>O<sub>3<\/sub><\/li>\n<\/ol>\n<p>Question 5.<br \/>\nHow does ammonia react with a solution of Cu<sup>2+<\/sup>?<br \/>\nAnswer:<br \/>\nWhen ammonia (aqueous solution is ammonium hydroxide) reacts with a solution of Cu<sup>2+<\/sup>, a deep blue solution is obtained due to the formation of tetraamine copper (II) ion.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23230\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-25.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 25\" width=\"395\" height=\"94\" \/><\/p>\n<p>Question 6.<br \/>\nWhat is the covalence of nitrogen in N<sub>2<\/sub>O<sub>5<\/sub>?<br \/>\nAnswer:<br \/>\nStructural formula of N<sub>2<\/sub>O<sub>5<\/sub><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23231\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-26.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 26\" width=\"184\" height=\"120\" \/><br \/>\nSince, N atom has 4 shared electron pairs; the valence of N is 4.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 7.<br \/>\nBond angle in PH<sub>4<\/sub><sup>+<\/sup> is higher than that in PH<sub>3<\/sub>. Why ?<br \/>\nAnswer:<br \/>\nIn both PH<sub>4<\/sub><sup>+<\/sup> and PH<sub>3<\/sub>, phosphorus atom is sp<sup>3<\/sup> hybridized. In PH<sub>4<\/sub><sup>+<\/sup>, all the four orbitals are bonded whereas in PH<sub>3<\/sub>, there is a lone pair of electrons too. Due to lone pair-bonded pair repulsion in PH<sub>3<\/sub>, the bond angle is less than 109.5\u00b0.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23232\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-27.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 27\" width=\"414\" height=\"212\" \/><\/p>\n<p>Question 8.<br \/>\nWhat happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO<sub>2<\/sub>? .<br \/>\nAnswer:<br \/>\nWhite phosphorus dissolves in boiling NaOH in an inert atmosphere of CO<sub>2<\/sub> producing phosphine (PH<sub>3<\/sub>) gas and sodium hypophosphite (NaH<sub>2<\/sub>PO<sub>2<\/sub>).<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23233\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-28.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 28\" width=\"557\" height=\"73\" \/><\/p>\n<p>Question 9.<br \/>\nWhat happens when PCl<sub>5<\/sub> is heated ?<br \/>\nAnswer:<br \/>\nIn PCl<sub>3<\/sub>, there are 5P &#8211; Cl bonds, out of which three are equatorial (longer) and two are axial (shorter). When PCl<sub>5<\/sub> is heated strongly, two less stable axial bonds break and phosphorus trichloride (PCl<sub>3<\/sub>) is formed.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23234\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-29.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 29\" width=\"479\" height=\"302\" \/><br \/>\nStructure of PCl<sub>5<\/sub> showing axial and equatorial bonds.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 10.<br \/>\nWrite a balanced equation for the hydrolytic reaction of PCl<sub>5<\/sub> in heavy water.<br \/>\nAnswer:<br \/>\nReaction of PCl<sub>5<\/sub> in heavy water (D<sub>2<\/sub>O)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23235\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6a-Group-15-Elements-30.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements 30\" width=\"532\" height=\"130\" \/><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 6(a) Group-15 Elements Textbook Questions and Answers. AP Inter 2nd Year Chemistry Study Material Lesson 6(a) Group-15 Elements Very Short Answer Questions Question 1. Why does the reactivity of nitrogen differ from phosphorus? Answer: Nitrogen gas exists as diatomic molecule. Dut to the presence &#8230; <a title=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements\" class=\"read-more\" href=\"https:\/\/apboardsolutions.in\/ap-inter-2nd-year-chemistry-study-material-chapter-6a\/\" aria-label=\"More on AP Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements\">Read more<\/a><\/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.in\/wp-json\/wp\/v2\/posts\/22210"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=22210"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22210\/revisions"}],"predecessor-version":[{"id":22211,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22210\/revisions\/22211"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=22210"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=22210"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=22210"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}