Answer to Question #84648 in Inorganic Chemistry for Sk Abidur Rahaman




 &quot;BF3,PF3 and ClF3 are AX3 type of molecules, but their structure and bond angles are different&quot;- Explain
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there are 3 types of hybridisation for carbon in organic compounds, sp, sp2 and sp3 depending on the number of sigma and pi bonds between the carbon atoms. more the ‘s’ character more is the electronegativity of the carbon as ‘s’ orbitals are closer to the nucleus, hence electrons held more closely. (i) adsorption is a surface phenomenon that causes the accumulation of molecules of a substance at the surface of a solid or liquid rather than in the bulk. in adsorption, the substance gets concentrated at the surface only. it does not penetrate through the surface to the bulk of the solid or liquid. for example, when a chalk stick is dipped into an ink solution, only its surface becomes coloured.


on the other hand, the process of absorption is a bulk phenomenon. in absorption, the absorbed substance gets uniformly distributed throughout the bulk of the solid or liquid. for example, when a sponge is dipped in water the whole sponge gets wets.


(ii)according to the schulze-hardy rule, the effectiveness of the salt causing flocculation depends on the charge on the ion of opposite sign to the charge on the sol particles. the greater the magnitude of the opposite charge, the higher the ability of a salt to coagulate the sol.  
thus, trivalent salt alcl3 is more effective in causing the coagulation of a negatively charged sol than divalent salt mgcl2.


(iii)out of sulphur sol and proteins, sulphur sol forms multimolecular colloids. sulphur sol consists of particles containing a thousand or more s8 molecules. on the other hand, proteins form macromolecular colloids. ch3nh2 will appear as a stronger base since n for (ch3)2nh has bulkier substituents that will block n from accepting extra h (bl base = proton acceptor)[readmore]since ch3nh2  is the stronger base than (ch3)2nh, its c.acid will be weaker compared to (ch3)2nhstronger base → ch3nh2 stronger c.acid → (ch3)2nh3+ recall that there are several types of intermolecular forces:1. ion-dipole interaction – occurs between an ion and a polar covalent compound2. hydrogen bonding – occurs in compounds where hydrogen is directly connected to an electronegative element such as n, o, or f3. dipole-dipole interaction – occurs between two polar covalent compounds4. induced-dipole interaction – occurs when a nonpolar compound interacts with a polar compound5. dispersion forces – occurs in all compounds. this is the primary intermolecular force exhibited by nonpolar compounds.[readmore]although hcl is a polar molecule, it still exhibits dispersion forces because it occurs in all compounds according to ncert answer would be option (1)and

but according to general chemistry answer would be (3)andoption 1)and





option 2)andoption 3)andoption 4)and cao a) he helium (he) is an element by itself. he atom can only exhibit dispersion forces.b) hclh          1a                   1 × 1 e– = 1 e–cl         7a                   1 × 7 e– = 7 e–   total: 8 valence e–[readmore]cl is more electronegative than h, so hcl is a polar covalent compound. hcl exhibits dipole-dipole interaction.c) h2oh          1a                   1 × 2 e– = 2 e–o          6a                   1 × 6 e– = 6 e–   total: 8 valence e– since hydrogen is bonded directly to oxygen.h2o exhibits hydrogen bonding.d) cl2cl         7a                   1 × 7 e– = 14 valence e–composed of the same atom. there's no electronegativity difference.cl2 can only exhibit dispersion forces. a) van der waals (present, all molecule/atoms contain it)[readmore](b) dipole-dipole (present, c=o creates a partial charge on c and o)(c) hydrogen bonding (not present, h-bonds are only present in compounds with h attached on o, n or f)(d) ion-ion (not present, only appears on compound with full charges) NaBr   

  

SO3   

  

CaCl2   

  

Mg3N2 we’re being asked to determine which of the given solutions is a good buffer system.recall that a buffer can be composed of the following:• weak base and its conjugate acid• weak acid and its conjugate base• weak acid and a strong base ▪ concentration of weak acid > concentration of strong base• weak base and a strong acid ▪ concentration of weak base > concentration of strong acid[readmore]a. 0.19 m potassium hydroxide + 0.27 m potassium bromide.potassium hydroxide → koh → group 1a k+ + oh- → strong basepotassium bromide → kbr breaks up in the solution: kbr → k+ + br-▪ k+ is a neutral ion → will not contribute to ph  hbr       +      h2o       →      br-      +         h3o+(weak acid)                            (conjugate base)•  strong base + weak base• not a good buffer systemb. 0.28 m hydroiodic acid + 0.18 m sodium iodide. hydroiodic acid → hi → weak acidsodium iodide → nai → breaks up in the solution: nai → na+ + i-hi       +      h2o       →      i-      +         h3o+(weak acid)                            (conjugate base)• weak acid + conjugate base = buffer• the solution is a good buffer systemc. 0.30 m ammonia + 0.37 m ammonium nitrate. ammonia → nh3 → weak baseammonium nitrate → nh4no3 → breaks up in the solution: nh4no3 → nh4+ + no3-nh3       +      h2o       →      nh4+      +         oh-(weak base)                            (conjugate acid)• weak base + conjugate acid = buffer• the solution is a good buffer systemd. 0.17 m hypochlorous acid + 0.12 m hydrobromic acid. hypochlorous acid → hclo → weak acidhydrobromic acid → hbr → weak acid•  weak acid + weak acid• not a good buffer systeme. 0.38 m calcium bromide + 0.30 m barium bromide.calcium bromide → cabr → breaks up in the solution: cabr → ca+ + br-barium bromide → babr → breaks up in the solution: babr → ba+ + br-•  conjugate base + conjugate base• not a good buffer systemanswer: the following aqueous solutions are good buffer systems b. 0.28 m hydroiodic acid + 0.18 m sodium iodide and c. 0.30 m ammonia + 0.37 m ammonium nitrate we are being asked to classify the following reactions based on the following categories:combination/synthesis → multiple reactants combine to form a single productw/o change in oxidation numbersa + b → abdecomposition →reactant splits into multiple products with no change in oxidation numbersab → a + bcombustion reaction: reaction with oxygen that produces carbon dioxide and waterhydrocarbon + o2 → co2 + h2osingle displacement →single element displaces another element in a compoundab + c → ac + b[readmore]let’s analyze the given reactionsa. fe2(co3)2 (s) → fe2o3(s) + 2co2(g) break down of fe2(co3)2(s) to its component ions → decomposition reactionanswer: this is a decomposition reaction.b. 2 c6h11oh(l) + 17 o2(g) → 12co2 (g) + 12h2o(l) reaction with oxygen that produces carbon dioxide and water → combustion reactionanswer: this is a combustion reaction.c. p4o10(s) + 6h2o(l) → 4h3po4(aq) two or more substances combine to form a single compound → combination reactionanswer: this is a combination reaction.d. 2c(s) + mno2(s) → mn(s) + 2co(g) substitute one element for another element in a compound →single-displacement reaction answer: this is a single-displacement reactione. 2naclo3 (s) → 2nacl(s) + 3o2(g)break down of compound into its simpler substances → decomposition reactionanswer: this is a decomposition reaction