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85951.

Account for the following: Although the electron gain enthalpy of fluorine is less negative than that of chlorine, fluorine is a stronger oxidising agent than chlorine.

Answer»

Solution :Besides ELECTRON GAIN enthalpy, the other factors that influence the oxidising power are BOND dissociation enthalpy and HYDRATION energy of the anion. The LAST two factors are in favour of `F_2` to act as oxidising agent.
Discussion
85952.

Account for the following: Although Fluorine has less negative electron gain enthalpy yet F_(2)is strong oxidizing agent.

Answer»

Solution :DUE to small size and LOW BOND DISSOCIATION enthalpy
Discussion
85953.

Account for the following : Although amino group is o- and p- directing in aromatic, electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.

Answer»

Solution :Nitration is usually carried out with a mixture of conc. `HNO_(3)` + conc. `H_(2)SO_(4)`. In presence of these acids, aniline gets protonated to form anilinium ION. Therefore, in presence of acids, the reaction mixture consists of aniline and anilinium ion. While `-NH_(2)` group in aniline is o, p-directing and activating, the `NH_(3)` group in anilinium ion is m-directing and deactivating. Now nitration of aniline mainly gives p-nitroaniline DUE to steric hindrance at o-position, the nitration of anilinium ion givesm-nitroaniline. In actual PRACTICE, approx. a 1 : 1 mixture of p-nitroaniline and m-nitroaniline is obtained, as per the reactions given below :
Discussion
85954.

Account for the following Although amino group is o - and p - directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m - nitroaniline.

Answer»

SOLUTION :Direct nitration of aniline gives o and p - nitro aniline but in strong ACID MEDIUM aniline is protonated to form anilinium ion which is m - directing and HENCE m - nitro aniline is ALSO formed.
Discussion
85955.

Account for the following: Acidic character decreases from N_(2)O_(3) to Bi_(2)O_(3) in group 15.

Answer»

Solution :As the SIZE INCREASES, ELECTRONEGATIVITY DECREASES/non-metallic character decreases
Discussion
85956.

Account for the following: (a) Sulphurous acid is a reducing agent. (b) Fluorine forms only one oxoacid. (c ) Boiling point of noble gases increases from He to Rn.

Answer»

Solution :(a) Sulphurous acid is a REDUCING agent
`H_(2)SO_(3) +H_(2)O +2Fe^(3+) to 2Fe^(2+)+SO_(4)^(2-)+4H^(+)`
`Fe^(3+)` ions are reduced to `Fe^(2+)` ions.
(b) Fluorine forms only one oxoacid.
Due to high electronegativity and small size, fluorine forms only one oxoacid.
(c ) Boiling point of noble GASES increases from He to Rn.
Size of the atom increases from He to Rn. Therefore, the VAN der Waals force of attraction, which is proportional to surface area ALSO increases from He to Rn. Thus, boiling point increases from He to Rn.
Discussion
85957.

Account for the following: (a) Molecular nitrogen N_2is not particularly reactive. (b) H_3PO_3 is diprotic.(c) Nitrogen forms no pentahalides like phosphorus. (d) Water has higher boiling point than H_2S .

Answer»

Solution :(a) Molecular nitrogen is not reactive. Molecular nitrogen has TRIPLE bond `N_ (N -= N)`and has non-polar CHARACTER. Due to the presence of triple bond, it has very high bond dissociation energy (945 kJ `"mol"^(-1)` ) and, therefore, it does not react with other elements under normal conditions. However, it may react at high temperatures.
(b) `H_3PO_3`has the structure given below:

In this structure, it has only two - OH groups and therefore, gives only two ionisable `H^+`ions in aqueous solution :
`H_2PO_3 iff H^(+) + H_2PO_(3)^(-)`
`H_2PO_(3)^(-) iff H^(+) + HPO_(3)^(2-)`
It gives only two series of salts `NaH_2PO_3` and `Na_2 HPO_3`and behaves as dibasic.
(c) The electronic configuration of nitrogen is
`1s^(2) 2S^(2) 2p_(x)^(1) 2p_(y)^(1) 2p_(z)^(1)`
The outermost shell does not contain d-orbitals and therefore, nitrogen cannot extend its octet. Hence, it can form maximum of four bonds and cannot form pentahalides. On the other hand, P has vacant d-orbitals in its outermost shell and therefore, can extend its octet. Hence, pentahalides of phosphorus are known.
(d) In `H_2O` , hydrogen is bonded to electronegative element oxygen and therefore, it can form hydrogen bonds. As a result, WATER exists as associated molecules and therefore, exists as a liquid.

On the other hand, `H_2S`has negligible tendency to form hydrogen bonding because of low electronegativity of S. The molecules are, therefore, held up by only weak van der Waals” forces. Therefore, it exists as gas. Therefore, water has higher boiling point than `H_2 S` .
Discussion
85958.

Account for the following : (a) Silver chloride dissolves in excess of NH_(3)

Answer»

Solution :(a)AgCIformssoluble complexwith `NH_(3)`
AgCI `+2NH_(3) to [Ag(NH_(3)) ]C]`
Discussion
85959.

Account for the following : (a) Europlum (II) is more stable than cerium (II). (b) Transition metal have high enthalpies of atomisation . (c ) Actinoid ions are generally coloured.

Answer»

Solution :(a) Europlum has stable electronic configuration , i.e.,`[Xe] 4F^(7) 5d^(0)6s^(0)`.
(b) Due to LARGE number of unpaired electrons in their atom, there is larger and stronger METALLIC bonding.
(c ) Unpaired electron are present in their ions WHCIH undergo f-f TRANSITION.
Discussion
85960.

Account for the following: (a) Lower members of alcohols are soluble in water but higher members are not. (b) Alcohols cannot be used as solvent for Grignard reagent.

Answer»

Solution :(a) Alcohols are soluble in water because they form intermolecular hydrogen BONDING with water. Lower membrane are completely miscible with water and the higher membrane are not. This is because of the INCREASE in size of hydrophobic (water repealing) alkyl group in the alcohol.
(ii) Strong basic substances like organo METALLIC compounds (RMgX - Grignard reagent reagent) are decomposed by alcohol.
`R-OH+ CH_(3)MGBR to R-O-Mg-Br + CH_(4)`
Hence, alcohols cannot be used as a solvent for Grignard reagent.
Discussion
85961.

Account for the following: (a) Europium(II) is more stable than cerium(II). (b) Transition metals have high enthalpies of atomisation. (c) Actinoid ions are generally coloured.

Answer»

Solution :(a) Europium(II) has stable electronic configuration [Xe] `4f^(7)5d^(0)6s^(0)` while ccrium(II) has comparatively LESS stable configuration [Xe] `4f^(2)5d^(0)6s^(0)`.
(b) TRANSITION metals have high values of enthalpies of atomisation. This is due to a large number of electrons in their atoms. They give rise to strong interatomic interactions and hence strong bonding between their molecules.
(c) Actinoid IONS are generally COLOURED. This can be explained in terms of UNPAIRED electrons undergoing f-f transitions.
Discussion
85962.

Account for the following : (a)CuCl_(2)is more stable than Cu_(2)Cl_(2). (b) Atomic radii of 4d and 5d series elements are nearly same. (c ) Hydrochloric acid is not used in permanganate titrations. or Account for the following : (a) Eu^(2+) is a strong reducing agent. (b) Orange colour of dichromate ion changes to yellow in alkaline medium. (c ) E^(@) ( M^(2+) // M ) valuesfor transiton metals show irregular variation.

Answer»

SOLUTION :(a) `Eu^(2+)`can lose electron to FORM more stable `Eu^(3+)` PROPERTY )
Discussion
85963.

Account for the following: (a) Carboxylic acids with five or fewer carbon atoms are water soluble but, higher ones are insoluble. (b) Highly branched carboxylic acids are less acidic than unbranched acids. (c) Acetic acid can be halogenated in prescnce of phos-phorus and chlorine but formic acid cannot be halogenated in the same way. (d) carbon-oxygen bonds length in formic acid are 1.23 Å and 1.36 Å but in sodium formate both carbon oxygen bonds have same value, i.e., 1.27 Å. Formic acid is stronger than acetic acid. (f) Fluoroacetic acid is stronger than chloroacetic acid (g) Formic acid shows reducing properties. (h) beta-Keto acids undergo decarboxylation easily. (i) o-Hydroxybenzoic acid is stronger than p-hydroxy benzoic acid. (ii) Oxidation of toluene by acidic KMnO_(4) gives poor yield of benzoic acid while oxidation of p-nitro toluene gives good yield of p-nitrobenzoic acid.

Answer»

Solution :(a) The solubility of RCOOH in water is due to hydrogen BONDING of-COOH group and water. R-portion being lyophobic resists solubility. As R gets large, this factor prevails over the first factor and thus, higher acids become insoluble.
(b) The-COOH group of the branched acid is shielded from solvent molecules and cannot be stabilized by solvation as effectively as unbranched acid like the acetate anion.
(c )This is HVZ reaction. It occurs only in those carboxylicacids which have `alpha`-hydrogen atoms. Acetic acid possesses three `alpha`-hydrogen atoms but formic acid does not have even a single `alpha`- hydrogen atom. Thus, formic acis does not undergo this reaction]
The formate ion present in sodium formate is a resonance hybrid of the following two structures :
(d) `H-underset(O)underset(||)C=O^(-)harrH-underset(O^(-))underset(|)C=O`
Thus, in resonance hybrid the bond LENGTH of C-O are identical. In formic acid, no such resonance exists and thus, the bond lengths are different.
(E ) n acetic acid, methyl group is present which exerts +I negative charge on the carboxylate ion and destabilise it. The loss of proton becomes comparatively difficult in comparison to formic acid. Hence, acetic acid is a weaker acid than formic acid.
(f) Both fluorine and chlorine are electron withdrawing substituents. However, the capacity of fluorine is more than chlorine as it is more electronegative than chlorine. The fluoroacetate ion is more stabilised and thus, fluoroacetic acid has higher tendency to lose its proton. Hence, it is a stronger acid than chloroacetic acid.
(g) Formic acid is easily oxidised to CARBON dioxide and water and thus acts as a reducing agent.
`HCOOH+[O]toH_(2)O+CO_(2)`
(h) `beta`-Keto acids are unstable acids. These readily undergo decarboxylation through a cyclic transition state.

The anion from salicylic acid in o-isomer is stabilized by intramolecular hydrogen bonding which does not exist in p-hydroxybenzoic acid.

(ii) OXIDATION is an electrophile, it can DESTROY the ring in case of toluene. But in p-nitrotoluene, the `-NO_(2)` group deactivates the benzene ring and thus increases the yield of p-nitrobenzoic acid
Discussion
85964.

Account for the following: (a) CuCl_(2) is more stable than Cu_(2)Cl_(2). (b) Atomic radii of 4d and 5d series elements are nearly same. (c) Hydrochloric acid is not used in permanganate titrations.

Answer»

Solution :(a) `CuCl_(2)` is more stable than `Cu_(2)Cl_(2)`. This is because in `CuCl_(2)`, Cu is in + 2 oxidation state which is more stable due to HIGH hydration enthalpy COMPARED to `Cu_(2)Cl_(2)` in which Cu is in + 1 oxidation state.
(b) Atomic radii of 4d and 5d series elements are NEARLY same. This happens because of lanthanide contraction.
(c) Hydrochloric acid is not used in permanganate titration. This is because hydrochloric acid is oxidised to chlorine.
`2HCl+O to Cl_(2)+H_(2)O`
Discussion
85965.

Account for the following : (a) Chlorine water has both oxidising and bleaching properties. (b) H_(3)PO_(2) and H_(3)PO_(3) act as good reducing agents while H_(3)PO_(4) does not. (c) On addition of ozone gas to KI solution, violet vapours are obtained.

Answer»

Solution :(a) In presence of moisture or water, `Cl_(2)` gives nascent oxygen which is responsible for its oxidising and bleaching properties as SHOWN below :
`Cl_(2) + H_(2)O rarr ,HCl + HClO' rarr 2HCl + underset("Nascent oxygen")([O])`
(i) It oxidises acidified ferrous sulphate to ferric sulphate
`{:(""Cl_(2)+H_(2)O rarr 2HCl + O),(" "FeSO_(4)+H_(2)SO_(4) + O rarr Fe_(2)(SO_(4))_(3)+H_(2)O),(bar (underset(("Green"))(2FeSO_(4))+H_(2)SO_(4) + Cl_(2) rarr underset(("Brown"))(Fe_(2)(SO_(4))_(3))+2HCl)):}`
(ii) It bleaches vegetable and ORGANIC colouring matter to colourless SUBSTANCES by oxidation
`Cl_(2) + H_(2)O rarr 2HCl + [O]`
Vegetable colouring matter `+ O rarr` Oxidised colourless substances.
(b) The structures of `H_(3)PO_(2), H_(3)PO_(3) and H_(3)PO_(4)` are :
Due to the presence of P-H bonds, both `H_(3)PO_(2) and H_(3)PO_(3)` act as reducing agents. In contrast, `H_(3)PO_(4)` does not have any P-H bonds and hence it does not act as a reducing agent.

(c) `O_(3)` is a POWERFUL oxidising agent. Therefore, it oxidises aq. KI to violet vapours of `I_(2)`
`{:(""O_(3)(g) rarr O_(2)(g) + O(g)),(""2KI (aq) + H_(2)O (l) + O(g) rarr 2KOH (aq) + I_(2)(s)),(bar(underset(("Colourless"))underset("Potassium iodide")(2KI(aq))+H_(2)O (l) + O_(3)(g) rarr 2KOH (aq) + O_(2)(g)+underset(("Violet"))underset("Iodine")(I_(2)(s)))):}`
Discussion
85966.

Account for the following : (a) Aspirin drug helps in the prevention of heart attack. (b) Diabetic patients are advised to take artificial sweeteners instead of natural sweeteners. (c) Detergents are non-biodegradable while soaps are biodegradable.

Answer»

Solution :(a) Aspirin acts as a blood thinner. Therefore, the chances of heart attack are reduced.
(b) Artificial sweeteners have ALMOST zero calorific value. They don.t require insulin which is in restricted quantity in diabetic persons. On the other HAND, if NATURAL sweetener is consumed by a diabetic, it will raise the level of blood sugar.
(c) Soaps are decomposed by microorganism PRESENT in sewage WHEREAS detergents are not.
Discussion
85967.

Account for the following : (a) Aniline does not undergo Friedel Crafts alkylation. (b) Although -NH_(2) group is an ortho and para-directing group, nitration of aniline gives alongwith ortho and para-derivatives meta-derivative also.

Answer»

Solution :(a) Aluminium CHLORIDE used in Friedel Crafts reaction is a LEWIS acid and aniline is a Lewis base. The two COMPOUNDS form an adduct. That is why Friedel Crafts alkylation does not proceed further.
(b) `-NH_(2)` group GETS protonated to `-NH_(3)`. This group is electron WITHDRAWING which is why we get the meta-derivative also.
Discussion
85968.

Account for the following (a) Aluminium undergo slow corrosion than iron. (b) H_2 - O_2 fuel cell is more useful than other cells .

Answer»

Solution :(a) Aluminium , copper silver also undergo corrosion but at a slower RATE than iron.
For eg., let US consider the REDUCTION of Aluminium
`Al_((s)) to Al_((aq))^(3+) + 3e^(-)`
`Al^(3+)` which reacts with oxygen in air to form a protective coating of `Al_2O_3`. This coating act as a protective film for the inner surface . So further corrosion is PREVENTED.
(b) `H_2 - O_2` fuel cell is more advanced. Because it is highly efficient. It is POLLUTION free. In this , `H_2 - O_2` fuel cell, energy of combustion of fuel is directly converted to electrical energy.
Discussion
85969.

Account for the following : (a) All Scandium salts are white.

Answer»

SOLUTION :(a) Sc has only + 3 oxidation state, there is no UNPAIRED ELECTRON.
Discussion
85970.

Account for the following : 2-methyl-2-nitro propane has neither of the properties.

Answer»

Solution :2-methyl - 2 - NITRO propane neither soluble in alkali nor reacts with NITROUS ACID. Because it contains a tertiary nitro group where aciform is not possible and it does not CONTAIN `alpha`- H atoms. So it has neither of the properties.
Discussion
85971.

Account for the followig : CO_(2) is used in referigeration

Answer»

Solution :SOLID `CO_(2)`, PRODUCE cooling and sublimes directly into vapour state, HENCE used for REFERIGERATION.
Discussion
85972.

Account for the followig : CO is poisonous

Answer»

Solution :CO forms carboxy-haemoglobin COMPLEX with haemoglobin of blood which is about 300 TIMES more stable than oxygen - haemoglobin complex and thus it STOPS the supply of oxygen and hence, leads to death of the person.
Discussion
85973.

Account for the fact that the charge of the colloidal particles is due to the selective adsorption of the ions. Give one example.

Answer»

Solution :The colloidal solutions GENERALLY contain traces of electrolytes. The charge on the colloidal particles is due to the preferentially adsorption of positive or negative ions from the electrolyte. If the colloidal particles adsorb positive ions, they get positive charge. Similarly, if the colloidal particles have preference to adsorb negative charge, they acquire negative charge. In general, the particles constituting the dispersed phase adsorb only those ions preferentially which are common with their own lattice ions. LET us consider an example The `FE(OH)_3 `sol prepared by the hydrolysis of `FeCl_3` . has positive charge because it preferentially adsorbs` Fe^(3+)`ions on its surface from the solution.
`FeCl_3 iff Fe^(3+) + 3Cl^(-)`
`Fe(OH)_3 + Fe^(3+) iff Fe(OH)_3. Fe^(3+)` (sol)
This colloidal sol may also be REPRESENTED as
` [Fe(OH)_3] Fe^(3+) : 3Cl^(-)`
where `[Fe(OH)_3] ` implies precipitate of FERRIC hydroxide on which `Fe^(3+)`are adsorbed. The ions on the right of dotted line represent ions present in excess in dispersion medium.
Discussion
85974.

Account for the followig : CO is used in the extraction of metals.

Answer»

Solution :CO being a good reducing AGENT, reduces SEVERAL metal OXIDES into crude metal. HENCE it is used in extraction of metals.
Discussion
85975.

Accountfor thefactthat pine oil is used in thefroth floatation process.

Answer»

Solution :Pine oil enhancesthe NON - wettabilityof ORE particlesby water. In other words, ore particlesarepreferentially WETTED by pineoil and HENCE become lighter and rise tothe surfacealong withthefroth.
Discussion
85976.

Account for the fact that [Ni(CO)_(4)] has tetrahedral geometry whereas [Ni(CN)_(4)]^(2-) has square planar geometry.

Answer»

Solution :In `[Ni(CO)_(4)]`, Ni is `sp^(3)` hybridised but in `[Ni(CN)_(4)]^(2-)` it is `dsp^(2)` hybridised.
Discussion
85977.

Account for the fact that ammonia is a good complexing agent.

Answer»

Solution :Ammonia is a GOOD COMPLEXING agent due to the presence of a lone pair of ELECTRONS on NITROGEN.
Discussion
85978.

Account for the acidity of L-ascorbic acid (pK_(a)=4.21). Which of the following is most acidic H? (Marked in the structure as H^(a),H^(b),H^(c ) "and" H^(d))

Answer»

`H^(a)`
`H^(B)`
`H^(c )`
`H^(d)`

Solution :The anion FORMED by the REMOVAL of enolic `H` is stable because the negative charge is STABILISED by resonance, i.e., by delocalisation of the negative charge to the `O` of the `(C=O)` through `(C=C)`, as shown below.
Discussion
85979.

Account for the acidic nature of HClO_4 In terms of Bronsted-Lowry Theory, Identify its conjugate base.

Answer»

Solution :`HClO_4 leftrightarrow H^(+) +ClO_4^(-)`
According to LOWRY -Bronsted concept, a strong acid has weak conjugate base and a weak acid has a strong conjugate base.
Let us consider the stabilities of the conjugate bases `ClO_4^(-),ClO_3^(-),ClO_2^(-) and ClO^(-)` formed from these acid `HClO_4,HClO_3,HClO_2,HOCl` respectively. these anions are stabilized to GREATEST extent, it has lesseer attraction for proton and therefore,will behave as weak base.
Consequently the corresponding acid will be strongest because weak conjugate base has strong acid and strong conjugate base has weak acid.
The charge stabilization increases in the order, `CIO^(-) LT ClO_2^(-) lt ClO_3^(-) lt ClO_4^(-)`
This means `ClO_4^(-)` will have maximum stability and therefore will have minimum attraction for `H^+` . Thus `ClO_4^(-)` will be weakest base and its conjugate acid `HClO_4` is the strongest acid.
`ClO_4^(-)` is the conjugate base of the acid `HClO_4`.
Discussion
85980.

Account for Reduction of CH_3CN " gives "CH_3CH_2NH_2 while CH_3NC " gives "(CH_3)_2NH.

Answer»

SOLUTION :`CH_3CN` (METHYL cyanide) on REDUCTION gives `CH_3CH_2NH_2` (ethylamine) because addition of hydrogen takes palce at `equiv CN`
`{:(CH_3-CequivNunderset(4[H])overset(LiAH_4)(to)CH_3-CH_2NH_2),("Ethylamine (Primary amine) "):}`
Whereas `CH_3NC` (Methyl ISOCYANIDE) on reduction with `LiAlH_4` gives secondary amine

In methyl cyanide, -CN group is attached to alkyl group and by reduction it gives a primary amine whereas in methyl isocyanide -NC group is attached to alkyl group and by reduction it gives a secondary amine.
Discussion
85981.

Account for the acidic nature of HCIO_4. In terms of Bronsted - Lowry theory, identify its conjugate base.

Answer»

Solution :(i) The acidic character of oxy of chlorine increases with increase in oxidation number of HALOGEN ATOM. The oxidation number of chlorine in `HCIO_4` (per chloric acid) is 7. Increase in oxidation number weakness the `O-H` bond strength and increases the acidity. Therefore `HCIO_4` is a strong acid.
(ii) According to Bronsted - Lowry theory. removal of HYDROGEN ion or a proton froman acid is its conjugate BASE.
(iii) So the conjugate base of `HCIO_4` is `CIO_4^(-1)`
Discussion
85982.

Account for reducing nature of Formic acid.

Answer»

Solution :(i) Formic acid (HCOOH) is unique because it contains both an aldehyde group and carboxyl group also.

(ii) Hence it can ACT as a reducing agent. It reduces Fehling's solution Tollen's reagent and DECOLOURISES pink coloured `KMnO_(4)` solution.
(iii) Whereas in acetic acid, there is no aldehyde group and it cannot act as reducing agent.
(iv) Formic acid reduces ammoniacal silver NITRATE solution (Tollen's reagent) to metallic silver.
`HCOOH+Ag_(2)OtoH_(2)O+CO_(2)+2Agdarr` (metallic silver)
(v) Formic acid reduces Fehling's solution. It reduces blue coloured cupric IONS to red coloured cuprous ions.
`HCOO^(-)+underset(("blue"))(2Cu^(2+))+5OH^(-)toCO_(3)^(2-)+underset("(red)")(Cu_(2)O)+3H_(2)O`
Discussion
85983.

Account for Nitration of aniline with "conc."HNO_3 may end up with same meta nitro product.

Answer»

SOLUTION :Nitration of ANILINE with con. `HNO_3` RESULTS in the formation of m-nitro aniline because nitric ACID is a strong acid. It protonates aniline FORMING anilinium ion `C_6H_5NH_3^+` because of positive charge on nitrogen, it is meta directive and `-NO_2` group is substituted at meta position.
Discussion
85984.

Account for p-toluidine is a stronger base than p-nitroaniline.

Answer»

SOLUTION :p-Toluidine contains a methyl GROUP which has + I EFFECT (electon withdrawing group) and due to this, p-toluidine is a stronger base than p-nitro aniline in which the nitro group is less reactive and it deactivate the benzene ring make a less basic.
Discussion
85985.

Account for : (i) Amino group is aniline iso- and p- directing in aromatic electrophilic substitution reactions. Aniline on nitration gives a substantial amount of m - nitroaniline. (ii) Aniline does not go Friedel Crafts reaction.

Answer»

Solution :(i) It is because aniline is protonated to form ANILINIUM cation, in which` NH_(3)` group is meta - directing.
(ii) It is because aniline is BASIC, can form ADDUCT with `AICI_(3)`,ELECTROPHILE cannot be GENERATED.
Discussion
85986.

Account for following : Chloromethane reacts with KCN to form ethane nitrile as main product and with AgCNto form methyl isocyanide as chief product.

Answer»

Solution :`CN^(-)`is an ambident nucleophile. In KCN, C acts as the nucleophile because of the ionic nature of K-C bond. In AgCN, N acts as the nucleophile because of covalent nature of Ag - Cbond. Hence, chloromethane reacts with KCN to form ethane nitrile as the main product while with AgCN, methyl isocyanide is the chief product.
`CH_(3)CL + KCN tounderset("Ethane nitrile")(CH_(3)CN + KCL)`
`CH_(3)Cl + AgCN to underset("isocyanide")underset("Methyl")(CH_(3)NC) + AgCl `
Discussion
85987.

Account for following : Use of DDT was banned in United States in 1973.

Answer»

SOLUTION :DDT is CHEMICALLY stable and is fat soluble. It is not metabolised easily by animals and, THEREFORE, it was BANNED in US in 1973.
Discussion
85988.

Account for following : Benzylic halides show high reactivity towards S_(N)1reaction.

Answer»

Solution :Benzyl CARBOCATION is STABILISED through resonance. Hence, it shows high REACTIVITY TOWARDS `S_(N)1`reaction.
Discussion
85989.

Account for different magnetic behaviour of hexacyanoferrate (III) and hexafluoroferrate (III).

Answer»

Solution :In `[Fe(CN)_(6)]^(3-),CN^(-)` is a STRONG ligand. Hence, electrons in the d-subshell pair up and there is only one unpaired electron left. In `[FeF_(6)]^(3-)`, `F^(-)` is a weak ligand. Hence, electrons in the 3d subshell do not pair up. There are five unpaired electrons in the 3d-subshell. (The HYBRIDISATION involved is `SP^(3)d^(2)` OUTER ORBITAL complex is formed). Hence `[FeF_(6)]^(3-)` has greater magnetic moment than `[Fe(CN)_(6)]^(3-)`.
Discussion
85990.

Account for (CH_3)_2NH requires two moler proportion of CH_2I to give the same crystalline product formed by (CH_3)_2N with one mole of CH_3I.

Answer»

Solution :`(CH_3)_2NH` (Secondary AMINE) REQUIRES 2 MOLES of `CH_3I`to give a quatermary salt (crystalline PRODUCT) whereas `(CH_3)_2N` (tertiary amine) require only one mole of `CH_3I` to give the same quaternary salt. It is due to the number of alkyl groups present in amines.
`{:((CH_3)_2-NH+2CH_2I to [(CH_3)_4]^+I^-),("Secondary amineTetramethyl ammonium iodide "):}`.
Discussion
85991.

Account for any two of the following : (a) Amines are basic substances while amides are neutral. (b) Aromatic amines are weaker bases than aliphatic amines.

Answer»

Solution :(a) In amines, alkyl group is ELECTRON releasing which increases electron density on NITROGEN making them basic whereas in amides `R-overset(overset(C)(||))(C)-"group"` is electron WITHDRAWING, therefore they are neutral.
(b) It is because aryl group is electron withdrawing which decreases electron density on nitrogen making them LESS basic whereas alkyl group is electron is electron releasing which makes alkylamines more basic. MOREOVER, the lone pair of electrons on N in aromatic amine is delocalised due to resonance with benzene ring.
Discussion
85992.

Account for any two of the following: (a) Amines are basic substances while amides are neural. (b) Aromatic amines are weaker bases than aliphatic amines.

Answer»

Solution :(a) In amines alkyl GROUP is an electron releasing group which increases the electron DENSITY on nitrogen thus making them basic whereas in amides `R-overset(O)overset(||)(C)-` group is electron withdrawijg, therefore they are neutral.
(b) It is because aryl group is an electron withdrawing group which DECREASES electron density on nitrogen atom, making them less basic whereas alkyl group is electron releasing whcih makes alkylamines more basic.
Discussion
85993.

According VSEPR theory, in the electronic geometry of the molecule(s), electron pair tend to minimise repulsion and follow the repulsion order in presence of lone pair of electrons with bond pair of electrons. lp-lpgt lp-bp gtbp-bp and similarly double bond pair of electrons follow the repulsion order with angle bond pair of electrons. double bond -double bond gt double bond -single bond gtsingle bond -single bond Which of the following statement is CORRECT about XeO_(2)F_(2)

Answer»

`XeO_(2)F_(2)` is `sp^(3)d` hybridized and has lone pair of ELECTRON at axial position of its electronic GEOMETRY
`XeO_(2)F_(2)` has maximum 10-lone pair of electrons
`XeO_(2)F_(2)` has distored SEE -saw shape.
`XeO_(2)F_(2)` contains `dpi-dpi` bond.

Answer :A
Discussion
85994.

According VSEPR theory, in the electronic geometry of the molecule(s), electron pair tend to minimise repulsion and follow the repulsion order in presence of lone pair of electrons with bond pair of electrons. lp-lpgt lp-bp gtbp-bp and similarly double bond pair of electrons follow the repulsion order with angle bond pair of electrons. double bond -double bond gt double bond -single bond gtsingle bond -single bond which of the following molecular geometry is distored geometry from their ideal geometry.

Answer»

`PCl_(3)F_(2)`
`SOF_(4)`
`XeF_(5)^(-)`
`XeO_(3)F_(2)`

ANSWER :A
Discussion
85995.

According to which one of the following laws it is indicated that when two or more gases react with one another, their volumes bear a simple ratio ?

Answer»

LAW of mass action
Law of multiple proportions
Law of RECIPROCAL proportions
Law of COMBINING volumes

Solution :Law of multiple PROPORTION.
Discussion
85996.

According to werner, the arrangement of 6 groups around the metal centre within the co-ordination sphere can lead to three possible geometries (For Q. 1 and Q. 2) : planar hexagon, trigonal prism and octahedral. For a particular geometry, if the groups (i.e. ligands) are arranged in different ways then the possibility of geometrical isomerism will arise. A complex [PtCl_(2)(NH_(3))_(4)]^(2+) in RENREW'S universe has a total 3 stereo-isomers possible. If Werner was present, then, according to him, what can be the probable shape of the complex ?

Answer»

PLANAR HEXAGON
TRIGONAL prison
octahedron
both (a) and (B)

SOLUTION :
Discussion
85997.

According to werner, the arrangement of 6 groups around the metal centre within the co-ordination sphere can lead to three possible geometries (For Q. 1 and Q. 2) : planar hexagon, trigonal prism and octahedral. For a particular geometry, if the groups (i.e. ligands) are arranged in different ways then the possibility of geometrical isomerism will arise. If in RENREW'S universe, [PtBrCl(NH_(3))Py] complex was found to have two stereo-isomers, then according to Werner, which inference can be correct ? (P) Complex must be square planar (Q) Complex may be tetrahedral (R ) Complex may show geometrical isomerism (S) Complex must exhibit optical isomerism

Answer»

<P>P, Q
Q, R
Q, S
P,S

Solution :Mabcd can SHOW G.I. if PLANAR. ALSO, it can show O.I. if tetrahedral.
Discussion
85998.

According to Werner's theory, the primary valencies of the central metal atom

Answer»

are satisfied by negative IONS are neutral molecules
are satisfied by negative ions
are equal to its coordination number
decide the geometry of the complex.

Solution :According to Werner.s theory, the primary VALENCIES of the central metal ATOM are satisfied by negative ions.
Discussion
85999.

According to VSEPR theory. The most probable shape of the molecule having 4 electron pairs around the central atom is:

Answer»

nexahedral
tetrahedral
octahedral
linear

Answer :B
Discussion
86000.

According to VSEPR theory, in which species do all the atoms lie in the same plane? 1CH_(3)^(+) 2 CH_(3)^(-)

Answer»

1 only
2 only
both 1 and 2
NEITHER 1 or 2

Solution :
Discussion