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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1951. |
Complete the folllowing. `XeF_(2)+H_(2)O to` |
| Answer» `2XeF_(2)+2H_(2)O to2Xe+4HF+O_(2)` | |
| 1952. |
Which of the following properties does correspond to the order? `HI lt HBr lt HCl lt HF`A. Thermal stabilityB. Reducing powerC. Ionic characterD. Dipole moment |
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Answer» Correct Answer - B Reducing power increases in the order as H-X bong length increases from HF to HI. `HF lt HCl lt HBr lt HI ` |
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| 1953. |
Complete the folllowing. `NaF+XeF_(6)to` |
| Answer» `NaF+XeF_(6)toNa^(+)[XeF_(7)]^(-)` | |
| 1954. |
Complete the folllowing. `XeF_(2)+PF_(5)to` |
| Answer» `XeF_(2)+PF_(5)to[XeF]^(+)A["F"_(6)]^(-)` | |
| 1955. |
Complete the folllowing. `XeF_(6)+AsF_(5)to` |
| Answer» `2XeF_(6)+ASF_(5)to[Xe_(2)F_(11)]^(+)[ASF_(6)]^(-)` | |
| 1956. |
Which of the following is not true about helium ?A. It has the lowest boiling point.B. It has the highest first ionization energy.C. It can diffuese through rubber an dplastic material.D. It can form clathrate compound. |
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Answer» Correct Answer - D Due to small size of He. It escapes from interstitial spaces `//` voids of molecular lattice of quinols. |
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| 1957. |
Complete the folllowing. `XeF_(4)+SbF_(5)to` |
| Answer» `XeF_(4)+SbF_(5)to[XeF_(3)]^(+)[SbF_(6)]^(-)` | |
| 1958. |
`SbF_(5)` reacts with `XeF_(4)` to form an adduct. The shapes of cation and anion in the adduct are respectively `:`A. square planar , trigonal bipyramidalB. `T-` shaped , octahedralC. square pyramidal, octahedralD. square planar, octahedral |
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Answer» Correct Answer - 2 `XeF_(4)+SbF_(5)rarr[XeF_(3)]^(+)[SbF_(6)]^(-)rarr underset(sp^(3)d "bent" T- "shape")([XeF_(3)]^(+))+underset(sp^(3)d^(2)"octahedral")([SbF_(6)]^(-))` |
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| 1959. |
The order `HF lt HCl lt HBr lt HI ` corresponds to which of the following propertiesA. Bond lengthB. Thermal stabilityC. Ionic characterD. Dipole moment |
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Answer» Correct Answer - A A bond length increases in the order all the other three properties decreases in this order. |
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| 1960. |
Consider the following statements , `(I)` Amongst `HCl,HBr,HI` and `HOCl,HCl` is most stable to heat. `(II)` Chlorine gas if evolved when potassium chloride reacts with iodine. `(III)` The basicity of `F^(-),Cl^(-),Br^(-)` , and `I^(-)` follows the order `F^(-)gt Cl^(-) gtBr^(-) gtI^(-)` `(IV)` Sodium hypochlorite is used as bleaching and sterilising agent of these,A. `(I),(II),` and `(III)` are correctB. `(I),(II)`, and `(IV)` are correctC. `(I),(III)`, and `(IV)` are correctD. All of these |
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Answer» Correct Answer - 3 `(I)HCl` because of high bond enthalpy of it. `(II)2KCIO_(3)+I_(2)rarr 2KIO_(3)+Cl_(2)` `(III)` Correct order , `F^(-)` because of high electron density on account of small size easily donate the electron pair, is strongest base. `(IV)` It is used as bleaching as well sterilising agent ( as it acts strong oxidising agent ). |
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| 1961. |
Match List I with List II and select the correct answer using the codes given below the lists `:` `{:("List "I,,"List "II),(a.(SiH_(3))_(3)N,,i. "3 centre -2-electron bond"),(b.BF_(2),,ii. sp^(3)-"hybridization"),(c. SiO_(2),,iii.p pi-p pi " bond"),(d.B_(2)H_(6),,iv.p pi-p pi " bond"):}` Code `:`A. `{:(,a,b,c,d),((1),iv,iii,i,ii):}`B. `{:(,a,b,c,d),((2),ii,iii,iv,i):}`C. `{:(,a,b,c,d),((3),i,ii,iii,iv):}`D. `{:(,a,b,c,d),((4),iv,iii,ii,i):}` |
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Answer» Correct Answer - 4 `{:(a.(SiH_(3))_(3)N,(p pi-d pi " bond"),b.BF_(3),(p pi-p pi "bond"),),(c.SiO_(2),(sp^(3)-"hybridization"),d.B_(2)H_(6),(3"centre"-2-"electron bond"),):}` |
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| 1962. |
The structure of `N(CH_(3))` is `"………….."` while that of `N(SiH_(3))_(3)` is `"………….."`. |
| Answer» Correct Answer - pyremidal, planar | |
| 1963. |
Why `(SiH_(3))_(3)` is a weaker base than `(CH_(3))_(3)N` ? |
| Answer» This is because the lone pair of electron of `N` (present in the2p orbital ) is transferred to the empty d-orbital of Si (`ppi-dpi` overlapping) and hence it is not available for protonation . As a result, `N(SiMe_(3))_(3)` is a weaker base than `N(CH_(3))_(3)` or `N(SiH_(3))_(3)` is a stronger base than `N(CH_(3))_(3)`. | |
| 1964. |
`+2` oxidation state of lead is more stable than `+4` oxidation state . Give reasons. |
| Answer» Among the elements of group 14, carbon does not have d- or f-electrons . Therefore , it does not show inert pair effect. Consequently, it shows an oxidation state of`+4`due to the presence of two electrons in the s- and two electrons in the p-orbital of the valence shell. In contrast , all other elements from Ge to Pb contain either dor both d- and f-electron increases, the inert pair effect becomes more and more prominent. In other words, as we move dow the group from Ge to Pb, the stability of +2 oxidation state increases while that of `+4` oxidation state decreases. Therefore , the tendency of Ge, Sn and Pb to exhibit +2 oxidation state increases with increasing atomic number in group 14. | |
| 1965. |
What is the major source of helium ? |
| Answer» The major source of helium is natural gas. | |
| 1966. |
Why do noble gases form compounds with fluorine and oxygen only ? |
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Answer» Noble gases form compounds with flourine and oxygen only. Reason : Oxygen and Fluorine are most electronegative elements. |
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| 1967. |
Which of the following structure is smaller to graphite?A. `B`B. `B_(4)C`C. `B_(2)H_(6)`D. `BN` |
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Answer» Correct Answer - D `BN` has hexagonal form like layer structure of graphite. Boron nitride `(BN)` is also called inorganic grphite. |
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| 1968. |
Solid `PCI_(5)` exits asA. `PCl_(5)`B. `PCl_(4)^(+)`C. `PCl_(6)^(-)`D. `PCl_(4)^(+)` and `PCL_(6)^(-)` |
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Answer» Correct Answer - D Solid `PCl_(5)` exists as `PCl_(4)^(+)` and `PCl_(6)^(-)`. |
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| 1969. |
When on excess of chlorine is treated with ammonia ,the products formed areA. `N_(2) and NCl_(3)`B. `N_(2) and HCl `C. `N_(2) and NH_(4)Cl `D. `NCl_(3) and HCl ` |
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Answer» Correct Answer - D Ammonia on reaction with excess of chlorine gives nitrogen trichloride . `NH_3+underset("Excess")(3Cl_(2))toNCl_(3)+3HCl` |
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| 1970. |
The reaction `3ClO^(-)(aq)rarrClO_(3)^(-)(aq)+2Cl^(-)(aq)` an example of :A. (A) oxidation reactionB. (B) reduction reactionC. (C) disproportionation reactionD. (D) decomposition reaction |
| Answer» Correct Answer - C | |
| 1971. |
The reaction `3ClO^(-)(aq)rarrClO_(3)^(-)(aq)+2Cl^(-)(aq)` an example of :A. oxidation reactionB. reduction oxidationC. disproportionationD. decomposition reaction |
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Answer» Correct Answer - C |
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| 1972. |
Which one is the anhydride of `HCIO_(4)` ?A. `Cl_(2)O`B. `ClO_(2)`C. `Cl_(2)O_(6)`D. `Cl_(2)O_(7)` |
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Answer» Correct Answer - D |
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| 1973. |
From `B_(2)H_(6)`, all the following can be prepared exceptA. `B_(2)O_(3)`B. `H_(3)BO_(3)`C. `B_(2)(CH_(3))_(6)`D. `NaBH_(4)` |
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Answer» Correct Answer - C `{:(B_(2)H_(6)+3O_(2)toB_(2)O_(3)+3H_(2)O+Heat),(B_(2)H_(6)+6H_(2)OtoH_(3)BO_(3)+6H_(2)),(2NaH+B_(2)H_(6)overset(ether)toNaBH_(4)):}` |
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| 1974. |
HI can be prepared by all the following methods except :A. `Pl_3 + H_2O`B. `Kl + H_2SO_4`C. `H_2+I_2 overset(Pt)to`D. `I_2+H_2S` |
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Answer» Correct Answer - B |
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| 1975. |
From `B_(2)H_(6)`, all the following can be prepared exceptA. `H_(3)BO_(3)`B. `[BH_(2)(NH_(3))_(2)]^(+)[BH_(4)]^(-)`C. `B_(2)(CH_(3))_(6)`D. `NaBH_(4)` |
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Answer» Correct Answer - 3 `CH_(3)` group being larger can not form a bridge between two small sized boron atoms. |
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| 1976. |
What happens when `H_(2)PO_(3)` is heated ? |
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Answer» The oxidation state of P in `H_(3)PO_(3)" is "+3`. Since this value is intermediate between the highest (+5) and lowest(-3) oxidation states of P, therefore, when `H_(3)PO_(3)` is heated it undergoes disproportionation to form `PH_(3)andH_(3)PO_(4)` with oxidation states of -3 and +5 respectively. `underset("Orthophosphorous acid")overset(+3)(4H_(3)PO_(3))overset("Heat")tounderset("Phosphine")overset(-3)(PH_(3))+underset("Orthophosphoric acid")overset(+5)(H_(3)PO_(4))` |
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| 1977. |
`H_(3)PO_(3)` undergoes disprotionation reaction but `H_(3)PO_(4)` does not ? Explain. |
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Answer» The maximum and minimum oxidation states of P are -3 and +5. But the oxidation state of P in `H_(3)PO_(3)` is +3. Therefore, it can increase its oxidation state to +5 in `H_(3)PO_(4)` and decrease its oxidation state to -3 in `PH_(3)`. Thus, `H_(3)PO_(3)` shows disproportionation reaction. `underset("Orthophoshorus acid")overset(+3)(4H_(3)PO_(3))overset("Heat")tounderset("Phoshine")overset(-3)(PH_(3))+underset("Orthophosphoric acid")overset(+5)(3H_(3)PO_(4))` In contrast, the oxidation state of P in `H_(3)PO_(4)" is "+5`, therefore, it cannot increase its oxidation state beyond +5 and hence it does not show disproportionation reaction. |
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| 1978. |
White phosphours reacts with chlorine and the product hydrolyses in the presence of water. Calculate the mass of HCl obtained by the hydrolysis of the product formed by the reaction of 62 g of white phosphorus with chlorine in the presence of water. |
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Answer» The chemical equations for the reactions in involved are : `{:(,P_(4)+6Cl_(2) to 4PCl_(3)),(,underline(PCl_(3) + 3H_(2)O to H_(3) PO_(3)+3HCl]xx4),(,underset(4xx31=124 g)(P_(4))+ 6Cl_(2)+12H_(2)O to 4H_(3)PO_(3)+underset (12 xx 36.5g)(12HCl)):}` Now, 124 g of white P give `HCl=12xx36.5g` `:.` 62 g of white P will give `HCl=(12xx36.5)/(124)xx62=219g` |
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| 1979. |
How does ammonia react with a solution of `Cu^(2+)`? |
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Answer» `Cu^(2+)` ions react with excess of ammonia to form a deep blue coloured complex. `Cu^(2+)(aq)+4NH_(4)OH(aq)tounderset("Tetramminecopper (II) ion (Deep blue)")([Cu(NH_(3))_(4)]^(2+))+4H_(2)O` |
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| 1980. |
Which of the following is parpmagnetic?A. `K_(2)MnO_(4)`B. `KMnO_(4)`C. `VCl_(3)`D. `K_(2)Cr_(2)O_(7)` |
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Answer» Correct Answer - A `K_(2)overset(+6)MnO_(4)` has one unpaired electron. |
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| 1981. |
Ammonia can not be prepared by :A. reduction of nitriteB. reduction of nitrateC. reduction of nitrideD. hydrolysis of amides |
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Answer» Correct Answer - C Nitride cannot be reduced any further |
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| 1982. |
Which of the following compound has maximum`S-S` linkage?A. `Na_(2)S_(2)O_(3)`B. `Na_(2)S_(2)O_(4)`C. `Mna_(2)S_(4)O_(6)`D. `Na_(2)S_(6)O_(6)` |
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Answer» Correct Answer - D `Na^(+ -)O-underset(O) underset(||)overset(O) overset(||)S-(S)_(4)-underset(O) underset(||)overset(O) overset(||)S-O^(-)Na^(+)` |
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| 1983. |
Which of the following does not have S-S linkage but have O-O linkage ?A. `S_(2)O_(8)^(2-)`B. `S_(2)O_(6)^(2-)`C. `S_(2)O_(5)^(2-)`D. `S_(2)O_(3)^(2-)` |
| Answer» Correct Answer - A | |
| 1984. |
Hydrides of oxygen and sulphur differ in physical state due toA. presence of intermolecular hydrogen bonding in the hydrides of oxygenB. more electronegativity of oxygenC. stronger S- S bonds compared to O-O bondsD. repulsion of lone pair of electrons on oxygen atoms |
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Answer» Correct Answer - A `H_2O` has intermolecular H-bonding , whereas `H_2S` does not. |
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| 1985. |
In `O_2` (O-O) the number of electrons which are paired isA. 14B. 16C. 8D. 7 |
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Answer» Correct Answer - A `O_2` has two unpaired `e^-` in antibonding molecular orbital. Therefore number of paired `e^-` is 14 |
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| 1986. |
All the elements of group 14 form tetranaldes of general formula `MX_(4)` which are compounds (except `SnF_(4)` and `PbF_(4)`) and have tetrahedral structure. The stability of these tetrahaldes dec reases as we move from C to Pb. The of `MX_(2)` increases as we move from C to Pb. Tetrahaldes are convalent compounds while dhalides are ionic. Which of the following is not acidic?A. `PCl_(3)`B. `SbCl_(3)`C. `BICl_(3)`D. `C Cl_(4)`. |
| Answer» `C Cl_(4)` is not acidic. All other chlorides form acids upon hyrdolysis. | |
| 1987. |
All the elements of group 14 form tetranaldes of general formula `MX_(4)` which are compounds (except `SnF_(4)` and `PbF_(4)`) and have tetrahedral structure. The stability of these tetrahaldes dec reases as we move from C to Pb. The of `MX_(2)` increases as we move from C to Pb. Tetrahaldes are convalent compounds while dhalides are ionic. `SiCl_(4)` is easily hydrolysed but `C Cl_(4)` is not. this is becauseA. bonding in `SiCl_(4)` is ionicB. silicon is non-metallicC. silicon can extand its coordination number beyond four but carbon cannotD. silicon can form hydrogen bonds but carbon cannot. |
| Answer» It is the correct asnwer. | |
| 1988. |
All the elements of group 14 form tetranaldes of general formula `MX_(4)` which are compounds (except `SnF_(4)` and `PbF_(4)`) and have tetrahedral structure. The stability of these tetrahaldes dec reases as we move from C to Pb. The of `MX_(2)` increases as we move from C to Pb. Tetrahaldes are convalent compounds while dhalides are ionic. Which of the following does not exist?A. `C Cl_(4)`B. `SiF_(40`C. `PbBr_(4)`D. `SnCl_(4)` |
| Answer» `PbBr_(4)` does not exist, intact, Br ions reduce `Pb^(4+)` ions to `Pb^(2+)` which are more stable according to inert pair effect also. | |
| 1989. |
Assertion (A) Heavier elements of group 15 do not form `ppi-ppi` bonds. Reason (R) Their atomic orbitals cannot have effective overiapping due to their large size. Choose the most suitable option.A. Both A and R are correct, R is the correct explanation of AB. Both A and R are correct, R is not the correct explanation of AC. A is correct, R is incorrectD. R is correct, A is incorrect |
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Answer» Correct Answer - A Heavire element of group 15 do not from `ppi-ppi` bonds as their atomic orbital are so large and diffius that they cannot have effecitve overlapping |
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| 1990. |
`{:(,"Column I",,"Column II"),((A),Bi^(3+) rarr (BiO)^(+),(p),"Heat"),((B),[AlO_(2)]^(-) rarr Al(OH)_(3),(q),"Hydrolysis"),((C ),SiO_(4)^(4-) rarr Si_(2)O_(7)^(6-),(r ),"Acidification"),((D),(B_(4)O_(7))^(2-)rarr [B(OH)_(3)],(s),"Dilution by water"):}` |
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Answer» Correct Answer - (A-q; B-s ; C-p,r; D-q,r) A-q `Bi^(3+) +H_(2)O overset("Hydrolysis")hArr underset("Bismuth oxycation")(BiO^(+)) + 2H^(+)` `B-s` `AlO_(2)^(-) + 2H_(2)O overset("Dilution")underset("with water")rarr Al(OH)_(3) + OH^(-)` C-p, r `underset("Orthosilicate")(2SiO_(4)^(4-)) + 2H^(+) overset((i)Acidification)underset((ii) Delta)rarrunderset("Pyrosilicate")(Si_(2)O_(7)^(6-)) + H_(2)O` D-q, r `B_(4)O_(7)^(2-)+ 2H^(+) overset("Acidification")rarr H_(2)B_(4)O_(7)` `H_(2)B_(4)O_(7) + 5H_(2)O overset("Hydrolysis")rarr 4H_(3)BO_(3)` |
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| 1991. |
Zeolites is the name given to some aluminesllicates of metals with the genral formula `[M_(x//n )(AlO_(2))_(x)(SiO_(2))_(3)] zH_(2)O`, where n is the charge on the metal cation while z is the number of molecules of water hydration present. The metal cations participating in the zeolite formation are usually `Na^(+),K^(+)` or `Ca^(2+)`. the zeolites in the hydrated form are used as ion-exchanges in removing hardness of water. However. if these are heated in vacuum for sometime, they low the molecules of water and get dehydrated . pores or cavities are formed at the places wehre `H_(2)O` molecules were present. Thus, they become highly porous and can act as catalysis in certain reactions. (i) Why are zeolites called shape-selective catalysis? (ii) What is the value associated with zeolites? |
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Answer» (i) Zeolites are known as shape selective catalysts because their activity is linked with pore size. In other words, they act as molecular sleves. For example, zeolite sodium alumino silicate can adsorb straight cahin hydrocarbons since their molecules can fit into the pores. the isomeric branched hydrocarbons donot fit into the pores becaue of smaller size. (ii) Zeolites act as useful chain catalysts for petro-chemical industry for carrying out cracking of certain hydrocarbons and isomerisation. For example, in the presence of zeolite `ZSM-5` (Zeolite sleve of porosity5) methanol is converted into gasoline of high octane number. |
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| 1992. |
Silicones are the synthetic organois licone polymers which contains Si-O-Si linkages. These are represented by the general formula `(R_(2) SiO)_(x)`, Here R may be alkyl or phenyl group. In general, silicones are formed by reacting Grignar reagent with `SiCl_(4)` and this is followed to hydrolysis. (i) How are linear silicones formed? (ii) How are cross- linked silicones prepared? (iii) What is the value associated with silicones? |
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Answer» (i) Linear silicones are formed by the hydrolysis of dichlorosilances followed by polymerisation. (ii) Cross-linked silicones are formed by the hydrolysis of trichlorosilanes followed by polymerisation. (iii) Silicones have wide range of applications. Silicone-rubbers retain their elasticity over a range of temperature These are used as insulating materials for electrical motorws and other appliances. Silicones are also used inmaking greases that act as lubircants in acroplnaes. |
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| 1993. |
Compound `(X)` on reduction with `LiAlH_(4)` gives a hydride `(Y)` containing `21.72 %` hydrogen along with other products. The compound `(Y)` reacts with air explosively resulting in formation of boron trioxide. Identify `(X)` and `(Y)`. Give balanced reactions involved in the formation of `(Y)` and its reaction with air. Give the structure of `(Y)`. |
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Answer» According to availbale information Compound (X) `overset(LiAIH)rarr"Hydride"(Y)` (contains 21.72% hydrogen)+other produce Hydride (Y) `overset(O_(2)("air"))rarrB_(2)O_(3)`+other products. The data shows that the compound (X) is a boron compound an composed (Y) is a hydride of boron which reacts explosively with air to form `B_(2)O_(3)` . The boron (Y) is most probably diborane `(B_(2)H_(6))` because it contains in it nearly 21.72% hydrogen which can be calculated as follows: `B_(2)H_(6)-=6H` `(2xx10.8xx6xx1)` parts ( 6xx1) parts 27.6g 6.0g % of hydrogen (H) in `B_(2)H_(6)=((6.0 g))/((27.6g))xx100=21.74%` Thus, the compound (X) is a halde of boron (e.g. `BF_(3)` or `BCl_(3)`) and the reaction involved is: ltbr. `4BCl_(3)+3LiAIH_(4)rarr3AICl_(3)+3LiCl+2B_(2)H_(6)` ltbr. [X] [Y] `B_(2)H_(6)+3O_(2)overset("Heat")rarrB_(2)O_(3)+3H_(2)O` For the structure of diborane, please consult section 11.4. |
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| 1994. |
Thermodynamically most stable allotrope of phosphorus is :A. Red B. WhiteC. BlackD. Yellow |
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Answer» Correct Answer - C |
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| 1995. |
The type of hybridisation of boron is diborane is (a) `sp` , (b) `sp^(2)` , (c ) `sp^(3)`, (d) `dsp^(2)` |
| Answer» In `B_(2)H_(6), B` is `sp^(3)`-hybridized. Therefore , option (c ) is correct. | |
| 1996. |
Boric acid is polymeric due to (a) its acidic nature , (b) the presence of hydrogen bonds (c ) its monobasic nature ,(d) its geomtry |
| Answer» Boric acid is polymeric due to the presence of H-bonds. Therefore , option b its correct. | |
| 1997. |
Thermodynamically the most stable form of carbon is (a) diamond, (b) graphite, (c ) fullerees , (d) coal |
| Answer» Thermodynamically the most stable form of carbon is graphite, i.e, option (b) is correct. | |
| 1998. |
Assertion Hydrolysis of `(CH_(3))-(2)SiCl_(2)` results in linear chain silicates. Reason Adding of some `(CH_(3))_(3)SiCl_(2)` control the molecular weight of silicones.A. Both assertion and reason are correct and reason is the correct explanation of the assertion.B. Both assertion and reason are correct but reason is not the correct explanation of assertion.C. Assertion is correct but reason is incorrectD. Assertion is incorrect but reason is correct. |
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Answer» Correct Answer - B |
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| 1999. |
Find the number of species which undergo hydrolysis at room temperature. ` C Cl_(4),SiCl_(4),BCl_(3),NF_(3),PCl_(5),PCl_(3),SF_(6),SO_(2)Cl_(2)` |
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Answer» Correct Answer - `0006` `SiCl_(4),BCl_(3),PCl_(5),PCl_(3),SO_(2)Cl_(2),AlCl_(3)` |
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| 2000. |
Find the number of reagents which would lead to unsymmetrical cleavage of diborane. `CH_(3)NH_(2),NH_(3),(CH_(3))_(2)NH,(CH_(3))_(3)N, , N_(2)H` |
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Answer» Correct Answer - `0003` `CH_(3)NH_(2),NH_(3),(CH_(3))_(2)NH` |
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