InterviewSolution
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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.
| 101. |
On fusion of a mixture of `Na-2CO_3` and ` CaCO_3` with silica at `1500^@C`, a liquid consisting of silicatees of sdium and calcium is formed. On cooling . Liquid become viscous and eventually ceasesto flow . It becomes solid and is known as flass. By varying the proportions of the three basic ingredients and by adding other substances, the properties of glass can be altered, Glass can be represented as `R_2 O. MO.6SiO_2`, where `R= Na` or `K, M = CaBa, Zn` or `pbSiO_2` may be replaced by `Akl_2 O_3 , b_2 O_3` or `P_2 O_5`. Colourd glasses are obtained by adding certain metallic oxdes or salts in the fused mass, Glass is attached by `HF` and this property is used to make marking on the galss , This is known as etching . The glass on rapid coolling becomes brittle and fragile . The artciles of glass are cooled neither slowly nor very rapidly . The process of graducal cooling of flass iscalled annealing . The acid that cannot be stroed in galss is .A. `HF`B. `Hcl`C. HBr`D. Hl` |
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Answer» Correct Answer - A `6HF + Na_2SiO_3 rarr Na_2SiF_6 + 3H_2O`. |
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| 102. |
Water gas is amisture of ………………….. And ………………….. , |
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Answer» Correct Answer - ` CO` and N_2` `CO` amd `H_2` . |
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| 103. |
what is the bond order of carbon momoxide ? |
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Answer» Correct Answer - 3 `(sigma1s^(2) sigma^(*) 1s^2 sigma2s^(2) sigma^(*) 2s^(2) pi 2p_(x)^(2) -= pi2 p_(z)^(2) sigma_y^(2) sigma2 p_(z)^()2)` Bond order ` = (1)/(2) (N_b-N_a) = (1)/(2) (10- 4) = 3.0`. |
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| 104. |
Bond energy is highest for :A. ` Sn-Sn`B. ` C-C`C. `Si -Si`D. `Ge-Ge` |
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Answer» Correct Answer - B Due to maxium corpaticbitlity in size orbitals involved in `C-C` nond ( i.e., 2p -2p overlap ). It is maximum stable and hence maxium enrgy . |
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| 105. |
The average value of C-C bond order in graphite isA. 4/3B. 3/4C. 3/2D. 1 |
| Answer» xiv. Bond order is ` 4//3 = 1.33`. | |
| 106. |
Oftern agroundglas stropper gets stuck in the neck of a glass blttlecontinng`NaOH` solution . This is due to ,A. The presence of dirt particles in bhetweemnB. The formation of solid silicate sin-between by the reaction sof ` SiO_2` of galss with `NaOH` .C. The formation of `Na_2 CO_3` in-between by the reaction of `CO_2` of aire and ` NaOH`.D. Glass contains a born compound which forms a precipotte with the ` NaOH` solution . |
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Answer» Correct Answer - B `SiO_(2)+2NaOHrarrunderset("silicate")underset("Sodium")(Na_(2)SiO_(3))+H_(2)O` |
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| 107. |
`C-C` bond length is maximum inA. diamondB. graphiteC. naphthaleneD. fullerene |
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Answer» Correct Answer - A In diamond onlu ` C-C` bonds are present , whicle in other `C-C` and `C=C` nonds are present . Hence `C-C` bond length in maximum in diamond . |
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| 108. |
Complete the reaction ltbr. `SnCl_4 + C_2 H_5 Cl+ Na rarr` |
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Answer» Butane is formed . `Sn Cl_4 + 2 C_2 H_5 Cl + 2 Na rarr Na_2 Sn Cl-6 +C_4H_(10)`. |
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| 109. |
a. `SnCl_4` + HCl+I_2 rarr (X) +(Y) ` |
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Answer» `SnCl_2 + HCl+I_2 rarr (X) +(Y)` `SnCl_2` acts as a reducing agent and `I_2` as an osidasing agent , hence `(X)` and `(Y)` are `SnCl_4` and `HI` respectivley . `Sn Cl_2 + 2 HCl + L_2 rarr underset ((X)) (SnCl_4) + underset ((Y)) (2HI)`. |
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| 110. |
A colourless gas which burns with blue flame and reduction `Cuo` to `Cu` is.A. ` N_2`B. `CO`C. `CO_2`D. `NO_2` |
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Answer» Correct Answer - B `CuO+CO rarr Cu + CO_2`. `CO` burns with a blue falme. |
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| 111. |
Elemental carbon appears in many structural forms or allotropes . Three of these formssare crystalline -diamond ,graphite and the recentluy discovered fullerne (bucky ball ) -while more than ` 40` others including coke and carbon black ae amorphous . Now there seems to be set afourth crystalline allotrope of carbon. reported in `1995` by ` Lagow at the University of Texas . Newly discovered allotrope if carbon has the form :A. polyyneB. fullereneC. bucky ballD. none of these |
| Answer» Correct Answer - A | |
| 112. |
The basic structural unit of silicates isA. `SiO_3^(2-)`B. `SiO_4^(2-)`C. `SiO^(O+)`D. `SiO_4^(4-)` |
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Answer» Correct Answer - D `SiO_4^(4-)`. |
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| 113. |
`C` and `Si` are always tetravelent , but Ge, Sn and Pb show divalency Gvie reason . |
| Answer» Anong forup `14` elements `C` and `Si` have noble gas caore heneath the valence shell electrons wherwseas in `Ga, Sn` and `Pb` fully filled lesser shielding `d` and (or) f-orbitals are present in-=between the noble gas coare and valence shell electron . As a result inerst pair effect is exhibited by Ga, Sn and Pb and not ny `C` and `Si`. Hence `C` and (are alwyas tentravelent but `1` Ga Sn and Pb are always divlent in nature . | |
| 114. |
Give one chemical reaction to show that tin (II) is a reducing agent whereas `Pb` (II) is not . |
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Answer» Both tin and lead exhilbit two oxidation states i.e, and `+4 +2` due to inert paire effect . As inert paire effect increase down the froup `(narr)` stabillity of `+2` oxidation state increases and that of `+4` oxidation states decerases . Hence `+2` oxidations tate is mores stable in tin than in lead. Therefore tin (II) is a reducing agent as tin (II) oxidatises to tine (IV) i.e., more stabel oxdiations tate and `Pb` (II) being more stable oxidations tate will not behave as reducing agent . Example : `Sn^(2+) + 2 Fe^(3+) rarr Sn^(4+) + 2 Fe^(2+)` `Sn` (II) behaves as reducing agnet it self gets oxidised to `Sn` (IV) and reduces `Fe` (III) to `fe` (II) . |
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| 115. |
Assertion (A) : Diamond in the haredst possible substance and is a networek covalent sokid . Reason (R) : All the `C` atoms in diamond are `sp^2` hybrdised .A. If both (A) and (R) are correct and (R) is correct explanation of (A)B. If both (A) and (R) are correct and (R) is not correct explanation of (A)C. If (A) is correct , but (R) is incorrectD. If (A) is incrorect byt (R) is incroroect . |
| Answer» Correct Answer - A | |
| 116. |
`CO` is isostructural with :A. `SnCl_2`B. `HgCl_2`C. `SCl_2`D. `NO_2` |
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Answer» Correct Answer - B::C Same explanation as in `Q. 6` above . |
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| 117. |
No form fo elecmental silicon is comparabnle to graphite Give reason . |
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Answer» In graphite carbon is `sp^2` hybridese and each carbon forms convalent bonds with three carbon atoms and forms extende hexaogonal rign -Like structure in the layer . Each carbon is now left with one unpaird electron in unhyridsed `2p` orbital which undrgoes sidways ovelap to form `p pi- p pi` double bonds . Thus , graphite has a two-dimensional sheet-like (alyered) stuture . On the other hand Si is not capable of forming structure like fraphite. Due to its bigger size. exteny of overlap , which cannot form `Si = Si`. Rather. Si prefes to undrgo `sp^3` hdycridisation and hence silicon has a diamond-like three-dimensional netword structure ` (##KSV_INORG_CHM_P1_C07_S01_026_S01.png" width="80%">. |
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| 118. |
x. The hubrid state fo carbon atoms in ` C_(60)` molecule is :A. spB. `sp^(2)`C. `sp^(3)`D. `dsp^(2)` |
| Answer» ix. `C` in `C_(60)` molecule is `sp^2` hybrdised . | |
| 119. |
Compounds wich readily unfergo hydrolysis are :A. `CCl_4`B. `BCl_3`C. `SiCl_4`D. `CF_4` |
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Answer» Correct Answer - B::C ` BC l_3` and ` SiCl_4` have vacant orbital of suitable enrgy to acept a pair of electron from `H_2O` molecule and thus easily undergo hydroysis, whereas in ` CC l_4` and ` CF_(4+)` due to the absence of low lying d-orbitals , hdrolysis is not possible . ` BC l_3 + 3 H_2O rarr B(OH)_3 +3HCl` `S iCl_4+ 4H_2O rarr H_4SiO_4 +4HCl`. |
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| 120. |
Give reason : `CO` is readily absorbed by ammonitcal cuprous chloride , but not ` CO_2` . |
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Answer» Carbon monoxide `(CO)` due to the presence fo lone pair on `C` acts as a Lewis base and thus forms soluble complex in ammoniacal cuprous chloride . `Cu Cl + NH_3 : CO rarr underset ("Soluble comlex") ([Cu (CO) (NH_3) ]^(o+) Cl^(Θ))`. However ` CO_2` does not behave as a lewis base since it does not have lone paire of electron on the carbon atom and hence does not dissolve inammoniacal cuprous chlaroide . |
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| 121. |
Lead solution may be titrated with satndard ` EDTA` at `ph=6` using which indicator ?A. Methylnthymol blueB. Eriochrome Bloack `T`C. Methyl orangeD. Eosion |
| Answer» Correct Answer - A | |
| 122. |
When a lead salt is heated with sodium carbonate in charcoal cavity it . Give .A. yellow incrustationB. brownC. blackD. blue |
| Answer» Correct Answer - A | |
| 123. |
One reacently discovered allotrope of carbon `(e.g. C_(60))` is commonly known as ……………………. , |
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Answer» Correct Answer - B::C One recently discosverd allotrope of carbon `(e.g. C_(60))` is commonly known as Buckminsterfullerene . |
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