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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.
| 251. |
Inhalation of carbon monoxide may cause death becauseA. it produces poisonous `HCN` inside the bodyB. it damages proteins of the body by cleaving peptide bondsC. it combines with the iron in hemoglobin and blocks the intake of oxygen by blood.D. it burns the lungs inside |
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Answer» Correct Answer - C `CO` is very poisonous because it has a `300 `- fold greater affinity for blood hemoglobin than does oxygen: Hemoglobin + `CO rarr` Carboxyhemoglobin `"Hemoglobin"+O_(2)underset(larr)(rarr)"Oxyhemoglobin"` Thus, quite low concentrations of `CO` in air are sufficient to prevent oxygen absorption in the lungs. Without a continuous supply of oxygen, the brain loses consciousness and death follows (due to asphyxiation) unless the supply of oxygenated hemoglobin is restored. |
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| 252. |
Carbon monoxide can be estimated by reaction withA. ammoniacal solution of cuprous chlorideB. (2) ammoniacal solution of silver nitrateC. `PdCl_(2)`D. `I_(2)O_(5)` |
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Answer» Correct Answer - D `5CO(g) + I_(2)O_(5) rarr I_(2)(s) + 5CO_(2)(g)` The iodine which is produced quantitatively is titrated with standard sodium thiosulphate solution. `CO` is readily absorbed by an ammoniacal solution of `CuCl` to give `CuCl . CO . 2H_(2)O`. It reduces an ammoniacal solution of `AgNO_(3)` to silver (black). In the absence of other gaseous reducing agents, this serves as a test for the gas. It also reduces `PdCl_(2)` to `Pd` `PdCl_(2) + CO + H_(2)O rarr Pd + CO_(2) + 2HCl` |
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| 253. |
Ammonia can be dried by `:`A. Conc. `H_(2)SO_(4)`B. `P_(4)O_(10)`C. CaOD. Anhydrous `CaCl_(2)` |
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Answer» Correct Answer - C Ammonia can be dried over CaO. |
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| 254. |
`H_(2)S` is a stronger acid than `H_(2)O`. Explain |
| Answer» Hydrogen sulphide `(H_(2)S)` is a stronger acid `(K_(a)=1.0xx10^(-7))` than water `(K_(a)=1.0xx10^(-14))` because S-H bond is longer than O-H bond. Its bond dissociation enthalpy is therfore, less as compared to that of O-H bond in `H_(2)O`. | |
| 255. |
What happens when aluminium is treated with conc. `HNO_(3)`? |
| Answer» Protective oxide layer | |
| 256. |
Which has the maximum electropositive character ?A. `Ba gt AI gt Ga gt In gt TI`B. `B lt AI lt Ga lt In lt TI`C. `B lt AI gt Ga lt In gt TI`D. `B lt AI gt Ga gt In gt TI` |
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Answer» Correct Answer - D As we move from B to AI, the sum of `Delta_(i) H_(1) + Delta_(i) H_(2) + Delta_(i) H_(3)` decreases substantially (6887 kJ `mol^(-1)` to 5137 kJ `mol^(-1)`)due to increase in the atomic size and hence AI has a high tendency to lose electrons. Since the electrode potentials increases from AI to TI therefore, their electropositive character decreases, i.e., AI (-1.66 V) to Ga (-0.56 V) to In (-0.34 V) to TI (+ 1.26 V) accordingly. |
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| 257. |
Which of the following is the strongest oxidising agent?A. `Br_(2)`B. `I_(2)`C. `Cl_(2)`D. `F_(2)` |
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Answer» Correct Answer - 4 The correct order of oxidizing power is `F_(2) gt Cl_(2) gt Br_(2) gt I_(2)` Fluorine is the strongest oxidising agent . There are two main reason for this, 1. `F_(2)` has a low enthalpy of dissociation which arises from the weakness of the `F-F` bond. 2. `F^(-)` has a high enthalpy of hydration which arises from the high charge density of `F^(-)` ion due to its smaller size. |
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| 258. |
The strongest oxidising agent among the following isA. `ClO_(4)^(-)`B. `BrO_(4)^(-)`C. `lO_(4)^(-)`D. `ClO_(3)` |
| Answer» Correct Answer - B | |
| 259. |
Which of the following statement given below is incorrect?A. `ONF` is isoelectronic with `O_(2)N^(-)`B. `OF_(2)` is an oxide of fluorineC. `Cl_(2)O_(7)` is an anhydride of perchloric acidD. `O_(3)` molecule is bent |
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Answer» Correct Answer - 2 Fluorine forms two binary compounds with oxygen I.e, `OF_(2)` and `O_(2)F_(2)` However they are called as oxygen fluroides rather than oxides of fluoride since the electronegativity fo `F` is higher than that of `O`. (1) Both `ONF (8+7+9=24)` and `NO_(2)^(-) (7+8+8+1=24)` are isoelectronic. (2) as they have some `MO` configuration due to same total number of electrons. (3) `Cl_(2)O_(7)` is an anhydride of perchloric acid `Cl_(2)O_(7)+H_(2)Orarr2HClO_(4) overset(-H_(2)O)underset(Delta)rarrCl_(2)O_(7)` `O_(3)` molecules (steric number : `2+1`) is bent `//`angular in shape. |
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| 260. |
Which of the following trihalides of `P`is least acidic?A. `PI_(3)`B. `PBr_(3)`C. `PCl_(3)`D. `PF_(3)` |
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Answer» Correct Answer - 1 The trihalides of `P, As` and `Sb` (esecially the fluorides and chlorides ) show acceptor porperties due to the presence of d-orbitals i.e. they can accept an electrons pair and can act as Lewis acids: `PF_(3)+F_(2)rarrPF_(5)` `PCl_(3)+Cl_(2)rarrPCl_(3)` `SbF_(3)+2F^(-)rarr[SbF_(5)]^(2-)` However, the Lewis acid strength decreases in the order: `PCl_(3) gt AsCl_(3) gt SbCl_(3)` (due to increasing size of `d` orbitals).`PF_(3) gt PCl_(3) gt PBr_(3) gt PI_(3)` (due to increasing siza of halogen). |
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| 261. |
Which of the following trihalides has the maximum bond angle?A. `PI_(3)`B. `PBr_(3)`C. `PCl_(3)`D. `PF_(3)` |
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Answer» Correct Answer - 4 Like ammonia, trihalides have pyramidal shapes, i.e. the central atom is `sp^(3)` hybridized. There of the four `sp^(3)`-orbitals from `sigma`-bonds with halgon atoms while the fourth `sp^(3)`-orbital contains the lone pair of electrons. The bond angles of the trihalides of an element decrease as the electronegativity of the halogen increasese, i.e. `underset((96.3^(@)))(PF_(3)) lt underset((100.4^(@)))(PCl_(3)) lt underset((101^(@)))(PBr_(3)) lt underset((102^(@)))(PI_(3))` As the electronegativity of halgen increases , the bond pairs move away from the central `P` atom. As a result lone pairs -bond hair replusions increase while bond pair-bond pair repulsions decreases .Consequently the bond angles increase. |
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| 262. |
Give two use of `Cl_(2)` |
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Answer» (i) In manufacturing `C Cl_(4),DDT,CHCl_(3)` etc. (ii) Sterilising drinking water. |
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| 263. |
Which of the following is not correct about graphite?A. Graphite cleaves easily between the layers.B. The density if graphite is higher than that if diamond.C. Graphite is an excellent lubricant.D. In graphite, the conduction of electricity can occur in a sheet, but not from one sheet to another. |
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Answer» Correct Answer - B The wide spacing of sheets in graphite implies that the atoms do not pack together to fill space very effectively. Thus, the density of graphite `(2.22 g cm^(-3))` is lower than that of diamond `(3.51 g cm^(-3))`. `(1)` Bonding between the different layers is weak. Thus, graphite cleaves easily between the layers which accounts for the remarkable softness of the crystals. `(3)` By virtue of the ability of sheets of `C` atoms to slide over one another, graphite is use as a lubricant, either on its own or in graphite oil. `(4)` The conductivity in the plane of the sheets is about `5000` times greater than that at right angles to the sheets. |
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| 264. |
A lead pencil contains____mixture.A. lead-clayB. charcoal-clayC. graphite-clayD. carbon back-clay |
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Answer» Correct Answer - C Graphite is used as graphite - clay mixtures in lead pencils. The higher the proportion of clay, the "harder" the pencil. The common mixture is designated `'HB'`. the higher - clay (harder) mixtures are designated by various `'H'` numbers, for example, `'2H'`, the higher - graphite (softer) mixtures are designated by various `'B'` numbers. |
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| 265. |
Bond angle in `H_(2)O(104.5^(@))` is higher than the bond angle of `H_(2)S (92.1^(@))`. The difference is due to A. O is diatomic and S is tetra-atomicB. difference in electronegativity of S and OC. difference in oxidation states of S and OD. difference in shapes of hybrid orbitals of Sand O. |
| Answer» Correct Answer - B | |
| 266. |
Graphite can be converted into synthetic diamond underA. very high pressure and at high temperatureB. very high pressure and at low temperatureC. low pressure and at high temperatureD. low pressure and at low temperature |
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Answer» Correct Answer - A The density of diamond `(3.5 g cm^(-3))` is much greater than that of graphite `(2.2 g cm^(-3))`. So a simple application of the Le Chatelier principle indicates that diamond formation from graphite is favored under the conditions of high pressure because on increasing pressure, the equilibrium shifts in that direction which results in a decrease of volume or an increase of density. Further more, to overcome the considerable activation energy barrier accompanying the rearrangement of covalent bonds, high temperatures are also required. |
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| 267. |
Name of the synthetic radioactive element of group 16 having atomic number 116 isA. LivermoriumB. TennessineC. LivernoriumD. Moscovium |
| Answer» Correct Answer - A | |
| 268. |
Fill in the blanks by choosing an appropriate option. (i) "--------" is a synthetic radioactive element of group 15 having electronic configuration (ii) "----------"A. `{:(""(i)," "(ii)),(""_(155)Mc, [Rn]5f^(14)6d^(10)7s^(2)7p^(3)):}`B. `{:(""(i)," "(ii)),(""_(155)Mc, [Xe]5f^(14)6d^(10)7s^(2)7p^(3)):}`C. `{:(""(i)," "(ii)),(""_(116)Lv, [Rn]5f^(14)6d^(10)7s^(2)7p^(4)):}`D. `{:(""(i)," "(ii)),(""_(114)Fl, [Rn]5f^(14)6d^(10)7s^(2)7p^(2)):}` |
| Answer» Correct Answer - A | |
| 269. |
Which of the following group -14 elements is a radioactive element?A. FleroviumB. GermaniumC. NihoniumD. Gallium |
| Answer» Correct Answer - A | |
| 270. |
Carbon forms a large number of compounds due to :A. tetravalency of carbonB. strong catenation property of carbonC. allotropic property of carbonD. non-matallic character of carbon |
| Answer» Correct Answer - B | |
| 271. |
Carbon shows a maximum covalency of four whereas other members can expand their cavalence due toA. absence of d-orbitals in carbonB. ability of carbon to form `p pi - p pi` multiple bondsC. small size of carbonD. catenation of carbon. |
| Answer» Correct Answer - A | |
| 272. |
Which of the following bonds will be most polar?A. `N-F`B. `N-N`C. `N-Cl`D. `O-F` |
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Answer» Correct Answer - 1 Because electronegaivity difference is maximum in `N-F` bond. |
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| 273. |
Maximum strength of gtdrogen bonding is shown byA. `H_(2)S`B. `HF`C. `H_(2)O`D. `NH_(3)` |
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Answer» Correct Answer - 2 Because `F` atom has the smallest size and highest electronegativity. |
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| 274. |
which of the following quickely absorbs oxygen?A. Turpentine oilB. Cinnamon oilC. Alkaline solution of pyragollolD. Carbon tetrachlorde |
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Answer» Correct Answer - 3 Oxygen can be removed from a mixture of gases by dissolving it in an alkaline solution of pyrogallol. Other solvents inculding galcial acetic acid readily dissolve ozone gas. |
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| 275. |
Nitrogen is used to fill electric bulbs becauseA. it is lighter than airB. it makes the bulb to glowC. it does not support combustionD. it is non-toxic. |
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Answer» Correct Answer - C `N_(2)` does not support combustion and thus, it is filled in electric bulbs. |
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| 276. |
Ammonia is used in detection of `Cu^(2+)` ion becauseA. aqueous solution of `NH_(3)` reacts with `Cu^(2+)` ion to form deep blue coloured complexB. `NH_(3)` reacts with `Cu^(2+)` ion to give blue precipitate of CuOC. aqueous solution of `NH_(3)` reacts with `Cu^(2+)` ion to form white coloured complexD. `NH_(3)` reacts with `Cu_(2+)` ion to give green precipitate. |
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Answer» Correct Answer - A `Cu_((aq))^(2+)+4NH_(3(aq))rarrunderset("Deep blue")([Cu(NH_(3))_(4)]_((aq))^(2+))` |
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| 277. |
When orthophosphoric acid is heated to `600^(@)C` the product formed isA. `HPO_(3)`B. `H_(3)PO_(3)`C. `P_(2)O_(5)`D. `PH_(3)` |
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Answer» Correct Answer - 1 When heated at `250^(@)C` it yields pyrophosphoric acid: `2H_(3)PO_(4) overset(250^(@)C)rarrH_(4)P_(2)O_(7)+H_(2)O` On further heating, it yields metaphosphoric acid. `H_(3)PO_(4)overset(Delta)underset(600^(@)C)rarrHPO_(3)+H_(2)O` or `H_(4)P_(2)O_(7) overset(Delta)rarr2HPO_(3)+H_(2)O` It yields `P_(4)O_(10)`, when strongly heated at red heat: `4H_(3)PO_(4)overset(Delta)rarrP_(4)O_(10)+6H_(2)O` |
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| 278. |
Reduction of hydrazine with zinc and `HCI` givesA. `N_(2)O`B. `N_(2)`C. `NH_(2)OH`D. `NH_(3)` |
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Answer» Correct Answer - 4 In acidic solutions, hydrazine usually behaves as a mild reducing agent, though powerful reducing agents can reduce to `N_(2)H_(4)` to `NH_(3)` ,thus causing `N_(2)H_(4)` to be oxidized. `overset(-2)N_(2)H_(4)+Zn+HClrarroverset(3-)(2NH_(3))+ZnCl_(2)` |
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| 279. |
Which of the following compound gives a black precipitate with ammonia solution?A. `AgCl`B. `HgCl_(2)`C. `CuSO_(4)`D. `Hg_(2)Cl_(2)` |
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Answer» Correct Answer - 4 Ammonia forms a grey precipitate withmercurous chloride: `Hg_(2)Cl_(2)+2NH_(4)OHrarrubrace(Hg+HgNH_(2)Cl)_("Grey")+NH_(4)Cl+2H_(2)O` it forms a white precipitate with mercuric chloride: `Hg_(2)Cl_(2)+2NH_(4)OHrarr underset("mercuric chloride")underset("Amido")(HgNH_(2)Cl)+NH_(4)Cl` |
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| 280. |
which of the following compounds is formed when `ZnSO_(4)`is treated with liquor ammonia?A. `Zn(OH)_(2).Zn(NH_(3))_(4)SO_(4)`B. `Zn(OH)_(2).ZnSO_(4).NH_(3)`C. `[Zn(NH_(3))_(4)]SO_(4)`D. `[Zn(NH_(3))_(6)]SO_(4)` |
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Answer» Correct Answer - 3 Zinc hydroxide is formed which dissolved in the form of complex in excess of `NH_(4)OH.` `ZnSO_(4)+2NH_(4)OHrarrZn(OH)_(2)+(NH_(4))_(2)SO_(4)` `Zn(OH)_(2)+(NH_(4))_(2)SO_(4)+2NH_(4)OHrarr[Zn(NH_(3))_(4)]SO_(4)+4H_(2)O` |
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| 281. |
Producer gas is made by blowing____through red hot cokeA. `N_(2)`B. airC. `O_(2)`D. steam |
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Answer» Correct Answer - B `ubrace(C+O_(2)+4N_(2))_("Air")rarrCO_(2)underset("Producer gas")(underset(2CO+4N_(2))underset(darr+C)(+)4N_(2))` |
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| 282. |
The molecular having the smaleest bond angle isA. `NCl_(3)`B. `AsCl_(3)`C. `SbCl_(3)`D. `PCl_(3)` |
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Answer» Correct Answer - C As the electronegativity of the central atom decreases, the bond angle also decreases. |
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| 283. |
A very wide range of oxidation states is shown by the elementA. `Sb`B. `As`C. `P`D. `N` |
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Answer» Correct Answer - 4 In the case of nitrogen , a very wide ranfe of oxidation state exisit as it can form compounds with less electronegative as well as more electronegative elements: `-3` in `NH_(3), -2` in `N_(2)H_(4), -1 NH_(2)OH, 0` in `N_(2), +1` in `N_(2)O +2` in `NO, +3` in `HNO_(2), +4` in `NO_(2)` and `+5` in `HNO_(3)` |
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| 284. |
Which of the following solids sublimes on heating?A. PhosphorusB. BismuthC. AntimonyD. Arsenic |
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Answer» Correct Answer - 4 Arsentic sublimes at `615^(@)C`. The boiling points, in general increases from top to bottom in the group but the melting point increases upto arsenic and then decreases up to Bismuth ,Thus arsenic has the highest boiling point while bismuth has the highest boiling point. Also note that `Sb` and `Bi` have the fairly charcateristic long liquid range. (`Sb: 631^(@)C` to `1387^(@)C` while `Bi: 271^(2)C` to `1564^(@)C`). |
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| 285. |
Which of the following Group `15` elements do not show allotropy?A. `N` and`As`B. `N` and `Sb`C. `N` and `Bi`D. `As` and `Sb` |
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Answer» Correct Answer - 3 Except `N` and `Bi` , all the group `15` elements exhibits allotropy. The allotroph of phosphorous is rather complex but essentially , there are three allotropic forms known as white, red and black phosphorous . Arsenic and antimony also exhibit allotrophy , an unstable form of each element ,structurally similar to that of white phosphorus being formed by rapid condensation of their capours. this unstable form is readily transformed into a denser allotrope which is metallic and similar in structure to that of black phosphorus. Bismuth does not exhibit allotroph and adopts this latter structure(i.e. the layer structure). Bismuth is unusual, because the expands when it forms the silid. This unsual behaviour is also found with `Ga` and `Ge` . |
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| 286. |
which of the following is not true about structure of carbon dioxide?A. In `CO_(2)` carbon is sp-hybridisedB. C forms two sigma bonds one with each oxygen atom and two `p pi - p pi` bondsC. `CO_(2)` is a liner covalent compoundD. it is a polar molecule |
| Answer» Correct Answer - D | |
| 287. |
Carbon dioxide gas is detected by its action onA. lime waterB. baryta waterC. both (1) and (2)D. blue litmus |
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Answer» Correct Answer - C To identify `CO_(2)`, the lime water `[Ca(OH)_(2)]` test or baryta water `[Ba(OH)_(2)]` test is used. In this test, a gas is bubbled into a saturated solution of calcium hydroxide (lime water) or barium hydroxide (baryta water). When the gas is `CO_(2)`, the lime water or baryta water turns milky as a white precipitate of `CaCO_(3)` or `BaCO_(3)` is formed: `CO_(2)(g) + Ca(OH)_(2)(aq.) rarr CaCO_(3)(s) + H_(2)O(l)` `CO_(2)(g) + Ba(OH)_(2)(aq.) rarr BaCO_(3)(s) + H_(2)O(l)` The additon of more `CO_(2)` results in the disappearance of precipitate as soluble calcium hydrogen carbonate or barium hydrogen carbonate forms: `CO_(2)(g) + CaCO_(3)(s) + H_(2)O(l) rarr Ca^(2+)(aq.) + 2HCO_(3)^(-)(aq.)` |
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| 288. |
Which of the following is the electron-deficient molecule?A. `C_(2)H_(6)`B. `SiH_(4)`C. `SnCl_(4)`D. `B_(2)H_(6)` |
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Answer» Correct Answer - D `B_(2)H_(6)` is electron deficient because `16 e^(-)`s are required to form `8` normal covalent bonds but only `12 e^(-)s` are available from the `2 B` and `6 H` atoms. |
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| 289. |
Which of the following molecule has trigonal planner geometry?A. `BF_(3)`B. `BCl_(3)`C. `BI_(3)`D. All of these |
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Answer» Correct Answer - D In boron trihalides `(BX_(3))`, the steric number of central `(B)` atom is three `(3 + 0)`. Therefore, the molecule has trigonal planar geometry. |
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| 290. |
What is the number of valence electrons in iodine monochloride ?A. 12B. 14C. 16D. 18 |
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Answer» Correct Answer - B `: overset(* *)underset(* *)I : overset(* *)underset(* *)Cl : 7+7=14` |
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| 291. |
Which of the following compounds does not give `N_(2)`on heating?A. `NaN_(3)`B. `(NH_(4))_(2)Cr_(2)O_(7)`C. `NH_(4)NO_(2)`D. `KNO_(3)` |
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Answer» Correct Answer - 4 `2KNO_(3)overset(Delta)rarr2KNO_(2)+O_(2)` `2NaN_(3)overset(Delta)rarr2Na+3Na_(2)` `Ba(N_(3))_(2)overset(Delta)rarrBa+3N_(2)` `(NH_(4))_(2)Cr_(2)O_(7)rarrN_(2)+4H_(2)O+Cr_(2)O_(3)` `NH_(4)NO_(2)rarrN_(2)+2H_(2)O` |
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| 292. |
convalent and ionic (in a particular state )radii increase in size down that group. Only a small increase in covalent radius is observed as one moves fromA. `N` to`P`B. `P` to`As`C. `As` to`Sb`D. `Sb` to`Bi` |
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Answer» Correct Answer - 4 There is a considerable increase in covalent radius from `N`to `P` (`70` pm to `110`pm).However ,from `Sb`to `Bi`only a small increase in covalent radius is observed (`141`pm to `148`pm.) This is due to the presence of completely filled `d` and `f` subshells in `Bi` |
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| 293. |
If the atomic number if `C` is `6`, then the atomic number of `Sn` will beA. `50`B. `45`C. `35`D. `40` |
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Answer» Correct Answer - A It is `6 + 8 + 18 + 18 = 50`. |
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| 294. |
The minimum increase in covalent radius is fromA. `C` to `Si`B. `Si` to `Ge`C. `Ge` to `Sn`D. `Sn` to `Pb` |
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Answer» Correct Answer - D Due to the presence of full `4f` subshell (in `Pb`) which shields the nuclear charge ineffectively. |
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| 295. |
Which fo the following metal is rendered passive by the action of highly concentrated nitric acid (~80%)?A. `Fe`B. `Al`C. `Ni`D. All of these |
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Answer» Correct Answer - 4 Nitric acid of any concentration has no effect on very unreactive metals such as gold and platinum. The concentrated acid has no effect either on aluminimum ,iron or nickel ,this is unexpected .it is thought that the surface of these three metals is first attacked,a thin impervious oxide film being formed which stops further action, the phenomenon is referred to as passivity. Other metals which become passive are cobalt and chromium |
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| 296. |
Oleum isA. fuming `H_(2)SO_(4)`B. fuming `HNO_(3)`C. fuming `HCl`D. fuming `H_(2)CO_(3)` |
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Answer» Correct Answer - 1 Oleum (also called fumisng sulphuric acid ) is sulphuric acid cotaining and excess of sulphur trioxide. For example `20%` oleum contains `20% SO_(3)` and `80% H_(2)SO_(4)` .It is extermely corrosive and contains some disulphuric `(Vi)` acid or pyrosulphuric acid .It is used for sulphonation and nitration. |
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| 297. |
Which is used in the laboratory for last drying of neutral gases?A. Anhyd `CaCl_(2)`B. `Na_(3)PO_(4)`C. `P_(2)O_(5)`D. Activated charcoal |
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Answer» Correct Answer - 3 It is the most effective drying or dehydrating agent below `100^(@)C` |
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| 298. |
The reasonating structure of `NO`are represented byA. `.overset(..)N=overset(..)underset(..)O:harr.overset(+)overset(..)N equiv overset(-)overset(..)O`B. `:overset(.)N=overset(..)underset(..)Oharr:overset(-)overset(..)N=overset(+)overset(.)O:`C. `N=overset(..)O:harr:N=overset(..)O`D. `:N=O:harr:Nequivoverset(..)O` |
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Answer» Correct Answer - 2 The structure of the molecules is a reasonance hybrid of the two forms: `:overset(.)N=overset(..)O:harr:overset(-)overset(..)N=overset(+)overset(.)O:` The bond length `N-O` is `115`pm, which is intermediate between a double and a triple bond. Nitric oxide readily loses its unpaired electron from the antibonding orbital to form the nitrosyl ion `(NO^(+))`, which is diamagnetic and has a shorter`N-O` bond length `(106 p m)` then that of the parent molecules `(115 p m)`. This triple-bonded (bond order is `3`) ion is isoelectronic with `CO,` and `N_(2)`. It forms many analofous metal complexes. |
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| 299. |
The number of valence electrons in nitric oxide is`11`. Among these,one unpaired electron occupaies a`//`anA. antibonding `sigma` orbitalB. bonding `pi` orbitalsC. bonding `sigma` orbitalsD. antibonding `pi` orbitals |
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Answer» Correct Answer - 4 it is colourless neutral paramagnetic gas its molecular orbital diagram resembles that of carbon monoxide ,but with one additional electron that occupies an antibounding orbital. Hence the prediated net bond order is `(2.5)`. |
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| 300. |
Laboratory grade concentrated nitric acid is a `"____"` azeotropic mixtureA. `50%`B. `68%`C. `98%`D. `33%` |
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Answer» Correct Answer - 2 Ostwald process gives a `HNO_(3)` solution of concentration `60%` by weight. The product can be concentrated to`68%` by distillation ,when a constant boiling mixture is obtained. More concentrated acid, which is needed for certain uses such as nitration can be made by distilling this mixture with sulphuric acid (a dehydrating agent). Laboratory grade nitric acid contains `68%` of the `HNO_(3)` by mass and has a density of `1.504 g cm^(-3)` |
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