InterviewSolution
Saved Bookmarks
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.
| 1551. |
Hydrogen bromide is dried by passing the gas through:A. (A) quick lineB. anhydrous calcium chlorideC. (C) potassium hydroxide pelletD. (D)con. `H2SO_4` |
| Answer» Correct Answer - B | |
| 1552. |
`F_2` reacts with `H_2O` as follows: `F_2+H_2O to H^(+) + F^(-) +O_2` Which of the following halogens shows same reaction but In opposite direction?A. `Br_2`B. `Cl_2`C. `I_2`D. All |
| Answer» Correct Answer - C | |
| 1553. |
Chlorine water on standing loses Its colour and forms:A. (1) HCI onlyB. (2) HCI and `HCIO_2`C. (3) HCI and HOCID. (4) HOCI and `HOCl_2` |
| Answer» Correct Answer - C | |
| 1554. |
Chlorine gas is dried over:A. CaOB. NaOHC. conc. `H_2SO_4`D. dil. `H_2SO_4` |
| Answer» Correct Answer - C | |
| 1555. |
Each of the following is true for white and red phosphorus except that theyA. can be oxidised by heating in airB. are both soulube in `CS_2`C. consists of same kind of atomsD. can be converted into one another |
|
Answer» Correct Answer - B White phosphorus , is soluble in `CS_(2)` whereas red phosphorus is insoluble in it . |
|
| 1556. |
Nitrogen shows different oxidation states in the range:A. `0` to `5`B. `-3` to `+5`C. `3` to `-5`D. `-5` to `+3` |
|
Answer» Correct Answer - B Nitrogen shows difference oxidation states in the range `-3` to `5:` |
|
| 1557. |
Bond dissociation enthalpy of E-H (E=element) bonds is given below. Which of the compounds will act as strongest reducing agent ? Compound `{:("Compound",NH_(3),PH_(3),AsH_(3),SbH_(3)),(Delta_("diss")(E-H)//kJmol^(-1),389,322,297,255):}`A. `NH_(3)`B. `PH_(3)`C. `AsH_(3)`D. `SbH_(3)` |
|
Answer» Correct Answer - D Weaker the E-H bond, stronger the reducing agent i.e., `SbH_(3)`. |
|
| 1558. |
On treating `PCl_(5)` with `H_(2)SO_(4)`, sulphuryl chloride `(SO_(2)Cl_(2))` is formed as the final product. This shows that `H_(2)SO_(4)`A. has two hydroxyl groups in its structureB. is a derivative of sulphur dioxideC. is a dibasic acidD. has greater affinity for water |
|
Answer» Correct Answer - A `PCl_5+HO-underset(O)underset(||)overset(O)overset(||)S-OH toCl-underset(O)underset(||)overset(O)overset(||)S-Cl+POCl_(3)+H_(2)O` `PCl_(5)` attacks -OH group and replace it by -Cl group . Hence , reaction of `PCl_(5)` with `H_2SO_(4)` shows the presence of two -OH groups in `H_2SO_4` . |
|
| 1559. |
`NH_3` gas is dried overA. CaOB. `HNO_(3)`C. `P_(2)O_(5)`D. `CuSO_(4)` |
|
Answer» Correct Answer - A `:. NH_3 ` reacts with `P_(2)O_(5) or CaCl_(2) or "Conc" H_2SO_4`. Therefore , it cannot be dreid over either of them . `CuSO_(4) and HNO_(3) ` are not drying agent . `NH_3` can be dried over CaO only because it is a drying agent with which it does not react . |
|
| 1560. |
Which of the following acids possesses oxidising, reducing, and complex forming properties ?A. HClB. `HNO_(2)`C. `H_(2)SO_(4)`D. `HNO_(3)` |
|
Answer» Correct Answer - B `HNO_(2)`, in it oxidation number of nitrogen is `+3`(i.e.in between `-3 " to " 5`) |
|
| 1561. |
Nitrogen shows different oxidation states in the rangeA. `0 to +5`B. `-3 to +5`C. `-5 to +3`D. `-3 to +3` |
|
Answer» Correct Answer - B `-3 "to" +5` |
|
| 1562. |
Which one of the following compounds has the smallest bond angle in its molecule ?A. `OH_(2)`B. `H_(2)S `C. `NH_(3) `D. `SO_(2)` |
|
Answer» Correct Answer - B Bond angles are `OH_(2)(104^(@)),H_2S(92^(@)),NH_(3)(107^(@)) and SO_(2)(119.5^(@))`. Hence , `H_2S` has smallest bond angle among the given compound . |
|
| 1563. |
The molecule havaing smallest bond angle isA. `AsCI_(3)`B. `SbCl_(3)`C. `PCl_(3)`D. `NCl_(3)` |
|
Answer» Correct Answer - B `SbCl_(3)` has the smallest bond angle because amongst N,P, As and Sb,Sb has the lowest electronegativity. |
|
| 1564. |
Which of the following statements is wrong?A. The stability of hydrides increases from `NH_(3)` to `BiH_(3)` in group 15 of the periodic tableB. Nitrogen cannot from `dpi-ppi` bondC. N-N single bond is weaker than the P-P single bondD. `N_(2)O_(4)` has two resonance structures |
|
Answer» Correct Answer - A `underset ("Least stable")(BiH_(3)) lt SbH_(3)lt AsH_(3) ltPH_(3) ltunderset("Most stable")(NH_(3))` |
|
| 1565. |
The molecule having smallest bond angle isA. `NCl_(3)`B. `AsCl_(3)`C. `SbCl_(3)`D. `PCl_(3)` |
|
Answer» Correct Answer - C As we move down the group , the repulison of bond pair and lone pair of centre atom decreases. |
|
| 1566. |
Assertion. Silicons are hydrophobic in nature. Reason. `Si-O-Si` linkage are moisture sensitive.A. If both assertion and reason are true, and reason is the true explanation of the assertion.B. If both assertion and reason are true, but reason not is the true explanation of the assertion.C. If assertion is true, but reason is false.D. If both assertion and reason are false. |
|
Answer» Correct Answer - C Correct reason. Si atoms in silicones are surrounded by non-polar alkyl groups, which repel water molecules. |
|
| 1567. |
Dangling valency is not exhibited by which of the following allotrope of carbon ?A. GraphiteB. DiamondC. `C_(60)`D. All of these |
|
Answer» Correct Answer - B `C_(60)` is the purest allotropic form of carbon & it has a smooth structure without any dangling valency. |
|
| 1568. |
The oxidation state of iron in `[Fe(H_(2)O)_(5)NO]^(2+)` is |
|
Answer» Correct Answer - C `{:(,+2,+2,-2,+2,-2),("["Fe(H_2O)_5NO"]"^(2+)=,"["Fe,"("H_2,O")"_5,N,O"]"^(2+)):}` |
|
| 1569. |
The reaction of `P_4` with X leads selectively top `P_4O_10` . The X isA. dry `O_2`B. a mixture of `O_2` and `N_2`C. moist `O_2`D. `O_2` in the presence of aqueous NaOH |
|
Answer» Correct Answer - B `P_4+5O_2 "(excess)" underset((N_2+O_2)) overset"excess of air"toP_4O_10` |
|
| 1570. |
The reaction of `P` with `X` leads selectively to `P_4 O_6. X` isA. dry `O_(2)`B. a mixture of `O_(2)` and `N_(2)`C. moist `O_(2)`D. `O_(2)` in the presence of aqueous `NaOH` |
|
Answer» Correct Answer - B |
|
| 1571. |
`BF_(3)` isA. electron-deficient compoundsB. Lewis acidC. used as rocket fuelD. ionic compound. |
| Answer» Correct Answer - A::B | |
| 1572. |
Which one of the following is pyrophosphoric acid ?A. `H_(3)PO_(4)`B. `H_(4)P_(2)O_(7)`C. `H_(4)P_(2)O_(5)`D. `H_(3)PO_(4)` |
| Answer» Correct Answer - B | |
| 1573. |
Write a balanced chemical equation for the reaction showing catalytic oxidation of `NH_(3)` by atmosoheric oxygen. |
|
Answer» Ammonia `(NH_(3))` on catalytic oxidation by atmospheric oxygen in presence of Rh/Pt gauge at 500 k under pressure of 9 bar produces nitrous oxide. Balanced chemical reaction can be written as `4NH_(3)+underset("From air")(5O_(2))overset("Pt//Rh gauge catalyst")underset(500K,9 "bar")to4NO+6H_(2)O` |
|
| 1574. |
The gases produced in the reaction, `Pb(NO_(3))_(2)overset(Delta)toandNH_(4)NO_(3)overset(Delta)to` are respectivelyA. `N_(2)O,NO`B. `N_(2)O,NO_(2)`C. `NO,NO_(2)`D. `NO_(2),N_(2)O` |
|
Answer» Correct Answer - D `2Pb(NO_(3))_(2)overset(Delta)to2PbO+4BO_(2)+O_(2)` `NH_(4)NO_(3)overset(Delta)toN_(2)O+2H_(2)O` |
|
| 1575. |
Which of the following properties is not shown by NO ?A. Its bond order is 2.5B. It is diamagnetic in the gaseous state.C. It is a neutral oxide.D. It combines with oxygen to form nitrogen dioxide. |
|
Answer» Correct Answer - B NO exists it has odd number of electrons (7+8=15) and hence is paramagnetic. All othe statements are corrent. |
|
| 1576. |
Why the elements of the second row (first short period) show a number of differences in properties from other members of their respective families ? |
|
Answer» The differences in the properties of the first member of a group from those of the other members are due to (i) the smaller size of the atom. (ii) high ionization enthalpy. (iii) absence of d-obritals. |
|
| 1577. |
In qualitative analysis when `H_(2)S` is passed through an aqueous solution of salt acidified with dil. HCL, a block precipitate is obtainned .On boiling the precipitate with dil. `HNO_(3)` it forms a solution of blue colour . Addition of excess of aqueous solution of ammonia to this solution given ....... .A. deep blue precipitate of Cu `(OH)_(2)`B. deep blue solution of `[Cu(NH_(3))_(4)]^(2+)`C. deep blue solution of Cu `(NO_(3))_(2)`D. deep blue solution of Cu `(OH)_(2)`. Cu `(ON_(3))_(2)` |
|
Answer» Correct Answer - B In qualitative analysis when `H_(2)S` is passed through an aqueous sloution of salt acidified with dil . HCl a black ppt. of CuS is obtained . `CuSO_(4)+H_(2)Soverset(dil. HCl)tounderset("blackppt")( CuS +)H_(2)SO_(4)` On boiling CuS with dil. `HNO_(3)` it forms a blue coloured solution and the following reactions occur `3CuS+8HNO_(3)to3Cu(NO_(3))_(2)+2NO+3S+4H_(2)O` `S+2HNO_(3)toH_(2)SO_(4)+NO` `2Cu^(2+)+SO_(4)^(2-)+2NH_(3)+2H_(2)OtoCu(OH)_(2)*CuSO_(4)+2NH_(4)OH` `Cu(OH)_(2)*CuSO_(4)+8NH_(3)tounderset("Tetraammine copper"(||)("Deep blue solution"))(2[Cu(NH_(3))_(4)]SO_(4)+2OH^(-)+SO_(4)^(2-)` |
|
| 1578. |
`BF_(3) and NF_(3)` both molecules, are covalent, but `BF_(3)` is non - polar and `NF_(3)` is polar.Its reason isA. boron is a metal and nitrogen is a gas in uncombined stateB. `BF_(3)` bonds have no dipole moment whereas `NF_(3)` bond have dipole momentC. atomic size of boron is smaller than that of nitrogenD. `BF_(3)` is symmetrical molecule whereas `NF_(3)` is unsymmetrical |
|
Answer» Correct Answer - D `BF_(3)` is symmetrical planar , although it has polar bonds but resultant dipole moment is zero . In `NF_(3)` , Ione pair cause distortion , hence polarity arises . |
|
| 1579. |
A colourless inorganic salt (A) decomposes completely at about `250^(@)C` to give only two products, (B) and (C), leaving no residue. The oxide (C) is a liquid at room temperature and neutral to moist litmus paper while the gas (B) is a neutral oxide. White phosphours burns in excess of (B) to produce a strong white dehydrating agent. Write balanced equations for the reactions involved in the above process. |
|
Answer» (i) Since a colourless inorganic salt (A) on heating to `250^(@)C` decomposes to form two oxides (B) and (C) leaving no residue, therefore, compound (A) must be ammonium nitrate `(NH_(4)NO_(3))`. `underset((A))(NH_(4)NO_(3))overset(250^(@)C)tounderset((B))(N_(2)O)+underset((C))(2H_(2)O)` (ii) The fact that inorganic salt (A) is actually ammonium nitrate is confirmed by the observation that oxide (C), i.e., `H_(2)O` is a liquid at room temperature. (iii) If the oxide (C) is `H_(2)O` then other neutral oxide (B) must be nitrous oxide, `N_(2)O`. (iv) The fact that the neutral oxide (B) is actually `N_(2)O` is confirmed by the observation that white phosphorus burns in `N_(2)O` to form phosphorus pentoxide which is a strong dehydrating agent. `underset("Nitrous oxide (B)")(10N_(2)O)+P_(4)tounderset("Phosphorus pentoxide")(P_(4)O_(10))+10N_(2)` Thus, inorganic salt (A) is ammonium nitrate `(NH_(4)NO_(3))`, (B) is nitrous oxide `(N_(2)O)` and (C) is water `(H_(2)O)`. |
|
| 1580. |
(a) Classify following oxides as neutral, acidic , basic or amphoteric `CO, B_(2)O_(3), SiO_(2), CO_(2), Al_(2)O_(3), PbO_(2), Tl_(2)O_(3)` (b) Write suitable chemcial equations to show their nature. |
|
Answer» (a) Neutral oxides : CO : Acidic oxides : `B_(2)O_(3), SiO_(2), CO_(2)` Amphoteric oxides : `Al_(2)O_(3), PbO_(2)` Basic oxide : `Tl_(2)O_(3)` (b) (i) Being acidic `B_(2)O_(3), SiO_(2)` and `CO_(2)` react with alkalis to form salts. `underset("anhydride")underset("Boric")(B_(2)O_(3)) + 2NaOH rarr underset("Sod. metaborate")(2NaBO_(2) + H_(2)O) , underset("Silica")(SiO_(2))+2NaOH overset(Delta)rarrunderset(" Sod. silicates ")(Na_(2)SiO_(3))+H_(2)O` `underset(" Carbon dioxide ")(CO_(2)) + 2NaOH rarr underset(" Sod. carbonate ")(Na_(2)CO) + H_(2)O` (ii) Being amphoteric, `Al_(2)O_(3)` and `PbO_(2)` react with both acids and bases. `underset("Alumina")(Al_(2)O_(3)) + 2NaOH overset("Fuse")rarr underset(" Sod. meta-aluminate ")(NaAlO_(2)) + H_(2)O , Al_(2)O_(3) + 3H_(2)SO_(4) rarr Al_(2)(SO_(4))_(3) + 3H_(2)O` `underset(" Lead dioxide ")(PbO_(2)) + 2NaOH rarr underset(" Sod. plumbate ")(Na_(2)PbO_(3)) + H_(2)O , 2PbO_(2) + 2H_(2)SO_(4) rarr 2PbSO_(4) +2H_(2)O + O_(2)` (iii) Being basic, `Tl_(2)O_(3)` dissolves in acids. `Tl_(2)O_(3) + 6 HCl rarr 2TlCl_(3) + 3H_(2)O`. |
|
| 1581. |
In some of the reactions, thallium resembles aluminium whereas in others it resembles with group `1` metals. Support this statement by giving some evidence. |
| Answer» Aluminium (Al) generally exhibits +3 oxidation state in its compoudns. Thallium (Tl) the last member of the group -13, is expected to show oxidation states of +3 and +1 (due to inert pair effect). Thus, the metal resembles aluminium in exhibiting +3 oxidation state. Both form trichlorides `TiCl_(3)` and `AlCl_(3)` at the same time., it also resembles alkali metals of group 1 in exhibing +1 oxidation state (e.g., both thallium and sodium form monochlorides TiCL and NaCl) For further details, consult 11.2. | |
| 1582. |
a. Classify following oxides as neutral, acidic, basic or amphoteric: `CO,B_(2)O_(3),SiO_(2),CO_(2),Al_(2)O_(3),PbO_(2),Tl_(2)O_(3)`. b. Write suitable chemical reaction to show their nature. |
|
Answer» Neutral oxide: CO Acidic oxides : `SiO_(2),CO_(2),B_(2)O_(3)` Basic oxides : `Tl_(2)O_(3),PbO_(2)` Amphoteric oxide : `Al_(2)O_(3)`. |
|
| 1583. |
Acid strength of oxy acids of chlorine follows the orderA. `HClOltHClO_(2)ltHClO_(3)ltHClO_(4)`B. `HClO_(4)ltHClO_(3)ltHClO_(2)ltHClO`C. `HClO_(4)ltHClO_(3)ltHClOltHClO_(2)`D. None of these |
|
Answer» Correct Answer - B `Hoverset(+7)(ClO_(4))gtHoverset(+5)(ClO_(3))gtHoverset(+3)(ClO_(2))gtHoverset(+1)(ClO)` |
|
| 1584. |
In the contact process for industrial manufacture of sulphuric acid, some amount of sulphuric acid is used a starting material. Explain briefly. What is the catalyst used in the oxidation of `SO_2` ? |
| Answer» Correct Answer - A::B::C::D | |
| 1585. |
The number of S-S bonds, in sulpher trioxide trimer `(S_(3)O_(9))` is :A. threeB. twoC. oneD. zero |
|
Answer» Correct Answer - D |
|
| 1586. |
Explain the following : (a) Boron has high melting and boiling points. (b) The `p pi - p pi` back bonding occurs in the halides of boron and not in those of aluminimum. ( c) Boron and aluminium halides behave as Lewis acids. (d) Aluminium forms `[AlF_(6)]^(3-)` ion, but boron does not form `[BF_(6)]^(3-)` ion. |
|
Answer» (i) The `Delta_(1)H_(1)` value of `Ga(578kJ"mol"^(-1))` is slightly higher than that of `Al(577kJ "mol"^(-1))` wheras, it is expected to be less. Acutally, gallium atom has ten 3d electrons which do not exert as much screening efect as is done by a and p electrons in case of alumimum. Therefore there is an unexpected inc rease in the magnitude of effective nuclear charge as we move from aluminum to gallium and as a result, the `Delta_(1)H_(1)` of gallium is little more than that of aluminium, inspita=e of the fact that atomic size of aluminium is less. (ii) The size of boron atom is very small, in order to form `B^(3+)` ion, the element needs a very high ionisation enthalpy `(Delta_(1)H_(1)+Delta_(1)H_(2)+Delta_(1)H_(3))` This is not availab le easily. therefore, It does not form `B^(3+)` ion or the compounds of the element are not ionic, therese are generally covalent in nature. (iii) For answer, Consult concept based Queston (Q.1) |
|
| 1587. |
The correct stability order for boron halides isA. `BF_(3)gtBCl_(3)gtBBr_(3)gtBl_(3)`B. `BCl_(3)gtBF_(3)gtBBr_(3)gtBr_(3)`C. `Bl_(3)gtBBr_(3)gtBCl_(3)gtBF_(3)`D. `BBr_(3)gtBCl_(3)gtBl_(3)gtBF_(3)` |
| Answer» (a) The stability order for boron halides is explained in terms of back-bonding. | |
| 1588. |
Assign a reason for each of the following statements : (i) Ammonia is a stronger bass than pjhosphine. (ii) Sulpher in vapour state exhibits a paramagnetic behaviour : |
|
Answer» (i) Both `NH_(3)` and `PH_(3)` have a lone pair of electrons on the central atom. Hence, they act as Lewis bases. When `NH_(3)`or `PH_(3)` accepts a proton to form `NH_(4)^(+)PH_(4)^(+)`, an additional `N-H` or `P-H` bond is formed. `H_(3)N : +H^(-) to NH_(4)^(+)` `H_(3)P : +H^(-) to PH_(4)^(+)` Now, because of the smaller size and the unavailability of d-orbitals in N, the lone pair on N is less delocalised than the lone pair present on P. Thus the N-H bond is much stronger than the `P-H` bond. As a result, the ability of `NH_(3)` to accept a proton is much more than that of `PH_(3)`. Therefore, `NH_(3)` is a stronger base than `PH_(3)`. (ii) In vapour state, sulphur exists as `S_(2)` molecule. `S_(2)` molecule, like `O_(2)`, has two unpaired electrons in the anti-bonding `n*`orbitals. Hence, like `O_(2)`,it exhibits paramagnetism. |
|
| 1589. |
Aluminium forms `[AIF_(6)]^(3-)` ion but boron does not form `[BF_(6)]^(3-)` ion. Explain. |
| Answer» Boron can show a maximum valency as well as co-ordination number of four. Therefore it cannot form `[BF_(6)]^(3-)` ion where it is to exhibit co-ordination number six. On the other and, aluminium can extend its valency as well as co-ordination number to six due to the presence of vacant 3d orbitals. Therefore, it can form `[AIF_(6)]^(3-)` ion. | |
| 1590. |
`(p pi-p pi)` back bonding occurs in the halides of boron but not in those of aluminium. Explain. |
| Answer» `p pi- p pi` back bond is linked with the size of the central atom in the halide and decreases with the increase in size. Sinc e the size of aluminium is quite big as compared to that of boron. `p pi-p pi` back bonding is possible in boron and not in aluminium. | |
| 1591. |
The oxidation number of `S` in `S_(8),S_(2)F_(2)`, and `H_(2)S`, respectively, areA. `0,+1` and `-2`B. `+2,+1` and `-2`C. `0,+1` and `+2`D. `-2, +1` and `-2` |
|
Answer» Correct Answer - A Oxidation state of molecular sulphur `S_(8)` is zero Oxidation state of sulphur in `S_(2)F_(2)=2x +2(-1)=0 , 2x=+2` or `x=1` Oxidation state of sulphur in `H_(2)S=2(+1)+x=0` or `x=-2` |
|
| 1592. |
Hydrolysis of one mole of peroxy disulphuric acid producesA. two moles of sulphuric acidB. two moles of peroxymonosulphuric acid.C. one mole of sulphuric acid and one mole of peroxy monosulphuric acidD. one mole of sulphuric acid, one mole of hydrogen peroxide. |
| Answer» Correct Answer - C | |
| 1593. |
The products of the chemical reaction between `Na_(2)S_(2)O_(3),Cl` and `H_(2)O` are :A. `HCI+Na_(2)S`B. `HCI+NaHSO_(4)`C. `HCI+Na_(2)SO_(3)`D. `NaHCIO_(3)+H_(2)O` |
|
Answer» Correct Answer - B `Na_(2)S_(2)O_(3)+4Cl+5H_(2)O to 2NaHSO_(4)+8HCl` |
|
| 1594. |
When sulphur is boiled with `Na_(2)SO_(4)` solution, the compound formed is :A. sodium sulphideB. sodium sulphateC. sodium persulphateD. sodium thiosulphate |
|
Answer» Correct Answer - D `Na_(2)SO_(3)+Soverset("Alkaline solution")underset("Boil")(to)Na_(2)S_(2)O_(3)` |
|
| 1595. |
Which ten is boiled with alkali solution, the product is -A. `SnO_(2)`B. `Sn(OH)_(2)`C. `Sn(OH)_(4)`D. `SnO_(3)^(2-)` |
| Answer» Correct Answer - D | |
| 1596. |
Sindoor used by women is an oxide of lead with the formula -A. `PbO`B. `PbO_(2)`C. `Pb_(3)O_(4)`D. `Pb_(2)O_(3)` |
| Answer» Correct Answer - C | |
| 1597. |
Lead `(IV)` oxide is obtained by:A. heating lead (II) oxide strongly in airB. heatind lead strongly in pure oxygenC. heating `Pb_(3)O_(4)` with conc. `HNO_(3)`D. oxidizing lead with conc. `HNO_(3)` |
|
Answer» Correct Answer - C `Pb_(3)O_(4)+4HNO_(3)overset(Delta)toPb(NO_(3))_(2)+2H_(2)O+2PbO_(2)` |
|
| 1598. |
Select the incorrect statement.A. Silicones are hydrophobic in nature.B. `Si-O-Si` linkage are moisture senstive.C. `SnI_(4)` is an orange solid on account of charge transfer.D. Silicones are resistant to most chemicals due to high strength of the `Si-C` bond and stable silica like strucurre of `Si-O-Si-O-Si`. |
| Answer» Correct Answer - B | |
| 1599. |
Which of the followng oxides of nitrogen is a coloured gas ?A. `N_(2)O`B. `NO`C. `N_(2)O_(4)`D. `NO_(2)` |
|
Answer» Correct Answer - D |
|
| 1600. |
Which of the following oxides of nitrogen is a coloured gas ?A. `N_2O`B. `NO`C. `N_2O_5`D. `NO_2` |
|
Answer» Correct Answer - D The colour of `NO_2` is reddish brown.All others are colourless. |
|