Class 12 Chemistry · 72 questions · 72 with answers
On addition of conc. H2SO4 to a chloride salt, colourless fumes are evolved but in case of iodide salt, violet fumes come out. This is because
(iii) HI gets oxidised to I2
In qualitative analysis when H2S is passed through an aqueous solution of salt acidified with dil. HCl, a black precipitate is obtained. On boiling the precipitate with dil. HNO3, it forms a solution of blue colour. Addition of excess of aqueous solution of ammonia to this solution gives _________.
(ii) deep blue solution of [Cu (NH3)4]
In a cyclotrimetaphosphoric acid molecule, how many single and double bonds are present?
(i) 3 double bonds; 9 single bonds
Which of the following elements can be involved in pπ–dπ bonding?
(iii) Phosphorus
Which of the following pairs of ions are isoelectronic and isostructural? 2– –
(i) CO3 , NO3 – 2–
Affinity for hydrogen decreases in the group from fluorine to iodine. Which of the halogen acids should have highest bond dissociation enthalpy?
(i) HF
Bond dissociation enthalpy of E—H (E = element) bonds is given below. Which of the compounds will act as strongest reducing agent? Compound NH3 PH3 AsH3 SbH3 –1 ∆diss (E—H)/kJ mol 389 322 297 255
(iv) SbH3
On heating with concentrated NaOH solution in an inert atmosphere of CO2, white phosphorus gives a gas. Which of the following statement is incorrect about the gas?
(iii) It is more basic than NH3.
Which of the following acids forms three series of salts?
(iii) H3PO4
Strong reducing behaviour of H3PO2 is due to
(iii) Presence of one –OH group and two P–H bonds
On heating lead nitrate forms oxides of nitrogen and lead. The oxides formed are ______.
(ii) NO2, PbO
Which of the following elements does not show allotropy?
(i) Nitrogen
Maximum covalency of nitrogen is ______________.
(iii) 4
Which of the following statements is wrong?
(i) Single N–N bond is stronger than the single P–P bond.
A brown ring is formed in the ring test for NO3 ion. It is due to the formation of 2+
(i) [Fe(H2O)5 (NO)]
Elements of group-15 form compounds in +5 oxidation state. However, bismuth forms only one well characterised compound in +5 oxidation state. The compound is
(ii) BiF5
On heating ammonium dichromate and barium azide separately we get
(i) N2 in both cases
In the preparation of HNO3, we get NO gas by catalytic oxidation of ammonia. The moles of NO produced by the oxidation of two moles of NH3 will be ______.
(i) 2
The oxidation state of central atom in the anion of compound NaH2PO2 will be ______.
(iii) +1
Which of the following is not tetrahedral in shape? +
(iii) SF4
Which of the following are peroxoacids of sulphur?
(i) H2SO5 and H2S2O8
Hot conc. H2SO4 acts as moderately strong oxidising agent. It oxidises both metals and nonmetals. Which of the following element is oxidised by conc. H2SO4 into two gaseous products?
(iii) C
A black compound of manganese reacts with a halogen acid to give greenish yellow gas. When excess of this gas reacts with NH3 an unstable trihalide is formed. In this process the oxidation state of nitrogen changes from _________.
(i) – 3 to +3
In the preparation of compounds of Xe, Bartlett had taken O2 Pt F6 as a base compound. This is because
(iii) both O2 and Xe have almost same ionisation enthalpy.
In solid state PCl5 is a _________.
(iv) ionic solid with [PCl4] tetrahedral and [PCl6] octahedra
Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidising power. – – – Ion ClO4 IO4 BrO4 V V V Reduction E =1.19V E =1.65V E =1.74V potential E /V – – –
(iii) BrO4 > IO4 > ClO4 – – –
Which of the following is isoelectronic pair?
(ii) BrO2 , BrF2
If chlorine gas is passed through hot NaOH solution, two changes are observed in the oxidation number of chlorine during the reaction. These are ________ and _________.
(i) 0 to +5
(iii) 0 to –1
Which of the following options are not in accordance with the property mentioned against them?
(ii) MI > MBr > MCl > MF Ionic character of metal halide.
(iii) F2 > Cl2 > Br2 > I2 Bond dissociation enthalpy.
Which of the following is correct for P4 molecule of white phosphorus?
(ii) It has six P–P single bonds.
(iv) It has four lone pairs of electrons.
Which of the following statements are correct?
(i) Among halogens, radius ratio between iodine and fluorine is maximum.
(iii) Among interhalogen compounds maximum number of atoms are present in iodine fluoride.
(iv) Interhalogen compounds are more reactive than halogen compounds.
Which of the following statements are correct for SO2 gas?
(i) It acts as bleaching agent in moist conditions.
(iii) It’s dilute solution is used as disinfectant.
Which of the following statements are correct?
(iii) P4 molecule in white phohsphorus have angular strain therefore white phosphorus is very reactive.
(iv) PCl is ionic in solid state in which cation is tetrahedral and anion is octahedral.
Which of the following orders are correct as per the properties mentioned against each?
(i) As2O3 < SiO2 < P2O3 < SO2 Acid strength.
(iv) H2O > H2S > H2Se > H2Te Thermal stability. 95 p-Block Elements
Which of the following statements are correct?
(i) S–S bond is present in H2S2O6.
(ii) In peroxosulphuric acid (H2SO5) sulphur is in +6 oxidation state.
In which of the following reactions conc. H2SO4 is used as an oxidising reagent?
(ii) 2HI + H2SO4 → I2 + SO2 + 2H2O
(iii) Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O
Which of the following statements are true?
(i) Only type of interactions between particles of noble gases are due to weak dispersion forces.
(ii) Ionisation enthalpy of molecular oxygen is very close to that of xenon.
In the preparation of H2SO4 by Contact Process, why is SO3 not absorbed directly in water to form H2SO4?
Acid fog is formed, which is difficult to condense. Pt/Rh gauge catalyst
Write a balanced chemical equation for the reaction showing catalytic oxidation of NH3 by atmospheric oxygen.
4NH3 + 5O2 500K, 9 bar → 4NO + 6H2O (From air)
Write the structure of pyrophosphoric acid.
Pyrophosphoric acid
PH3 forms bubbles when passed slowly in water but NH3 dissolves. Explain why?
NH3 forms hydrogen bonds with water therefore it is soluble in it but PH3 cannot form hydrogen bond with water so it escapes as gas.
In PCl5, phosphorus is in sp3d hybridised state but all its five bonds are not equivalent. Justify your answer with reason.
[Hint : It has trigonal bipyramidal geometry]
Why is nitric oxide paramagnetic in gaseous state but the solid obtained on cooling it is diamagnetic?
In gaseous state NO2 exists as monomer which has one unpaired electron but in solid state it dimerises to N2O4 so no unpaired electron is left hence solid form is diamagnetic.
Give reason to explain why ClF3 exists but FCl3 does not exist.
Because fluorine is more electronegative as compared to chlorine.
Out of H2O and H2S, which one has higher bond angle and why?
Bond angle of H2O is larger, because oxygen is more electronegative than sulphur therefore bond pair electron of O–H bond will be closer to oxygen and there will be more bond-pair bond-pair repulsion between bond pairs of two O–H bonds. –
SF6 is known but SCl6 is not. Why?
Due to small size of fluorine six F ion can be accomodated around sulphur whereas chloride ion is comparatively larger in size, therefore, there will be interionic repulsion. 101 p-Block Elements
On reaction with Cl2, phosphorus forms two types of halides ‘A’ and ‘B’. Halide A is yellowish-white powder but halide ‘B’ is colourless oily liquid. Identify A and B and write the formulas of their hydrolysis products. – 2+
A is PCl5 (It is yellowish white powder) P4 + 10Cl2 → 4PCl5 B is PCl3 (It is a colourless oily liquid) P4 + 6Cl2 → 4PCl3 Hydrolysis products are formed as follows : PCl3 + 3H2O → H3PO3+3HCl PCl5 + 4H2O → H3PO4 + 5HCl – 2+ + 3+
In the ring test of NO3 ion, Fe ion reduces nitrate ion to nitric oxide, which 2+ combines with Fe (aq) ion to form brown complex. Write the reactions involved in the formation of brown ring.
NO3 + 3Fe + 4H → NO + 3Fe + 2H2O 2+ 2+ [Fe(H2O)6] + NO → [Fe(H2O)5(NO)] + H2O (brown complex)
Explain why the stability of oxoacids of chlorine increases in the order given below: HClO < HClO2 < HClO3 < HClO4
Oxygen is more electronegative than chlorine, therefore dispersal of – – negative charge present on chlorine increases from ClO to ClO4 ion because number of oxygen atoms attached to chlorine is increasing. Therefore, stability of ions will increase in the order given below : – – – – ClO < ClO2 < ClO3 < ClO4 Thus due to increase in stability of conjugate base, acidic strength of corresponding acid increases in the following order HClO < HClO2 < HClO3 < HClO4
Explain why ozone is thermodynamically less stable than oxygen.
See the NCERT textbook for Class XII, page 186.
P4O6 reacts with water according to equation P4O6 + 6H2O → 4H3PO3. Calculate the volume of 0.1 M NaOH solution required to neutralise the acid formed by dissolving 1.1 g of P4O6 in H2O.
P4O6 + 6H2O → 4H3PO3 H3PO3 + 2NaOH → Na2 HPO3 + 2H2O] × 4 (Neutralisation reaction) P4O6 + 8NaOH → 4Na2 HPO4 + 2H2O 1 mol 8 mol Product formed by 1 mol of P4O6 is neutralised by 8 mols of NaOH 1.1 1.1 ∴ Product formed by mol of P4O6 will be neutralised by × 8 mol 220 220 of NaOH Molarity of NaOH solution is 0.1M ⇒ 0.1 mol NaOH is present in 1 L solution 1.1 1.1 × 8 88 4 ∴ × 8 mol NaOH is present in L = L = L = 0.4 L = 220 220 × 0.1 220 10 400 mL of NaOH solution.
White phosphorus 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.
P4 + 6Cl2 → 4PCl3 PCl3 + 3H2O → H3PO3 + 3HCl] × 4 P4 + 6Cl2 + 12H2O → 4H3PO3 + 12HCl 1 mol of white phosphorus produces 12 mol of HCl 62 1 62g of white phosphorus has been taken which is equivalent to = mol. 124 2 Therefore 6 mol HCl will be formed. Mass of 6 mol HCl = 6 × 36.5 = 219.0 g HCl
Name three oxoacids of nitrogen. Write the disproportionation reaction of that oxoacid of nitrogen in which nitrogen is in +3 oxidation state.
Three oxoacids of nitrogen are (i) HNO2, Nitrous acid (ii) HNO3 , Nitric acid (iii) Hyponitrous acid, H2N2O2 Disproportionation 3HNO2 → HNO3 + H2O + 2NO
Nitric acid forms an oxide of nitrogen on reaction with P4O10. Write the reaction involved. Also write the resonating structures of the oxide of nitrogen formed.
4HNO3 + P4O10 → 4HPO3 + 2N2O5
Phosphorus has three allotropic forms —
(i) white phosphorus
(ii) red phosphorus and
Give an example to show the effect of concentration of nitric acid on the formation of oxidation product.
Dilute and concentrated nitric acid give different oxidation products on reaction with copper metal. 3Cu + 8HNO3 (dil.) → 3Cu(NO3)2 + 2NO + 4H2O Cu + 4HNO3 (Conc.) → 3Cu(NO3)2 + 2NO + 2H2O
PCl5 reacts with finely divided silver on heating and a white silver salt is obtained, which dissolves on adding excess aqueous NH3 solution. Write the reactions involved to explain what happens.
PCl5 + 2Ag → 2AgCl + PCl3 + – AgCl + 2NH3(aq) → [Ag(NH3)2] Cl (soluble complex)
Phosphorus forms a number of oxoacids. Out of these oxoacids phosphinic acid has strong reducing property. Write its structure and also write a reaction showing its reducing behaviour.
Structure of phosphinic acid (Hypophosphorous acid) is as follows : Reducing behaviour of phosphinic acid is observable in the reaction with silver nitrate given below : 4AgNO3 + 2H2O + H3PO2 → 4Ag + 4HNO3 + H3PO4 103 p-Block Elements IV. Matching Type
Match the compounds given in Column I with the hybridisation and shape given in Column II and mark the correct option. Column I Column II (A) Xe F6 (1) sp3d3 – distorted octahedral (B) Xe O3 (2) sp3d2 - square planar (C) Xe OF4 (3) sp3 - pyramidal (D) Xe F4 (4) sp3 d2 - square pyramidal 97 p-Block Elements Code :
(i) A (1) B (3) C (4) D (2)
Match the formulas of oxides given in Column I with the type of oxide given in Column II and mark the correct option. Column I Column II (A) Pb3O4 (1) Neutral oxide (B) N2O (2) Acidic oxide (C) Mn2O7 (3) Basic oxide (D) Bi2O3 (4) Mixed oxide Code :
(ii) A (4) B (1) C (2) D (3)
Match the items of Columns I and II and mark the correct option. Column I Column II (A) H2SO4 (1) Highest electron gain enthalpy (B) CCl3NO2 (2) Chalcogen (C) Cl2 (3) Tear gas (D) Sulphur (4) Storage batteries Code :
(i) A (4) B (3) C (1) D (2)
Match the species given in Column I with the shape given in Column II and mark the correct option. Column I Column II (A) SF4 (1) Tetrahedral (B) BrF3 (2) Pyramidal – (C) BrO 3 (3) Sea-saw shaped + (D) NH 4 (4) Bent T-shaped Code :
(ii) A (3) B (4) C (2) D (1)
Match the items of Columns I and II and mark the correct option. Column I Column II (A) Its partial hydrolysis does not (1) He change oxidation state of central atom (B) It is used in modern diving apparatus (2) XeF6 (C) It is used to provide inert atmosphere (3) XeF4 for filling electrical bulbs (D) Its central atom is in sp 3d hybridisation (4) Ar Code :
(iii) A (2) B (1) C (4) D (3)
Assertion (A): N2 is less reactive than P4.
Reason (R): Nitrogen has more electron gain enthalpy than phosphorus.
Assertion (A): HNO3 makes iron passive.
Reason (R): HNO3 forms a protective layer of ferric nitrate on the surface of iron. 99 p-Block Elements
Assertion (A): HI cannot be prepared by the reaction of KI with concentrated H2SO4
Reason (R): HI has lowest H–X bond strength among halogen acids.
Assertion (A): Both rhombic and monoclinic sulphur exist as S8 but oxygen exists as O2.
Reason (R): Oxygen forms pπ – pπ multiple bond due to small size and small bond length but pπ – pπ bonding is not possible in sulphur.
Assertion (A): NaCl reacts with concentrated H2SO4 to give colourless fumes with pungent smell. But on adding MnO2 the fumes become greenish yellow.
Reason (R): MnO2 oxidises HCl to chlorine gas which is greenish yellow.
Assertion (A): SF6 cannot be hydrolysed but SF4 can be.
Reason (R): Six F atoms in SF6 prevent the attack of H2O on sulphur atom of SF6.
An amorphous solid “A” burns in air to form a gas “B” which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aqueous KMnO4 solution and reduces Fe3+ to Fe2+. Identify the solid “A” and the gas “B” and write the reactions involved.
‘A’ is S8 ‘B’ is SO2 gas ∆ → 8SO S8 + 8O2 2 – 2– + 2MnO4 + 5SO2 + 2H2O → 5 SO4 + 4H + 2Mn2+ (violet) (colourless) 2– + 2Fe3+ + SO2 + 2H2O → 2Fe2+ + SO4 + 4H ∆
On heating lead (II) nitrate gives a brown gas “A”. The gas “A” on cooling changes to colourless solid “B”. Solid “B” on heating with NO changes to a blue solid ‘C’. Identify ‘A’, ‘B’ and ‘C’ and also write reactions involved and draw the structures of ‘B’ and ‘C’.
Pb(NO3)2 2PbO + 4NO2 673K (A) (Brown colour) On cooling 2NO2 Heating N2O4 (B) (Colourless solid) ∆ 250 K 2NO + N2O4 → 2 N2O3 (C) (Blue solid) (Structure of N2O4) (Structure of N2O3)
On heating compound
(A) gives a gas
(B) which is a constituent of air. This gas when treated with 3 mol of hydrogen (H2) in the presence of a catalyst gives another gas
(C) which is basic in nature. Gas C on further oxidation in moist condition gives a compound
(D) which is a part of acid rain. Identify compounds (A) to (D) and also give necessary equations of all the steps involved.