Test-5/P Block/Set-S Mittal Sir

Test-5/P Block/Set-S Mittal Sir

Test-5/p_block/Set-S Mittal Sir

III-A/172 , Nehru Nagar Ghaziabad. Ph : 9811212090, 9868502091.

email : , website :

Type : JEE MAIN 2015-16

TEST-5/12th & 12th Passed/Set-S

Topic : ‘p block elements’

Date : 27th NOV2015

Max Marks : 120 Time : 50 Minutes

Section A

Single Correct Answer Type(+2/-0.5)

This section contains 20 questions. Each question has 4 choices (a), (b), (c) and (d) for its answer, out of which ONLY ONE is correct.

  1. In view of the sings of rG for the following reactions :

PbO2 + Pb  2PbO, rG < 0

SnO2 + Sn  2SnO, rG > 0

Which oxidation states are more characteristic for lead and tin?

(a) for lead + 2, for tin + 2(b) for lead + 4, for tin + 2

(c) for lead + 2, for tin + 4(d) for lead + 4, for tin + 4

  1. Total number of lone pairs present in one molecule of ozone

(a) 3(b) 4(c) 5(d) 6

  1. The number of P – O – P bonds in cyclic metaphosphoric acid is

(a) zero(b) two(c) three(d) four

  1. Ammonia can be dried by

(a) conc.H2SO4(b) CaO(c) P4O10(d) anhydrous CaCl2

  1. Extra pure N2 can be obtained by heating

(a) NH3 with CuO(b)Ba(N3)2 (c) (NH4)2Cr2O7(d) NH4NO3

  1. Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid is

(a) Cl2O6(b) Cl2O7(c) ClO2(d) Cl2O

  1. In the brown ring test of NO3–, the formula of brown ring formed is

(a) [Fe(H2O)5 NO]2+(b) [Fe(H2O)5N2O]+ (c) [Fe(H2O)5 N2O]2+ (d) [Fe(H2O)5 NO]+

  1. Which of the following element does not show allotropism

(a) S(b) P(c) O(d) N

  1. H3BO3 is :

(a) Monobasic and strong Lewis acid(b) Tribasic and weak Bronsted acid

(c) Monobasic and weak lewis acid (d) Monobasic and weak Bronsted acid

  1. Blue liquid which is obtained on reacting equimolar amounts of two gases at –30C is

(a) N2O5(b) N2O4(c) N2O3(d) N2O

  1. Name of the structure of silicates in which three oxygen atoms of [SiO4]4– are shared

(a) Pyrosilicate(b) Sheet silicate

(c) Linear chain silicate(d) Three dimensional silicate

  1. Which reduces AgNO3 into Ag

(a) HNO2(b) H3PO4(c) H3PO2(d) PH3

  1. Nitrous acid (HNO2) is usually produced in situ because it is unstable and in aqueous solution it disproportionate into

(a) HNO3 & NH3(b) HNO3 & N2(c) HNO3 & NO (d) HNO3 & N2O

  1. In which compound all the bonds are of same bond length

(a) N2O5(b) XeF4(c) B2H6(d) Al2Cl6

  1. A gas that cannot be collected over water is :

(a) O2(b) N2(c) SO2(d) PH3

  1. Identify the incorrect statement among the following

(a) Br2 reacts with hot and strong NaOH solution to give NaBr and H2O

(b) Ozone reacts with SO2 to give SO3

(c) Silicon reacts with NaOH(aq) in the presence of air to give Na2SiO3 and H2O

(d) Cl2 reacts with excess of NH3 to give N2 and HCl

  1. Lowest bond enthalpy is of

(a) I – I(b) Cl – Cl(c) Br – Br(d) F – F

  1. Highest boiling point is of

(a) H2S(b) H2O(c) H2Se(d) NH3

  1. SO2 and Cl2 both are bleaching agent but they differ in –

(a) SO2 bleach by oxidation while Cl2 bleach by reduction

(b) SO2 bleach by reduction while Cl2 bleach by oxidation

(c) SO2 works in dry state while Cl2 works in aqueous medium

(d) SO2 works in aqueous medium while Cl2 works in dry state

  1. The most reactive species in aquaregia is

(a) NOCl(b) Cl+(c) NO(d) [Cl]

Section B

Single Correct Answer Type (+4/-1)

This section contains 20 questions. Each question has 4 choices (a), (b), (c) and (d) for its answer, out of which ONLY ONE is correct.

  1. The gas (es) produced when a mixture of Zn & NaCl is treated with concentrated H2SO4 is –

(i) H2, (ii) Cl2, (iii) HCl, (iv) SO2

(a)(iii) & (iv)(b) (ii) & (iv)(c)(i) & (iv)(d) (i), (ii) & (iv)

  1. A yellowish white powder ‘X’ which on heating form a colourless oily liquid ‘Y’ along with a gas ‘Z’. The oily liquid is prepared by the reaction of a tetrahedral compound ‘A’ with thionyl chloride. ‘Y’ is

(a) ICl3(b) ICl5(c) PCl3(d) PCl5

  1. Among O2, NO, NO2, N2O3, S2 (vapours), F2 & Cl2 total number of paramagnetic gases are

(a) 2(b) 3(c) 4(d) 5

  1. 2 mols of H3PO4 were treated with three mols of NaOH, the product will carry

(a) 1 mol NaH2PO4 & 1 mol Na2HPO4(b) 2 mols Na2HPO4 and 1 mol H3PO4

(c) 1 mol Na3PO4 & 1 mol H3PO4(d) 2 mols NaH2PO4 & 1 mol Na2HPO4

  1. 0.1 mol Zn 0.1 mol Cu 0.1 mol Zn 0.1 mol Cu

Conc Conc dil dil

HNO3 HNO3 HNO3 HNO3

[A] [B] [C] [D]

If in all four cases metal is limiting reagent in which case lowest amount of gas will

be released (consider H2O as liquid)

(a) A(b) B(c) C(d) D

  1. Column ‘A’Column ‘B’

(i) Number of moles of SOCl2(P) 3

required to react one mol of

P4 to prepare PCl3

(ii) Number of P – P bonds in one(Q) 4

molecule of whitephosphorus

(iii) Covalency of nitrogen in N2O5(R) 6

(iv) In hydrolysis of XeF4

mols of XeF4 required (S) 8

to obtain two mols of xenon

  1. M – O – M bond is found in how many of these compound

M  Metal/Non metal (other than oxygen)

Pyrophosphoric acid, Phosphorus trioxide, Phosphorus pentaoxide, Nitrogen (III) oxide,

Nitrogen (v) oxide, cyclotrimeta phosphoric acid, Pyrosilicate, orthosilicate, Borax

(a) 5(b) 6(c) 7(d) 8

  1. Among the following the process (s) which is (are) endothermic

(I) e– + He(g) He(g)(II) 2O3(g)  3 O2(g)

(III) e– + F(g) F(g)(IV) 2e– + O(g) O–2(g)

(a) I & IV(b) III & IV(c) I, II & IV(d) I, II, III & IV all

  1. Among the following, the correct statement is

(a) Between NH3 and PH3, NH3 is a better electron donor because the lone pair of electrons occupies

sp3 orbital and is more directional

(b) Between NH3 and PH3, PH3 is a better electron donor because the lone pair of electrons occupies

sp3 orbital and is more directional

(c) Between NH3 and PH3, NH3 is a better electron donor because the lone pair of electrons occupies

spherical s-orbital and is less directional

(d) Between NH3 and PH3, PH3 is a better electron donor because the lone pair of electrons occupies

spherical s-orbital and is less directional

  1. 10 litre of an aqueous solution containing 1 mol of I2 was taken. In this solution 5 mols of Cl2 were passed. If no reactant was left, then number of mols of NaOH which will be required to neutralize completely are –

(a) 5(b) 6(c) 10(d) 12

  1. Among the following the case(s) in which product is NH3 gas

(I) NaNO3 + Zn (II) CaNCN

(III) NH4NO3(IV) (NH4)2Cr2O7

(a) I & II both (b) III & IV both(c) II only (d) I, II, III & IV all

  1. Quantitative method for estimation of O3 is –

(a)Reaction of O3 with KI in acidic medium and then using Na2S2O3 to estimate I2

(b) Reaction of O3 with KI in basic medium and then using Na2S2O3 to estimate I2

(c) Reaction of O3 with PbS and then O2 evalved is measured at S.T.P

(d) Reaction of O3 with H2O2

  1. In which case bond angle increases

(I) NH3 + H+ NH4(II) PH3 + H+ PH4

(III) NO2 NO2+ + e–(IV) NO2 + e– NO2–

(a) II & III(b) I & III(c) I, II & III(d) All I, II, III & IV

  1. When dissolved in water disproportionation take place

(I) XeF2(II) XeF4(III) XeF6(IV) NO2 (V) SO2

(a) III & IV(b) II & IV (c) I & IV(d) III, IV & V

  1. 25 ml of household solution was mixed with 30 ml of 0.50 M KI and 10 ml of 4N acetic acid. In the titration of the liberated iodine, 48 ml of 0.25 N Na2S2O3 was used to reach the end point. The molarity of the household bleach solution is

(a) 0.48 M(b) 0.96 M(c) 0.024 M(d) 0.24 M

  1. Among the following total number of molecules which show hydrolysis when dissolved in water XeF2, XeF4, XeF6, CCl4, SiCl4, PCl3, PCl5, NF3, SF4

(a) 5(b) 6(c) 7(d) 8

  1. Orthophosphorus acid disproportionate on heating. One molecule of orthophosphorus acid form

(a) ½ mol of phosphoric acid and ½ mols of hypophosphorus acid

(b) 2/3 mols of phosphoric acid and 1/3 mol of phosphine

(c) ¾ mols of phosphoric acid and ¼ mols of phosphine

(d) ¾ mols of phosphoric acid and ¼ mol of hypophosphorus acid

  1. Among the following total number of mono proteic acids are –

Hypophosphorous acid, Hypophosphoric acid, Boric acid, Auric acid,

Chloroplatinic acid, Chromic acid, Nitrous acid, Hypochlorus acid, Chloric acid

(a) 3(b) 4(c) 5(d) 6

  1. M – F bond length is shorter than expected in –

(a) BF3(b) SF6(c) CF4(d) PF5

  1. E x2/x – is positive for all halogens the value depend upon the H of the process

½ X2(g) X(g) X–(g) X–(aq)

The highest positive value of E for F2 is due to

(i) Low bond energy of F – F bond , (ii) High electron gain enthalpy of Flourine

(iii) High hydration enthalpy of F– , (iv) High electronegativity of Fluorine

(a) (i) & (iii)(b) (i), (iii) & (iv)(c) (i), (ii) & (iii)(d) all (i), (ii), (iii) & (iv)

III-A/172 , Nehru Nagar Ghaziabad. Ph : 0120-4102287, 9811212090

email : , website :

Type : JEE MAIN 2015-16

TEST-5/12th & 12th Passed/Set-S

Topic : p block elements

Date : 27th NOV 2015

Max Marks : 120 Time : 50 Minutes

Note : Write your name and roll no. above in box before filling the answers.

Use only blue/black ball pen, use of pencil is strictly prohibited.

Section – A Section - B

Straight Objective Type (+2/-0.5) Straight Objective Type (+4/-1)

A B C D A B C D A B C D A B C D

1 11 21 31.

212 22 32.

313 23 33.

414 24 34.

515 25 35.

6 16 26 36.

717 27 37.

818 28 38.

919 29 39.

1020 30 40.

BMC CLASSES Page No -1

III-A/172, Nehru Nagar Ghaziabad. Ph : 0120-4102287, 9811212090 email: web: