{"id":17616,"date":"2022-05-24T06:00:58","date_gmt":"2022-05-24T00:30:58","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=17616"},"modified":"2022-05-16T15:22:55","modified_gmt":"2022-05-16T09:52:55","slug":"mcq-questions-for-class-11-chemistry-chapter-11","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/mcq-questions-for-class-11-chemistry-chapter-11\/","title":{"rendered":"MCQ Questions for Class 11 Chemistry Chapter 11 The p-Block Elements with Answers"},"content":{"rendered":"

Students are advised to practice the\u00a0NCERT MCQ Questions for Class 11 Chemistry Chapter 11 The p-Block Elements with Answers Pdf free download is available here. MCQ Questions for Class 11 Chemistry with Answers<\/a> are prepared as per the Latest Exam Pattern. Students can solve these The p-Block Elements Class 11 MCQs Questions with Answers and assess their preparation level.<\/p>\n

The p-Block Elements Class 11 MCQs Questions with Answers<\/h2>\n

Solving the The p-Block Elements Multiple Choice Questions of Class 11 Chemistry Chapter 11 MCQ can be of extreme help as you will be aware of all the concepts. These MCQ Questions on The p-Block Elements Class 11 with answers pave for a quick revision of the Chapter thereby helping you to enhance subject knowledge. Have a glance at the MCQ of Chapter 11 Chemistry Class 11 and cross-check your answers during preparation.<\/p>\n

Question 1.
\nConsider the following statement about Ozone I. O3<\/sub> is formed by the interaction of fluorine. II. It turns tetramethyl base paper as violet. III. It turns benzidine paper as brown. The correct set of true statement is
\n(a) I and II
\n(b) I, II and III
\n(c) I and III
\n(d) II and III<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) I, II and III
\nExplanation:
\nOzone is formed by the interaction of fluorine. It turns tetramethyl base paper and benzidine paper as violet and brown respectively.
\nHence, the correction option is (2).<\/p>\n<\/details>\n

\n

Question 2.
\nIn the compound of type ECl3<\/sub>, where E = B, P, As, or Bi, the angle Cl \u2013 E \u2013 Cl for different E are ion the order:
\n(a) B = P = As = Bi
\n(b) B > P > As > Bi
\n(c) B < P = As = Bi
\n(d) B < P < As < Bi<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) B > P > As > Bi
\nExplanation:
\nBCl3<\/sub> is trigonal planar in structure and bond angles are 120\u00b0 each. PCl3<\/sub>, AsCl3<\/sub>, and BiCl3<\/sub> are pyramidal in shape with sp\u00b3-hybridization.
\nIn all of them, the bond angles are less than the normal tetrahedral angle of 109.28, and also these bond angles decrease down the group.
\nTherefore, the correct order of bond angles is as follows:
\nB > P > As > Bi<\/p>\n<\/details>\n

\n

Question 3.
\nIn white phosphorous(P4<\/sub>) molecule, which one is not correct:
\n(a) 6P-P single bonds are present
\n(b) 4P-P single bonds are present
\n(c) 4 lone pair of electrons is present
\n(d) P-P-P bond angle is 60\u00b0<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 6P-P single bonds are present
\nExplanation:
\nIt has total four lone pairs of electrons situated at each P – atom.
\nIt has six P_P single bond
\n\"MCQ<\/p>\n<\/details>\n

\n

Question 4.
\nAll the elements of oxygen family are
\n(a) Non metals
\n(b) Metalloids
\n(c) Radioactive
\n(d) Polymorphic<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) Polymorphic
\nExplanation:
\nGroup 16 elements are called polymorphic elements because all elements show allotropy except Te.<\/p>\n<\/details>\n

\n

Question 5.
\nWhich of the following will not produce hydrogen gas ?
\n(a) Reaction between Fe and dil. HCl
\n(b) Reaction between Zn and NaOH
\n(c) Reaction between Zn and conc. H2<\/sub>SO4<\/sub>
\n(d) Electrolysis of NaCl in Nelsons cell<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) Reaction between Zn and conc. H2<\/sub>SO4<\/sub>
\nExplanation:
\nConcentrated sulphuric acid reacts with Zn to give SO2<\/sub> and not H2<\/sub><\/p>\n<\/details>\n

\n

Question 6.
\nAmorphous form of Silica is
\n(a) Tridymite
\n(b) Kieselguhr
\n(c) Cristobalite
\n(d) Quartz<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) Cristobalite
\nExplanation:
\nSilicon Dioxide\/ Silica\/ Quartz –
\nCovalent, three dimensional solid network in which each silicon is covalently bond to four oxygen atoms (sp\u00b3 hybridisation) forming a tetrahedral structure.
\nFunction of quartz –
\nAs piezoelectric material in clocks, radio, television broadcasting and mobile communication.
\nQuartz, tridymite, cristobalite is crystalline form, and kieselguhr is an amorphous form of silica.<\/p>\n<\/details>\n

\n

Question 7.
\nGraphite is a soft solid lubricant extremely diffcult to melt. The reason for this anomalous behaviour is that graphite.
\n(a) Has carbon atoms arranged in large plates of rings of strongly bound carbon atoms with weak interplate bonds
\n(b) Is a non – crystalline substance
\n(c) Is an allotropic from of carbon
\n(d) Has molecules of variable molecular masses like polymers.<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) Has carbon atoms arranged in large plates of rings of strongly bound carbon atoms with weak interplate bonds
\nExplanation:
\nC-atoms oi graphite form covalently bonded plates (layers) These layers are held together by weak forces of attraction. i.e., one layer can slide over other to cause lubricacy. lt cannot be melted easily as a large number of atoms being bonded strongly in the layer to form big entity.<\/p>\n<\/details>\n

\n

Question 8.
\nBorax is used as a cleansing agent because on dissolving in water, it gives
\n(a) Alkaline solution
\n(b) Acidic solution
\n(c) Bleaching solution
\n(d) Amphoteric solution.<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) Alkaline solution
\nExplanation:
\nBorax dissolves in water to give an alkaline solution.
\nNa2<\/sub>B4<\/sub>O7<\/sub> + 7H2<\/sub>O \u21d4 2NaOH + 4H3<\/sub>BO3<\/sub>.<\/p>\n<\/details>\n

\n

Question 9.
\nAmong the C-X bond (where, X = Cl, Br, I) the correct decreasing order of bond energy is
\n(a) C\u2212I > C\u2212Cl > C\u2212Br
\n(b) C\u2212I > C\u2212Br > C\u2212Cl
\n(c) C\u2212Cl > C\u2212Br > C\u2212I
\n(d) C\u2212Br > C\u2212Cl > C\u2212I<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) C\u2212Cl > C\u2212Br > C\u2212I
\nExplanation:
\nAmong the C-X bond (where, X = Cl, Br, I), the correct decreasing order of bond energy is
\nC\u2212Cl > C\u2212Br > C\u2212l<\/p>\n<\/details>\n

\n

Question 10.
\nOn heating boron with caustic potash, the pair of products formed are
\n(a) Potassium Borate + Dihydrogen
\n(b) Potassium Borate + Water
\n(c) Potassium Borate + H2<\/sub>
\n(d) Borax + Dihydrogen.<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) Potassium Borate + Dihydrogen
\nExplanation:
\n2B + 2KOH + 2H2<\/sub>O \u2192 2KBO2<\/sub> + 3H2<\/sub>
\nBoron react with potassium hydroxide and water to produce potassium metaborate and hydrogen.<\/p>\n<\/details>\n

\n

Question 11.
\nWhich of the following statements regarding ozone is not correct?
\n(a) The oxygen-oxygen bond length in ozone is identical with that of molecular oxygen
\n(b) The ozone is response hybrid of two structures
\n(c) The ozone molecule is angular in shape
\n(d) Ozone is used as a germicide and disinfectant for the purification of air.<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) The oxygen-oxygen bond length in ozone is identical with that of molecular oxygen
\nExplanation:
\nThe oxygen\u2013oxygen bond length in ozone is identical with that of molecular oxygen<\/p>\n<\/details>\n

\n

Question 12.
\nThere is no S-S bond in
\n(a) S2<\/sub>O2-<\/sup>4<\/sub>
\n(b) S2<\/sub>O2-<\/sup>5<\/sub>
\n(c) S2<\/sub>O2-<\/sup>3<\/sub>
\n(d) S2<\/sub>O2-<\/sup>7<\/sub><\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) S2<\/sub>O2-<\/sup>7<\/sub>
\nSolution :
\nThere is no S-S bond in S2<\/sub>O2-<\/sup>7<\/sub>
\n\"MCQ<\/p>\n<\/details>\n

\n

Question 13.
\nWhich is strongest Lewis acid?
\n(a) BF3<\/sub>
\n(b) BCl3<\/sub>
\n(c) BBr3<\/sub>
\n(d) BI3<\/sub><\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) BF3<\/sub>
\nExplanation:
\nLarger the size of halogen atom less is the back donation of electrons into empty 2p orbital of B.<\/p>\n<\/details>\n

\n

Question 14.
\nFertilizer having the highest nitrogen percentage is:
\n(a) Calcium cyanamide
\n(b) Urea
\n(c) Ammonium nitrate
\n(d) Ammonium sulphate<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) Urea
\nExplanation:
\nEvery compound has 2N atoms (i.e., same mass of N), thus compound with the lowest molecular mass (i.e., urea) will have the highest N percentage.<\/p>\n<\/details>\n

\n

Question 15.
\nIn general, the Boron Trihaides act as
\n(a) Strong reducing agent
\n(b) Lewis Acids
\n(c) Lewis Bases
\n(d) Dehydrating Agents<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) Lewis Acids
\nExplanation:
\nThe boron atom in trihaldies has only six electrons in the valence shell and hence can accept a pair of electrons in the vacant p-orbital to complete its octet. As a result, boron trihaldies act as a Lewis acids.<\/p>\n<\/details>\n

\n

Question 16.
\nWhich of the following is not a mineral of boron?
\n(a) Colemanite
\n(b) Kernite
\n(c) Boric Anhydride
\n(d) Borax<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) Boric Anhydride
\nExplanation:
\nThe most important boron minerals in commercial terms are; Tincal, Colemanite, Kernite, Ulexite, Pandermite, Boracite, Szaybelite and Hydroboracite. The main boron minerals transformed by Eti Maden, the World Boron Leader, into high value added products in international quality standards are; Tincal, Colemanite and Ulexite.<\/p>\n<\/details>\n

\n

Question 17.
\nWhich phosphorus is used as a rat poison?
\n(a) White
\n(b) Violet
\n(c) Red
\n(d) Black<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) White
\nExplanation:
\nWhite phosphorous is least stable and most toxic of all allotropes. Upon coming on contact with air it is toxic and causes severe liver damage on digestion so it is used as rat poison.<\/p>\n<\/details>\n

\n

Question 18.
\nThe structure of diBorane contains
\n(a) Four 2c – 2e bonds and two 3c – 2e bonds
\n(b) Two 2c – 2e bonds and two 3c – 2e bonds
\n(c) Two 2c – 2e bonds and two 3c – 3e bonds
\n(d) Four 2c – 2e bonds and four 3c – 2e bonds<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) Four 2c – 2e bonds and two 3c – 2e bonds
\nExplanation:
\nAccording to molecular orbital theory, each of the two boron atoms is in sp\u00b3 hybrid state. Of the four hybrid orbitals, three have one electron each while the fourth is empty. Two of the four orbitals of each of the boron atom overlap with two terminal hydrogen atoms forming two normal B \u2013 H \u03c3-bonds. One of the remaining hybrid orbital (either filled or empty) of one of the boron atoms, 1s orbital of hydrogen atoms (bridge atom) and one of hybrid orbitals of the other boron atom overlap to form a delocalised orbital covering the three nuclei with a pair of electrons. Such a bond is known as three centre two electron (3c – 2e) bonds
\n\"MCQ<\/p>\n<\/details>\n

\n

Question 19.
\nNitrogen (I) oxide is produced by:
\n(a) Thermal decomposition of ammonium nitrate
\n(b) Disproportionation of N2<\/sub>O4<\/sub>
\n(c) Thermal decomposition of ammonium nitrite
\n(d) None of the above<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) Thermal decomposition of ammonium nitrite
\nExplanation:
\nNitrous Oxide (N2<\/sub>O) can be produced by thermal decomposition of ammonium nitrate:
\nNH4<\/sub>NO3<\/sub>(s) \u2192 N2<\/sub>O (g) + 2H2<\/sub>O(l)<\/p>\n<\/details>\n

\n

Question 20.
\nRed phosphorus is chemically less reactive because
\n(a) It does not contain P – P bonds
\n(b) It dos not contain tetrahedral P4<\/sub> molecules
\n(c) It does not catch fire in air even upto 400\u00b0C
\n(d) It has a polymeric structure<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) It has a polymeric structure
\nExplanation:
\nRed phosphorus is less reactive because of its gaint polymeric structure.<\/p>\n<\/details>\n

\n

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Students are advised to practice the\u00a0NCERT MCQ Questions for Class 11 Chemistry Chapter 11 The p-Block Elements with Answers Pdf free download is available here. MCQ Questions for Class 11 Chemistry with Answers are prepared as per the Latest Exam Pattern. Students can solve these The p-Block Elements Class 11 MCQs Questions with Answers and …<\/p>\n

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