{"id":17580,"date":"2022-05-24T10:00:56","date_gmt":"2022-05-24T04:30:56","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=17580"},"modified":"2022-05-16T15:24:53","modified_gmt":"2022-05-16T09:54:53","slug":"mcq-questions-for-class-11-chemistry-chapter-4","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/mcq-questions-for-class-11-chemistry-chapter-4\/","title":{"rendered":"MCQ Questions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure with Answers"},"content":{"rendered":"

Students are advised to practice the\u00a0NCERT MCQ Questions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure 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 Chemical Bonding and Molecular Structure Class 11 MCQs Questions with Answers and assess their preparation level.<\/p>\n

Chemical Bonding and Molecular Structure Class 11 MCQs Questions with Answers<\/h2>\n

Solving the Chemical Bonding and Molecular Structure Multiple Choice Questions of Class 11 Chemistry Chapter 4 MCQ can be of extreme help as you will be aware of all the concepts. These MCQ Questions on Chemical Bonding and Molecular Structure 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 4 Chemistry Class 11 and cross-check your answers during preparation.<\/p>\n

Question 1.
\nBased on VSEPR theory, the number of 90\u00b0 F-Br-F angles in BrF5<\/sub> is
\n(a) 0
\n(b) 2
\n(c) 4
\n(d) 8<\/p>\n

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

Answer: (a) 0
\nExplanation:
\nBrF5<\/sub> has octahedral geometry and square pyramidal shape.
\nIt has one lone pair and five bond pairs. So, geometry will be octahedral. But since lone pair repels too much, all the four bond pairs that had to be on a square planar surface, will tilt. This will result in a destructed geometry.<\/p>\n<\/details>\n

\n

Question 2.
\nThe hybrid state of sulphur in SO2<\/sub> molecule is :
\n(a) sp\u00b2
\n(b) sp\u00b3
\n(c) sp
\n(d) sp\u00b3d<\/p>\n

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

Answer: (a) sp\u00b2
\nExplanation:
\nThe hybridisation of sulphur in SO2<\/sub>\u200b is sp\u00b2. Sulphur atom has one lone pair of electrons and two bonding domains. Bond angle is <120\u00b0 and molecular geometry is V-shape, bent or angular<\/p>\n<\/details>\n

\n

Question 3.
\nIn allene (C3<\/sub>H4<\/sub>), the type(s) of hybridisation of the carbon atoms is (are)
\n(a) sp and sp\u00b3
\n(b) sp and sp\u00b2
\n(c) Only sp\u00b2
\n(d) sp\u00b2 and sp\u00b3<\/p>\n

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

Answer: (b) sp and sp\u00b2
\nExplanation:
\nAllene: H2<\/sub>\u200bC = C = CH2<\/sub>. The central carbon is attached to other carbons by two sigma and two pi bonds so its hybridisation will be sp. The terminal carbon is attached to other carbon and hydrogen by 3 sigma and 2 pi bonds and hence, its hybridisation will be sp\u00b2<\/p>\n<\/details>\n

\n

Question 4.
\nThe state of hybridization of the central atom and the number of lone pairs over the central atom in POCl3<\/sub> are
\n(a) sp, 0
\n(b) sp\u00b2, 0
\n(c) sp\u00b3, 0
\n(d) dsp\u00b2, 1<\/p>\n

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

Answer: (c) sp\u00b3, 0
\nExplanation:
\nThe central atom of POCl3<\/sub> that is P has sp3 hybridization and the number of lone pairs over the central atom in POCl3<\/sub> is zero.<\/p>\n

The central atom is P, which has 5 valence electrons. Out of them, two valence electrons are used to form a P = O double bond, while the other 3 valence electrons are used to form 3 P-Cl bonds. The molecular geometry of POCl3<\/sub> is tetrahedral with asymmetric charge distribution around the central atom. Therefore this molecule is polar. The structure of this compound is tetrahedral and hybridisation of P is sp\u00b3.<\/p>\n

On the central atom P, there are 4 bonding electron clouds (1 P = O double bond and 3 P-Cl bonds) but no lone pair of electrons.<\/p>\n<\/details>\n

\n

Question 5.
\nThe charge\/size ratio of a cation determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizing order of the polarizing power of the cationic species, K+<\/sup>, Ca2+<\/sup>, Mg2+<\/sup>, Be2+<\/sup>?
\n(a) Ca2+<\/sup> < Mg2+<\/sup> < Be+<\/sup> < K+<\/sup>
\n(b) Mg2+<\/sup> < Be2+<\/sup> < K+<\/sup> < Ca2+<\/sup>
\n(c) Be2+<\/sup> < K+<\/sup> < Ca2+<\/sup> < Mg2+<\/sup>
\n(d) K+<\/sup> < Ca2+<\/sup> < Mg2+<\/sup> < Be2+<\/sup><\/p>\n

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

Answer: (d) K+<\/sup> < Ca2+<\/sup> < Mg2+<\/sup> < Be2+<\/sup>
\nExplanation:
\nHigh charge and small size of the cations increases polarisation.
\nAs the size of the given cations decreases as
\nK+<\/sup> > Ca2+<\/sup> > Mg2+<\/sup> > Be2+<\/sup>
\nHence, polarising power decreases as K+<\/sup> < Ca2+<\/sup> < Mg2+<\/sup> < Be2+<\/sup><\/p>\n<\/details>\n

\n

Question 6.
\nWhich one of the following does not have sp\u00b2 hybridised carbon?
\n(a) Acetone
\n(b) Acetic acid
\n(c) Acetonitrile
\n(d) Acetamide<\/p>\n

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

Answer: (c) Acetonitrile
\nExplanation:
\nAcetonitrile does not contain sp\u00b2 hybridized carbon.<\/p>\n<\/details>\n

\n

Question 7.
\nWhich one of the following is paramagnetic?
\n(a) NO+<\/sup>
\n(b) CO
\n(c) O2<\/sub>–<\/sup>
\n(d) CN<\/p>\n

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

Answer: (c) O2<\/sub>–<\/sup>
\nExplanation:
\nO2<\/sub>–<\/sup>(17)Super oxide has one unpaired electron. Since Paramagnetism is shown by those molecules which have at least one unpaired electron, hence, O2<\/sub>–<\/sup> is paramagnetic.
\nOption 1)
\nNO+<\/sup>
\nThis solution is incorrect
\nOption 2)
\nCO
\nThis solution is incorrect
\nOption 3)
\nO2<\/sub>–<\/sup>
\nThis solution is correct
\nOption 4)
\nCN–<\/sup>
\nThis solution is incorrect<\/p>\n<\/details>\n

\n

Question 8.
\nWhich of the following structures will have a bond angle of 120\u00b0 around the central atom?
\n(a) Linear
\n(b) Tetrahedral
\n(c) Triangular
\n(d) Square planar<\/p>\n

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

Answer: (c) Triangular
\nExplanation:
\nWhen three electrons pairs get as far apart from each other, a trigonal planar structure is formed. The bond angle in this structure will be 120\u00b0.<\/p>\n<\/details>\n

\n

Question 9.
\nAn atom of an element A has three electrons in its outermost orbit and that of B has six electrons in its outermost orbit. The formula of the compound between these two will be
\n(a) A3<\/sub> B6<\/sub>
\n(b) A2<\/sub> B3<\/sub>
\n(c) A3<\/sub> B2<\/sub>
\n(d) A2<\/sub> B<\/p>\n

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

Answer: (b) A2<\/sub> B3<\/sub>
\nExplanation:
\nA has 3 electrons in outermost orbit and B has 6 electrons in its outermost orbits. So A can give three electrons to complete its octet and B needs 2 electrons to complete its octet. So 2 atoms of A will release 6 electrons and 3 atoms of B will need six electrons to complete their octet
\nSo, the formula will be A2<\/sub>\u200bB3<\/sub><\/p>\n<\/details>\n

\n

Question 10.
\nIn which of the following , the angle around the central atom is largest?
\n(a) CS2<\/sub>
\n(b) SF4<\/sub>
\n(c) SO2<\/sub>
\n(d) BBR3<\/sub><\/p>\n

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

Answer: (b) SF4<\/sub>
\nExplanation:
\nBond angle is determined by the hybridization.
\nSO2<\/sub> \u200b undergo sp3 hybridization and exhibits 109\u00b0 bond angle
\nBBr3<\/sub>\u200b undergo sp2 hybridization and exhibits 120\u00b0 bond angle
\nCS2<\/sub>\u200b undergo sp hybridization and exhibits 180\u00b0 bond angle.
\nSF4<\/sub>\u200b undergo sp3 d hybridization and exhibits different bond angles.
\nSo, the least bond angle is exhibited by CS2<\/sub><\/p>\n<\/details>\n

\n

Question 11.
\nBased on lattice enthalpy and other considerations which one the following alkali metals chlorides is expected to have the higher melting point?
\n(a) RbCl
\n(b) KCl
\n(c) NaCl
\n(d) LiCl<\/p>\n

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

Answer: (c) NaCl
\nExplanation:
\nThe highest melting point will be NaCl, it is because, the lattice energy decreases as the size of alkali metal increases so going down the group the melting point decreases, but due to the covalent bonding in LiCl, its melting point is lower than NaCl and so NaCl is expected to have maximum melting point in the alkali chlorides.\u200b<\/p>\n<\/details>\n

\n

Question 12.
\nIn which of the following substances, the intermolecular forces are hydrogen bonds?
\n(a) Hydrogen Chloride
\n(b) Hydrogen Sulphide
\n(c) Dry Ice
\n(d) Ice<\/p>\n

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

Answer: (d) Ice
\nExplanation:
\nIce is held together by hydrogen bonds.<\/p>\n<\/details>\n

\n

Question 13.
\nWhich one of the following pairs of species have the same bond order?
\n(a) CN\u2212<\/sup> and NO+<\/sup>
\n(b) CN\u2212<\/sup> and CN+<\/sup>
\n(c) O2<\/sub>\u2212<\/sup> and CN\u2212<\/sup>
\n(d) NO+<\/sup> and CN+<\/sup><\/p>\n

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

Answer: (d) NO+<\/sup> and CN+<\/sup>
\nExplanation:
\nCN\u2212<\/sup> and NO+<\/sup> are isoelectronic 14 and have same bond order.<\/p>\n<\/details>\n

\n

Question 14.
\nDipole-induced dipole interactions are present in which of the following pairs?
\n(a) H2<\/sub>O and alcohol
\n(b) Cl2<\/sub> and CCl4<\/sub>
\n(c) HCl and He atoms
\n(d) SiF4<\/sub> and He atoms<\/p>\n

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

Answer: (c) HCl and He atoms
\nExplanation:
\nHCl is polar (\u03bc \u2260 0) and He is non-polar (\u03bc = 0) gives dipole-induced dipole interaction.<\/p>\n<\/details>\n

\n

Question 15.
\nIn allene (C3<\/sub>H4<\/sub>), the type(s) of hybridisation of the carbon atoms is (are)
\n(a) sp and sp\u00b3
\n(b) sp and sp\u00b2
\n(c) Only sp\u00b2
\n(d) sp\u00b2 and sp\u00b3<\/p>\n

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

Answer: (b) sp and sp\u00b2
\nExplanation:
\nAllene: H2<\/sub>\u200bC = C = CH2<\/sub>. The central carbon is attached to other carbons by two sigma and two pi bonds so its hybridisation will be sp. The terminal carbon is attached to other carbon and hydrogen by 3 sigma and 2 pi bonds and hence, its hybridisation will be sp\u00b2<\/p>\n<\/details>\n

\n

Question 16.
\nThe weakest interparticle forces are found in which of the following?
\n(a) Ionic Solids
\n(b) Metallic Solids
\n(c) Molecular Solids
\n(d) All types of solids are equal in terms of interparticle forces.<\/p>\n

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

Answer: (c) Molecular Solids
\nExplanation:
\nMolecular solids are the weakest.
\nBecause Ionic solids and metallic solids are a type of bond. Bonds are stronger than Van der Waals forces.<\/p>\n<\/details>\n

\n

Question 17.
\nWhich of the following types of hybridisation leads to three dimensional geometry of bonds around the carbon atom ?
\n(a) sp
\n(b) sp\u00b2
\n(c) sp\u00b3
\n(d) None of these<\/p>\n

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

Answer: (b) sp\u00b2
\nExplanation:
\nsp\u00b2 hybrid structures have trigonal planar geometry, which is two dimensional.<\/p>\n<\/details>\n

\n

Question 18.
\nIf the bond length and dipole moment of a diatomic molecule are 1.25 A and 1.0 D respectively, what is the percent ionic character of the bond?
\n(a) 10.66
\n(b) 12.33
\n(c) 16.66
\n(d) 19.33<\/p>\n

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

Answer: (c) 16.66
\nExplanation:
\n\u03bcionic<\/sub> = (1.6 \u00d7 10\u221219<\/sup> C)(1.25 \u00d7 10\u221210<\/sup> m)\/(3.335 \u00d7 10\u221230<\/sup>) (Cm\/D)
\n= 5.99 D
\n%ionic character = (100 \u00d7 \u03bcobs<\/sub>)\/ (\u03bcionic<\/sub>)
\n=100 \u00d7 15.99 = 16.66%<\/p>\n<\/details>\n

\n

Question 19.
\nThe number of types of bonds between two carbon atoms in calcium carbide is
\n(a) Two sigma, two pi
\n(b) One sigma, two pi
\n(c) One sigma, one pi
\n(d) Two sigma, one pi<\/p>\n

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

Answer: (b) One sigma, two pi
\nExplanation:
\nA single bond between two atoms is always considered as sigma bond.
\nA double bond between two atoms is always considered as one sigma and one pi bond
\nA triple bond between two atoms is always considered as one sigma bond and two pi bonds.
\nSo according to the given structure CaC2<\/sub> (Calcium carbide) has 1 sigma and 2 pi bonds<\/p>\n<\/details>\n

\n

Question 20.
\nWhich of the following species contain non-directional bonds ?
\n(a) NCl3<\/sub>
\n(b) BeCl2<\/sub>
\n(c) BCl3<\/sub>
\n(d) RbCl<\/p>\n

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

Answer: (b) BeCl2<\/sub>
\nExplanation:
\nIonic or electrovalent bonds are called non-directional bonds, the meaning of non-directional is that these type of bonds does not have any special type of geometry, that is only attraction of positive and negative charge as we know ionic bonds made between metal[positively charged] and non-metal[negatively charged]<\/p>\n

RbCl is ionic compound and non-directional<\/p>\n<\/details>\n

\n

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Students are advised to practice the\u00a0NCERT MCQ Questions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure 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 Chemical Bonding and Molecular Structure Class 11 MCQs …<\/p>\n

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