{"id":17572,"date":"2022-05-24T11:00:05","date_gmt":"2022-05-24T05:30:05","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=17572"},"modified":"2022-05-16T15:25:21","modified_gmt":"2022-05-16T09:55:21","slug":"mcq-questions-for-class-11-chemistry-chapter-2","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/mcq-questions-for-class-11-chemistry-chapter-2\/","title":{"rendered":"MCQ Questions for Class 11 Chemistry Chapter 2 Structure of Atom with Answers"},"content":{"rendered":"

Students are advised to practice the\u00a0NCERT MCQ Questions for Class 11 Chemistry Chapter 2 Structure of Atom 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 Structure of Atom Class 11 MCQs Questions with Answers and assess their preparation level.<\/p>\n

Structure of Atom Class 11 MCQs Questions with Answers<\/h2>\n

Solving the Structure of Atom Multiple Choice Questions of Class 11 Chemistry Chapter 2 MCQ can be of extreme help as you will be aware of all the concepts. These MCQ Questions on Structure of Atom 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 2 Chemistry Class 11 and cross-check your answers during preparation.<\/p>\n

Question 1.
\nWhich of the following pair of ions have same paramagnetic moment?
\n(a) Cu+2<\/sup>, Ti+3<\/sup>
\n(b) Mn+2<\/sup>, Cu+2<\/sup>
\n(c) Ti+4<\/sup>, Cu+2<\/sup>
\n(d) Ti+3<\/sup>, Ni+2<\/sup><\/p>\n

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

Answer: (a) Cu+2<\/sup>, Ti+3<\/sup>
\nExplanation:
\n29<\/sub>Cu = [Ar] 3d10<\/sup> 4s1<\/sub>
\nCu2+<\/sup> = [Ar] 3d9<\/sup> (n = 1)
\n\"MCQ
\n22<\/sub>Ti = [Ar] 3d\u00b2 4s\u00b2
\nTi3+<\/sup> = [Ar] 3d\u00b2 (n = 1)
\n\"MCQ
\nBoth of these ions have one unpaired electron, hence these have same paramagnetic moment.<\/p>\n<\/details>\n

\n

Question 2.
\nThe charge to mass ratio of \u03b1 – particles is approximately \u2026\u2026 the charge to mass ratio of protons
\n(a) Twice
\n(b) Half
\n(c) Four times
\n(d) Six times<\/p>\n

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

Answer: (b) Half
\nExplanation:
\nLet charge of proton be +e, then charge of alpha particle will be + 2e. Similarly let mass of proton be m, then mass of alpha particle will be 4m.
\nNow, specific charge = (charge) \/ (mass of the substance.)
\nFor proton, specific charge = (e\/m)
\nFor alpha particle, specific charge = (2e)\/ (4m)
\nTherefore there ratio is: (e\/m) \u00d7 (4m)\/ (2e) = 2 : 1<\/p>\n<\/details>\n

\n

Question 3.
\nThe frequency of a wave of light is 12 \u00d7 1014<\/sup>s-1<\/sup>. The wave number associated with this light
\n(a) 5 \u00d7 10-7<\/sup>m
\n(b) 4 \u00d7 10-8<\/sup>cm-1<\/sup>
\n(c) 2 \u00d7 10-7<\/sup>m-1<\/sup>
\n(d) 4 \u00d7 104<\/sup>cm-1<\/sup><\/p>\n

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

Answer: (d) 4 \u00d7 104 <\/sup>cm-1<\/sup>
\nExplanation:
\nFrequency \u03bd = 12 \u00d7 1014<\/sup>s-1<\/sup> and
\nvelocity of light c = 3 \u00d7 1010 <\/sup>cms-1<\/sup>.
\nWe know that the wave number \u03bd–<\/sup> = (v\/c)
\n= (12 \u00d7 1014<\/sup>)\/(3 \u00d7 1010<\/sup>)
\n= 4 \u00d7 104<\/sup> cm-1<\/sup><\/p>\n<\/details>\n

\n

Question 4.
\nIn a multi – electron atom, which of the following orbitals described by the three quantum numbers will have the same energy in the absence of magnetic acid and electric fields? (a) n = 1, l = 0, m = 0 (b) n = 2, l = 0, m = 0 (c) n = 2, l = 1, m = 1 (d) n = 3, l = 2, m = 1 (e) n = 3, l = 2, m = 0
\n(a) (a) and (b)
\n(b) (b) and (c)
\n(c) (c) and (d)
\n(d) (d) and (e)<\/p>\n

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

Answer: (d) (d) and (e)
\nExplanation:
\nIn the absence of magnetic and electric fields, the orbitals defined by magnetic quantum number are degenerate (of same energy). So, energy of the orbital, in the absence of magnetic and electric fields depends on the (n + l) value.
\nHigher the (n + l) value, larger the energy of the orbital.
\nOrbitals with same n and n + 1 values are degenerate and have same energy.
\nIn the given combinations, d and e have same n and n + l value and so, have same energy in the absence of magnetic and electric fields.<\/p>\n<\/details>\n

\n

Question 5.
\nThe electronic transitions from n = 2 to n = 1 will produce shortest wavelength in (where n = principal quantum state)
\n(a) Li2+<\/sup>
\n(b) He+<\/sup>
\n(c) H
\n(d) H+<\/sup><\/p>\n

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

Answer: (a) Li2+<\/sup>
\nExplanation:
\n(1\/\u03bb) = Z\u00b2RH<\/sub> [(1\/n1<\/sub>\u00b2) \u2013 (1\/n2<\/sub>\u00b2)], n1<\/sub> = 1, n2<\/sub> = 2
\nTherefore, (1\/ \u03bb) = Z\u00b2RH<\/sub> [(1\/1) \u2013 (1\/4)]
\n= (3\/4) Z\u00b2RH<\/sub>
\n\u03bb \u221d (1\/Z)\u00b2
\nTherefore, Li2+<\/sup> will produce shortest wave length.<\/p>\n<\/details>\n

\n

Question 6.
\nIn a hydrogen atom, if energy of an electron in ground state is 13.6 eV, then that in the 2nd excited state is
\n(a) 1.51 eV
\n(b) 3.4 eV
\n(c) 6.04 eV
\n(d) 13.6 eV<\/p>\n

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

Answer: (a) 1.51 eV
\nExplanation:
\nEn = (13.6)\/(n\u00b2) eV
\nor E = (13.6)\/(9)eV
\n= 1.51 eV<\/p>\n<\/details>\n

\n

Question 7.
\nThe credit of discovering neutron goes to
\n(a) Rutherford
\n(b) Thomson
\n(c) Goldstein
\n(d) Chadwick<\/p>\n

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

Answer: (d) Chadwick
\nExplanation:
\nThe essential nature of the atomic nucleus was established with the discovery of the neutron by James Chadwick in 1932 and the determination that it was a new elementary particle, distinct from the proton.<\/p>\n<\/details>\n

\n

Question 8.
\nThe maximum number of electrons that can be accommodated in fifth energy level is
\n(a) 10
\n(b) 25
\n(c) 50
\n(d) 32<\/p>\n

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

Answer: (c) 50
\nExplanation:
\nThe maximum number of electrons that can occupy a given energy level n is given by
\nmax. no. of electrons = 2n\u00b2
\nSo the number of orbitals that are present in an energy level n is given by
\nof orbitals = n\u00b2
\nAlso a given orbital can hold a maximum of 2 electrons, which is why the maximum number of electrons that can be added to a given energy level n is twice the number of orbitals present on said energy level.
\nAccording to given,
\nn = 5, it refers to the fifth energy level, holds
\nof orbitals = 5\u00b2
\nof orbitals = 25
\nThis means that the maximum number of electrons that can be added to the fifth energy level is
\nmax no. of electrons = 2 \u00d7 25
\n= 50 electrons<\/p>\n<\/details>\n

\n

Question 9.
\nAccording to Aufbaus principle, which of the three 4d, 5p and 5s will be filled with electrons first
\n(a) 4d
\n(b) 5p
\n(c) 5s
\n(d) 4d and 5s will be filled simultaneously<\/p>\n

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

Answer: (c) 5s
\nExplanation:
\nAccording to the Aufbaus principle, electron will be first enters in those orbital which have least energy. So decreasing order of energy is 5p > 4d > 5s.<\/p>\n<\/details>\n

\n

Question 10.
\nA hydrogen atom in its ground state absorbs 10.2 eV of energy. The orbital angular momentum is increased by (Given Planck constant h = 6.6 \u00d7 10-34<\/sup> Jsec)
\n(a) 1.05 \u00d7 10-34<\/sup> Jsec
\n(b) 3.16 \u00d7 10-34<\/sup> Jsec
\n(c) 2.11 \u00d7 10-34<\/sup> Jsec
\n(d) 4.22 \u00d7 10-34<\/sup> Jsec<\/p>\n

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

Answer: (a) 1.05 \u00d7 10-34<\/sup> Jsec
\nExplanation:
\nElectron after absorbing 10.2 eV energy goes to its first excited state (n = 2) from ground state (n = 1).
\nTherefore, Increase in momentum = (h)\/(2\u03c0)
\n= (6.6 \u00d7 10-34<\/sup>)\/ (6.28)
\n= 1.05 \u00d7 10-34<\/sup> Js.<\/p>\n<\/details>\n

\n

Question 11.
\nThe ionization enthalpy of hydrogen atom is 1.312 \u00d7 106<\/sup> J mol-1<\/sup>. The energy required to excite the electron in the atom from n = 1 to n = 2 is
\n(a) 8.51 \u00d7 105<\/sup> Jmol-1<\/sup>
\n(b) 6.56 \u00d7 105<\/sup> Jmol-1<\/sup>
\n(c) 7.56 \u00d7 105<\/sup> Jmol-1<\/sup>
\n(d) 9.84 \u00d7 105<\/sup> Jmol-1\/sup><\/sup><\/p>\n

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

Answer: (d) 9.84 \u00d7 105<\/sup>J mol-1<\/sup>
\nExplanation:
\nEnergy required when an electron makes transition from n = 1 to n = 2
\nE2<\/sub> = \u2212(1.312 \u00d7 106<\/sup> \u00d7 (1)\u00b2)\/(2\u00b2)
\n= \u22123.28 \u00d7 105<\/sup>J mol-1<\/sup>
\nE1<\/sub> = \u22121.312 \u00d7 106<\/sup>J mol-1<\/sup>
\n\u0394E = E2<\/sub> \u2212 E1<\/sub>
\n= \u22123.28 \u00d7 105<\/sup> – (\u221213.2 \u00d7 106<\/sup>)
\n\u0394E = 9.84 \u00d7 105<\/sup>J mol-1<\/sup><\/p>\n<\/details>\n

\n

Question 12.
\nFor principal quantum number n = 4, the total number of orbitals having l = 3 is
\n(a) 3
\n(b) 7
\n(c) 5
\n(d) 9<\/p>\n

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

Answer: (b) 7
\nExplanation:
\nFor n = 4 and l = 3, the orbital is 4f.
\nNumber of values of m= no. of orbitals= (2l + 1) = 7.<\/p>\n<\/details>\n

\n

Question 13.
\nMaximum number of electrons in a subshell with l = 3 and n = 4 is
\n(a) 10
\n(b) 12
\n(c) 14
\n(d) 16<\/p>\n

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

Answer: (c) 14
\nExplanation:
\nn = 4, so 4th shell and l = 3 so it is f subshell.
\nThus n = 4, l = 3 indicates 4f orbitals.
\nIn f subshell there are 7 orbitals and each orbital can accommodate a maximum of 2 electrons. So, maximum no. of electrons in 4f subshell = 7 \u00d7 2 = 14.<\/p>\n<\/details>\n

\n

Question 14.
\nWhich hydrogen-like species will have same radius as that of Bohr orbit of hydrogen atom?
\n(a) n = 2, Li2+<\/sup>
\n(b) n = 2, Be3+<\/sup>
\n(c) n = 2, He+<\/sup>
\n(d) n = 3, Li2+<\/sup><\/p>\n

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

Answer: (b) n = 2, Be3+<\/sup>
\nExplanation:
\nr = (r0<\/sub>) \u00d7 (n\u00b2\/Z) \u00c5
\nwhereas,
\nr0<\/sub> = radius of 1st Bohrs orbit of hydrogen atom = 0.529 \u00c5
\nFor r = r0<\/sub>–<\/sup>
\n(n\u00b2\/Z) = 1
\nn\u00b2 = Z (1)
\nBecause for n = 2 , Be+3<\/sup>
\nZ = 4, which satisfies equation (1).
\nHence, n = 2,Be+3<\/sup> will have the same radius of 1 st Bohrs orbit of a hydrogen atom.<\/p>\n<\/details>\n

\n

Question 15.
\nThe magnetic quantum number specifies
\n(a) Size of orbitals
\n(b) Shape of orbitals
\n(c) Orientation of orbitals
\n(d) Nuclear Stability<\/p>\n

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

Answer: (c) Orientation of orbitals
\nExplanation:
\nThe magnetic quantum number specifies orientation of orbitals.<\/p>\n<\/details>\n

\n

Question 16.
\nIn Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inner-orbit jumps of the electron for Bohr orbits in an atom of hydrogen?
\n(a) 3 \u2192 2
\n(b) 5 \u2192 2
\n(c) 4 \u2192 1
\n(d) 2 \u2192 5<\/p>\n

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

Answer: (b) 5 \u2192 2
\nExplanation:
\nSince the line in hydrogen spectrum lies within visible region that is at red end, therefore it is corresponds to the Balmer series.
\nThe line at the red end suggests that
\nThe first line of Balmer series is n = 3 to n =2
\nThe second line of Balmer series is n = 4 to n = 2
\nThe third line of Balmer series is n = 5 to n = 2<\/p>\n<\/details>\n

\n

Question 17.
\nIn chromium atom, in ground state, the number of occupied orbitals is
\n(a) 14
\n(b) 15
\n(c) 7
\n(d) 12<\/p>\n

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

Answer: (b) 15
\nExplanation:
\nThe configuration of
\n24<\/sub>Cr is 1s\u00b2, 2s\u00b2, 2p6<\/sup>, 3s\u00b2, 3p6<\/sup>, 3d5<\/sup>, 4s1<\/sup>
\nTherefore, total s-orbitals = 4
\ntotal p-orbitals = 6
\ntotal d-orbitals = 5 and thus
\nThus, total orbitals = 4 + 6 + 5 = 15.<\/p>\n<\/details>\n

\n

Question 18.
\nA sub-shell with n = 6 , l = 2 can accommodate a maximum of
\n(a) 12 electrons
\n(b) 36 electrons
\n(c) 10 electrons
\n(d) 72 electrons<\/p>\n

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

Answer: (c) 10 electrons
\nExplanation:
\nn = 6, \u2113 = 2 means 6d \u2192 will have 5 orbitals.
\nTherefore max 10 electrons can be accommodated as each orbital can have maximum of 2 electrons.<\/p>\n<\/details>\n

\n

Question 19.
\nWhich of the following sets of quantum numbers represents the highest energy of an atom?
\n(a) n = 3, l = 0, m = 0, s = +1\/2
\n(b) n = 3, l = 1, m = 1, s = +1\/2
\n(c) n = 3, l = 2, m = 1, s = +1\/2
\n(d) n = 4, l = 0, m = 0, s = +1\/2<\/p>\n

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

Answer: (c) n = 3, l = 2, m = 1, s = + 1\/2
\nExplanation:
\nn = 3, l = 0 represents 3s orbital n = 3, l = 1 represents 3p orbital n = 3, l = 2 represents 3d orbital n = 4, l = 0 represents 4s orbital The order of increasing energy of the orbitals is 3s < 3p < 4s < 3d.<\/p>\n<\/details>\n

\n

Question 20.
\nThe value of Plancks constant is 6.63 \u00d7 10-34<\/sup>Js. The velocity of light is 3.0 \u00d7 108<\/sup>ms-1<\/sup>. Which value is closest to the wavelength in nanometres of a quantum of light with frequency of 8 \u00d7 1015<\/sup>s-1<\/sup>
\n(a) 3 \u00d7 107<\/sup>
\n(b) 2 \u00d7 10-25<\/sup>
\n(c) 5 \u00d7 10-18<\/sup>
\n(d) 4 \u00d7 101<\/sup><\/p>\n

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

Answer: (d) 4 \u00d7 101<\/sup>
\nExplanation:
\n\u03bb = (c\/v) = (3 \u00d7 108<\/sup>)\/(8 \u00d7 1015<\/sup>)
\n= 3.75 \u00d7 10-8<\/sup>
\n= 3.75 \u00d7 10-8<\/sup> \u00d7 109<\/sup> nm
\n= 4\u00d7101<\/sup> nm.<\/p>\n<\/details>\n

\n

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Students are advised to practice the\u00a0NCERT MCQ Questions for Class 11 Chemistry Chapter 2 Structure of Atom 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 Structure of Atom Class 11 MCQs Questions with Answers and …<\/p>\n

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