MVM Gateway Training Program Chapter Wise Test - Chemistry PDF

**MVM Gateway Training Program** **Chapter Wise Test -- Chapter 2 -- XI -- Chemistry** **Chapter: *Atomic Structure*** 1. The radius of second Bohr\'s orbits for hydrogen atoms is: (h = 6.6262×10^-34^ J s, m~e~ = 9.109 × 10^-31^ kg, e = 1.6021 × 10^-19^C) (A)1.65 Å (B) 4.76 Å (C) 0.529...

**MVM Gateway Training Program** **Chapter Wise Test -- Chapter 2 -- XI -- Chemistry** **Chapter: *Atomic Structure*** 1. The radius of second Bohr\'s orbits for hydrogen atoms is: (h = 6.6262×10^-34^ J s, m~e~ = 9.109 × 10^-31^ kg, e = 1.6021 × 10^-19^C) (A)1.65 Å (B) 4.76 Å (C) 0.529 Å (D)2.12 Å **2.** Total number of orbitals associated with third shell will be \_\_\_\_\_\_\_\_\_\_. \(A) 2. (B) 4. (C) 9. (D) 3. \(A) 50 (B) 75 (C) 10 (D) 25 **4.** Who was the first scientist to propose a model for the structure of an atom? \(A) J.J. Thomson (B) Dalton (C) Ernest Rutherford (D) E. Goldstein **5.** Number of angular nodes for 4d orbital is \_\_\_\_\_\_\_. \(A) 4 (B) 3 (C) 2 (D) 1 **6. The mass number of a nucleus is:** **(A) Always less than its atomic number.** **(B) Always more than its atomic number.** **(C) Sometimes equal to its atomic number.** **(D) Sometimes equal and sometimes more than its atomic number** ***In the following questions, two statements are given---one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:*** ***(a) Both Assertion (A) and Reason (R) are correct statements, and Reason (R) is the correct explanation of the Assertion (A).*** ***(b) Both Assertion (A) and Reason (R) are correct statements, but Reason (R) is not the correct explanation of the Assertion (A).*** ***(c) Assertion (A) is correct, but Reason (R) is incorrect statement.*** ***(d) Assertion (A) is incorrect, but Reason (R) is correct statement.*** **7.** **Assertion(A):** Black body is an ideal body that emits and absorbs radiations of all frequencies. **Reason(R):** The frequency of radiation emitted by a body goes from a lower frequency to higher frequency with an increase in temperature. **Reason(R):** The path of an electron in an atom is clearly defined. **9.** **Assertion(A):** All isotopes of a given element show the same type of chemical behaviour. **Reason(R):** The chemical properties of an atom are controlled by the number of electrons in the atom. **10.** **Assertion(A):** Atoms are electrically neutral. **Reason(R):** Number of protons and electrons are different. **11.** How many electrons will be present in the subshells having m~s~ value of for n = 4? **12.** Prove that if the uncertainty in position of a moving electron is equal to its de-Broglie wavelength then its velocity is completely uncertain. **13.** The Vividh Bharati station of All India Radio, Delhi, broadcasts on a frequency of 1,368 kHz. Calculate the wavelength of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to? **[Section -- C (SA II)] (2×3= 6)** **14. (i) An orbital has n = 2, what are the possible values of** [*l*]{.math.inline} **and**[ *m*]{.math.inline}**?** **15.** (i) What is common between [*d*~xy ~]{.math.inline} and [*d*~*x*^2^ − *y*^2^~]{.math.inline} orbitals? **16.** Orbitals are region or space where there is maximum probability of finding electrons. Qualitatively, these orbitals can be distinguished by their size, shape and orientation. An orbital of small size means there is more chance of finding the electron near the nucleus. Shape and orientation means the direction in which probability of finding electron is the maximum. Atomic orbitals can be distinguished by quantum numbers. Each orbital is designated by three quantum numbers n, l and m~l~ (magnetic quantum number) which define energy, shape and orientation but these are not sufficient to explain spectra of multi-electrons atoms. Spin quantum number (m~s~) determines the spin of electron. Spin angular momentum of electron has two orientations relative to chosen axis which are distinguished by spin quantum numbers m~s~ which can take values +1/2 and -1/2. **17.** Bohr\'s model explained electrons can revolve only in certain permitted orbits whose angular momentum is integral multiple of h/2π, associated with fixed amount of energy. Bohr theory could successfully explain stability of atoms and spectrum of unielectron species. Hydrogen spectrum consist of Lyman, Balmer, Paschen, Brackett and Pfund series. Bohr\'s theory could not explain spectrum of multi-electron species, Stark effect, Zeeman effect, dual nature of matter, de Broglie equation and Heisenberg uncertainty principle which lead to orbital concept. Electrons were filled in orbitals according to Aufbau\'s principle, Hund\'s Rule and Pauli\'s exclusion principle. Each electron is identified by four quantum numbers n, l, m~l~ and m~s~ out which n, I, m~l ~ were derived from Schrodinger\'s wave equation. Half filled and completely filled orbitals are more stable due to exchange energy and symmetrical distribution of electrons.\ (a) Heat treatment of muscular pain involves radiation of wavelength 900 nm which spectral line of H-atom is suitable for this purpose.\ (b)Arrange 4d, 3d, 4p and 3p in increasing order of energy.\ (c) What is name of spectrum of radiation emitted by substance that has absorbed radiation?\ (d) What rules out the probability of existence of definite path of electrons? **18.** (i). Show that the circumference of Bohr\'s orbit for the H-atom is an integral multiple of the de-Broglie wavelength of electron revolving around the orbit. (ii). Explain that the effect of the Heisenberg uncertainty principle is significant only for motion of microscopic objects and is negligible for that of macroscopic objects. \[h = 6.626 ×10^-34^Js, Mass of electron = 9.1 ×10^-31^ ~kg~\]. (iii). State Hund\'s Rule of maximum multiplicity. **19**. A photon of wavelength 4 × 10^-7^ m strikes on metal surface, the work function of the metal being 2.13eV. Calculate: (i). The energy of the photon (eV), (ii). The kinetic energy of the emission, (iii). The velocity of the photoelectron (1 eV = 1.6020 × 10^-19^ J).