(c) 6.6 x 10-31 cm So, a 4p-orbital has (4 - 2) = 2 radial nodes. υ2 = 65.9 x 104 m2 s-2 n = 3, n = 4. Solution: (a) A and R are correct and R is the correct explanation of A. If l = 0, number of electrons = (2l + 1) Answer: Question 8. Answer: = \(\sqrt{(1(1+1)}) \mathrm{h} / 2 \pi\) Uncertainty in position = ∆x Answer: ∆x = 10-5m; ∆v =  = 5.27 x 10-28mv According to de Brogue equation, A = \(\frac {h}{mv}\) Answer: or = \(\frac{0.529 x 9}{3}\) Which of the following pairs of d-orbitals will have electron density along the axes ? Splitting of spectral lines in an electric field is called ……………. 1. represents the directional orientation of orbital Question 30. Planck’s quantum hypothesis: (d) 0.4, Question 23. So, 4p orbital has 2 radial nodes and 1 angular node. E6 = -13.6 / 62 ; E5 = – 13.6 / 52 If l = 0,  4s orbital = 1 orbital Answer: Wave length in cm = 1.065 x 10-34 x 100 (c) A is correct but R is wrong. Which of the following is not used in writing electronic configuration of an atom? l = azimuthal quantum number =1 Calculate the total number of angular nodes and radial nodes present in 3p orbital. Assumptions of Bohr atom model. (a) 4 (d) m . 2:38 33.2k LIKES Assign the symbol to the element. The diagram above shows: 1. sin2x has identical nodes to si… Albert Einstein proposed that light has dual nature, i.e. Number of angular nodes = 3 because l = 3 4f2 : It means that the element has 2 electrons in outermost 4f shell. Illustrate the significance of de Broglie equation with an iron ball and an electron. \(\widehat{\mathrm{H}}\) Ψ = EΨ ……………(1) Each orbital can hold two electrons (with opposite spins), giving the d orbitals a total capacity of 10 electrons. Energy of an electron in hydrogen atom in ground state is -13.6 eV. (b) \(\frac{\sqrt{2 \mathrm{h}}}{2 \pi}\) = 0 – \(\left(-\frac{2.18 \times 10^{-18}}{4} \mathrm{J}\right)\) = 5.45 x 10-19 J. Answer: Energy of an electron in ground state = -13.6 eV. What is exchange energy? For 4d orbital l = 2 n = 3, main shell is m. En = \(\frac{(-1312.8) Z^{2}}{n^{2}}\) kJ mol-1 Number of angular nodes in 3d orbital = ? Students can Download Chemistry Chapter 2 Quantum Mechanical Model of Atom Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Chemistry Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations. (c) 3d7 4s0 Question 46. (b) [Xe] 4f7, 6s2, [Xe] 4f7 5d1 6s2 and [Xe] 4f9 6s2 (d) 1s2 2s2 2p6 3s2 3p6 3d5 4s3 λ = 3 x 10-25 m. Question 37. If n = 4, l values are 0, 1, 2, 3 Find out the difference in number of angular nodes and number of radial nodes in the orbital to whichlast electron of chromium present. Therefore, here Planar orbital = 2 So the planar node of 4d-orbital is2. Mass of an electron = m = 9.1 x 10-31 kg. (b) n = 3, l = 0. = 1.2 x 10-11 m asked Mar 6, 2018 in Class XI Chemistry by vijay Premium ( 539 points) structure of atom The uncertainty in the position of a moving bullet of mass 10 g is 10 s m. Calculate the uncertainty in its velocity? Answer: Question 3. The spin quantum number represents the spin of the electron and is denoted by the letter ‘m. ∴ Equation (1) can be written as, (c) 2 l – l Solution: Number of angular nodes = l (c) n = 5 to n = 3 (b) 2, 1, +1, – ½ ∆v = Uncertainty in velocity = \(\frac {0.1}{100}\) x 2.2 x 103 ms-1 . Answer: Question 8. (d) \(\frac{\sqrt{6} h}{2 \pi}\) Energy of an electron in 2nd main shell = \(\frac{(-13.6) Z^{2}}{n^{2}}\); Z = 1, n = 2 f orbital. The region where the probability density function of electron reduces to zero is called During exchange process, the energy is released and the released energy is called exchange energy. 1. Cr3+ electronic configuration is Is2 2s2 2p6 3s23p6 3d3. More number of exchanges are possible only in the case of half filled and fully filled configurations. Orbitals in Physics and Chemistry is a mathematical function depicting the wave nature of an electron or a pair of electrons present in an atom. Question 8. Answer: Question 15. (b) – 82.05 k J mol-1 If l = 3 orbital = -3,-2, -1, 0, +1, +2, +3 = 7 A subdivision of the available space within an atom for an electron to orbit the nucleus. (b) 3, 1,-l, +½ (a) nickel For the electron, the de Broglie wavelength is significant and measurable while for the iron ball it is too small to measure, hence it becomes insignificant. Answer: (b) – 82.05 k J mol-1 Using Aufbau principle, write the ground state electronic configuration of following atoms. (a) 4, 2, -2, +½ If l = 0 orbital = 4s =1 Calculate the Bohr radius of the third orbit and calculate the energy of an electron in 4th orbit. λ = \(\frac{h}{mv}\) where h = 6.62 x 10-34 JS (a) Zeeman effect. (iv) for l = 0; m = -1 not possible \(\frac{1}{2}\) mv2 = 1.609 x 10-16V Momentum of electron, mv = \(\frac {h}{λ}\) de Broglie equation is …………… Total number of nodes in 4f orbital = 3. (d) s The angular nodes are either planar or conical in shape. (d) [Xe] 4f8 5d1 6s2[Xe] 4f7 5d1 6s2 and [Xe] 4f9 6s2 (a) equal to the number of protons For 4f orbital: Number of angular nodes = 3 because l = 3 Number of radial nodes = n – l – l =4 – 3 – 1 = 0 Total number of nodes in 4f orbital = 3. 1. Assertion : The spectrum of He+ is expected to be similar to that of hydrogen Calculate the energy required to remove an electron completely from the n = 2 orbit. (a) A and R are correct and R is the correct explanation of A. = 2.238 x 10-25 m. Question 38. (b) 1 Uncertainty in momentum = 0.8792 x 10-24 kg ms-1 (or) = 8.792 x 10-25 kg ms-1, Question 49. (2) v = 100 cm s-1 = 100 x 10-2 m s-1 So in the case of a 4f orbital, we have n=4. = 1.274 x 10-27 kg ms-1 ………(i) Louis de Broglie extended this concept and proposed that all forms of matter showed dual character. What Is The Orbital Angular Momentum Of An Electron In A 4f Orbital? The radial nodes for the nd orbital are (n − 3) nodes. Answer: To find n, solve the equation: nodes=n-1; in this case, 5=n-1, so n=6. s = spin quantum number = +\(\frac {1}{2}\) – \(\frac {1}{2}\). They allowed the accelerated beam of electrons to fall on a nickel crystal and recorded the diffraction pattern. For ns orbital, (n-1) nodes are found in it. The formula for getting the number of radial nodes is n - 2. (b) (i) and (ii). Question 12. (1) When n = 3 (c) Uncertainty principle Question 7. The f orbital is far more complex than the d orbital and observed in heavy elements. Now, Mass no. All p orbitals have a planar nodal surface. One electron is present in hydrogen atom, the four quantum numbers are n = 1, l = 0, m = 0 and s = + \(\frac {1}{2}\). He combined the following two equations of energy of which one represents wave character (hυ) and the other represents the particle nature (mc. Mn2+ electronic configuration is 1s 1s2 2s2 2p6 3s2 3p6 3d5, 2. (a) both assertion and reason are true and reason is the correct explanation of assertion. Answer: (c) – E / 9 Answer: The set of quantum numbers is not possible because for l = 0, m. The set of quantum numbers is not possible because, for n = 3, l = 3. Question 6. Number of radial nodes in 3d orbital = ? \(\frac{-13.6}{9}\) = -E Question 43. The stabilization of a half filled d-orbital is more pronounced than that of the p-orbital why? ∆x.m∆υ = \(\frac {h}{4π}\) or ∆υ = \(\frac {h}{4πm∆x}\); E = \(\frac{-13.6}{2^{2}}\) = \(\frac{-13.6}{4}\) = -3.4 atom-1. (c) 6 (c) rn= \(\frac{(-1312.8) Z^{2}}{n^{2}}\) kJ mol-1 The energies E1 and E2 of two radiations are 25 eV and 50 eV respectively. Classification of elements and periodicity in properties, General principles and process of Isolation of metals, S - block elements - alkali and alkaline earth metals, Purification and characteristics of organic compounds, Some basic principles of organic chemistry, Principles related to practical chemistry. Question 22. (a) λ = \(\frac {h}{mv}\). (b) dxz, dyz pp hybridisation (conjectured) In late period-8 elements a hybrid of 8p 3/2 and 9p 1/2 is expected to exist, where "3/2" and "1/2" refer to the total angular momentum quantum number. Number of electrons in the main shell = 2n2 n = 4, for N shell. (b) 2 (d) – 3.4 eV atom-1 (b) 22 Mention the shape of s, p, d orbitals. m  = 0.1 kg. (a) (i), (ii), (iii) and (iv) = 30 x 10-26 According to de Brogue equation, λ = \(\frac {h}{mv}\) (b) λ1 = 2 λ2 Question 27. ∆x = \(\frac{h}{∆v.4π}\) …………. (b) ∆E.∆v ≥ h/4πm (b) ii and iii i. (As the number of neutrons are 11.1% more than the number of electrons) (d) Stark effect Copper (Z = 29) of element = No. = 5400 x 10-10 m. Question 14. Solution: Question 21. (b) Neils Bohr ∆E = hv = hc / λ. Question 10. Calculation of the velocity of electron What is the maximum numbers of electrons that can be associated with the following set of quantum numbers ? 1 angular node means ℓ=1 which tells us that we have a p subshell, specifically the p z orbital because the angular node is on the xy plane. (c) \(\sqrt{2 \times 4}\) 1. What are quantum numbers? So, it contain 5 unpaired electrons. (6) 14 Enjoy :) (d) – 3.4 eV atom-1 Question 24. The exactly half filled orbitals have greater stability. (c) 1s2 2s2 2p6 3s2 3p6 3d6 Samacheer Kalvi 11th Chemistry Quantum Mechanical Model of Atom 2 – Marks Questions. The seeds in the atomic number only a thin gold foil ( d ) a is.. 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