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The arrangement of orbitals on the basis of energy is based upon their (n + l) value. Lower the value of (n + l), lower is the energy. For orbitals having same values of (n + l), the orbital with lower value of n will have lower energy.
Based upon the above information, arrange the following orbitals in the increasing order of energy.
5f, 6d, 7s, 7p
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The arrangement of orbitals on the basis of energy is based upon their (n + l) value. Lower the value of (n + l), lower is the energy. For orbitals having same values of (n + l), the orbital with lower value of n will have lower energy.
Based upon the above information, solve the questions given below:
Which of the following orbitals has the lowest energy?
4d, 4f, 5s, 5p
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The arrangement of orbitals on the basis of energy is based upon their (n + l) value. Lower the value of (n + l), lower is the energy. For orbitals having same values of (n + l), the orbital with lower value of n will have lower energy.
Based upon the above information, solve the questions given below:
Which of the following orbitals has the lowest energy?
5p, 5d, 5f, 6s, 6p
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The electronic configuration of valence shell of Cu is 3d104s1 and not 3d94s2. How is this configuration explained?
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What is the difference between the terms orbit and orbital?
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Match the following species with their corresponding ground state electronic configuration.
| Atom / Ion | Electronic configuration |
| (i) \[\ce{Cu}\] | (a) 1s2 2s2 2p6 3s2 3p6 3d10 |
| (ii) \[\ce{Cu^{2+}}\] | (b) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 |
| (iii) \[\ce{Zn^{2+}}\] | (c) 1s2 2s2 2p6 3s2 3p6 3d10 4s1 |
| (iv) \[\ce{Cr^{3+}}\] | (d) 1s2 2s2 2p6 3s2 3p6 3d9 |
| (e) 1s2 2s2 2p6 3s2 3p6 3d3 |
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Match the quantum numbers with the information provided by these.
| Quantum number | Information provided |
| (i) Principal quantum number | (a) orientation of the orbital |
| (ii) Azimuthal quantum number | (b) energy and size of orbital |
| (iii) Magnetic quantum number | (c) spin of electron |
| (iv) Spin quantum number | (d) shape of the orbital |
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Match the following
| (i) Photon | (a) Value is 4 for N shell |
| (ii) Electron | (b) Probability density |
| (iii) ψ2 | (c) Always positive value |
| (iv) Principal quantum number n | (d) Exhibits both momentum and wavelength |
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Match species given in Column I with the electronic configuration given in Column II.
| Column I | Column II |
| (i) \[\ce{Cr}\] | (a) [Ar]3d84s0 |
| (ii) \[\ce{Fe^{2+}}\] | (b) [Ar]3d104s1 |
| (iii) \[\ce{Ni^{2+}}\] | (c) [Ar]3d64s0 |
| (iv) \[\ce{Cu}\] | (d) [Ar] 3d54s1 |
| (e) [Ar]3d64s2 |
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Which of the following property of water can be used to explain the spherical shape of rain droplets?
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How does the surface tension of a liquid vary with increase in temperature?
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The magnitude of surface tension of liquid depends on the attractive forces between the molecules. Arrange the following in increasing order of surface tension:
water, alcohol \[\ce{(C2H5OH)}\] and hexane \[\ce{[CH3(CH2)4CH3)]}\].
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Name two phenomena that can be explained on the basis of surface tension.
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Why does sharp glass edge become smooth on heating it upto its melting point in a flame? Explain which property of liquids is responsible for this phenomenon.
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Explain the term ‘laminar flow’. Is the velocity of molecules same in all the layers in laminar flow? Explain your answer.
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Consider the reactions given below. On the basis of these reactions find out which of the algebric relations given in options (i) to (iv) is correct?
(a) \[\ce{C (g) + 4 H (g) -> CH4 (g); ∆_rH = xkJ mol^{-1}}\]
(b) \[\ce{C (graphic) + 2H2 (g) -> CH4 (g); ∆_rH = ykJ mol^{-1}}\]
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One mole of acetone requires less heat to vapourise than 1 mol of water. Which of the two liquids has higher enthalpy of vapourisation?
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The enthalpy of atomisation for the reaction \[\ce{CH4(g) -> C(g) + 4H(g)}\] is 1665 kJ mol–1. What is the bond energy of \[\ce{C – H}\] bond?
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Use the following data to calculate \[\ce{∆_{lattice}H^Θ}\] for \[\ce{NaBr}\].
\[\ce{∆_{sub}H^Θ}\] for sodium metal = 108.4 kJ mol–1
Ionization enthalpy of sodium = 496 kJ mol–1
Electron gain enthalpy of bromine = – 325 kJ mol–1
Bond dissociation enthalpy of bromine = 192 kJ mol–1
\[\ce{∆_fH^Θ}\] for \[\ce{NaBr (s)}\] = – 325 kJ mol–1
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The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction.
\[\ce{H2(g) + Br(g) -> 2HBr(g)}\]
Given that Bond energy of \[\ce{H2, Br2}\] and \[\ce{HBr}\] is 435 kJ mol–1, 192 kJ mol–1 and 368 kJ mol–1 respectively.
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