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(a)
Interpretation:
The orbital which is higher in energy should be identified in the given pairs of hydrogen orbitals.
Concept Introduction:
The energies of orbitals in the hydrogen atom depend on the value of the principal quantum number (n). When n increases, energy also increases. For this reason, orbitals in the same shell have the same energy in spite of their subshell. The increasing order of energy of hydrogen orbitals is
1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f
In the case of one 2s and three 2p orbitals in the second shell, they have the same energy. In the third shell, all nine orbitals (one 3s, three 3p and five 3d) have the same energy. All sixteen orbitals (one 4s, three 4p, five 4d and seven 4f) in the fourth shell have the same energy.
The energy levels of the different orbitals in hydrogen atom are easily explained by considering the given diagram. Here, each box represents one orbital. Orbitals with the same principal quantum number (n) have the same energy.
Principal Quantum Number (n)
The principal quantum number (n) assigns the size of the orbital and specifies the energy of an electron. If the value of n is larger, then the average distance of an electron in the orbital from the nucleus will be greater. Therefore the size of the orbital is large. The principal quantum numbers have the integral values of 1, 2, 3 and so forth and it corresponds to the quantum number in
To find: Identify the orbital which is higher in energy in the given pair 1s, 2s orbitals of hydrogen.
Find the value of ‘n’
(b)
Interpretation:
The orbital which is higher in energy should be identified in the given pairs of hydrogen orbitals.
Concept Introduction:
The energies of orbitals in the hydrogen atom depend on the value of the principal quantum number (n). When n increases, energy also increases. For this reason, orbitals in the same shell have the same energy in spite of their subshell. The increasing order of energy of hydrogen orbitals is
1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f
In the case of one 2s and three 2p orbitals in the second shell, they have the same energy. In the third shell, all nine orbitals (one 3s, three 3p and five 3d) have the same energy. All sixteen orbitals (one 4s, three 4p, five 4d and seven 4f) in the fourth shell have the same energy.
The energy levels of the different orbitals in hydrogen atom are easily explained by considering the given diagram. Here, each box represents one orbital. Orbitals with the same principal quantum number (n) have the same energy.
Principal Quantum Number (n)
The principal quantum number (n) assigns the size of the orbital and specifies the energy of an electron. If the value of n is larger, then the average distance of an electron in the orbital from the nucleus will be greater. Therefore the size of the orbital is large. The principal quantum numbers have the integral values of 1, 2, 3 and so forth and it corresponds to the quantum number in Bohr’s model of the hydrogen atom. If all orbitals have the same value of ‘n’, they are said to be in the same shell (level). The total number of orbitals for a given n value is n2. As the value of ‘n’ increases, the energy of the electron also increases.
To find: Identify the orbital which is higher in energy in the given pair 2p, 3p orbitals of hydrogen
Find the value of ‘n’
(c)
Interpretation:
The orbital which is higher in energy should be identified in the given pairs of hydrogen orbitals.
Concept Introduction:
The energies of orbitals in the hydrogen atom depend on the value of the principal quantum number (n). When n increases, energy also increases. For this reason, orbitals in the same shell have the same energy in spite of their subshell. The increasing order of energy of hydrogen orbitals is
1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f
In the case of one 2s and three 2p orbitals in the second shell, they have the same energy. In the third shell, all nine orbitals (one 3s, three 3p and five 3d) have the same energy. All sixteen orbitals (one 4s, three 4p, five 4d and seven 4f) in the fourth shell have the same energy.
The energy levels of the different orbitals in hydrogen atom are easily explained by considering the given diagram. Here, each box represents one orbital. Orbitals with the same principal quantum number (n) have the same energy.
Principal Quantum Number (n)
The principal quantum number (n) assigns the size of the orbital and specifies the energy of an electron. If the value of n is larger, then the average distance of an electron in the orbital from the nucleus will be greater. Therefore the size of the orbital is large. The principal quantum numbers have the integral values of 1, 2, 3 and so forth and it corresponds to the quantum number in Bohr’s model of the hydrogen atom. If all orbitals have the same value of ‘n’, they are said to be in the same shell (level). The total number of orbitals for a given n value is n2. As the value of ‘n’ increases, the energy of the electron also increases.
To find: Identify the orbital which is higher in energy in the given pair 3dxy, 3dyz orbitals of hydrogen
Find the value of ‘n’
(d)
Interpretation:
The orbital which is higher in energy should be identified in the given pairs of hydrogen orbitals.
Concept Introduction:
The energies of orbitals in the hydrogen atom depend on the value of the principal quantum number (n). When n increases, energy also increases. For this reason, orbitals in the same shell have the same energy in spite of their subshell. The increasing order of energy of hydrogen orbitals is
1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f
In the case of one 2s and three 2p orbitals in the second shell, they have the same energy. In the third shell, all nine orbitals (one 3s, three 3p and five 3d) have the same energy. All sixteen orbitals (one 4s, three 4p, five 4d and seven 4f) in the fourth shell have the same energy.
The energy levels of the different orbitals in hydrogen atom are easily explained by considering the given diagram. Here, each box represents one orbital. Orbitals with the same principal quantum number (n) have the same energy.
Principal Quantum Number (n)
The principal quantum number (n) assigns the size of the orbital and specifies the energy of an electron. If the value of n is larger, then the average distance of an electron in the orbital from the nucleus will be greater. Therefore the size of the orbital is large. The principal quantum numbers have the integral values of 1, 2, 3 and so forth and it corresponds to the quantum number in Bohr’s model of the hydrogen atom. If all orbitals have the same value of ‘n’, they are said to be in the same shell (level). The total number of orbitals for a given n value is n2. As the value of ‘n’ increases, the energy of the electron also increases.
To find: Identify the orbital which is higher in energy in the given pair 3s, 3d orbitals of hydrogen
Find the value of ‘n’
(e)
Interpretation:
The orbital which is higher in energy should be identified in the given pairs of hydrogen orbitals.
Concept Introduction:
The energies of orbitals in the hydrogen atom depend on the value of the principal quantum number (n). When n increases, energy also increases. For this reason, orbitals in the same shell have the same energy in spite of their subshell. The increasing order of energy of hydrogen orbitals is
1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f
In the case of one 2s and three 2p orbitals in the second shell, they have the same energy. In the third shell, all nine orbitals (one 3s, three 3p and five 3d) have the same energy. All sixteen orbitals (one 4s, three 4p, five 4d and seven 4f) in the fourth shell have the same energy.
The energy levels of the different orbitals in hydrogen atom are easily explained by considering the given diagram. Here, each box represents one orbital. Orbitals with the same principal quantum number (n) have the same energy.
Principal Quantum Number (n)
The principal quantum number (n) assigns the size of the orbital and specifies the energy of an electron. If the value of n is larger, then the average distance of an electron in the orbital from the nucleus will be greater. Therefore the size of the orbital is large. The principal quantum numbers have the integral values of 1, 2, 3 and so forth and it corresponds to the quantum number in Bohr’s model of the hydrogen atom. If all orbitals have the same value of ‘n’, they are said to be in the same shell (level). The total number of orbitals for a given n value is n2. As the value of ‘n’ increases, the energy of the electron also increases.
To find: Identify the orbital which is higher in energy in the given pair 4f, 5s orbitals of hydrogen
Find the value of ‘n’
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Chapter 3 Solutions
Chemistry: Atoms First
- Could you please solve the first problem in this way and present it similarly but color-coded or step by step so I can understand it better? Thank you!arrow_forwardCould you please solve the first problem in this way and present it similarly but (color-coded) and step by step so I can understand it better? Thank you! I want to see what they are doingarrow_forwardCan you please help mne with this problem. Im a visual person, so can you redraw it, potentislly color code and then as well explain it. I know im given CO2 use that to explain to me, as well as maybe give me a second example just to clarify even more with drawings (visuals) and explanations.arrow_forward
- Part 1. Aqueous 0.010M AgNO 3 is slowly added to a 50-ml solution containing both carbonate [co32-] = 0.105 M and sulfate [soy] = 0.164 M anions. Given the ksp of Ag2CO3 and Ag₂ soy below. Answer the ff: Ag₂ CO3 = 2 Ag+ caq) + co} (aq) ksp = 8.10 × 10-12 Ag₂SO4 = 2Ag+(aq) + soy² (aq) ksp = 1.20 × 10-5 a) which salt will precipitate first? (b) What % of the first anion precipitated will remain in the solution. by the time the second anion starts to precipitate? (c) What is the effect of low pH (more acidic) condition on the separate of the carbonate and sulfate anions via silver precipitation? What is the effect of high pH (more basic)? Provide appropriate explanation per answerarrow_forwardPart 4. Butanoic acid (ka= 1.52× 10-5) has a partition coefficient of 3.0 (favors benzene) when distributed bet. water and benzene. What is the formal concentration of butanoic acid in each phase when 0.10M aqueous butanoic acid is extracted w❘ 25 mL of benzene 100 mL of a) at pit 5.00 b) at pH 9.00arrow_forwardCalculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0 Group of answer choices 0.0269 kJ/mole 2610 kJ/mole 27.6 kJ/mole 0.215 kJ/mole 20.8 kJ/molearrow_forward
- Calculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0 choices: 0.0269 kJ/mole 2610 kJ/mole 27.6 kJ/mole 0.215 kJ/mole 20.8 kJ/molearrow_forwardCalculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0arrow_forwardDon't used hand raiting and don't used Ai solutionarrow_forward
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