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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.
Figure 1
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.
Figure 1
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.
Figure 1
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.
Figure 1
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.
Figure 1
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 7 Solutions
Connect 1 Semester Access Card for General Chemistry: The Essential Concepts
- The following 'H NMR spectrum was taken with a 750 MHz spectrometer: 1.0 0.5 0.0 10.0 9.0 8.0 7.0 6.0 5.0 4.0 3.0 ' 2.0 1.0 0.0 (ppm) What is the difference Av in the frequency of RF ac Δν ac radiation absorbed by the a and c protons? (Note: it's not equal to the difference in chemical shifts.) Round your answer to 2 significant digits, and be sure it has an appropriate unit symbol. = O O a will shift left, c will shift right. O a will shift right, c will shift left. a and c will both shift left, with more space between them. Suppose a new spectrum is taken with a 500 MHz spectrometer. What will be true about this new spectrum? O a and c will both shift left, with less space between them. O a and c will both shift right, with more space between them. O a and c will both shift right, with less space between them. Which protons have the largest energy gap between spin up and spin down states? O None of the above. ○ a Ob Explanation Check C Ar B 2025 McGraw Hill LLC. All Rights Reserved.…arrow_forwardWhat mass of Na2CO3 must you add to 125g of water to prepare 0.200 m Na2CO3? Calculate mole fraction of Na2CO3, mass percent, and molarity of the resulting solution. MM (g/mol): Na2CO3 105.99; water 18.02. Final solution density is 1.04 g/mL.arrow_forward(ME EX2) Prblms Can you please explain problems to me in detail, step by step? Thank you so much! If needed color code them for me.arrow_forward
- Experiment #8 Electrical conductivity & Electrolytes. Conductivity of solutions FLINN Scientific Scale RED LED Green LED LED Conductivity 0 OFF OFF 1 Dim OFF 2 medium OFF 3 Bright Dim Low or Nowe Low Medium High 4 Very Bright Medium nd very high AA Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ SE=Strong Electrolyte, FE = Fair Electrolyte CWE = Weak Electrolyte, NE= Noni Electrolyte, #Solutions 1 0.1 M NaCl 2/1x 102 M NaCl, 3/1X103 M Nall Can Prediction M Observed Conductivity Very bright red Bright red Dim red you help me understand how I'm supposed to find the predictions of the following solutions? I know this is an Ionic compound and that the more ions in a solution means it is able to carry a charge, right? AAAA Darrow_forward(SE EX 2) Prblsm 4-7: Can you please explain problems 4-7 and color code if needed for me. (step by step) detail explanationsarrow_forward(SE EX 2) Problems 8-11, can you please explain them to me in detail and color-code anything if necessary?arrow_forward
- (ME EX2) Problems 15-16 Could you please explain problems 15 through 16 to me in detail, step by step? Thank you so much! If necessary, please color-code them for me.arrow_forward1.)show any electrophilic aromatic substitution, identify the electriphile, nucleophile and transition statearrow_forward(SE EX 2) Problems 15-16, can you please explain them to me in detail and color-code anything if necessary?arrow_forward
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