Videos
(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
GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST
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- Can u show the process as to how to get these?arrow_forwardSketch the expected 'H NMR spectra for the following compound. Label all of the H's in the structure and the corresponding signal for the spectra you sketch. Make sure you include the integration value and the splitting pattern for each signal Indicate how many signals you would expect in the 13C NMRarrow_forwardUse IUPAC naming rules to name the following hydrocarbon compounds: CH2-CH3 | a) CH-CH-CH2-CH-CH-CH3 b) | CH2 CH3 | CH3 CH3 \ / C=C H 1 H CH2-CH3 c) d) CH=C-CH3 e) CH3-CH2-CH2-CH=CH-CH3 f) CH2=CH-CH2-CH=CH-CH3 g) CH3-CH2-C = C-CH2-CH3 h)arrow_forward
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- Jaslev Propose a synthesis of the following starting from benzene and any other reagents and chemicals. No mechanisms are required. Indicate the condition for each step plus the major product for each step. More than two steps are required. Step 1 Step 2 مہد Brarrow_forwardPart C: The line formula for another branched alkane is shown below. i. In the IUPAC system what is the root or base name of this compound? ii. How many alkyl substituents are attached to the longest chain? iii. Give the IUPAC name for this compound.arrow_forwardPart D: Draw the Structural Formula for 4-ethyl-2-methylhexane Part E. Draw the Structural Formula for 1-chloro-3,3-diethylpentane (Chloro = Cl)arrow_forward
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