Energies of Orbitals
CHXI02:STRUCTURE OF ATOM

307317 Arrange the following orbitals in decreasing order of energy.
\({\rm{A}}.{\rm{n}} = 3,l = 0,\;{\rm{m}} = 0\)
\({\rm{B}}.{\rm{n}} = 4,l = 0,\;{\rm{m}} = 0\)
\({\rm{C}}.{\rm{n}} = 3,l = 1,\;{\rm{m}} = 0\)
\({\rm{D}}.{\rm{n}} = 3,l = 2,\;{\rm{m}} = 1\)

1 D \(>\) B \(>\) C \(>\) A
2 \(\mathrm{A}>\mathrm{C}>\mathrm{B}>\mathrm{D}\)
3 B \(>\) D \(>\) C \(>\) A
4 D \(>\) B \(>\) A \(>\) C
CHXI02:STRUCTURE OF ATOM

307318 The energy of an electron of \({\rm{2}}{{\rm{p}}_{\rm{y}}}\) orbital is

1 Greater than \({\rm{2}}{{\rm{p}}_{\rm{x}}}\) orbital
2 Less than \({\rm{2}}{{\rm{p}}_{\rm{z}}}\) orbital
3 Equal to \({\rm{2s}}\) orbital
4 Same as that of \({\rm{2}}{{\rm{p}}_{\rm{x}}}\) and \({\rm{2}}{{\rm{p}}_{\rm{z}}}\) orbitals
CHXI02:STRUCTURE OF ATOM

307319 Compare the energies of following sets of quantum numbers for multielectron system.
\[\begin{array}{l}
{\rm{(I)n}} = 4,l = 1\\
{\rm{(II)n}} = 4,l = 2\\
{\rm{(III)n}} = 3,l = 1\\
{\rm{(IV)n}} = 3,l = 2\\
{\rm{(V)n}} = 4,l = 0
\end{array}\]
Choose the correct answer from the options given below

1 (V) \({\mathrm{>}}\) (III) \({\mathrm{>}}\) (I) \({\mathrm{>}}\) (IV) \({\mathrm{>}}\) (II)
2 (III) \({\mathrm{ < }}\) (V) \({\mathrm{ < }}\) (IV) \({\mathrm{ < }}\) (I) \({\mathrm{ < }}\) (II)
3 (II) \({\mathrm{>}}\) (I) \({\mathrm{>}}\) (III) \({\mathrm{>}}\) (V) \({\mathrm{>}}\) (IV)
4 (V) \({\mathrm{>}}\) (III) \({\mathrm{>}}\) (IV) \({\mathrm{>}}\) (I) \({\mathrm{>}}\) (II)
CHXI02:STRUCTURE OF ATOM

307320 The electrons identified by quantum numbers \(\mathrm{n}\) and l , are as follows
\[\begin{array}{l}
{\rm{I}}{\rm{.n}} = 4,l = 1\,\,\,\,\,\,\,\,\,\,\,{\rm{II}}{\rm{.n}} = 4,l = 0\\
{\rm{III}}{\rm{.n}} = 3,l = 2\,\,\,\,\,\,\,{\rm{IV}}.{\rm{n}} = 3,l = \end{array}\]
If we arrange them in order of increasing energy, i.e. from lowest to highest, the correct order is

1 IV \( < \) II \( < \) III \( < \) I
2 II \( < \) IV \( < \) I \( < \) III
3 I \( < \) III \( < \) II \( < \) IV
4 III \( < \) I \( < \) IV \( < \) II
CHXI02:STRUCTURE OF ATOM

307317 Arrange the following orbitals in decreasing order of energy.
\({\rm{A}}.{\rm{n}} = 3,l = 0,\;{\rm{m}} = 0\)
\({\rm{B}}.{\rm{n}} = 4,l = 0,\;{\rm{m}} = 0\)
\({\rm{C}}.{\rm{n}} = 3,l = 1,\;{\rm{m}} = 0\)
\({\rm{D}}.{\rm{n}} = 3,l = 2,\;{\rm{m}} = 1\)

1 D \(>\) B \(>\) C \(>\) A
2 \(\mathrm{A}>\mathrm{C}>\mathrm{B}>\mathrm{D}\)
3 B \(>\) D \(>\) C \(>\) A
4 D \(>\) B \(>\) A \(>\) C
CHXI02:STRUCTURE OF ATOM

307318 The energy of an electron of \({\rm{2}}{{\rm{p}}_{\rm{y}}}\) orbital is

1 Greater than \({\rm{2}}{{\rm{p}}_{\rm{x}}}\) orbital
2 Less than \({\rm{2}}{{\rm{p}}_{\rm{z}}}\) orbital
3 Equal to \({\rm{2s}}\) orbital
4 Same as that of \({\rm{2}}{{\rm{p}}_{\rm{x}}}\) and \({\rm{2}}{{\rm{p}}_{\rm{z}}}\) orbitals
CHXI02:STRUCTURE OF ATOM

307319 Compare the energies of following sets of quantum numbers for multielectron system.
\[\begin{array}{l}
{\rm{(I)n}} = 4,l = 1\\
{\rm{(II)n}} = 4,l = 2\\
{\rm{(III)n}} = 3,l = 1\\
{\rm{(IV)n}} = 3,l = 2\\
{\rm{(V)n}} = 4,l = 0
\end{array}\]
Choose the correct answer from the options given below

1 (V) \({\mathrm{>}}\) (III) \({\mathrm{>}}\) (I) \({\mathrm{>}}\) (IV) \({\mathrm{>}}\) (II)
2 (III) \({\mathrm{ < }}\) (V) \({\mathrm{ < }}\) (IV) \({\mathrm{ < }}\) (I) \({\mathrm{ < }}\) (II)
3 (II) \({\mathrm{>}}\) (I) \({\mathrm{>}}\) (III) \({\mathrm{>}}\) (V) \({\mathrm{>}}\) (IV)
4 (V) \({\mathrm{>}}\) (III) \({\mathrm{>}}\) (IV) \({\mathrm{>}}\) (I) \({\mathrm{>}}\) (II)
CHXI02:STRUCTURE OF ATOM

307320 The electrons identified by quantum numbers \(\mathrm{n}\) and l , are as follows
\[\begin{array}{l}
{\rm{I}}{\rm{.n}} = 4,l = 1\,\,\,\,\,\,\,\,\,\,\,{\rm{II}}{\rm{.n}} = 4,l = 0\\
{\rm{III}}{\rm{.n}} = 3,l = 2\,\,\,\,\,\,\,{\rm{IV}}.{\rm{n}} = 3,l = \end{array}\]
If we arrange them in order of increasing energy, i.e. from lowest to highest, the correct order is

1 IV \( < \) II \( < \) III \( < \) I
2 II \( < \) IV \( < \) I \( < \) III
3 I \( < \) III \( < \) II \( < \) IV
4 III \( < \) I \( < \) IV \( < \) II
CHXI02:STRUCTURE OF ATOM

307317 Arrange the following orbitals in decreasing order of energy.
\({\rm{A}}.{\rm{n}} = 3,l = 0,\;{\rm{m}} = 0\)
\({\rm{B}}.{\rm{n}} = 4,l = 0,\;{\rm{m}} = 0\)
\({\rm{C}}.{\rm{n}} = 3,l = 1,\;{\rm{m}} = 0\)
\({\rm{D}}.{\rm{n}} = 3,l = 2,\;{\rm{m}} = 1\)

1 D \(>\) B \(>\) C \(>\) A
2 \(\mathrm{A}>\mathrm{C}>\mathrm{B}>\mathrm{D}\)
3 B \(>\) D \(>\) C \(>\) A
4 D \(>\) B \(>\) A \(>\) C
CHXI02:STRUCTURE OF ATOM

307318 The energy of an electron of \({\rm{2}}{{\rm{p}}_{\rm{y}}}\) orbital is

1 Greater than \({\rm{2}}{{\rm{p}}_{\rm{x}}}\) orbital
2 Less than \({\rm{2}}{{\rm{p}}_{\rm{z}}}\) orbital
3 Equal to \({\rm{2s}}\) orbital
4 Same as that of \({\rm{2}}{{\rm{p}}_{\rm{x}}}\) and \({\rm{2}}{{\rm{p}}_{\rm{z}}}\) orbitals
CHXI02:STRUCTURE OF ATOM

307319 Compare the energies of following sets of quantum numbers for multielectron system.
\[\begin{array}{l}
{\rm{(I)n}} = 4,l = 1\\
{\rm{(II)n}} = 4,l = 2\\
{\rm{(III)n}} = 3,l = 1\\
{\rm{(IV)n}} = 3,l = 2\\
{\rm{(V)n}} = 4,l = 0
\end{array}\]
Choose the correct answer from the options given below

1 (V) \({\mathrm{>}}\) (III) \({\mathrm{>}}\) (I) \({\mathrm{>}}\) (IV) \({\mathrm{>}}\) (II)
2 (III) \({\mathrm{ < }}\) (V) \({\mathrm{ < }}\) (IV) \({\mathrm{ < }}\) (I) \({\mathrm{ < }}\) (II)
3 (II) \({\mathrm{>}}\) (I) \({\mathrm{>}}\) (III) \({\mathrm{>}}\) (V) \({\mathrm{>}}\) (IV)
4 (V) \({\mathrm{>}}\) (III) \({\mathrm{>}}\) (IV) \({\mathrm{>}}\) (I) \({\mathrm{>}}\) (II)
CHXI02:STRUCTURE OF ATOM

307320 The electrons identified by quantum numbers \(\mathrm{n}\) and l , are as follows
\[\begin{array}{l}
{\rm{I}}{\rm{.n}} = 4,l = 1\,\,\,\,\,\,\,\,\,\,\,{\rm{II}}{\rm{.n}} = 4,l = 0\\
{\rm{III}}{\rm{.n}} = 3,l = 2\,\,\,\,\,\,\,{\rm{IV}}.{\rm{n}} = 3,l = \end{array}\]
If we arrange them in order of increasing energy, i.e. from lowest to highest, the correct order is

1 IV \( < \) II \( < \) III \( < \) I
2 II \( < \) IV \( < \) I \( < \) III
3 I \( < \) III \( < \) II \( < \) IV
4 III \( < \) I \( < \) IV \( < \) II
CHXI02:STRUCTURE OF ATOM

307317 Arrange the following orbitals in decreasing order of energy.
\({\rm{A}}.{\rm{n}} = 3,l = 0,\;{\rm{m}} = 0\)
\({\rm{B}}.{\rm{n}} = 4,l = 0,\;{\rm{m}} = 0\)
\({\rm{C}}.{\rm{n}} = 3,l = 1,\;{\rm{m}} = 0\)
\({\rm{D}}.{\rm{n}} = 3,l = 2,\;{\rm{m}} = 1\)

1 D \(>\) B \(>\) C \(>\) A
2 \(\mathrm{A}>\mathrm{C}>\mathrm{B}>\mathrm{D}\)
3 B \(>\) D \(>\) C \(>\) A
4 D \(>\) B \(>\) A \(>\) C
CHXI02:STRUCTURE OF ATOM

307318 The energy of an electron of \({\rm{2}}{{\rm{p}}_{\rm{y}}}\) orbital is

1 Greater than \({\rm{2}}{{\rm{p}}_{\rm{x}}}\) orbital
2 Less than \({\rm{2}}{{\rm{p}}_{\rm{z}}}\) orbital
3 Equal to \({\rm{2s}}\) orbital
4 Same as that of \({\rm{2}}{{\rm{p}}_{\rm{x}}}\) and \({\rm{2}}{{\rm{p}}_{\rm{z}}}\) orbitals
CHXI02:STRUCTURE OF ATOM

307319 Compare the energies of following sets of quantum numbers for multielectron system.
\[\begin{array}{l}
{\rm{(I)n}} = 4,l = 1\\
{\rm{(II)n}} = 4,l = 2\\
{\rm{(III)n}} = 3,l = 1\\
{\rm{(IV)n}} = 3,l = 2\\
{\rm{(V)n}} = 4,l = 0
\end{array}\]
Choose the correct answer from the options given below

1 (V) \({\mathrm{>}}\) (III) \({\mathrm{>}}\) (I) \({\mathrm{>}}\) (IV) \({\mathrm{>}}\) (II)
2 (III) \({\mathrm{ < }}\) (V) \({\mathrm{ < }}\) (IV) \({\mathrm{ < }}\) (I) \({\mathrm{ < }}\) (II)
3 (II) \({\mathrm{>}}\) (I) \({\mathrm{>}}\) (III) \({\mathrm{>}}\) (V) \({\mathrm{>}}\) (IV)
4 (V) \({\mathrm{>}}\) (III) \({\mathrm{>}}\) (IV) \({\mathrm{>}}\) (I) \({\mathrm{>}}\) (II)
CHXI02:STRUCTURE OF ATOM

307320 The electrons identified by quantum numbers \(\mathrm{n}\) and l , are as follows
\[\begin{array}{l}
{\rm{I}}{\rm{.n}} = 4,l = 1\,\,\,\,\,\,\,\,\,\,\,{\rm{II}}{\rm{.n}} = 4,l = 0\\
{\rm{III}}{\rm{.n}} = 3,l = 2\,\,\,\,\,\,\,{\rm{IV}}.{\rm{n}} = 3,l = \end{array}\]
If we arrange them in order of increasing energy, i.e. from lowest to highest, the correct order is

1 IV \( < \) II \( < \) III \( < \) I
2 II \( < \) IV \( < \) I \( < \) III
3 I \( < \) III \( < \) II \( < \) IV
4 III \( < \) I \( < \) IV \( < \) II