Part A Classify each orbital diagram for ground-state electron configurations by

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Part A

Classify each orbital diagram for ground-state electron configurations by the rule or principle it violates.

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Aufbau violation

Hund violation

Pauli violation

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Part B

The following sets of quantum numbers, listed in the order n, , m, and ms, were written for the last electrons added to an atom. Identify which sets are valid and classify the others by the rule or principle that is violated.

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Pauli violation

Other violation

Valid

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Part C

Rank the following ions in order of decreasing radius: F,Cl,Br,I, and At. Use the periodic table as necessary.

Rank from largest to smallest radius. To rank items as equivalent, overlap them.

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Smallest radius

Largest radius

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Part D

The following ions contain the same number of electrons. Rank them in order of decreasing ionic radii.

Rank from largest to smallest radius. To rank items as equivalent, overlap them.

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Smallest radius

Largest radius

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Part E

Based on position in the periodic table and electron configuration, arrange these elements in order of decreasing Ei1.

Rank the elements from highest to lowest ionization energy. To rank items as equivalent, overlap them.

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Na

F

Li

N

B

O

Lowest ionization energy

Highest ionization energy

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Successive ionization energies

Removal of successive electrons always requires more energy (e.g., Ei3>Ei2>Ei1) because a negative electron is being removed for a successively higher positive charge. This pattern of increasing values for successive ionization energies shows a drastic increase in magnitude when an electron is removed from a completely filled energy level or one that is identical to that of a noble gas.

Part F

Examine the following set of ionization energy values for a certain element. How many valence electrons does an atom of the neutral element possess?

Enter your answer numerically as an integer.

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electrons

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Part G

The trend in second ionization energy for the elements from lithium to fluorine is not a smooth one. Predict which of these elements has the highest second ionization energy and which has the lowest and explain your choices.

Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer.

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1s22s22p3

F+

C+

highest

lowest

N+

1s22s2

1s2

B+

Li+

Be+

O+

Based on the electron configuration of the 1+ ions, should have the second ionization energy since the removal of the second electron involves removing a core electron. The second ionization energy should have since removing the second electron takes you to the configuration.

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Aufbau violation

Hund violation

Pauli violation

3p 3s

Explanation / Answer

Part A

Firstly given figure is not following Aufbau's Principle as it is not according to increasing energies. Order should be 1s 2s 2p 3s 3p 4s 3d 4p and 5s.

Secondly in 3p case, two electrons have same spin that is not acceptable according to Pauli's exclusion principle.

None of them is violating Hund's rule of maximum multiplicity.

Part B

In the given case all configurations are valid except last one because if n is principle quantum number then l= 0 to n-1.

Here, n is 3 so l can have values upto 2 and not 3. So it is violating the Pauli's Exclusion principle.

Part C

On going downwards number of shells increases and so also radius increases and on moving from left to right effective nuclear charge increases and hence radius decreases. So order should be At>I>Br>Cl>F

Part D

Higher the number of protons, greater is the force on electrons and lesser will be the radius. So order should be N3->O2->F->Na+>Mg2+>Al3+

Part E

Ionization energy increases from left to right and decreases down the table. So order should be N>F>O>B>Li>Na. Nitrogen has higher ionization energy because of its half filled orbital and stable configuration.

Part F

Since last 2 energies are high it means they may have half filled or fulfilled configuration. That is if we are talking about p subshell then they may have 3 or 6 electrons.First 2 have very low energies so there may be only 1 or 4 electron and intermediate energy range cases may have 2 or 5 electrons.

Part G

After giving 1 electron, Li and B get fulfilled orbitals and O get half filled orbitals. But since Li+ has less number of shells so it is difficult to take out electron from it and hence it will have highest second ionization energy.

If we try to remove second electron F- will come to half filled configuration and others will loose their configurations, so it will be easy to take out electron from flouride ion.Hence F- will have lowest second ionization energy.

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