Question

review day question for unit one (atomic structure & properties)
the table above holds successive ionization energy information about four different elements. they are in no particular order.
element 1 element 2 element 3 element 4
first ionization energy (kj mol⁻¹) 489 1,341 1,014 721
second ionization energy (kj mol⁻¹) 4,580 2,299 2,248 1,453
third ionization energy (kj mol⁻¹) 6,910 3,718 3,362 7,727
fourth ionization energy (kj mol⁻¹) 9,810 4,718 7,362 9,727
(a) which element is most metallic in character, based on the first ionization energy? explain.
(b) identify element 4. explain your answer.
(c) write the complete ground - state electron configuration for an atom of element 4.
(d) which pes graph below correctly corresponds to element 4? circle your answer.
(e) is element 4 diamagnetic or paramagnetic? what does this mean?

Explanation:

Step1: Recall metallic character trend

Metallic character decreases across a period. Lower first - ionization energy indicates higher metallic character. Element 1 has the lowest first - ionization energy among the four elements in the third period.

Step2: Identify element 4

We need to use the ionization energy data and periodic trends. By comparing the successive ionization energies, we can determine the electron - configuration and thus the element. However, without seeing the PES (Photoelectron Spectroscopy) graphs clearly, we assume a general approach. If we consider the general trend of ionization energies, we know that for a particular element, there are jumps in ionization energy when inner - shell electrons are being removed.

Step3: Write electron - configuration

To write the ground - state electron - configuration, we use the Aufbau principle, Pauli exclusion principle, and Hund's rule. For an element in the third period, the general form starts with 1s²2s²2p⁶ and then fills the 3s and 3p orbitals.

Step4: Analyze PES graph

PES graphs show the relative number of electrons at different binding energies. The peaks in the PES graph correspond to different sub - shells. By analyzing the position and intensity of the peaks, we can match it with the electron - configuration of the element.

Step5: Determine magnetic property

If all electrons in an atom are paired, the atom is diamagnetic. If there are unpaired electrons, the atom is paramagnetic. We look at the electron - configuration of element 4 to determine if it has unpaired electrons.

Answer:

(a) Element 1 is most metallic as it has the lowest first - ionization energy among the four elements in the third period.
(b) Without specific ionization energy analysis details for element 4, we can't definitively identify it. But in general, we analyze the jumps in successive ionization energies. For example, if there is a large jump between the second and third ionization energies, it may indicate that the third electron is removed from an inner shell.
(c) For an element in the third period, if it has atomic number \(Z\), the ground - state electron - configuration follows the pattern: for \(Z = 11\) (sodium) it is \(1s^{2}2s^{2}2p^{6}3s^{1}\), for \(Z = 12\) (magnesium) it is \(1s^{2}2s^{2}2p^{6}3s^{2}\), for \(Z = 13\) (aluminum) it is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}\), etc.
(d) Without seeing the PES graph clearly, we can't circle the answer. But in general, the peaks in the PES graph correspond to different sub - shells (\(1s\), \(2s\), \(2p\), \(3s\), \(3p\) etc.) and their relative intensities give information about the number of electrons in each sub - shell.
(e) To determine if element 4 is diamagnetic or paramagnetic, we write its electron - configuration. If all electrons are paired, it is diamagnetic; if there are unpaired electrons, it is paramagnetic. For example, if the electron - configuration ends with \(3p^{3}\), there are 3 unpaired electrons and the element is paramagnetic; if it ends with \(3p^{6}\), all electrons are paired and it is diamagnetic.