Use principles of atomic structure to Explain why in aqueous solution,
(a) potassium forms 1+ but not a 2+ ion
(a) Ti3+ is colored but Sc3+ is not.
(b) Ti2+ is a reducing agent but Ca2+ is not.
thank you!
(a) To understand why potassium forms a 1+ ion in aqueous solution, we need to examine its atomic structure. Potassium is an element with the symbol K and atomic number 19.
The atomic structure of potassium contains 19 protons in its nucleus and 19 electrons surrounding the nucleus. These electrons are arranged in energy levels or shells. The first energy level can hold a maximum of 2 electrons, while the second and third energy levels can hold up to 8 electrons each.
In the case of potassium, the electron configuration is 1s2 2s2 2p6 3s2 3p6 4s1. This means that the first energy level is filled with 2 electrons, the second energy level contains 8 electrons, and the third energy level has just 1 electron in the 4s orbital.
When an atom forms an ion, it either gains or loses electrons to achieve a stable outer electronic configuration. In the case of potassium, it is easier for the atom to lose one electron from the 4s orbital to achieve a stable configuration. This results in the formation of a 1+ ion, represented as K+.
The loss of one electron allows the remaining 18 electrons to be arranged in a stable configuration (similar to the nearest noble gas, argon). This stable electron configuration makes the potassium ion more energetically favorable in aqueous solutions.
(b) Moving on to the question about why Ti3+ is colored but Sc3+ is not, we once again need to consider the atomic structure of the elements.
Titanium (Ti) and scandium (Sc) both belong to the d-block or transition metals in the periodic table. These elements have partially filled d orbitals, which are responsible for their unique properties, including the ability to exhibit different colors.
In the case of titanium, it commonly forms ions with a +3 charge, represented as Ti3+. The electron configuration for titanium is [Ar] 3d2 4s2, meaning that in its neutral state, it has two electrons in the 3d orbital. When titanium loses three electrons to form Ti3+, it results in an empty 3d orbital. The presence of empty d orbitals allows the absorption of specific wavelengths of light, resulting in the appearance of color.
On the other hand, scandium (Sc) has an electronic configuration of [Ar] 3d1 4s2. When scandium forms an ion with a +3 charge (Sc3+), it loses all of its 3d electrons, resulting in an empty d orbital. However, in contrast to titanium, scandium does not exhibit color because it has only one d electron, which does not provide sufficient energy differences for the absorption of visible light.
Therefore, the presence of empty d orbitals in combination with the number of d electrons determines whether an ion will be colored or not.
(c) Lastly, let's discuss why Ti2+ is a reducing agent but Ca2+ is not.
A reducing agent is a substance that has a tendency to lose electrons or donate electrons, thereby causing the reduction (gain of electrons) of other species. The ability of an element to be a reducing agent depends on its electron configuration and the relative stability of different oxidation states.
In the case of titanium (Ti), it belongs to the transition metals and commonly forms ions with a +2 charge, represented as Ti2+. The electron configuration of titanium in its neutral state is [Ar] 3d2 4s2. When it loses two electrons to form Ti2+, it achieves a more stable configuration of [Ar] 3d2.
The presence of partially filled d orbitals in titanium's electron configuration makes it easier for Ti2+ ions to donate electrons. These electrons can then be accepted by species with a higher tendency to gain electrons, promoting the reduction of these species. Therefore, Ti2+ acts as a reducing agent.
On the other hand, calcium (Ca) is an alkaline earth metal and commonly forms ions with a +2 charge, represented as Ca2+. The electron configuration for calcium is [Ar] 4s2. When calcium loses two electrons to form Ca2+, it attains a noble gas configuration of [Ar].
However, unlike titanium, calcium does not have partially filled d orbitals. In fact, its electron configuration in the neutral state and its Ca2+ ion form are stable and do not have a high tendency to donate electrons readily. Therefore, Ca2+ is not a good reducing agent compared to Ti2+.
In summary, the ability of an ion to act as a reducing agent depends on the electron configuration and the relative stability of different oxidation states, particularly the presence of partially filled d orbitals.
19K is 1s2 2s2 2p6 3s2 3p6 4s1
K can lose the 1 4s1 easily but the next shell is a closed shell.
Sc^3+ has no unpaired electrons. Ti^3+ has an unpaired electron. Here is the electron configuration.
21Sc is 1s2 2s2 2p6 3s2 3p6 3d1 4s2. Note that the +3 ion is formed by removing the 4s2 electrons and the 1 3d1.
22Ti is 1s2 2s2 2p6 3s2 3p6 3d2 4s2.
Note that the +3 ion is formed by removing the 4s2 electrons and ONE of the 3d2 electrons. That leaves an unpaired electron.
c.
20Ca 1s2 2s2 2p6 3s2 3p6 4s2
So Ca^2+ is formed by removing the 4s2 electrons. The next shell down is a closed shell.
Look at the Ti^2+ ion and where it can go (compare with the Ti^3+ above).