Suppose that a stable element with atomic number 119, symbol Wr, has been discovered.
(a) Write the ground-state electron configuration for Wr, showing only the valence-shell electrons.
(b) Would Wr be a metal or a nonmetal? Explain in terms of electron configuration.
(c) On the basis of periodic trends, would Wr have the largest atomic radius in its group or would it have the smallest? Explain in terms of electronic structure.
(d) What would be the most likely charge of the Wr ion in stable ionic compounds?
(e) Write a balanced equation that would represent the reaction of Wr with water.
(f) Assume that Wr reacts to form a carbonate compound.
(i) Write the formula for the compound formed between Wr and the carbonate ion, CO32–.
(ii) Predict whether or not the compound would be soluble in water. Explain your reasoning.
Look at the periodic table here. www.webelements.com. More than likely #119 will be the first element in row 8.
a. 8s1
b. Now that you know where it is you should be able to answer the others
thankk you!
(a) To determine the valence-shell electron configuration of an element with atomic number 119, we can look at its position in the periodic table. Since atomic number 119 is not a currently discovered element, we'll make an assumption based on the elements in the same group.
Element 117, Tennessine (Ts), is the heaviest element currently discovered. Since Wr is the element with atomic number 119, it would be in the 7th period of the periodic table. The element preceding it, Tennessine (Ts), has an electron configuration of [Rn] 5f14 6d10 7s2 7p5.
Assuming Wr follows a similar trend, the valence-shell electron configuration of Wr would be [Rn] 5f14 6d10 7s2 7p7.
(b) Based on the electron configuration, Wr would be a nonmetal. Nonmetals tend to have valence shells that are closer to being filled with electrons, making it more energetically favorable for them to gain or share electrons to achieve a stable electron configuration.
(c) On the basis of periodic trends, Wr would be expected to have the largest atomic radius in its group. Going down a group in the periodic table, the atomic radius generally increases due to the addition of new energy levels. Since Wr is in period 7, it will have more energy levels and a larger atomic radius compared to elements in lower periods.
(d) The most likely charge of the Wr ion in stable ionic compounds would depend on its tendency to gain or lose electrons. Since it is a nonmetal, it is more likely to gain electrons to achieve a stable electron configuration. Thus, Wr would typically form an ion with a charge of -1, gaining one electron to fill its valence shell.
(e) The reaction of Wr with water would depend on its behavior as a nonmetal. If Wr is a reactive nonmetal, it could potentially form a reaction with water. However, without specific information about Wr, it is not possible to write a balanced equation for this reaction.
(f) Assuming Wr reacts to form a carbonate compound:
(i) The formula for the compound formed between Wr and the carbonate ion, CO32–, would be WrCO3.
(ii) Whether or not the compound WrCO3 would be soluble in water would depend on the solubility rules for carbonates. Most carbonates are insoluble in water, except for those of Group 1 metals (Li+, Na+, K+, etc.) and ammonium ion (NH4+). Without further information, it is not possible to predict the solubility of WrCO3.
(a) To determine the ground-state electron configuration for Wr (atomic number 119), we first need to identify the electron configuration of the element preceding it. Since Wr is element 119, we know it follows the pattern of the periodic table where each successive element adds one electron. The element preceding Wr is element 118, which is oganesson (Og). The electron configuration of Og is [Rn]5f146d107s27p6.
To find the valence-shell electrons, we need to focus on the outermost energy level, which is the sixth energy level (n=6) for Wr. Based on the pattern in the periodic table, we can place the valence electrons as follows: 7s2, 6d10, 5f14. The valence configuration for Wr is therefore [Rn]5f146d107s27p6.
(b) To determine if Wr would be a metal or a nonmetal, we can look at its position on the periodic table. Wr is located in period 8, so it would be an element in the p-block. In general, elements on the left side of the periodic table are metals, while elements on the right side are nonmetals. Since Wr is further to the right side of the periodic table, it is more likely to be a nonmetal.
(c) The atomic radius generally decreases as we move across a period from left to right in the periodic table. This is due to increasing effective nuclear charge, which attracts the valence electrons more strongly, leading to a smaller atomic radius. Since Wr is located towards the right side of its period, it would have a smaller atomic radius compared to the other elements in its group.
(d) To predict the most likely charge of the Wr ion in stable ionic compounds, we need to consider its electron configuration and the tendency to gain/lose electrons. Since Wr is located in period 8 of the periodic table, it will likely gain electrons to achieve a stable electron configuration. The noble gas configuration it would aim to achieve is [Rn]. Therefore, Wr is likely to form an ion with a charge of -1 to complete its valence shell.
(e) The reaction of Wr with water would depend on the reactivity of Wr, but we can write a general equation for the reaction:
Wr + H2O -> WrO + H2
This equation represents the formation of a metal oxide (WrO) and hydrogen gas (H2) when Wr reacts with water.
(f)
(i) The carbonate ion has a charge of 2- (CO32-). To determine the formula for the compound formed between Wr and the carbonate ion, we need to balance the charges. Since Wr is likely to form an ion with a charge of -1, we would need two Wr ions to balance the charge of the carbonate ion. Therefore, the formula for the compound formed between Wr and the carbonate ion would be Wr2(CO3).
(ii) Whether or not the compound formed between Wr and the carbonate ion would be soluble in water depends on the solubility rules for carbonates. In general, carbonates are insoluble except for those of Group 1 metals and ammonium ions. Since Wr is not a Group 1 metal or an ammonium ion, the compound Wr2(CO3) is likely to be insoluble in water.