Highest Ionization Energy Quiz Questions

The valence shell electron configuration of the halogens is ns2np5. This means that in the outermost energy level (valence shell), there are 2 electrons in the s orbital (represented by ns2) and 5 electrons in the p orbital (represented by np5). This configuration allows the halogens to have a total of 7 valence electrons, which makes them highly reactive and likely to form compounds by gaining one electron to achieve a stable octet configuration.

Rb (Rubidium) is the largest atom among the given options. The size of an atom is determined by its atomic radius, which is the distance from the nucleus to the outermost electron shell. As you move down a group in the periodic table, the atomic radius generally increases due to the addition of more electron shells. Rb is located in Group 1 (alkali metals) and is lower in the periodic table compared to the other elements listed. Therefore, it has more electron shells and a larger atomic radius, making it the largest atom among the options provided.

Phosphorus has the highest first-ionization energy among the given elements. First-ionization energy refers to the energy required to remove the outermost electron from an atom in its gaseous state. Phosphorus has a smaller atomic radius and a higher nuclear charge compared to the other elements listed. These factors result in a stronger attraction between the nucleus and the outermost electron, making it more difficult to remove. Therefore, phosphorus has a higher first-ionization energy.

Metals are known for their high thermal conductivity, meaning they can easily transfer heat. However, the answer choice "poor thermal conductors" contradicts this property. Therefore, it is not a property of metals.

The first ionization energy is the energy required to remove one electron from an atom in its gaseous state. As we move across a period from left to right, the first ionization energy generally increases due to an increase in the effective nuclear charge. Among the given choices, the correct order of increasing first-ionization energy is Al, Si, S, Cl, Ar. This is because Al is the first element in the list and has the lowest ionization energy, while Ar is the last element and has the highest ionization energy.

Rb belongs to the group called the alkali metals. Alkali metals are highly reactive metals that belong to Group 1 of the periodic table. They have one valence electron and are known for their low melting points and softness. Rb, which stands for rubidium, is a highly reactive alkali metal that is silvery-white in color and has a low melting point. It is commonly used in research and in certain industries.

Cl has the most negative electron affinity because it is located in Group 7A of the periodic table, which is known as the halogens. The halogens have a strong tendency to gain an electron to achieve a stable electron configuration. Cl has a higher effective nuclear charge compared to Na, Ar, and I, which means that the attraction between the nucleus and the incoming electron is stronger. This results in a higher energy release when Cl gains an electron, leading to a more negative electron affinity.

SiO2 is the most acidic compound among the given options. This is because SiO2 is an oxide of a non-metal (silicon) and non-metal oxides tend to be acidic. When SiO2 reacts with water, it forms silicic acid (H2SiO3), which is a weak acid. On the other hand, Al2O3 and MgO are oxides of metals (aluminum and magnesium respectively) and metal oxides tend to be basic or amphoteric, not acidic. Therefore, SiO2 is the most acidic compound among the given options.

The first-ionization energy refers to the energy required to remove the outermost electron from an atom. Based on their positions in the periodic table, atoms with a higher atomic number tend to have a larger first-ionization energy. Ne (neon) has a higher atomic number than Mg (magnesium), so Ne is expected to have a larger first-ionization energy than Mg. Therefore, the pair Ne, Mg is the correct answer.

Potassium (K) would have the largest second-ionization energy among the given elements. This is because potassium has a relatively low atomic radius and a high effective nuclear charge, making it more difficult to remove a second electron from the atom compared to the other elements. As a result, potassium requires more energy to remove a second electron, leading to a higher second-ionization energy.

The atomic radius of an atom refers to the size of its electron cloud. In general, atomic radius increases as you move down a group on the periodic table, and decreases as you move across a period from left to right.

In this case, N is a nonmetal in Group 15, As is also a nonmetal in Group 15, K is a metal in Group 1, and Fr is a metal in Group 1. Since N and As are in the same group, N has a smaller atomic radius than As. K and Fr are also in the same group, but Fr is further down the group, so it has a larger atomic radius than K. Therefore, the correct order of increasing atomic radius is N

CO2 is the most acidic oxide among the given options. This is because it is a non-metallic oxide and non-metallic oxides tend to be acidic in nature. CO2 is formed by the combination of carbon and oxygen, and carbon is a non-metal. When CO2 dissolves in water, it forms carbonic acid, which further dissociates into H+ ions, making it acidic. On the other hand, Li2O, Al2O3, and CaO are all basic oxides because they are formed by the combination of metals and oxygen, and metal oxides tend to be basic in nature.

The effective nuclear charge is the net positive charge experienced by an electron in an atom or ion. It is determined by the number of protons in the nucleus and the shielding effect of inner electrons. In this case, all the given species have the same number of electrons, so the effective nuclear charge will be smallest for the species with the least number of protons. Li has the fewest protons (3) among the given species, so it will experience the smallest effective nuclear charge.

The atomic radius of an element is the distance from the nucleus to the outermost electron shell. As you move down a group in the periodic table, the atomic radius generally increases due to the addition of more electron shells. In the given set, Cl, S, O, the atomic radius should increase from Cl to S to O. However, the correct answer states that Cl, S, O is in the incorrect order of increasing atomic radius. Therefore, the correct answer is Cl, S, O.

Barium is the correct answer because it is the only element listed that starts with the letter "b" and has an atomic number greater than 50. Bismuth, bromine, boron, and beryllium all have atomic numbers less than 50. Therefore, barium is the only element that fits the given criteria.

The set of elements C, N, O is not in order of increasing atomic radius because the atomic radius generally increases as you move down a group in the periodic table. However, in this set, the atomic radius of N is smaller than that of O, which contradicts the trend.

Boron (B) would have a fourth-ionization energy much greater than the third. This is because as we move across a period in the periodic table, the ionization energy generally increases due to the increasing nuclear charge. However, Boron has an exception to this trend. It has a lower fourth-ionization energy compared to its third-ionization energy because removing a fourth electron from Boron would result in a stable half-filled p orbital, which is a more stable configuration. Therefore, it requires significantly more energy to remove the fourth electron from Boron compared to the third electron.

Phosphorus has the highest second-ionization energy among the given elements. Second-ionization energy refers to the energy required to remove a second electron from a positively charged ion. Phosphorus has a relatively small atomic radius and a high effective nuclear charge, which means that the second electron is strongly attracted to the nucleus. This strong attraction makes it more difficult to remove the second electron, resulting in a higher second-ionization energy compared to the other elements listed.

When calcium reacts with nitrogen, three calcium atoms combine with one nitrogen molecule to form calcium nitride. This is represented by the balanced equation 3Ca + N2 -> Ca3N2.

The given answer is correct because it arranges the atoms in order of increasing distance of the n=4 electron shell from the nucleus. As the atomic number increases, the size of the atom generally increases. Therefore, the larger atoms will have the n=4 electron shell farther from the nucleus. In this case, Ir (Iridium) has the smallest atomic number and thus the smallest size, followed by Zr (Zirconium), Rb (Rubidium), As (Arsenic), and Ca (Calcium), which has the largest atomic number and thus the largest size.

The 2p subshell is lowest in energy in the nitrogen (N) atom. This is because nitrogen has a higher atomic number than carbon (C), lithium (Li), beryllium (Be), and boron (B), which means it has more protons in its nucleus. The increased nuclear charge in nitrogen leads to a stronger attraction for its electrons, causing the 2p subshell to be lower in energy compared to the other atoms listed.

Rb (Rubidium) has the lowest melting point among the given options. This is because the melting point of an element is influenced by the strength of the metallic bonds between its atoms. Rubidium has weaker metallic bonds compared to Calcium (Ca) and Cesium (Cs), leading to a lower melting point.

When magnesium reacts with ammonia, it forms magnesium nitride and hydrogen gas. To find the mass of hydrogen formed, we need to determine the limiting reactant first.

The molar mass of Mg is 24.31 g/mol and the molar mass of NH3 is 17.03 g/mol.

Using the molar masses, we can calculate the number of moles of Mg and NH3 present in the given masses.

Number of moles of Mg = 9.47 g / 24.31 g/mol = 0.3896 mol
Number of moles of NH3 = 2.35 g / 17.03 g/mol = 0.1378 mol

The balanced chemical equation shows that 3 moles of H2 are produced for every 1 mole of Mg. Therefore, the number of moles of H2 produced can be calculated as:

Number of moles of H2 = 0.3896 mol Mg × 3 mol H2 / 1 mol Mg = 1.168 mol

Finally, we can calculate the mass of hydrogen formed by multiplying the number of moles by the molar mass of H2:

Mass of H2 = 1.168 mol × 2.02 g/mol = 2.36 g

Therefore, the mass of hydrogen formed in the reaction is 2.36 g.

In this reaction, the balanced equation shows that 1 mole of magnesium reacts with 2 moles of ammonia to produce 1 mole of hydrogen gas. To find the mass of hydrogen gas formed, we need to convert the given mass of magnesium to moles using its molar mass, and then use the mole ratio from the balanced equation to find the moles of hydrogen gas. Finally, we convert the moles of hydrogen gas to grams using its molar mass. The calculated mass of hydrogen gas formed is 0.228 grams.

The given answer arranges the atoms in order of increasing distance of the 3rd shell from the nucleus. The 3rd shell is further away from the nucleus compared to the 2nd shell. Therefore, the atom with the highest atomic number, Sr, will have the electrons in the 3rd shell farthest from the nucleus. Rb has a lower atomic number than Sr, so it will have the electrons in the 3rd shell closer to the nucleus. Se has a lower atomic number than Rb, so its electrons in the 3rd shell will be even closer to the nucleus. Ca has a lower atomic number than Se, so its electrons in the 3rd shell will be closer to the nucleus. Finally, Ar has the lowest atomic number among the given atoms, so its electrons in the 3rd shell will be closest to the nucleus.

The ionization potential of an element is the energy required to remove an electron from its outermost shell. In the case of Lithium, the nuclear charge (Z) is 3 and the principal quantum number (n) is 2 for the outermost electron. Using the formula given, we can calculate the energy as (-2.18 x 10-18 J)(3^2/2^2) = -2.18 x 10-18 J. However, we need to report the answer as a positive number, so the ionization potential of Lithium is 2.18 x 10-18 J, which is equivalent to 4.90e-18 J in scientific notation.