Strontium(Sr) lies to the right of rubidium(Rb) in the 5th period on the Periodic Table. Ionization energy (IE) increases as you go from left to right so Sr would have a higher IE than Rb. The actual reason for this is that both of them have their valence electrons in the 5th shell, so they are the same distance from the nucleus, however, Sr has more protons in the nucleus, and thus the attraction of these electrons is greater. Greater attraction means it takes more energy to remove them.
Rubidium has a larger atomic radius than lithium and iodine primarily because it is located further down the periodic table, in group 1. As you move down a group, additional electron shells are added, increasing the distance between the nucleus and the outermost electrons. Although iodine is larger than lithium, it has a higher nuclear charge, which draws its electrons closer, resulting in a smaller atomic radius compared to rubidium. Thus, the combination of more electron shells and weaker effective nuclear charge in rubidium leads to its larger atomic radius.
Smaller
In other words this question asks if the pores are bigger or smaller than the iodine molecules. So.... the answer is that the iodine are smaller than the pores and the pores are bigger, because the iodine molecules need room to pass through, and the pores are not selectively permeable because they let the iodine through.
An iodine atom has one more principal energy level than a bromine atom. Therefore the radius of an iodine atom is greater than the latter.
To calculate the energy needed to melt 25.4 grams of I2 (iodine), you can use the formula: energy = mass x heat of fusion. The heat of fusion for iodine is 15.52 kJ/mol. First, find the molar mass of I2 (253.8 g/mol) and then convert the mass to moles. Finally, multiply the moles by the heat of fusion to get the energy needed.
The ionization energy of fluorine is 1681 kJ/mol (the first) and the ionization energy of iodine is 1008,4 kJ/mol.
Because fluorine's size is lower than that of iodine, it has a greater ionization energy than iodine. Fluorine, on the other hand, appears to have a smaller shielding effect. As a result, fluorine's nucleus attracts more valence electrons than iodine's.
Rubidium and iodine form an ionic bond. Rubidium, a metal, donates an electron to iodine, a nonmetal, resulting in the transfer of electrons from rubidium to iodine to achieve stability. This forms the ionic compound rubidium iodide (RbI).
Yes. Rubidium is an alkali metal in the sodium group. It will react with iodine to form rubidium iodide:- 2Rb+ I2 -> 2RbI
The empirical formula for iodine is I2 and for rubidium is Rb.
A ionic bond forms between rubidium and iodine. Rubidium, a metal, donates its electron to iodine, a nonmetal, resulting in the formation of positively charged rubidium ions and negatively charged iodine ions that are then attracted to each other.
Rubidium metal would react with iodine to make rubidium iodide , according to the equation: 2 Ru + I2 -> 2 RuI
what is the balanced equation for Rubidium metal reacting with halogen iodine
Ionic bond is formed between rubidium and iodine, where rubidium donates its electron to iodine to complete its valence shell. Rubidium becomes a positively charged ion (cation) and iodine becomes a negatively charged ion (anion), resulting in the formation of an ionic compound, rubidium iodide.
Iodine has a higher electronegativity value than rubidium. Electronegativity increases across a period from left to right on the periodic table, so iodine, being on the right side of rubidium, has a higher electronegativity value.
Rubidium iodine
The lattice energy of potassium bromide is more exothermic than that of rubidium iodide because potassium and bromine have smaller atomic sizes and higher charges, which leads to stronger ionic bonding in potassium bromide. Rubidium and iodine have larger atomic sizes and lower charges, resulting in weaker ionic bonding in rubidium iodide. The stronger ionic bonding in potassium bromide requires more energy to break, resulting in a more exothermic lattice energy.