Question

An element's most stable ion forms an ionic compound with bromine, having the formula \(\mathrm{XBr}_{2}\). If the ion of element \(\mathrm{X}\) has a mass number of 230 and has 86 electrons, what is the identity of the element, and how many neutrons does it have?

Short answer

The element X is Polonium (Po) with a +2 charge when forming an ionic compound with bromine, and it has 146 neutrons.

Step-by-step solution

Determining the charge of element X

Since the ionic compound has the formula \(\mathrm{XBr}_{2}\), it means that the ion of element X has a charge that balances with the charges of two bromide ions. The bromide ion (\(\mathrm{Br}^-\)) has a charge of -1. Therefore, the ion of element X must have a charge of +2 to balance the charges. So, element X has lost 2 electrons to form this ionic compound.

Finding the number of protons

Element X has 86 electrons in its neutral state. When it loses 2 electrons and acquires a +2 charge, it has 84 electrons. Since the number of protons is equal to the number of electrons in a neutral atom, element X has 84 protons.

Identifying the element

Using the periodic table, we can identify element X by looking for the element with an atomic number (number of protons) of 84. The element with an atomic number of 84 is Polonium (Po).

Calculating the number of neutrons

The mass number of an element is the sum of the protons and neutrons in its nucleus. Since Polonium (X) has a mass number of 230 and 84 protons, we can find the number of neutrons using the formula: Number of Neutrons = Mass Number - Number of Protons Number of Neutrons = 230 - 84 Number of Neutrons = 146 So, the ion of element X is Polonium with a +2 charge (Po^2+) and has 146 neutrons.

Key concepts

Mass number

The mass number of an atom is a crucial concept in understanding the structure of elements. It represents the total number of protons and neutrons present in the nucleus of an atom. To calculate the mass number, add up the protons and neutrons of the atom. This addition results in the atomic mass, which essentially determines the identity and isotopic nature of an element.

• For example, the mass number of Polonium (Po) in the given ion is 230.
• This total signifies the combination of both protons and neutrons present within the atom's nucleus.

Understanding and calculating mass number allows scientists and students to gain insight into the stability and reactions of chemical elements. It is a foundational concept for atomic theory and order in the periodic table.

Electrons and protons count

Counting electrons and protons is vital for discerning an element's identity and its charge state. In a neutral atom, the number of electrons equals the number of protons. This balance ensures the overall neutral charge of the atom.

• Electrons are negatively charged particles orbiting around the nucleus.
• Protons are positively charged particles contained within the nucleus.

In the case of element X, initially, it has 86 electrons in the neutral state and a corresponding 86 protons, given by the context. When it forms an ion with a +2 charge, it loses 2 electrons, leaving it with 84 electrons as a positively charged ion. Hence, identifying an element in the periodic table often involves determining the atomic number, which equals the number of protons. For element X, having 84 protons points to Polonium, establishing its identity.

Charge of an ion

The charge of an ion is determined by the balance of electrons and protons in an atom. Ions form when atoms gain or lose electrons, resulting in a net positive or negative charge.

• A cation is a positively charged ion, formed by the loss of electrons.
• An anion is a negatively charged ion, formed by gaining electrons.

For element X, its ionic formation with bromine reveals its +2 charge. Since the XBr eformula indicates two bromide ( Br^- ) ions balance the charge of one X ion, element X must have lost 2 electrons to counteract the -2 total charge from the bromide ions. This loss of electrons results in a cationic state, Po^2+, for the ion of Polonium.