Chemistry, 13.08.2019 04:30 esperanzar5755

Identify each of the following elements: (a) a halogen whose anion contains 36 electrons, (b) a radioactive noble gas with 86 protons, (c) a group 6a element whose anion contains 36 electrons, (d) an alkali metal cation that contains 36 electrons, (e) a group 4a cation that contains 80 electrons.

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Answer from: Issaking0131

For a: The correct answer is Br^-

For b: The correct answer is Radon.

For c: The correct answer is $Se^{2-}$

For d: The correct answer is Rb^+

For e: The correct answer is $Pb^{2+}$

Explanation:

An ion is formed when an atom looses or gains electron.

When an atom looses electrons, it will form a positive ion known as cation.

When an atom gains electrons, it will form a negative ion known as anion.

Atomic number is defined as the number of protons or electrons that are present in a neutral atom.

Atomic number = number of protons = number of electrons

Atomic mass is defined as the sum of number of protons and neutrons that are present in an atom.

Atomic mass = Number of protons + Number of neutrons

For a:

Halogens are Fluorine (Z = 9), Chlorine (Z = 17), Bromine (Z = 35), Iodine (Z = 53) and Astatine (Z = 85)

Bromine is the 35th element of the periodic table having electronic configuration $[Ar]3d^{10}4s^24p^5$

This element will gain 1 electron and will form Br^- ion.

The ion formed has 36 electrons.

For b:

Noble gas having 86 protons means the element having atomic number 86 is Radon.

Thus, the correct answer is Radon.

For c:

Group 6A elements are Oxygen (Z = 8), Sulfur (Z = 16), Selenium (Z = 34), Tellurium (Z = 52) and Polonium (Z = 84)

Selenium is the 34th element of the periodic table having electronic configuration $[Ar]3d^{10}4s^24p^4$

This element will gain 2 electrons and will form $Se^{2-}$ ion.

The ion formed has 36 electrons.

For d:

Alkali metals are Lithium (Z = 3), Sodium (Z = 11), Potassium (Z = 19), Rubidium (Z = 37), Cesium (Z = 55) and Francium (Z = 87)

Rubidium is the 34th element of the periodic table having electronic configuration [Kr]5s^1

This element will loose 1 electron and will form $Rb^{+}$ ion.

The ion formed has 36 electrons.

For e:

Group 4A elements are Carbon (Z = 6), Silicon (Z = 14), Germanium (Z = 32), Tin (Z = 50) and Lead (Z = 82)

Lead is the 82th element of the periodic table having electronic configuration $[Xe]4f^{14}5d^{10}6s^26p^2$

This element will loose 2 electrons and will form $Pb^{2+}$ ion.

The ion formed has 80 electrons.

Answer from: Quest

The answer is b i think

Answer from: Quest

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