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2 years ago

What is the sum of the numbers of neutrons and electrons in the ion 208 Pb2+?

Chemistry

1 answer:

AysviL [449]2 years ago

3 0

Answer:

For ²⁰⁸Pb²⁺ cation, the sum of the number of neutrons and electrons = 206

Explanation:

Lead, chemical symbol Pb, is a chemical element which belongs to the group 14 of the periodic table. The atomic number of lead is 82 and it is a member of the p-block.

The isotope of lead with the mass number 208, has 126 neutrons.

Since, atomic number = number of protons = number of electrons for neutral atom

Therefore, for ²⁰⁸Pb: number of electrons= 82

So, for ²⁰⁸Pb²⁺ cation: number of electrons= 82 - 2 = 80

Therefore, for ²⁰⁸Pb²⁺ cation, the sum of the number of neutrons and electrons = number of electrons + number of neutrons = 80 electrons + 126 neutrons = 206.

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Give the name and formula of the compound formed from the following elements: (a) sodium and nitrogen; (b) oxygen and strontium;

lisov135 [29]

Answer:

For a: The chemical name and chemical formula formed is sodium nitride and $Na_3N$

For b: The chemical name and chemical formula formed is strontium oxide and $SrO$

For c: The chemical name and chemical formula formed is aluminium chloride and $AlCl_3$

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

The nomenclature of ionic compounds is given by:

1. Positive is written first.

2. The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.

For a: Sodium and nitrogen:

Sodium is the 11th element of periodic table having electronic configuration of $[Ne]3s^1$

To form $Na^{+}$ ion, this element will loose 1 electron.

Nitrogen is the 7th element of periodic table having electronic configuration of $[He]2s^22p^3$ .

To form $N^{3-}$ ion, this element will gain 3 electrons.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for sodium nitride is $Na_3N$

For b: Oxygen and strontium

Strontium is the 38th element of periodic table having electronic configuration of $[Kr]5s^2$ .

To form $Sr^{2+}$ ion, this element will loose 2 electrons.

Oxygen is the 8th element of periodic table having electronic configuration of $[He]2s^22p^4$ .

To form $O^{2-}$ ion, this element will gain 2 electrons.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for strontium oxide is $SrO$

For c: Aluminium and chlorine

Aluminium is the 13th element of periodic table having electronic configuration of $[Ne]3s^23p^1$

To form $Al^{3+}$ ion, this element will loose 3 electrons.

Chlorine is the 17th element of the periodic table having electronic configuration of $[Ne]3s^23p^5$

To form $Cl^{-}$ ion, this element will gain 1 electron.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for aluminium chloride is $AlCl_3$

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