Basics of Chemistry

Part-04 (Valancy, charge & ions)

In the previous three parts of this series, we have learnt about details of first 20 elements, calculated their atomic weight, drawn their atomic structures and learnt about shell & sub-shell electronic configurations.

Not seen Part-1, Part-2 & Part-3 yet??

Click here for Basics of Chemistry || Part-01 ||

Click here for Basics of Chemistry || Part-02 ||

Click here for Basics of Chemistry || Part-03 ||

Basics of Chemistry || Part-04 || Valancy & Radicals

In this part we will be discussing about valencies and radicals of the first 20 elements about which we have studied in previous parts. We will first learn to calculate valencies from shell and sub-shell electronic configurations and then we will be looking towards ions.

So, let us start without any delay...

After understanding shell electronic configuration, it is easy to predict valencies of the elements just by looking their last number.

For example, the Electronic configuration (E.C.) of Lithium is 2,1. So, its valancy will be 1. Similarly, the E.C. of Carbon is 2,4 so its valancy will be 4. Likewise, the E.C. of nitrogen is 2, 5 so according to our plan, the valancy must be 5 but nitrogen atom prefer to gain electrons rather than donating. So, its valancy becomes 8 - 5 = 3.

Note: We subtracted 5 by 8 because L-shell (2nd orbit) can accommodate maximum 8 electrons and tends to gain 8 electrons.

The valencies of first 20 elements is given below in the table.

Element	Shell electronic configuration	Valancy

Hydrogen (H)	1	1
Helium (He)	2	2 – 2 = 0
Lithium (Li)	2,1	1
Beryllium (Be)	2,2	2
Boron (B)	2,3	3
Carbon (C)	2,4	4
Nitrogen (N)	2,5	8-5 = 3
Oxygen (O)	2,6	8-6 = 2
Fluorine (F)	2,7	8-7 = 1
Neon (Ne)	2,8	8-8 = 0
Sodium (Na)	2,8,1	1
Magnesium (Mg)	2,8,2	2
Aluminum (Al)	2,8,3	3
Silicon (Si)	2,8,4	4
Phosphorous (P)	2,8,5	8-5 = 3
Sulphur (S)	2,8,6	8-4 = 2
Chlorine (Cl)	2,8,7	8-7 = 1
Argon (Ar)	2,8,8	8-8 = 0
Potassium (K)	2,8,8,1	1
Calcium (Ca)	2,8,8,2	2

Let us understand about charge of these elements.

Simply, the valancy indicates the charge of any element. You can calculate the charge on any ion just by looking to its valancy.

Note: 😂😂😂😂😂

1) After donation you becomes positive

2) After stealing from others you become negative

For example,

Lithium

E.C. = 2,1

Valancy = 1

So, it donates its outermost 1 electron and becomes positive.

Hence, ion will be Li⁺

Carbon

E.C. = 2,4

Valancy = 4

So, it can either donate its 4 electrons or can steal electrons from other.

Hence, its ion will be C⁺ ⁺ ⁺ ⁺ or C⁻ ⁻ ⁻ ⁻

Nitrogen

E.C. = 2,5

Valancy = 3

So, it is easier for nitrogen to steal 3 electrons for its outermost orbit rather than donating 5 electrons to others.

Hence, it has ion N⁻ ⁻ ⁻

Sodium

E.C. = 2,8,1

Valancy = 1

So, it is simple that sodium can donate its 1 extra electron and becomes positive.

Hence, its ion will be Na⁺

And so on....

Here is the summarized list of first 20 elements with their valencies and charge.

Element	Valancy	Charge

Hydrogen (H)	1	+
Helium (He)	0	0
Lithium (Li)	1	+
Beryllium (Be)	2	+ +
Boron (B)	3	+ + +
Carbon (C)	4	4+ or 4 -
Nitrogen (N)	3 or 5	- - -
Oxygen (O)	2	- -
Fluorine (F)	1	-
Neon (Ne)	0	0
Sodium (Na)	1	+
Magnesium (Mg)	2	+ +
Aluminum (Al)	3	+ + +
Silicon (Si)	4	4+ or 4 -
Phosphorous (P)	3 or 5	- - -
Sulphur (S)	2	- -
Chlorine (Cl)	1	-
Argon (Ar)	0	0
Potassium (K)	1	+
Calcium (Ca)	2	+ +

After understanding valancy, we , now, learn about ions. Simply, ions can be written by placing charge into power of symbol.

Just have a look to the given table.

Element	Valancy	Ions

Hydrogen (H)	1	H⁺
Helium (He)	0	He⁰
Lithium (Li)	1	Li⁺
Beryllium (Be)	2	Be^{+ +}
Boron (B)	3	B^{+ + +}
Carbon (C)	4	C^{4 +} or C^{4 -}
Nitrogen (N)	3 or 5	N^{- - -}
Oxygen (O)	2	O^{- -}
Fluorine (F)	1	F^-
Neon (Ne)	0	Ne⁰
Sodium (Na)	1	Na⁺
Magnesium (Mg)	2	Mg^{+ +}
Aluminum (Al)	3	Al^{+ + +}
Silicon (Si)	4	Si⁴⁺ or Si^4-
Phosphorous (P)	3 or 5	P^{- - -}
Sulphur (S)	2	S^{- -}
Chlorine (Cl)	1	Cl^-
Argon (Ar)	0	Ar⁰
Potassium (K)	1	K⁺
Calcium (Ca)	2	Ca^{+ +}

Hope that you liked it. In the next part (last part of this series) we will be trying to understand about molecular formula by criss-cross method.

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