SS1: CHEMISTRY – 1ST TERM Chemical Combination (Chemical Bonding) II | Week 10 Co-ordinate Covalent or Dative Bonding

Lesson 10, Topic 4

In Progress

← Previous Next→

Co-ordinate Covalent or Dative Bonding

Lesson Progress

0% Complete

Topic Content:

Co-ordinate Covalent or Dative Bonding
Formation of Ammonium ion (NH₄⁺)
Formation of Hydroxonium Ion or Oxonium Ion (H₂O⁺)
The Reaction between Ammonia and Boron Trifluoride, BF₃
Examples of Compounds that Exhibit Co-ordinate Covalent Bond
Properties of Co-ordinate Covalent Compounds

In co-ordinate covalent bonding, electrons are still shared but it is only one of the participating atoms that donate the pair of electrons to be shared. Such a pair of electrons is called the Lone pair.

Co-ordinate covalent bonding occurs between an atom that has a Lone-pair (unshared pair) of electrons in the outer shell and another atom that has an empty valence orbital and requires one pair of electrons to complete an octet or duplet structure.

The atom that contributes the pair of electrons to be shared is known as donor-atom, while the other is the acceptor atom.

Examples:

Formation of Ammonium ion (NH₄⁺):

Ammonia and water molecule possess Lone pair and they can enter into a co-ordinate covalent bonding. When Ammonia (NH₃) reacts with Hydrochloric Acid (HCl) the H⁺ from HCl accepts a Lone pair of electrons from Ammonia (NH₃) to form a co-ordinate covalent bond.

The H⁺ will attain a stable duplet structure of helium while the Nitrogen atom of the ammonia will also be stable.

When the ammonium ion, NH₄⁺, is formed, the fourth hydrogen is attached by a dative covalent bond, because only the hydrogen’s nucleus is transferred from the chlorine to the nitrogen. The hydrogen’s electron is left behind on the chlorine to form a negative chloride ion.

In simple diagrams, a co-ordinate bond is shown by an arrow. The arrow points from the atom donating the lone pair to the atom accepting it.

Formation of Hydroxonium Ion or Oxonium Ion (H₃O⁺):

Formation of Hydroxonium Ion or Oxonium Ion H3O

The Reaction between Ammonia and Boron trifluoride, BF₃:

Boron trifluoride, BF₃, is a compound which doesn’t have a noble gas structure around the boron atom. BF₃ is described as being electron deficient because boron only has 3 pairs of electrons in its bonding level, whereas there would be room for 4 pairs.

The lone pair on the nitrogen of an ammonia molecule can be used to overcome that deficiency, and a compound is formed involving a co-ordinate bond.

The Reaction between Ammonia and Boron trifluoride BF3

Using lines to represent the bonds, this could be drawn more simply as:

The second diagram shows another way that you might find co-ordinate bonds drawn. The nitrogen end of the bond has become positive because the electron pair has moved away from the nitrogen towards the boron – which has therefore become negative.

Examples of Compounds that Exhibit Co-ordinate Covalent Bond:

NH₄Cl – Ammonium Chloride
NH₄⁺ – Ammonium ion
H₃O⁺– Hydroxonium ion or oxonium ion
[CU(NH₃)₄]²⁺ – Tetraammine copper (II) ion
[Ag(NH₃)₂]^– – Silver diammine ion

Properties of Co-ordinate Covalent Compounds:

Covalent and co-ordinate covalent compounds have similar properties but co-ordinate covalent compounds are less volatile than covalent compounds.

Theory Questions:

1. (a) State four differences between ElectrovalentAn electrovalent or ionic bond is formed when a metal atom transfers one or more electrons to a non-metal atom. Ionic or electrovalent compounds are those compounds which are formed by... More and Covalent Compounds

(b) State the type of bonding formed in the following

(i) Carbon(IV) Oxide

(ii) Oxonium ion

(iii) Calcium Chloride

2. The Electronic configuration for the metal Calcium, the non-metals Silicon, and Chloride can be represented as

Ca – 2, 8, 8, 2

Si – 2, 8, 4

Cl – 2, 8, 7

Explain in terms of electrons, the formation of Calcium Chloride and Silicon Chloride.

Give two differences in physical properties you would expect between Calcium Chloride and Silicon Chloride

3. (a) Write down the electronic configuration of the Elements Q, R, and S, whose atomic numbers are as follows

Element	Atomic Number
Q	8
R	12
S	19

(b) Draw the diagrammatic representation of the compound formed between Q and R

(d) Is Q a metal or non-metal?

4. (a) (i) What is Dative bonding?

(ii) State one condition that must be met for Dative bonding to take place

(b) Given the following Atomic Numbers X = 12, Y = 17

(i) Write the formula of the Compound formed between X and Y

(ii) Name the type of Bond formed in 4b(i)

(iii) Give a simple definition of the type of Bond formed.

	Electrovalent Compounds	Covalent Compounds
1.	They have high melting and boiling points.	They have low melting and boiling points.
2.	They are solids that do not vaporize easily.	They are gases or volatile liquids.
3.	They are soluble in water or polar solvents but insoluble in non-polar solvents.	They are not soluble in water or polar solvents but soluble in non-polar solvents.
4.	They are electrolytes i.e they conduct electricity in solution or molten form.	They are non-electrolyte, i.e they do not conduct electricity.

SS1: CHEMISTRY - 1ST TERM

Quizzes

Follow