
The oxidation number of $[Co{{(N{{H}_{3}})}_{6}}]C{{l}_{2}}Br$ is:
a.) +6
b.) Zero
c.) +3
d.) +2
Answer
586.8k+ views
Hint: Remember that an ammonia ligand is a neutral ligand that contains no charge and is neutral in nature. Also remember that the oxidation numbers of all elements in a neutral coordination complex sum up to zero.
Complete answer:
Let us first look into what the concept of oxidation number really means before trying to solve the given question.
The oxidation state, sometimes referred to as oxidation number, describes the degree of oxidation (loss of electrons) of an atom in a chemical compound. Conceptually, the oxidation state, which may be positive, negative or zero, is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, with no covalent component. This is never exactly true for real bonds. Oxidation states are typically represented by integers which may be positive, zero, or negative
Now, let us try to see how oxidation numbers can be calculated for coordination compounds.
The oxidation state of the metal is determined based on the charges of each ligand and the overall charge of the coordination compound. For example, in\[\left[ Cr{{\left( {{H}_{2}}O \right)}_{4}}C{{l}_{2}} \right]Br\], the coordination sphere (in brackets) has a charge of 1+ to balance the bromide ion. The water ligands are neutral, and the chloride ligands are anionic with a charge of 1− each. To determine the oxidation state of the metal, we set the overall charge equal to the sum of the ligands and the metal: +1 = −2 + x, so the oxidation state (x) is equal to 3+.
Let us now apply these concepts to help calculate the oxidation number of the given compound.
As there are 6 ammonia ligands which are neutral and hence carry no charge. The chlorine ligands carry a charge of -1 each and thus they have -3 charge on them. But as the molecule in whole is a neutral the coordination sphere must carry a charge of +3. So, the oxidation state of Co is +3.
Thus, we can conclude that the answer is c).
Note:
Remember the following rules of the oxidation states for coordination compounds at all times:
1. The oxidation number of a free element is always 0.
2. The oxidation number of a monatomic ion equals the charge of the ion.
3. The oxidation number of H is +1, but it is -1 in when combined with less electronegative elements.
4. The oxidation number of O in compounds is usually -2, but it is -1 in peroxides.
5. The oxidation number of a Group 1 element in a compound is +1.
6. The oxidation number of a Group 2 element in a compound is +2.
7. The oxidation number of a Group 17 element in a binary compound is -1.
8. The sum of the oxidation numbers of all of the atoms in a neutral compound is 0.
9. The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.
Complete answer:
Let us first look into what the concept of oxidation number really means before trying to solve the given question.
The oxidation state, sometimes referred to as oxidation number, describes the degree of oxidation (loss of electrons) of an atom in a chemical compound. Conceptually, the oxidation state, which may be positive, negative or zero, is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, with no covalent component. This is never exactly true for real bonds. Oxidation states are typically represented by integers which may be positive, zero, or negative
Now, let us try to see how oxidation numbers can be calculated for coordination compounds.
The oxidation state of the metal is determined based on the charges of each ligand and the overall charge of the coordination compound. For example, in\[\left[ Cr{{\left( {{H}_{2}}O \right)}_{4}}C{{l}_{2}} \right]Br\], the coordination sphere (in brackets) has a charge of 1+ to balance the bromide ion. The water ligands are neutral, and the chloride ligands are anionic with a charge of 1− each. To determine the oxidation state of the metal, we set the overall charge equal to the sum of the ligands and the metal: +1 = −2 + x, so the oxidation state (x) is equal to 3+.
Let us now apply these concepts to help calculate the oxidation number of the given compound.
As there are 6 ammonia ligands which are neutral and hence carry no charge. The chlorine ligands carry a charge of -1 each and thus they have -3 charge on them. But as the molecule in whole is a neutral the coordination sphere must carry a charge of +3. So, the oxidation state of Co is +3.
Thus, we can conclude that the answer is c).
Note:
Remember the following rules of the oxidation states for coordination compounds at all times:
1. The oxidation number of a free element is always 0.
2. The oxidation number of a monatomic ion equals the charge of the ion.
3. The oxidation number of H is +1, but it is -1 in when combined with less electronegative elements.
4. The oxidation number of O in compounds is usually -2, but it is -1 in peroxides.
5. The oxidation number of a Group 1 element in a compound is +1.
6. The oxidation number of a Group 2 element in a compound is +2.
7. The oxidation number of a Group 17 element in a binary compound is -1.
8. The sum of the oxidation numbers of all of the atoms in a neutral compound is 0.
9. The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

