Advertisements
Advertisements
प्रश्न
The oxidation number of an element in a compound is evaluated on the basis of certain rules. Which of the following rules is not correct in this respect?
विकल्प
The oxidation number of hydrogen is always +1.
The algebraic sum of all the oxidation numbers in a compound is zero.
An element in the free or the uncombined state bears oxidation number zero.
In all its compounds, the oxidation number of fluorine is – 1.
उत्तर
The oxidation number of hydrogen is always +1.
Explanation:
Oxidation number of hydrogen is – 1 in metal hydrides like \[\ce{NaH}\].
APPEARS IN
संबंधित प्रश्न
Assign oxidation numbers to the underlined element in the following species:
H4P2O7
Assign oxidation numbers to the underlined elements in the following species:
H2S2O7
What are the oxidation numbers of the underlined elements in the following and how do you rationalise your results?
CH3CH2OH
Consider the elements: Cs, Ne, I and F
Identify the element that exhibits only negative oxidation state.
Consider the elements : Cs, Ne, I and F
Identify the element which exhibits neither the negative nor does the positive oxidation state.
In which of the following compounds, an element exhibits two different oxidation states.
The reaction \[\ce{Cl2 (g) + 2OH- (aq) -> ClO- (aq) + Cl- (aq) + H2O (l)}\] represents the process of bleaching. Identify and name the species that bleaches the substances due to its oxidising action.
Calculate the oxidation number of sulphur atom in the following compounds:
\[\ce{Na2S4O6}\]
The oxidation number and covalency of sulphur in sulphur molecules (Sg) are:
Oxidation state of sulphur in anions `"SO"_3^(2-)`, `"S"_2"O"_4^(2-)` and `"S"_2"O"_6^(2-)` increases in the orders: