Fluorine reacts with ice and results in the change:
H2O(s) + F2(g) → HF(g) + HOF(g)
Justify that this reaction is a redox reaction.
Let us write the oxidation number of each atom involved in the given reaction above its symbol as:
Here, we have observed that the oxidation number of F increases from 0 in F2 to +1 in HOF. Also, the oxidation number decreases from 0 in F2 to -1 in HF. Thus, in the above reaction, F is both oxidized and reduced. Hence, the given reaction is a redox reaction.
The first (ΔiH1) and the second (ΔiH) ionization enthalpies (in kJ mol–1) and the (ΔegH) electron gain enthalpy (in kJ mol–1) of a few elements are given below:
Elements | ΔiH1 | ΔiH | ΔegH |
I | 520 | 7300 | -60 |
II | 419 | 3051 | -48 |
III | 1681 | 3374 | -328 |
IV | 1008 | 1846 | -295 |
V | 2372 | 5251 | +48 |
VI | 738 | 1451 | -40 |
Which of the above elements is likely to be :
(a) the least reactive element.
(b) the most reactive metal.
(c) the most reactive non-metal.
(d) the least reactive non-metal.
(e) the metal which can form a stable binary halide of the formula MX2, (X=halogen).
(f) the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)?
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 4: Fluorine reacts with ice and results in the change: H2O(s) + F2(g) → HF(g) + HOF(g) Justi....
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Thanku so muchð¤
Being the most electronegative fluorine will never show positive oxidation state