Question

Which of the following atoms has been assigned only single oxidation number?

• A. H
• B. N
• C. F
• D. O

Answer 1

The correct option is C F

Analysing the options:
Option (A)
Fluorine (F) is the most electronegative element with electronic configuration
$1s^{2}2s^{2}2p^5$.
It readily accepts one electron to attain stable noble gas configuration and exist as $F^{-}$
$F+e^{-}\to F^{-}$

But due to its high electronegativity and small size, it cannot lose electron. Also it does not have vacant d−orbitals, so it can not form more than one $\sigma$ bond so it exhibits only one oxidation state i.e., -1.

Option (B)
Oxygen can exhibit more than one oxidation state. It has six valence electrons.
Electronic configuration of Oxygen is
$1s^{2}2s^{2}2p^4$.
E.g. : In $H_{2}O$, oxidation state of O is −2.
In $H_2{O}_2$, oxidation state of O is -1 (peroxy linkage).
In $O_2{F}_2$, oxidation state of O is +1.

Option (C)
Nitrogen can exhibit more than one oxidation state. It has five valence electrons.
Electronic configuration of Nitrogen:
$1s^{2}2s^{2}2p^3$.
E.g. : N in $N{H}_3$, has −3 oxidation state.
N in $N_3^{-}$ , has $-\frac{1}{3}$ oxidation state.
N in $HN{O}_2$, has +3 oxidation state.
N in $HN{O}_3$, has +5 oxidation state.

Option (D)
Hydrogen has $1s^1$ configuration thus it can gain one electron as well as it can lose one electron.
Thus, it can exhibit more than one oxidation state i.e., +1 and -1 and can exist as $H^{+}$ and $H^{-}$ respectively.
So, the correct answer is option (A).