
Hydrogen combines with other elements by
A. Losing an electron
B. Gaining and electron
C. Sharing an electron
D. Loosing, gaining, or sharing electrons
Answer
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Hint: Atomic number of Hydrogen is 1 and it is the lightest element. Hydrogen is considered non-metallic as it does not possess any metallic properties even though it is a group I element. It’s Electronic Configuration 1s1 so we can see that it has only 1 valence shell electron. It can only share or lose this one electron and also can accept one electron.
Complete Step by Step Answer:
The atomic no of hydrogen is 1 so its electronic configuration is:
In hydrogen, we can see that it has only one electron which is its valence electron. So, Hydrogen can easily lose one electron and become an H+ ion, and also it has a place to accept one electron so it accepts one electron and becomes an H- ion but it also shares one electron with other components.
Examples of each condition are:
1. In HF (Hydrogen Fluoride) Hydrogen loses its one electron and forms a bond with Fluorine this bond is called a polar covalent bond in this reaction fluorine accepts that electron and forms a covalent bond with hydrogen sharing a pair of electrons.
$\ H_2+F_2\longrightarrow HF$
Mechanism
$ H_2\longrightarrow 2H \bullet$
$2H\bullet\ \longrightarrow \ 2H^++\ 2e^-$
$F_2\longrightarrow 2F \bullet$
$2F\bullet+2e^-\ \longrightarrow {2F}^-$
$2F^-\ +\ {2H}^+\longrightarrow \ 2HF$
2. In NaH (Sodium Hydride) hydrogen gains one electron from sodium (Na) and forms an anion H- ion and then Na+ and H- combines together with an ionic bond.
$Na(s)+\frac{1}{2}H_2(g)\longrightarrow \ NaH(s)$
Mechanism
$Na\ \longrightarrow {\rm Na}^++e^-$
$H_2\longrightarrow 2H \bullet$
$H\bullet+e^-\ \longrightarrow H^-$
${\rm Na}^+\ +\ H^-\longrightarrow \ NaH$
3. In H2 which is a Hydrogen gas there exists a covalent bond between both the Hydrogen atoms and they share their single electrons with each other.
$2H\bullet\ \longrightarrow \ H_2$
Thus, Option (D) is correct
Note: Hydrogen does not exist in H- ion in the solutions it is a very unstable ion. Hydrogen has an important role in acid-base reaction as it acts as a proton, and acids are considered as proton donors and bases are considered as proton acceptors.
Complete Step by Step Answer:
The atomic no of hydrogen is 1 so its electronic configuration is:
1 |
1s1 |
In hydrogen, we can see that it has only one electron which is its valence electron. So, Hydrogen can easily lose one electron and become an H+ ion, and also it has a place to accept one electron so it accepts one electron and becomes an H- ion but it also shares one electron with other components.
Examples of each condition are:
1. In HF (Hydrogen Fluoride) Hydrogen loses its one electron and forms a bond with Fluorine this bond is called a polar covalent bond in this reaction fluorine accepts that electron and forms a covalent bond with hydrogen sharing a pair of electrons.
$\ H_2+F_2\longrightarrow HF$
Mechanism
$ H_2\longrightarrow 2H \bullet$
$2H\bullet\ \longrightarrow \ 2H^++\ 2e^-$
$F_2\longrightarrow 2F \bullet$
$2F\bullet+2e^-\ \longrightarrow {2F}^-$
$2F^-\ +\ {2H}^+\longrightarrow \ 2HF$
2. In NaH (Sodium Hydride) hydrogen gains one electron from sodium (Na) and forms an anion H- ion and then Na+ and H- combines together with an ionic bond.
$Na(s)+\frac{1}{2}H_2(g)\longrightarrow \ NaH(s)$
Mechanism
$Na\ \longrightarrow {\rm Na}^++e^-$
$H_2\longrightarrow 2H \bullet$
$H\bullet+e^-\ \longrightarrow H^-$
${\rm Na}^+\ +\ H^-\longrightarrow \ NaH$
3. In H2 which is a Hydrogen gas there exists a covalent bond between both the Hydrogen atoms and they share their single electrons with each other.
$2H\bullet\ \longrightarrow \ H_2$
Thus, Option (D) is correct
Note: Hydrogen does not exist in H- ion in the solutions it is a very unstable ion. Hydrogen has an important role in acid-base reaction as it acts as a proton, and acids are considered as proton donors and bases are considered as proton acceptors.
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