
Hydrogen loses its electron to form ${H^ + }$. In this respect it resembles:
A. Halogens
B. Alkali metals
C. Transition elements
D. Alkaline earth metal
Answer
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Hint: Hydrogen is the first element of the periodic table and is the lightest element. In a normal state, it exists in diatomic form. The name hydrogen has a Greek origin where hydro means water and gene means producers.
Complete step by step answer:
only one electron in its outermost shell so it easily loses the electron to form ${H^ + }$. If we talk about the halogen, its electronic configuration is $n{s^2}n{p^5}$ which means it contains seven electrons in its valence shell and requires only one electron to complete its octet. As it is quite easy to accept one electron instead of removing seven electrons so they can accept the one-electron instead of losing.
If we talk about alkali metals, their electronic configuration is $n{s^1}$. That means the last electron enters the s subshell. As they are larger and have one electron in its valence shell so alkali metals easily remove an electron. In the case of a transition metal, its electronic configuration is $(n - 1){d^{1 - 10}}n{s^{1 - 2}}$. That means the last electron enters the d orbital and they also exist in a variable oxidation state. And in the case of alkaline earth metal, the electronic configuration is $n{s^2}$. Since s orbital contains 2 electrons so it is a full-filled electronic configuration so it is quite difficult to remove an electron.
So, the correct answer is Option B.
Note: Hydrogen can exist in various forms such as isotopes, allotropes, molecular, and ionic. The Isotopes of hydrogen are protium, deuterium, and tritium and the Allotropes of hydrogen are orthohydrogen, parahydrogen.
Complete step by step answer:
only one electron in its outermost shell so it easily loses the electron to form ${H^ + }$. If we talk about the halogen, its electronic configuration is $n{s^2}n{p^5}$ which means it contains seven electrons in its valence shell and requires only one electron to complete its octet. As it is quite easy to accept one electron instead of removing seven electrons so they can accept the one-electron instead of losing.
If we talk about alkali metals, their electronic configuration is $n{s^1}$. That means the last electron enters the s subshell. As they are larger and have one electron in its valence shell so alkali metals easily remove an electron. In the case of a transition metal, its electronic configuration is $(n - 1){d^{1 - 10}}n{s^{1 - 2}}$. That means the last electron enters the d orbital and they also exist in a variable oxidation state. And in the case of alkaline earth metal, the electronic configuration is $n{s^2}$. Since s orbital contains 2 electrons so it is a full-filled electronic configuration so it is quite difficult to remove an electron.
So, the correct answer is Option B.
Note: Hydrogen can exist in various forms such as isotopes, allotropes, molecular, and ionic. The Isotopes of hydrogen are protium, deuterium, and tritium and the Allotropes of hydrogen are orthohydrogen, parahydrogen.
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