
Hydrogen resembles in many of its properties
A. Halogen
B. Alkali metals
C. Both a and b
D. None of these
Answer
162.9k+ views
Hint: The first and simplest element in the periodic table is the hydrogen atom having an electronic configuration $n{{s}^{1}}$. It contains one electron in its outermost valence shell, making it a half filled configuration. Hence, it has a tendency either to gain one electron to attain its nearest noble gas (He) configuration or to lose one electron and become a stable unipositive species.
Complete Step by Step Answer:
Alkali metals have an outer electronic configuration $n{{s}^{1}}$that resembles hydrogen. For example, sodium has an electronic configuration, $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{1}}$ i.e, it has one electron in its outermost shell. Thus, sodium tends to lose one electron to gain the most stable electronic configuration of the nearest noble gas, neon $(1{{s}^{2}}2{{s}^{2}}2{{p}^{6}})$.
$Na(1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{1}})\to N{{a}^{+}}(1{{s}^{2}}2{{s}^{2}}2{{p}^{6}})+{{e}^{-}}$
Like alkali metals, hydrogen forms oxides, halides, sulphides, etc.
Halogens have an outer electronic configuration $n{{s}^{2}}n{{p}^{5}}$ that also resembles hydrogen. For example, fluorine(F) has an electron configuration $1{{s}^{2}}2{{s}^{2}}2{{p}^{5}}$. It means fluorine needs one electron to attain the nearest noble gas configuration of neon (Ne).
$F(1{{s}^{2}}2{{s}^{2}}2{{p}^{5}})+{{e}^{-}}\to {{F}^{-}}(1{{s}^{2}}2{{s}^{2}}2{{p}^{6}})$
Like this, hydrogen resembles halogen and forms diatomic molecules like ${{H}_{2}},{{F}_{2}},C{{l}_{2,}}B{{r}_{2,}}{{I}_{2}}$molecules.
That’s why the hydrogen atom has been placed at the top separately for its unique properties. This special small-sized hydrogen atom is a strong reducing agent.
So, hydrogen resembles many of its properties alkali metals, and halogens both.
Thus, option (C) is correct.
Note: Hydrogen atom also shows differences with alkali metals and halogens. Like alkali metal atoms, hydrogen atoms are not metal, they are non-metal. Halogens have more than one shell and hydrogen has only one shell. Hydrogen forms neural oxides while alkali and alkaline earth metals form basic oxides.
Complete Step by Step Answer:
Alkali metals have an outer electronic configuration $n{{s}^{1}}$that resembles hydrogen. For example, sodium has an electronic configuration, $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{1}}$ i.e, it has one electron in its outermost shell. Thus, sodium tends to lose one electron to gain the most stable electronic configuration of the nearest noble gas, neon $(1{{s}^{2}}2{{s}^{2}}2{{p}^{6}})$.
$Na(1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{1}})\to N{{a}^{+}}(1{{s}^{2}}2{{s}^{2}}2{{p}^{6}})+{{e}^{-}}$
Like alkali metals, hydrogen forms oxides, halides, sulphides, etc.
Halogens have an outer electronic configuration $n{{s}^{2}}n{{p}^{5}}$ that also resembles hydrogen. For example, fluorine(F) has an electron configuration $1{{s}^{2}}2{{s}^{2}}2{{p}^{5}}$. It means fluorine needs one electron to attain the nearest noble gas configuration of neon (Ne).
$F(1{{s}^{2}}2{{s}^{2}}2{{p}^{5}})+{{e}^{-}}\to {{F}^{-}}(1{{s}^{2}}2{{s}^{2}}2{{p}^{6}})$
Like this, hydrogen resembles halogen and forms diatomic molecules like ${{H}_{2}},{{F}_{2}},C{{l}_{2,}}B{{r}_{2,}}{{I}_{2}}$molecules.
That’s why the hydrogen atom has been placed at the top separately for its unique properties. This special small-sized hydrogen atom is a strong reducing agent.
So, hydrogen resembles many of its properties alkali metals, and halogens both.
Thus, option (C) is correct.
Note: Hydrogen atom also shows differences with alkali metals and halogens. Like alkali metal atoms, hydrogen atoms are not metal, they are non-metal. Halogens have more than one shell and hydrogen has only one shell. Hydrogen forms neural oxides while alkali and alkaline earth metals form basic oxides.
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