In the second period, Li to Ne identify the elements in the following:
(a)- Most reactive element
(b)- Most non-reactive non-metal
(c)- Largest atomic radius
(d)- With valency equal to 4
Answer
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Hint: In the second period there are 8 elements in the order lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine, and neon in which the first three are metals and the rest are non-metals. First, while moving left to right the valency increases till 4 then decreases till 0.
Complete Solution :
The period in the periodic table is in which the number of the last shell remains the same but the number of electrons increases as we move left to right. In the second period, there are 8 elements in the order lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine, and neon in which the first three are metals and the rest are non-metals.
(a)- The reactivity of the elements can be decided on the factor of electronegativity. The element that has the highest electronegativity will have the highest reactivity, so fluorine has the highest electronegativity hence, it will be most reactive.
(b)- Carbon, nitrogen, oxygen, fluorine, and neon are non-metals, and the element which belongs to the noble gas group will be the least reactive because they have a complete octet. So, neon is the noble element therefore, it will be the most non-reactive non-metal.
(c)- Period in the periodic table is in which the number of the last shell remains the same but the number of electrons increases as we move left to right due to which the nuclear charge increases and the size becomes smaller. So, lithium will have the largest size.
(d)- First, while moving left to right the valency increases till 4 then decreases till 0. So, the valency of Li and F will be 1, the valency of Be and O will be 2, the valency of B and N will be 3, the valency of C will be 4, and the valency of Ne will be 0. Therefore, Carbon is the answer.
Note: The metallic character along the period from left to right decreases and the non-metallic character increases. The trend in electron gain enthalpy and ionization energy in the period is the same as that of electronegativity.
Complete Solution :
The period in the periodic table is in which the number of the last shell remains the same but the number of electrons increases as we move left to right. In the second period, there are 8 elements in the order lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine, and neon in which the first three are metals and the rest are non-metals.
(a)- The reactivity of the elements can be decided on the factor of electronegativity. The element that has the highest electronegativity will have the highest reactivity, so fluorine has the highest electronegativity hence, it will be most reactive.
(b)- Carbon, nitrogen, oxygen, fluorine, and neon are non-metals, and the element which belongs to the noble gas group will be the least reactive because they have a complete octet. So, neon is the noble element therefore, it will be the most non-reactive non-metal.
(c)- Period in the periodic table is in which the number of the last shell remains the same but the number of electrons increases as we move left to right due to which the nuclear charge increases and the size becomes smaller. So, lithium will have the largest size.
(d)- First, while moving left to right the valency increases till 4 then decreases till 0. So, the valency of Li and F will be 1, the valency of Be and O will be 2, the valency of B and N will be 3, the valency of C will be 4, and the valency of Ne will be 0. Therefore, Carbon is the answer.
Note: The metallic character along the period from left to right decreases and the non-metallic character increases. The trend in electron gain enthalpy and ionization energy in the period is the same as that of electronegativity.
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