
An element has the electronic configuration 2, 8, 4. It will be classified as:
(A) Metal
(B) non-metal
(C) metalloid
(D) none of these
Answer
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Hint: From taking the sum of electronic configuration, we can get the atomic number of element and thus we can identify the element. By observing where the element is occupied in a periodic table we can classify them into metal, nonmetals and metalloids.
Complete step by step answer:
- As we know, the maximum number of electrons present in a given shell is given by the formula $2{{(n)}^{2}}$ where n is given as the number of energy levels in the shell. The given electronic configuration 2, 8, 4. Hence there are two electrons in the K shell, eight electrons in the L shell and four electrons in the M shell. Thus the total number of electrons is 14.
- Since the atomic number is fourteen we can assume that the given element is silicon. From the electronic configuration, we could understand that the valency is four and thus it’s a group 14 element and belongs to period three in periodic table.
- Lets distinguish between metal, non-metal and metalloids. Metals are known for their lustrous, malleability, ductility and good conductivity whereas non-metals are non-lustrous, poor conductors and brittle. Metalloids will have intermediate properties between nonmetals and metals.
- The element silicon is a metalloid. They are better conductors of electricity than non-metals but not good as conductors of metals. They act as semiconductors and the silicon semiconductors are found in almost all electronic materials such as computers, mobile phones etc.
Therefore from the above discussions it’s clear that the element with the electronic configuration 2, 8, 4 can be classified into a metalloid.
So, the correct answer is “Option C”.
Note: The answer can be found through another approach too. As we know nonmetals have high electronegativities and metals have lower electronegativities. An element with the electronic configuration is 2, 8, 4 has electronegativity in between non-metal and metal. So it can be classified as a metalloid.
Complete step by step answer:
- As we know, the maximum number of electrons present in a given shell is given by the formula $2{{(n)}^{2}}$ where n is given as the number of energy levels in the shell. The given electronic configuration 2, 8, 4. Hence there are two electrons in the K shell, eight electrons in the L shell and four electrons in the M shell. Thus the total number of electrons is 14.
- Since the atomic number is fourteen we can assume that the given element is silicon. From the electronic configuration, we could understand that the valency is four and thus it’s a group 14 element and belongs to period three in periodic table.
- Lets distinguish between metal, non-metal and metalloids. Metals are known for their lustrous, malleability, ductility and good conductivity whereas non-metals are non-lustrous, poor conductors and brittle. Metalloids will have intermediate properties between nonmetals and metals.
- The element silicon is a metalloid. They are better conductors of electricity than non-metals but not good as conductors of metals. They act as semiconductors and the silicon semiconductors are found in almost all electronic materials such as computers, mobile phones etc.
Therefore from the above discussions it’s clear that the element with the electronic configuration 2, 8, 4 can be classified into a metalloid.
So, the correct answer is “Option C”.
Note: The answer can be found through another approach too. As we know nonmetals have high electronegativities and metals have lower electronegativities. An element with the electronic configuration is 2, 8, 4 has electronegativity in between non-metal and metal. So it can be classified as a metalloid.
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