Carbon belongs to the second period and 14 groups. Silicon belongs to the third period and group 14. If atomic number of carbon is 6, the atomic number of silicon is:
A. 7
B. 14
C. 24
D. 16
Answer
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Hint: Carbon belongs to second period and 14th group with atomic number 6 whose electronic configuration is ${\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{p}}^{\rm{2}}}$. There are four valence electrons in a carbon atom. All the elements of the 14th group will have the same number of valence electrons.
Complete step by step answer:
Carbon, silicon, germanium, tin, lead and flerovium are the 14th group elements. Therefore, all these elements will have 4 valence electrons. Elements of group 14 tend to adopt oxidation states of $ + 4$ and $ + 2$ for heavier elements due to inert pair effect. The atomic radii and metallic properties increase down the group, and the ionization energies decrease down the group.
Electronic configuration of the carbon atom with atomic number 6 will be ${\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{p}}^{\rm{2}}}$. And the electronic configuration of the silicon atom with atomic number 14 will be ${\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{p}}^{\rm{6}}}{\rm{3}}{{\rm{s}}^{\rm{2}}}{\rm{3}}{{\rm{p}}^{\rm{2}}}$. Both of them are non-metals. However, the element silicon is less non-metallic since the non-metallic character decreases down a group; it may be regarded as a semimetal.
So, the correct answer is Option B.
Note: During the filling of electrons, some rules need to be considered as the orbitals should be filled in the order of increasing energies. Therefore, we follow this order for the filling of the orbitals:
\[{\rm{1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, and 7p}}\]
Complete step by step answer:
Carbon, silicon, germanium, tin, lead and flerovium are the 14th group elements. Therefore, all these elements will have 4 valence electrons. Elements of group 14 tend to adopt oxidation states of $ + 4$ and $ + 2$ for heavier elements due to inert pair effect. The atomic radii and metallic properties increase down the group, and the ionization energies decrease down the group.
Electronic configuration of the carbon atom with atomic number 6 will be ${\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{p}}^{\rm{2}}}$. And the electronic configuration of the silicon atom with atomic number 14 will be ${\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{p}}^{\rm{6}}}{\rm{3}}{{\rm{s}}^{\rm{2}}}{\rm{3}}{{\rm{p}}^{\rm{2}}}$. Both of them are non-metals. However, the element silicon is less non-metallic since the non-metallic character decreases down a group; it may be regarded as a semimetal.
So, the correct answer is Option B.
Note: During the filling of electrons, some rules need to be considered as the orbitals should be filled in the order of increasing energies. Therefore, we follow this order for the filling of the orbitals:
\[{\rm{1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, and 7p}}\]
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