Elements X,Y,Z have atomic numbers 6,9 and 12 respectively.
Which one forms anions - negative ions;
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Solution
Let us first write the electronic configuration of elements X,Y and Z.
The electronic configuration of an element describes how electrons are distributed in its atomic orbitals.
The electrons are filled in the increasing number of shells viz. K,L,M,N,etc.
Number of electrons in a shell is given by 2n2 where n is the number of shells.
Putting n=1,2,3 and 4, we get the number of electrons 2,8,18,32 that can be filled in each shell respectively.
Element X-
The atomic number of Element X is 6.
2 electrons will fill the first shell i.e., the K shell.
4 electrons will fill the second shell i.e., the L shell.
Hence, its configuration will be 2,4.
Element Y-
The atomic number of Element Y is 9.
2 electrons will fill the first shell i.e., the K shell.
7 electrons will fill the second shell i.e., the L shell.
Hence, its configuration will be 2,7.
Element Z-
The atomic number of Element Y is 12.
2 electrons will fill the first shell i.e., the K shell.
8 electrons will fill the second shell i.e., the L shell.
2 electrons will enter the third shell i.e., the M shell.
Hence, its configuration will be 2,8,2.
Identification-
Elements that have 1,2 or 3 electrons in their outermost shell lose electrons to form positive ions i.e., cation and are metals.
Elements that have 4,5,6 or 7 electrons in their outermost shell gain electrons to form negative ions i.e., anions, and are non-metals.
It could gain four electrons forming C4– anion. But it would be difficult for the nucleus with six protons to hold on to ten electrons, that is, four extra electrons.
Here, element Y has 7 electrons in its outermost shell.
It will gain an electron to get 8 electrons in its outermost shell acquiring a negative charge due to an excess electron than the proton.