Question

Four elements A, B, C, D are given :

• A shows the presence of 20 neutrons, 17 protons and 17 electrons.
• B shows the presence of 18 neutrons, 17 protons and 17 electrons.
• C shows the presence of 10 neutrons, 9 protons and 10 electrons.
• D shows the presence of 4 neutrons, 3 protons and 2 electrons.

State which of the above is –

1. an anion
2. a cation
3. a pair of isotopes.

Write the formula of the compound formed between D and C.

Answer 1

For a. and b. and c. we need to find the valence electrons.
Element A has 17 electrons [2, 8, 7]
As there are 7 electrons in the outermost shell. It will gain 1 electron to attain a stable octet structure of the nearest gas – Neon [2, 8]
∴ \[\ce{A-}\] is Anion.
Element B also has 17 electrons [2, 8, 7] is short of 1 electron in the valence shell will gain 1 electron
∴ \[\ce{B-}\] Anion.
Element C has 10 electrons and electronic configuration [2, 8] has 8 electrons in the valence shell. Its octet is complete, but there are 9 protons.
∴ Element C must have 9 electrons
The 10th electron is gained to have a stable octet structure and hence 8 electrons in the valence shell.
∴ \[\ce{C-}\] is Anion.
Element D has protons 3 and electrons 2
It should have 3 electrons. This means it has lost 1 electron from valence shell to have a stable duplet structure of the nearest gas (inert) [2] Helium.
∴ \[\ce{D+}\] is Cation.
c. A and B are isotopes as both have the same atomic number but the different numbers of neutrons.
d. \[\ce{D+C^1-}\]
∴ The compound formed is [DC]