CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

An element A has an atomic number of 6. Another element B has 17 electrons in its one neutral atom.

(a) In which groups of the periodic table would you expect to find these elements?
(b) what type of bond is formed between A and B?
(c) Suggest a formula of the compound formed between A and B.

Open in App
Solution

(a) Element A has an atomic number of 6, so its electronic configuration is 2,4 (having 4 valence electrons).
Therefore, the group number for element A = valence shell + 10
= 4+10 =14.
Hence, the group number for element A is 14.

Element B has 17 electrons in its neutral atom. Its electronic configuration is 2,8,7.
Therefore, the group number of element B = valence shell + 10
= 7 + 10 = 17.
Hence, the group number for element B is 17.

(b) When two non-metals A and B combine, a covalent bond is formed, since group 14 elements and group 17 elements are all non-metallic in nature.

(c) When one atom of A combines with four atoms of B, a covalent compound with the formula AB4 is formed. This is because the element A has a valency of 4 and element B has a valency of 1.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Electronegativity
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon