The complete orbital notation diagram of an atom is shown. Five squares are shown aligned horizontally. Inside the first square
from the left is shown one upwards pointing arrow and one downwards pointing arrow. In the second square is another pair of upwards and downwards pointing arrow. Inside the third square is another pair of upwards and downwards pointing arrow. Inside the fourth square is another upwards and downwards pointing arrow. Inside the fifth square is an upwards pointing arrow. Based on the diagram, what values can be assigned to the angular momentum quantum number for the electrons in the atom? What information does this quantum number provide about the location of the electron?
1 answer:
Answer:
l = 0, 1, and 2; it is in a d orbital.
Explanation:
1. Angular momentum quantum number
One square means l = 0.
Three squares mean l = 1.
Five squares mean l = 2.
If an atom contains electrons with l = 2, it must also have electrons with l = 0 and 1.
Thus, the electrons in the atom can be assigned the angular momentum quantum numbers l = 0, 1, and 2.
Those in a set of five squares all have l = 2.
2. Location of electron
The last electron is the lone electron in the fifth square. Since l = 2, the electron is in a d orbital.
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Answer: Option (e) is the correct answer.
Explanation:
A bond that is formed when an electron is transferred from one atom to another results in the formation of an ionic bond.
For example, NaBr will be an ionic compound as there is transfer of electron from Na to Br.
Whereas a bond that is formed by sharing of electrons is known as a covalent bond.
For example, will be a covalent compound as there is sharing of electron between carbon and bromine atom.
Also, when electrons are shared between the combining atoms and there is large difference in electronegativity of these atoms then partial charges develop on these atoms. As a result, it forms a polar covalent bond.
For example, in a HBr compound there is sharing of electrons between H and Br. Also, due to difference in electronegativity there will be partial positive charge on H and partial negative charge on Br.
Thus, we can conclude that out of the given options HBr is the only compound that has polar covalent bonds.