
What is the number of electrons in an $A{l^{ + 3}}$ion?
Answer
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Hint :An atom's mass number tells us the total number of protons and of neutrons that atom has in its nucleus. Since we are dealing with an isotope of aluminium, it follows that this atom must have the exact same number of protons in its nucleus.
Complete Step By Step Answer:
This is an aluminum ion with a charge of $ + 3$. Having a $ + 3$ charge implies aluminum having an atomic number $13$ has lost three electrons.
The electron configuration of the neutral atom will be
$Al = 1{s^2},2{s^2},2{p^5}3{s^1}3{p^1}$
When aluminium forms $ + 3$ cations, $A{l^{ + 3}}$, it loses $3$electrons from its outermost energy shell.
$A{l^{ + 3}} = 1{s^2},2{s^2},2{p^6}.$
So the number of the electrons in an $A{l^{ + 3}}$ ion is $10$.
Also, since we realize that in common substance reactions, electrons are the particles that are included, this implies the number of protons are intact $13$. No doubt, $13$protons.
The atomic number represents the number of positively charged particles in the atomic nucleus. The aluminum nucleus contains $13$ protons, $13$ fundamental, positively charged particles; this is what identifies the nucleus as aluminum. To balance this charge (all matter is neutral after all), there must also be $13$ fundamental negatively charged particles associated with the aluminum nucleus.
Note :
Number of protons is equal to the number of electrons for simple atoms. But if we refer to an ion, and since we know that electrons are what participate in a chemical reaction, then the number of electrons might increase or decrease. An increase if a negative ion was formed and a decrease in electrons if a positive ion was formed.
Complete Step By Step Answer:
This is an aluminum ion with a charge of $ + 3$. Having a $ + 3$ charge implies aluminum having an atomic number $13$ has lost three electrons.
The electron configuration of the neutral atom will be
$Al = 1{s^2},2{s^2},2{p^5}3{s^1}3{p^1}$
When aluminium forms $ + 3$ cations, $A{l^{ + 3}}$, it loses $3$electrons from its outermost energy shell.
$A{l^{ + 3}} = 1{s^2},2{s^2},2{p^6}.$
So the number of the electrons in an $A{l^{ + 3}}$ ion is $10$.
Also, since we realize that in common substance reactions, electrons are the particles that are included, this implies the number of protons are intact $13$. No doubt, $13$protons.
The atomic number represents the number of positively charged particles in the atomic nucleus. The aluminum nucleus contains $13$ protons, $13$ fundamental, positively charged particles; this is what identifies the nucleus as aluminum. To balance this charge (all matter is neutral after all), there must also be $13$ fundamental negatively charged particles associated with the aluminum nucleus.
Note :
Number of protons is equal to the number of electrons for simple atoms. But if we refer to an ion, and since we know that electrons are what participate in a chemical reaction, then the number of electrons might increase or decrease. An increase if a negative ion was formed and a decrease in electrons if a positive ion was formed.
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