# Thread: How to calculate protons and electrons in NH3BF3?

1. As we know in NH3, N full fills its octet with three single bonds with H and a lone pair of electrons and so H. So this molecule is stable. In BF3, F full fills their octets by three single bonds with B but B is unable to fulfill its octet. It got 6 electrons by three single bonds with F. So, B needs 2 more electrons to fulfill its octet. So, BF3 receives the lone pair of electrons from the NH3 and make a coordinate covalent bond between NH3 and BF3, making a new molecule NH3BF3.
So my question is how to calculate the proton numbers in "NH3'' after giving the lone pair of electrons and ''BF3" after receiving the lone pair of electrons?

2.

3. Originally Posted by Indranil
As we know in NH3, N full fills its octet with three single bonds with H and a lone pair of electrons and so H. So this molecule is stable. In BF3, F full fills their octets by three single bonds with B but B is unable to fulfill its octet. It got 6 electrons by three single bonds with F. So, B needs 2 more electrons to fulfill its octet. So, BF3 receives the lone pair of electrons from the NH3 and make a coordinate covalent bond between NH3 and BF3, making a new molecule NH3BF3.
So my question is how to calculate the proton numbers in "NH3'' after giving the lone pair of electrons and ''BF3" after receiving the lone pair of electrons?
Eh? The protons are in the nuclei of the atoms, so they are not affected at all by what the arrangement of electrons is.

So how do you think you can calculate the number in NH3?

4. Originally Posted by exchemist
Originally Posted by Indranil
As we know in NH3, N full fills its octet with three single bonds with H and a lone pair of electrons and so H. So this molecule is stable. In BF3, F full fills their octets by three single bonds with B but B is unable to fulfill its octet. It got 6 electrons by three single bonds with F. So, B needs 2 more electrons to fulfill its octet. So, BF3 receives the lone pair of electrons from the NH3 and make a coordinate covalent bond between NH3 and BF3, making a new molecule NH3BF3.
So my question is how to calculate the proton numbers in "NH3'' after giving the lone pair of electrons and ''BF3" after receiving the lone pair of electrons?
Eh? The protons are in the nuclei of the atoms, so they are not affected at all by what the arrangement of electrons is.

So how do you think you can calculate the number in NH3?
Because NH3 and BF3 are connected by a coordinate bond, making this molecule NH3+ and BF3-.

5. Originally Posted by Indranil
Originally Posted by exchemist
Originally Posted by Indranil
As we know in NH3, N full fills its octet with three single bonds with H and a lone pair of electrons and so H. So this molecule is stable. In BF3, F full fills their octets by three single bonds with B but B is unable to fulfill its octet. It got 6 electrons by three single bonds with F. So, B needs 2 more electrons to fulfill its octet. So, BF3 receives the lone pair of electrons from the NH3 and make a coordinate covalent bond between NH3 and BF3, making a new molecule NH3BF3.
So my question is how to calculate the proton numbers in "NH3'' after giving the lone pair of electrons and ''BF3" after receiving the lone pair of electrons?
Eh? The protons are in the nuclei of the atoms, so they are not affected at all by what the arrangement of electrons is.

So how do you think you can calculate the number in NH3?
Because NH3 and BF3 are connected by a coordinate bond, making this molecule NH3+ and BF3-.
No, let's try to answer the question. Can you show me how you would calculate the number of protons in NH3? What answer do you get and why? You show me and we can discuss.

(You may think I am being difficult but we sometimes get people trying to get us to do their school work for them. We won't do that, but we will help them to learn.)

6. Originally Posted by exchemist
Originally Posted by Indranil
Originally Posted by exchemist
Originally Posted by Indranil
As we know in NH3, N full fills its octet with three single bonds with H and a lone pair of electrons and so H. So this molecule is stable. In BF3, F full fills their octets by three single bonds with B but B is unable to fulfill its octet. It got 6 electrons by three single bonds with F. So, B needs 2 more electrons to fulfill its octet. So, BF3 receives the lone pair of electrons from the NH3 and make a coordinate covalent bond between NH3 and BF3, making a new molecule NH3BF3.
So my question is how to calculate the proton numbers in "NH3'' after giving the lone pair of electrons and ''BF3" after receiving the lone pair of electrons?
Eh? The protons are in the nuclei of the atoms, so they are not affected at all by what the arrangement of electrons is.

So how do you think you can calculate the number in NH3?
Because NH3 and BF3 are connected by a coordinate bond, making this molecule NH3+ and BF3-.
No, let's try to answer the question. Can you show me how you would calculate the number of protons in NH3? What answer do you get and why? You show me and we can discuss.

(You may think I am being difficult but we sometimes get people trying to get us to do their school work for them. We won't do that, but we will help them to learn.)
N = 7 and H = 3 x 1 = 3, So the total proton number of NH3 = 10

7. Originally Posted by Indranil
Originally Posted by exchemist
Originally Posted by Indranil
Originally Posted by exchemist
Originally Posted by Indranil
As we know in NH3, N full fills its octet with three single bonds with H and a lone pair of electrons and so H. So this molecule is stable. In BF3, F full fills their octets by three single bonds with B but B is unable to fulfill its octet. It got 6 electrons by three single bonds with F. So, B needs 2 more electrons to fulfill its octet. So, BF3 receives the lone pair of electrons from the NH3 and make a coordinate covalent bond between NH3 and BF3, making a new molecule NH3BF3.
So my question is how to calculate the proton numbers in "NH3'' after giving the lone pair of electrons and ''BF3" after receiving the lone pair of electrons?
Eh? The protons are in the nuclei of the atoms, so they are not affected at all by what the arrangement of electrons is.

So how do you think you can calculate the number in NH3?
Because NH3 and BF3 are connected by a coordinate bond, making this molecule NH3+ and BF3-.
No, let's try to answer the question. Can you show me how you would calculate the number of protons in NH3? What answer do you get and why? You show me and we can discuss.

(You may think I am being difficult but we sometimes get people trying to get us to do their school work for them. We won't do that, but we will help them to learn.)
N = 7 and H = 3 x 1 = 3, So the total proton number of NH3 = 10
Exactly.

8. Originally Posted by exchemist
Originally Posted by Indranil
Originally Posted by exchemist
Originally Posted by Indranil
Originally Posted by exchemist
Originally Posted by Indranil
As we know in NH3, N full fills its octet with three single bonds with H and a lone pair of electrons and so H. So this molecule is stable. In BF3, F full fills their octets by three single bonds with B but B is unable to fulfill its octet. It got 6 electrons by three single bonds with F. So, B needs 2 more electrons to fulfill its octet. So, BF3 receives the lone pair of electrons from the NH3 and make a coordinate covalent bond between NH3 and BF3, making a new molecule NH3BF3.

Eh? The protons are in the nuclei of the atoms, so they are not affected at all by what the arrangement of electrons is.

So how do you think you can calculate the number in NH3?
Because NH3 and BF3 are connected by a coordinate bond, making this molecule NH3+ and BF3-.
No, let's try to answer the question. Can you show me how you would calculate the number of protons in NH3? What answer do you get and why? You show me and we can discuss.

(You may think I am being difficult but we sometimes get people trying to get us to do their school work for them. We won't do that, but we will help them to learn.)
N = 7 and H = 3 x 1 = 3, So the total proton number of NH3 = 10
Exactly.
So my question is how to calculate the proton numbers in "NH3'' after giving the lone pair of electrons and ''BF3" after receiving the lone pair of electrons?
But my question is that how many protons in "NH3+" of the molecule of NH3BF3 because this molecule (NH3 and BF3) is connected by a co-ordinating covalent bond, making NH3+ and BF3-?

9. Originally Posted by Indranil
But my question is that how many protons in "NH3+" of the molecule of NH3BF3 because this molecule (NH3 and BF3) is connected by a co-ordinating covalent bond, making NH3+ and BF3-?
Why do you think it would not be 10?

11. Now I understand. The proton numbers stay same whether it is an atom or a molecule or Ions. Only electron numbers change. Am I correct?

12. Originally Posted by PhDemon
I think it is about determining the formal charges of the atoms. For example, for F3BNH3, the boron atom has 6 electrons (4 in outer shell, 2 in inner shell), but has 5 protons. Therefore, the boron atom has a formal charge of –1. And the nitrogen atom also has 6 electrons (4 in outer shell, 2 in inner shell), but has 7 protons. Therefore, the nitrogen atom has a formal charge of +1.

13. Originally Posted by KJW
Originally Posted by PhDemon
I think it is about determining the formal charges of the atoms. For example, for F3BNH3, the boron atom has 6 electrons (4 in osauouterH shell, 2 in inner shell), but has 5 protons. Therefore, the boron atom has a formal charge of –1. And the nitrogen atom also has 6 electrons (4 in outer shell, 2 in inner shell), but has 7 protons. Therefore, the nitrogen atom has a formal charge of +1.
How did you say Boron has 6 electrons?

14. Originally Posted by Indranil
Now I understand. The proton numbers stay same whether it is an atom or a molecule or Ions. Only electron numbers change. Am I correct?
Yes. This is what I was trying to say to you in post 2 of this thread.

Chemical bonding is all about electrons. The nuclei of the atoms - and therefore the protons and neutrons they are composed of - stay just as they are.

15. Originally Posted by Indranil
Originally Posted by KJW
Originally Posted by PhDemon
I think it is about determining the formal charges of the atoms. For example, for F3BNH3, the boron atom has 6 electrons (4 in osauouterH shell, 2 in inner shell), but has 5 protons. Therefore, the boron atom has a formal charge of –1. And the nitrogen atom also has 6 electrons (4 in outer shell, 2 in inner shell), but has 7 protons. Therefore, the nitrogen atom has a formal charge of +1.
How did you say Boron has 6 electrons?
No. Read what he is saying carefully.

In Boron the element, the atom has the same number of electrons as protons, making it a neutral atom. That would be 5, 2 inner electrons and 3 in the 3 outer ("valence") shell. In compounds however this may not be so.

In the compound in question there is what is known as a "dative" (or "coordinate") covalent bond. Covalent bonds generally involve pairs of electrons. Normally each atom contributes one of these electrons. But, as you yourself pointed out, both electrons in the N-B bond come from N. Dative bonds are sometimes written like this: N->B, instead of N-B, in order to signify this.

This unbalances the electrical neutrality of the atoms, leading to a net deficit of electrons circulating round the N atom and a net excess of electrons circulating round the B atom, compared to the charges on the atomic nuclei (which themselves remain unchanged, as we have just said).

Hence he is saying the "formal" charge on B is -1 and that on N is +1, i.e. you can, very loosely speaking, view them as turned into "ions", of a sort, as a result of this process. Although I stress that they are not true ions as the charge separation is incomplete.

16. Originally Posted by exchemist
Originally Posted by Indranil
Originally Posted by KJW
Originally Posted by PhDemon
I think it is about determining the formal charges of the atoms. For example, for F3BNH3, the boron atom has 6 electrons (4 in osauouterH shell, 2 in inner shell), but has 5 protons. Therefore, the boron atom has a formal charge of –1. And the nitrogen atom also has 6 electrons (4 in outer shell, 2 in inner shell), but has 7 protons. Therefore, the nitrogen atom has a formal charge of +1.
How did you say Boron has 6 electrons?
No. Read what he is saying carefully.

In Boron the element, the atom has the same number of electrons as protons, making it a neutral atom. That would be 5, 2 inner electrons and 3 in the 3 outer ("valence") shell. In compounds however this may not be so.

In the compound in question there is what is known as a "dative" (or "coordinate") covalent bond. Covalent bonds generally involve pairs of electrons. Normally each atom contributes one of these electrons. But, as you yourself pointed out, both electrons in the N-B bond come from N. Dative bonds are sometimes written like this: N->B, instead of N-B, in order to signify this.

This unbalances the electrical neutrality of the atoms, leading to a net deficit of electrons circulating round the N atom and a net excess of electrons circulating round the B atom, compared to the charges on the atomic nuclei (which themselves remain unchanged, as we have just said).

Hence he is saying the "formal" charge on B is -1 and that on N is +1, i.e. you can, very loosely speaking, view them as turned into "ions", of a sort, as a result of this process. Although I stress that they are not true ions as the charge separation is incomplete.
First of all, Thank you for your amazing explanations. Now I got it but one point still I don't understand here. The point is below:
''Although I stress that they are not true ions as the charge separation is incomplete.''
Could you get this point easier a little bit please, so that I can understand the concept well?

17. Originally Posted by Indranil
Originally Posted by exchemist
Originally Posted by Indranil
Originally Posted by KJW
Originally Posted by PhDemon
I think it is about determining the formal charges of the atoms. For example, for F3BNH3, the boron atom has 6 electrons (4 in osauouterH shell, 2 in inner shell), but has 5 protons. Therefore, the boron atom has a formal charge of –1. And the nitrogen atom also has 6 electrons (4 in outer shell, 2 in inner shell), but has 7 protons. Therefore, the nitrogen atom has a formal charge of +1.
How did you say Boron has 6 electrons?
No. Read what he is saying carefully.

In Boron the element, the atom has the same number of electrons as protons, making it a neutral atom. That would be 5, 2 inner electrons and 3 in the 3 outer ("valence") shell. In compounds however this may not be so.

In the compound in question there is what is known as a "dative" (or "coordinate") covalent bond. Covalent bonds generally involve pairs of electrons. Normally each atom contributes one of these electrons. But, as you yourself pointed out, both electrons in the N-B bond come from N. Dative bonds are sometimes written like this: N->B, instead of N-B, in order to signify this.

This unbalances the electrical neutrality of the atoms, leading to a net deficit of electrons circulating round the N atom and a net excess of electrons circulating round the B atom, compared to the charges on the atomic nuclei (which themselves remain unchanged, as we have just said).

Hence he is saying the "formal" charge on B is -1 and that on N is +1, i.e. you can, very loosely speaking, view them as turned into "ions", of a sort, as a result of this process. Although I stress that they are not true ions as the charge separation is incomplete.
First of all, Thank you for your amazing explanations. Now I got it but one point still I don't understand here. The point is below:
''Although I stress that they are not true ions as the charge separation is incomplete.''
Could you get this point easier a little bit please, so that I can understand the concept well?
If the electrons in the N->B bond were distributed exactly symmetrically along the N-B axis, the charge separation would give N a charge of +1 and B a charge of -1.

But it won't be symmetrical, because as a net +ve charge starts to develop at one end and a net -ve charge starts to develop at the other, the pair of electrons in the bond will be pulled somewhat towards the + end and repelled somewhat from the - end. So you end up with a lopsided bond with the bonding electrons on average closer to N than to B. Thus the actual degree of charge separation will be rather less than a full unit.

18. Originally Posted by exchemist
Originally Posted by Indranil
Originally Posted by exchemist
Originally Posted by Indranil
Originally Posted by KJW
Originally Posted by PhDemon
I think it is about determining the formal charges of the atoms. For example, for F3BNH3, the boron atom has 6 electrons (4 in osauouterH shell, 2 in inner shell), but has 5 protons. Therefore, the boron atom has a formal charge of –1. And the nitrogen atom also has 6 electrons (4 in outer shell, 2 in inner shell), but has 7 protons. Therefore, the nitrogen atom has a formal charge of +1.
How did you say Boron has 6 electrons?
No. Read what he is saying carefully.

In Boron the element, the atom has the same number of electrons as protons, making it a neutral atom. That would be 5, 2 inner electrons and 3 in the 3 outer ("valence") shell. In compounds however this may not be so.

In the compound in question there is what is known as a "dative" (or "coordinate") covalent bond. Covalent bonds generally involve pairs of electrons. Normally each atom contributes one of these electrons. But, as you yourself pointed out, both electrons in the N-B bond come from N. Dative bonds are sometimes written like this: N->B, instead of N-B, in order to signify this.

This unbalances the electrical neutrality of the atoms, leading to a net deficit of electrons circulating round the N atom and a net excess of electrons circulating round the B atom, compared to the charges on the atomic nuclei (which themselves remain unchanged, as we have just said).

Hence he is saying the "formal" charge on B is -1 and that on N is +1, i.e. you can, very loosely speaking, view them as turned into "ions", of a sort, as a result of this process. Although I stress that they are not true ions as the charge separation is incomplete.
First of all, Thank you for your amazing explanations. Now I got it but one point still I don't understand here. The point is below:
''Although I stress that they are not true ions as the charge separation is incomplete.''
Could you get this point easier a little bit please, so that I can understand the concept well?
If the electrons in the N->B bond were distributed exactly symmetrically along the N-B axis, the charge separation would give N a charge of +1 and B a charge of -1.

But it won't be symmetrical, because as a net +ve charge starts to develop at one end and a net -ve charge starts to develop at the other, the pair of electrons in the bond will be pulled somewhat towards the + end and repelled somewhat from the - end. So you end up with a lopsided bond with the bonding electrons on average closer to N than to B. Thus the actual degree of charge separation will be rather less than a full unit.
As you said ''the pair of electrons in the bond will be pulled somewhat towards the + end and repelled somewhat from the - end'' Now my question is that if the '+ve' charge of N tries to pull the pair of electrons towards its + end, then how does N donate the pair of electron to B and the same question for B, how does B accept the pair of electrons as the end - ve repels the electrons? Could you explain it, please?

19. Originally Posted by exchemist
Originally Posted by Indranil
Originally Posted by exchemist
Originally Posted by Indranil
Originally Posted by KJW
Originally Posted by PhDemon
I think it is about determining the formal charges of the atoms. For example, for F3BNH3, the boron atom has 6 electrons (4 in osauouterH shell, 2 in inner shell), but has 5 protons. Therefore, the boron atom has a formal charge of –1. And the nitrogen atom also has 6 electrons (4 in outer shell, 2 in inner shell), but has 7 protons. Therefore, the nitrogen atom has a formal charge of +1.
How did you say Boron has 6 electrons?
No. Read what he is saying carefully.

In Boron the element, the atom has the same number of electrons as protons, making it a neutral atom. That would be 5, 2 inner electrons and 3 in the 3 outer ("valence") shell. In compounds however this may not be so.

In the compound in question there is what is known as a "dative" (or "coordinate") covalent bond. Covalent bonds generally involve pairs of electrons. Normally each atom contributes one of these electrons. But, as you yourself pointed out, both electrons in the N-B bond come from N. Dative bonds are sometimes written like this: N->B, instead of N-B, in order to signify this.

This unbalances the electrical neutrality of the atoms, leading to a net deficit of electrons circulating round the N atom and a net excess of electrons circulating round the B atom, compared to the charges on the atomic nuclei (which themselves remain unchanged, as we have just said).

Hence he is saying the "formal" charge on B is -1 and that on N is +1, i.e. you can, very loosely speaking, view them as turned into "ions", of a sort, as a result of this process. Although I stress that they are not true ions as the charge separation is incomplete.
First of all, Thank you for your amazing explanations. Now I got it but one point still I don't understand here. The point is below:
''Although I stress that they are not true ions as the charge separation is incomplete.''
Could you get this point easier a little bit please, so that I can understand the concept well?
If the electrons in the N->B bond were distributed exactly symmetrically along the N-B axis, the charge separation would give N a charge of +1 and B a charge of -1.

But it won't be symmetrical, because as a net +ve charge starts to develop at one end and a net -ve charge starts to develop at the other, the pair of electrons in the bond will be pulled somewhat towards the + end and repelled somewhat from the - end. So you end up with a lopsided bond with the bonding electrons on average closer to N than to B. Thus the actual degree of charge separation will be rather less than a full unit.
As you said ''the pair of electrons in the bond will be pulled somewhat towards the + end and repelled somewhat from the - end'' Now my question is that if the '+ve' charge of N tries to pull the pair of electrons towards its + end, then how does N donate the pair of electron to B and the same question for B, how does B accept the pair of electrons as the end - ve repels the electrons? Could you explain it, please?

20. Originally Posted by Indranil
Originally Posted by exchemist
Originally Posted by Indranil
Originally Posted by exchemist
Originally Posted by Indranil
Originally Posted by KJW
Originally Posted by PhDemon
I think it is about determining the formal charges of the atoms. For example, for F3BNH3, the boron atom has 6 electrons (4 in osauouterH shell, 2 in inner shell), but has 5 protons. Therefore, the boron atom has a formal charge of –1. And the nitrogen atom also has 6 electrons (4 in outer shell, 2 in inner shell), but has 7 protons. Therefore, the nitrogen atom has a formal charge of +1.
How did you say Boron has 6 electrons?
No. Read what he is saying carefully.

In Boron the element, the atom has the same number of electrons as protons, making it a neutral atom. That would be 5, 2 inner electrons and 3 in the 3 outer ("valence") shell. In compounds however this may not be so.

In the compound in question there is what is known as a "dative" (or "coordinate") covalent bond. Covalent bonds generally involve pairs of electrons. Normally each atom contributes one of these electrons. But, as you yourself pointed out, both electrons in the N-B bond come from N. Dative bonds are sometimes written like this: N->B, instead of N-B, in order to signify this.

This unbalances the electrical neutrality of the atoms, leading to a net deficit of electrons circulating round the N atom and a net excess of electrons circulating round the B atom, compared to the charges on the atomic nuclei (which themselves remain unchanged, as we have just said).

Hence he is saying the "formal" charge on B is -1 and that on N is +1, i.e. you can, very loosely speaking, view them as turned into "ions", of a sort, as a result of this process. Although I stress that they are not true ions as the charge separation is incomplete.
First of all, Thank you for your amazing explanations. Now I got it but one point still I don't understand here. The point is below:
''Although I stress that they are not true ions as the charge separation is incomplete.''
Could you get this point easier a little bit please, so that I can understand the concept well?
If the electrons in the N->B bond were distributed exactly symmetrically along the N-B axis, the charge separation would give N a charge of +1 and B a charge of -1.

But it won't be symmetrical, because as a net +ve charge starts to develop at one end and a net -ve charge starts to develop at the other, the pair of electrons in the bond will be pulled somewhat towards the + end and repelled somewhat from the - end. So you end up with a lopsided bond with the bonding electrons on average closer to N than to B. Thus the actual degree of charge separation will be rather less than a full unit.
As you said ''the pair of electrons in the bond will be pulled somewhat towards the + end and repelled somewhat from the - end'' Now my question is that if the '+ve' charge of N tries to pull the pair of electrons towards its + end, then how does N donate the pair of electron to B and the same question for B, how does B accept the pair of electrons as the end - ve repels the electrons? Could you explain it, please?
A covalent bond between two atoms forms because it leads to a more stable state, i.e. one of lower energy, that when the two atoms are separate. In this case there are two competing influences: one from the lowering of energy due to forming the bond and the other the raising of energy involved in charge separation. What actually happens will be at some balance point between these competing effects. One would expect that the stability conferred by forming a dative bond (the bond strength) will not be as great as that formed by a normal covalent bond in which no charge separation occurs.

21. Thank you all for your kind efforts.

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