2.93

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Sam Smoot 2L
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Joined: Fri Sep 29, 2017 7:04 am

2.93

Postby Sam Smoot 2L » Wed Nov 08, 2017 10:17 pm

I am confused by the solution to 2.93. It says that A = Na, B = Cl. I understand that Na+ is larger than Cl-, but what is the trend for determining the relative sizes of atoms in their neutral state? Thanks.

Sabrina Dunbar 1I
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Re: 2.93

Postby Sabrina Dunbar 1I » Wed Nov 08, 2017 10:20 pm

There is an error in the solutions manual, so this is what Dr. Lavelle has posted on his website in reference to this error:
2.93
In the picture, it shows A (smaller atom) + B (larger atom) --> C (larger ion) + D (smaller ion)
The solution manual says that A=Na and B=Cl, and it references Figure 2.20 which explicitly lists the atomic radius of Na to be 154 pm and the atomic radius of Cl to be 99 pm.
It also says that C=Na+ and D=Cl-, referencing Figure 2.22 which shows that the ionic radius of Na+ to be 102 pm and the ionic radius of Cl- to be 181 pm.
Error: A=Na; B=Cl; C=Na+; D=Cl- Correction: A=Cl ; B=Na; C= ; D=Na+

Haocheng Zhang 2A
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Re: 2.93

Postby Haocheng Zhang 2A » Wed Nov 08, 2017 10:36 pm

The size of atom in neutral state decrease from right to left, so A should be Cl and B should be Na. After the reaction, Na loses an electron and becomes Na+. It no longer have eletron in 3s orbit, so Na+ is actually smaller than Cl-.Thus, C should be Cl- and D should be Na+.

Sam Smoot 2L
Posts: 59
Joined: Fri Sep 29, 2017 7:04 am

Re: 2.93

Postby Sam Smoot 2L » Wed Nov 08, 2017 10:43 pm

Thanks for letting me know about the solution mistake. I also realize I made an error; I meant to say Na+ < Cl- in terms of ionic radii


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