# AR, IR, IE
**Atomic Radius**
* SInce we cannot pinpoint the positions of electrons, we cannot have a boundary as there will always be a probability of an electron being there
* Instead distance from adjacent Nuclei ( Internuclear Distance )
* Down a group --> AR increases $$\because$$ the # of E-levels increase
* More Mathematically: $$AR \propto N$$ ( where N is the Principle Energy Level )
* Across a period --> AR decreases $$\because N$$stays the same & $$Z$$increases --> More charge --> Atoms are more grouped together.
* More Mathematically: $$AR \propto \frac{1}{Z\_{eff}}$$--> They are inversely Proportional
**Ionic Radius**
* Where as Atomic Radius notes the distance between adjacent nucleus of **Neutral** atoms, **Ionic** Radius look at Cations & Anions
* Note Cations are smaller ( greater charge ) than a stable atom
* Note Anions are larger ( greater repulsion ) than a stable atom
* One can easily see that for [Isoelectrionic](https://www.google.com/search?q=isoelectronic+meaning\&rlz=1CAJIKU_enCA1075\&oq=isoelectronic+meaning\&aqs=chrome.0.0i131i433i512l3j0i512l7.2921j0j7\&sourceid=chrome\&ie=UTF-8\&safe=active\&ssui=on) Cations , the one with the greater sum charge has a smaller $$IR$$ ( more charge attracting )
* One can easily see that for [Isoelectrionic](https://www.google.com/search?q=isoelectronic+meaning\&rlz=1CAJIKU_enCA1075\&oq=isoelectronic+meaning\&aqs=chrome.0.0i131i433i512l3j0i512l7.2921j0j7\&sourceid=chrome\&ie=UTF-8\&safe=active\&ssui=on) Anions, the one with the least sum charge has a bigger $$IR$$( More charge repelling )
**Ionization Energy**
* **Mole** ( mol ) is an SL unit used to measure the amount of any substance: Typically used to measure large quantities of tiny entities like Atoms, Molecules, and particles. [here](https://knowmayus.gitbook.io/chemistry/unit-1/ar-ir-ie/ionization-energy-contd) for true definition ( IMPORTANT ).
* $$1 mol = 6.02214076 \* 10^{23}$$ or $$N\_A$$ Avogadro's number
* **First Ionization Energies** ( $$IE\_1$$) are a way to measure the attraction between the nucleus & the outer electron.
* More precisely its the amount of Energy to completely remove an electron from an neutral atom
* From this we can find some trends
* Down a group --> Increased distance from Nucleus to Valence
* IE becomes smaller
* More Mathematically: $$N \propto \frac{1}{IE}$$
* Across a period --> $$Z$$increases $$\therefore$$ charge increases
* IE becomes bigger
* More Mathematically: $$Z\_{eff} \propto IE$$
* Looking at the graph of Ionization Energies amongst the Elements, we can find evidence of why the First Energy Level has maximum of $$2 e^-1$$, evidence why s-orbitals contain only 2 electrons, etc..
**Screening Effect and Effective Nuclear Charge**
* $$Z\_{eff} = Z - S$$--> Approximate the attractive force Valence Electrons "Feel" from nucleus
With $$Z\_{eff}$$ being the *Effective Nuclear Charge.*
*With* $$Z$$ being the Nuclear Charge ( # of protons )
* They are tightly packed --> so act as one large charge
With $$S$$ being the # of *Core Electrons ,* that block/shield the valence electrons
**Electrons in Atoms**
* Lyman's Series --> From $$N\geq 2 \rightarrow N=1$$( UV )
* The elements $$H$$and $$He$$ have ground states at $$N = 1$$ ( why ? )
* Balmar Series --> From $$N \geq 3 \rightarrow N = 2$$ ( Visible )
* Paschen's Series --> From $$N \geq 4 \rightarrow N = 3$$
* Brackett Series --> From $$N \geq 5 \rightarrow N = 4$$
* Since the Energy Levels get closer and closer as N converges to $$\infin$$, eventually being so insignificantly close that it forms a [continuum](https://www.google.com/search?q=continuum+meaning\&rlz=1CAJIKU_enCA1075\&oq=contin\&aqs=chrome.0.69i59j69i57j46i131i199i433i465i512j0i131i433i512l2j69i61l3.982j0j7\&sourceid=chrome\&ie=UTF-8\&safe=active\&ssui=on) . Once reached there, the atom becomes an ion.
**Calculating IE from Spectral Frequency**
Goal: Relate Frequency & Energy
$$\therefore$$ we use the following equations
1. $$c = f \lambda$$ where c = $$3.00 \times 10^8$$ m/s
2. $$E= hf$$
We can now relate to result in the equation:
$$
E = \frac{hc}{\lambda}
$$
However note that [**IE**](https://knowmayus.gitbook.io/chemistry/unit-1/ar-ir-ie/ionization-energy-contd) is the amount of energy needed to remove 1 mole of $$e^-$$ from 1 mole of a gas $$\therefore$$ we must multiply by $$N\_A$$, Avogadro's Number. For convention, one must turn $$J$$ into $$kJ$$ for $$\frac{kJ}{mol}$$
**IE can be calculated from wavelength or frequency that corresponds to the energy of the electron transition.**
$$1 eV = 1.602 \times 10^{-19} J$$
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