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3 years ago

Arrange the elements in decreasing order of first ionization energy.

Chemistry

1 answer:

just olya [345]3 years ago

5 0

Answer:

The decreasing order of first ionization energy: Se > Ge > In > Cs

The decreasing order of first ionization energy: x > y > z

Explanation:

Ionization energy refers to the energy needed to completely pull out an electron from the valence shell of a neutral gaseous atom.

First ionization energy is the energy involved in the removal of first valence electron.

In the periodic table, down the group, as atomic radius of elements increases, the ionization energy decreases

Whereas, across a period, as atomic radius of elements decreases, the ionization energy increases.

PART (A):

Position of the given elements in the periodic table:

Indium (In): Group 13, period 5

Germanium (Ge): Group 14, period 4

Selenium (Se): Group 16, period 4

Caesium (Cs): Group 1, period 6

Thus, the increasing order of atomic radius: Se < Ge < In < Cs

Therefore, the decreasing order of first ionization energy: Se > Ge > In > Cs

PART (B):

Given elements:

element x: radius = 110 pm

element y: radius = 199 pm

element z: radius = 257 pm

Thus, the increasing order of atomic radius: x < y < z

Therefore, the decreasing order of first ionization energy: x > y > z

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Answer:

Approximately $2.47\times 10^{15}\; \rm Hz$ .

Explanation:

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$\displaystyle - \frac{k\, Z^2}{n^2}$ (note the negative sign in front of the fraction,)

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$Z$ is the atomic number of that atom. $Z = 1$ for hydrogen.
$n$ is the energy level of that electron.

The electron that produced the $n = 2$ line was initially at the

$\begin{aligned} &E_{n = 2} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{2^2} \cr & \approx -5.4475\times 10^{-19}\; \rm J\end{aligned}$ .

The electron would then transit to energy level $n = 1$ . Its energy would become:

$\begin{aligned} &E_{n = 1} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{1^2} \cr & \approx -2.179 \times 10^{-18} \; \rm J\end{aligned}$ .

The energy change would be equal to

$\begin{aligned}&\text{Initial Energy} - \text{Final Energy} \cr &= E_{n = 2} - E_{n = 1} \cr &= -5.4475 \times 10^{-19} - \left(-2.179 \times 10^{-18}\right) \cr & \approx 1.63425\times 10^{-18}\; \rm J \end{aligned}$ .

That would be the energy of a photon in that $n = 2$ spectrum line. Planck constant $h$ relates the frequency of a photon to its energy:

$E = h \cdot f$ , where

$E$ is the energy of the photon.
$h \approx 6.62607015\times 10^{-34}\; \rm J \cdot s$ is the Planck constant.
$f$ is the frequency of that photon.

In this case, $E \approx 1.63425 \times 10^{-18}\; \rm J$ . Hence,

$\begin{aligned} f &= \frac{E}{h} \cr &\approx \frac{1.63425\times 10^{-18}}{6.62607015\times 10^{-34}} \cr & \approx 2.47 \times 10^{15}\; \rm s^{-1}\end{aligned}$ .

Note that $1 \; \rm Hz = 1 \; \rm s^{-1}$ .

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