Atomic Structure - Section III

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The Quantum Mechanical Model of the Atom

Solution :

The uncertainty in the location is 0.1 Å i.e. $\Delta x = 0.1{\AA}$

By the Heisenberg’s uncertainty principle:

$\Delta x\Delta p \ge \frac{h}{{4\pi }}\,\,{\text{or}}\,\,\Delta x\,\,m\Delta {\text{v}} \ge \frac{h}{{4\pi }}$

$\Delta {\text{v}} \ge \frac{h}{{4\pi \Delta xm}}$

$\Delta {\text{v}} \ge \frac{{6.626 \times {{10}^{ - 34}}Js}}{{4 \times 3.14 \times 0.1 \times {{10}^{ - 10}}m \times 9.11 \times {{10}^{ - 31}}kg}}$

$= 5.79 \times {10^6}m\,\,{s^{ - 1}}$

Example - 20

Calculate the de Broglie wavelength of an electron traveling with a speed equal to 10 % of the speed of light.

Solution :

The speed of an electron traveling at 10% the speed of light is:

${\text{v}} = 3 \times {10^8}\,\,m{s^{ - 1}} \times \frac{{10}}{{100}} = 3 \times {10^7}\,\,m{s^{ - 1}}\,$ (Velocity of light $3{\text{ }} \times {\text{ }}108{\text{ }}m{s^{ - 1}}$ )

It’s de Broglie’s wavelength would be:

$\lambda = \frac{h}{{m{\text{v}}}} = \frac{{6.62 \times {{10}^{ - 34}}\,kg\,{m^2}{s^{ - 1}}}}{{9.1 \times {{10}^{ - 31}}\,\,kg\, \times 3 \times {{10}^7}\,\,m{s^{ - 1}}}} = 24.4\;{\text{pm}}$

Example - 21

Establish a relation between wavelength of a moving particle and its kinetic energy. The wavelength of a moving particle of mass $1.0{\text{ }} \times {\text{ }}{10^{ - 6}}{\text{ kg is }}3.313{\text{ }} \times {\text{ }}{10^{ - 29}}\;{\text{m}}{\text{.}}$ . Calculate its kinetic energy. $\left[ {h{\text{ }} = {\text{ }}6.626 \times {{10}^{ - 34}}{\text{ }}Js} \right]$

Solution :

The kinetic energy of $\\alpha$ particle of mass $m$ moving at a velocity $v$ is given by $K.E. = \frac{1}{2}m{{\text{v}}^2}$ . The de-Broglie wavelength of a particle of mass $m$ moving at a velocity $v$ is given by $\lambda = h/ {m{\text{v}}$ .

we can wrtie $\lambda = \frac{h}{{\sqrt {2K.E. \times m} }}$

The de Broglie wave length of a mass $1 \times {10^{ - 6}}kg$ is given to be Hence

$K.E. = \frac{{{h^2}}}{{2m{\lambda ^2}}} = \frac{{{{\left( {6.626 \times {{10}^{ - 34}}} \right)}^2}}}{{2 \times {{10}^{ - 6}} \times {{\left( {3.313{ \times ^{ - 29}}} \right)}^2}}}$

$= 2 \times {10^{ - 4}}J$

Example - 22

13.6 eV is needed for ionisation of a hydrogen atom. An electron in a hydrogen atom in its ground state absorbs 1.50 times as much energy as the minimum energy required for it to escape from the atom. What is the wavelength of the emitted electron ?

$({m_e}{\text{ }} = {\text{ }}9.109{\text{ }} \times {\text{ }}{10^{ - 31}}{\text{ kg}},{\text{ }}e{\text{ }} = {\text{ }}1.602{\text{ }} \times {\text{ }}{10^{ - 19}}{\text{ coulomb}},{\text{ }}h{\text{ }} = {\text{ }}6.63{\text{ }} \times {\text{ }}{10^{ - 34}}{\text{ }}J.s)$

Solution :

1.5 times of 13.6 eV, i.e. 20.4 eV, is absorbed by the hydrogen atom out of which 6.8 eV (20.4 – 13.6) is converted to kinetic energy of the electron.

$K.E.{\text{ }} = {\text{ }}6.8{\text{ }}eV{\text{ }} = {\text{ }}6.8{\text{ }}(1.602{\text{ }} \times {\text{ }}{10^{ - 19}}{\text{ coulomb}}){\text{ }}(1{\text{ volt}}){\text{ }} = {\text{ }}1.09{\text{ }} \times {\text{ }}{10^{ - 18}}{\text{ }}J$

${\text\;\;\;\;\;{Now,}}\,\,\,\,\,K.E. = \frac{1}{2}m{{\text{v}}^2}$

$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{v}}\, = \,\sqrt {\frac{{2K.E.}}{m}} = \sqrt {\frac{{2\left( {1.09 \times {{10}^{ - 18}}\,J\,} \right)}}{{\left( {9.109 \times {{10}^{ - 31}}\,\,kg} \right)}}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.55 \times {10^6}\,\,m/s$

$\therefore\;\;\;\;\;$ The de Broglie’s wavelength is:

$\lambda = \frac{h}{{m{\text{v}}}} = \frac{{\left( {6.63 \times {{10}^{ - 34}}\,\,J \cdot s} \right)}}{{\left( {9.109 \times {{10}^{ - 31}}\,\,kg} \right)\left( {1.55 \times {{10}^6}\,\,m/s} \right)}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4.70 \times {10^{ - 10}}\,\,{\text{metres}}{\text{.}}$

Example - 23

Two particles ‘A’ and ‘B’ are in motion. The momentum of particle ‘B’ is half of ‘A’ . If the wavelength associated with the particle ‘A’ is $5{\text{ }} \times {10^{ - 8}}{\text{ }}m$ , calculate the wavelength associated with the particle ‘B’.

Solution :

The de Broglie’s wavelength of a particle is given by:

$\begin \,\,\,\,\,\,\,\,\,\,\,\,\lambda = \frac{h}{p}. \hfill$

${\text{Thus}}\;\;\;\;\;\,\,\frac{{{\lambda _A}}}{{{\lambda _B}}} = \frac{{{p_B}}}{{{p_A}}} \hfill \end$

$\Rightarrow \;\;\;\;\; {\lambda _B} = {\lambda _A}\frac{{{p_A}}}{{{p_B}}} = 5 \times {10^{ - 8}} \times \frac{{{p_A}}}{{\frac{1}{2}{p_A}}} = {10^{ - 7}}m$

Example - 24

If an electron and proton when in motion have the same wavelength associated with each of them, which would be moving faster and why?

Solution :

As $\lambda \, = \,\frac{h}{{m{\text{v}}}}$

${\text{v}}\,\, \propto \,\,\frac{1}{m},\,\,$ If $\lambda$ is constant.

$\therefore\;\;\;\;\;$ Electron would be moving faster than proton due to its lesser mass.

Example - 25

Calculate the wavelength of an electron that has been accelerated in a particle accelerator through a potential difference of 100 million volts. $[1{\text{ }}eV{\text{ }} = {\text{ }}1.6{\text{ }} \times {\text{ }}{10^{ - 19}}{\text{ }}J,{\text{ }}me{\text{ }} = {\text{ }}9.1 \times {10^{ - 31}}{\text{ }}kg,{\text{ }}h{\text{ }} = {\text{ }}6.0{\text{ }} \times {\text{ }}{10^{ - 34}}{\text{ }}Js,{\text{ }}c{\text{ }} = {\text{ }}3.0{\text{ }} \times 108{\text{ }}m{s^{ - 1}}]$

Solution :

The kinetic energy gained is equal to the potential energy lost

$K.E. = e \times V\,\,;\,\,\,\,\,\,\,\,K.E. = \frac{{{P^2}}}{{2m}}$ (This will be understood after doing electrostatics in physics)

$\therefore \;\;\;\;\; P = \sqrt {2meV}$

Therefore the de Broglie’s wavelength is:

$\lambda = \frac{h}{{\sqrt {2{m_e}eV} }} = \frac{{6.6 \times {{10}^{-34}}}}{{\sqrt {2 \times 9.1 \times {{10}^{-31}} \times 1.6 \times {{10}^{ - 19}} \times {{10}^8}} }} = 0.123 \times {10^{ - 12}}m$

Example - 26

(a) Suppose the velocities of an electron and a rifle bullet of mass 0.03 kg are each measured with an uncertainty of $\Delta {\text{v}} = {10^{ - 3}}\,\,m{s^{ - 1}}$ . Calculate the minimum uncertainties in their positions. (Given : mass of an electron = $9.1 \times {10^{ - 31}}{\text{ }}kg$ )

(b) Comment on the calculated values of uncertainties in positions in the two cases.

Solution :

(a) According to Heisenberg’s uncertainty principle,

$\Delta x \times \Delta p \ge \frac{h}{{4\pi }}$

${\text{or}}\,\,\,\Delta x.m\Delta {\text{v}} \ge \frac{h}{{4\pi }}; \;\;\;\;\; \Delta x \ge \frac{h}{{4\pi m\Delta {\text{v}}}}$

$\Delta x\,\,{\text{for}}\,\,{\text{bullet}} \ge \,\frac{{6.625 \times {{10}^{ - 34}}}}{{4 \times 3.14 \times 0.03 \times {{10}^{ - 3}}}} = 1.76 \times {10^{ - 30}}m.$

(b) As the uncertainty in the position measurement of the bullet is negligibly small, it indicates that we can measure the position of a macroscopic object like a bullet very accurately but not of the microscopic object like an electron.

Example - 27

Show that the Heisenberg uncertainty principle is of negligible significance for an object of ${10^{ - 6}}$ kg mass. $\left( {h/4\pi = 0.528 \times {{10}^{ - 34}}kg\,\,{m^2}{s^{ - 1}}} \right)$

Solution :

According to Heisenberg’s uncertainty principle,

$\Delta x \times \Delta v \ge \frac{h}{{4\pi m}} \ge \frac{{0.528 \times {{10}^{ - 34}}kg\;{m^2}\,{s^{ - 1}}}}{{{{10}^{ - 6}}kg}} = 0.528 \times {10^{ - 28}}{m^2}{s^{ - 1}}$

As the product of uncertainty in position and velocity is very small, the uncertainty principle is of negligible significance for an object of mass ${10^{ - 6}}$ kg.

Example - 28

On the basis of the Heisenberg uncertainty principle, show that the electron can not exist within an atomic nucleus of radius ${10^{ - 15}}\;{\text{m}}$ .

(mass of electron = $9 \times {10^{ - 31}}{\text{ }}kg;{\text{ }}h{\text{/}}4\;\pi = {\text{ }}0.528 \times {10^{ - 34}}{\text{ }}kg{\text{ }}{m^2}{\text{ }}{s^{ - 1}}$ )

Solution :

The uncertainty principle is defined for an error only in one dimension ( $\Delta x$ is the error in measuring the $x$ coordinate). In this case when we speak of uncertainty in position, as the system is spherically symetric the dimension that has to be used is the radial location. If the electron is located in the nucleus the uncertainty in it’s radial position is given by $\Delta r = {10^{ - 15}}m$

According to uncertainty principle, $\Delta r \times \Delta p \ge \frac{h}{{4\pi }}$

Or $\;\;\;\;\;\Delta v \ge \frac{h}{{4\pi \times m \times \Delta r}} = \frac{{6.626 \times {{10}^{ - 34}}}}{{4 \times 3.142 \times 9.1 \times {{10}^{--31}} \times {{10}^{ - 15}}}} = 5.8 \times {10^{10}}m{s^{ - 1}}$

Thus the uncertainty in velocity is greater than the speed of light $(3 \times {10^8}{\text{ }}m{s^{ - 1}})$ , which is not possible. Hence an electron cannot be present inside the nucleus of an atom.

Example - 29

Uncertainty in velocity = 100 – 99.99 = 0.01%

$\therefore \;\;\;\;\; \Delta {\text{v}} = 50 \times \frac{{0.01}}{{100}} = 50 \times {10^{ - 4}} = 5 \times {10^{ - 3}}m{s^{ - 1}}$

By the Heisenberg’s uncertainty principle:

$\Delta x \times \Delta v \ge \frac{h}{{4m\pi }} \;\;\;\;{\text{or}}\;\;\;\;\Delta x = \frac{h}{{4m\pi \times \Delta v}}\; = \frac{{6.626 \times {{10}^{ - 34}}\;Js \times 7}}{{4 \times 9.1 \times {{10}^{ - 31}} \times 22 \times 5 \times {{10}^{ - 3}}}} = 1.15 \times {10^{ - 2}}\;m$

Example - 30

What is the maximum number of electrons in an atom in which the last electron, filled, has the following quantum numbers?

(a) $n{\rm{ }} = {\rm{ }}3,{\rm{ }}l{\rm{ }} = 1$ .	(b) $n{\rm{ }} = {\rm{ }}3,{\rm{ }}l{\rm{ }} = 2$
(c) $n{\rm{ }} = {\rm{ }}3,{\rm{ }}l{\rm{ }} = {\rm{ }}2{\rm{ and }}m{\rm{ }} = {\rm{ }} - 1$	(d) $n{\rm{ }} = {\rm{ }}3,{\rm{ }}l{\rm{ }} = {\rm{ }}2,{\rm{ }}m{\rm{ }} = {\rm{ }}0{\rm{ and }}s{\rm{ }} = {\rm{ }} - \frac{1}{2}$

Solution :

The solution to this problem has been presented in the form of a table.

Example - 31

The following incorrect set of quantum numbers in the order n, l, m, s are written for paired electrons or for one electron in an orbital. Correct them, assuming n values are correct and only one of the four quantum numbers is in correct.

(a) $1,0,0, + \frac{1}{2}, + \frac{1}{2}$

(b) $2,2,1, \pm \frac{1}{2}$

(c) $3,2,3, \pm \frac{1}{2}$

(d) 2, 1

Solution :

(a) Two electrons in the same orbital cannot have the same spin so the set of quantum numbers must be $1,0,0, \pm \frac{1}{2}$

(b) The highest value a azimuthal quantum number can take of a given principal quantum number $n$ is $n - 1$ . $\therefore\;$ the azimuthal quantum number needs correction and can be 0 or 1. As the magnetic quantum number is $m = 1, l$ can only be 1. Thus the corrected set is $2,1,1, \pm \frac{1}{2}$

(c) The magnetic quantum number for a given azimuthal quantum number can take values from –l to l. Therefore for $l = 2$ , $m$ cannot be 3. To correct one of them we may write $l \ge 3$ & $m = 3$ or l = 2 & $m = 2$ ,1,0,–1, –2 but would not be possible for $n = 3 l = 2$ and $m = 2$ ,1,0,–1, –2

$\therefore\;\;\;\;\;$ A corrected set would be

${\rm{3, 2,(2,1,0,}} - {\rm{1,}} - {\rm{2),}} \pm \frac{1}{2}$

(d) The spin quantum number cannot be zero it has to be either $+ \frac{1}{2}{\rm{ }} \;\;\; {\rm{or }} \;\;\; - \frac{1}{2}$

$\therefore \;\;\;$ The correct set should be

$2,1, - 1, + \frac{1}{2}{\rm{ or }} - \frac{1}{2}$

Example - 32

How many electrons can be placed (a) in the shell with $n = 2$ (b) in the shell with $n = 3$ (c) in the shell with $n = 3$ before the first electron enters the shell with $n = 4$ ?

Solution :

(a) This is a direct application of the formula derived for the maximum number of electrons in the nth shell. The principal quantum number is $n = 2$

$2{(n)^2}{\rm{ }} = {\rm{ }}2{(2)^2}{\rm{ }} = {\rm{ }}8$

(b) The principal quantum number is $n = 3$

$2{n^2}{\rm{ }} = {\rm{ }}2 \times {(3)^2}{\rm{ }} = {\rm{ }}18$

(c) We know that in the order of filling of atomic orbitals $4s$ comes before $3d$ . So the maximum no. of electrons in the $n = 3$ shell before the $n = 4$ shell has any electrons would be when the last electrons are filled in the $3 p$ sub shell. Then the configuration would be $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}.$

$\therefore\;\;\;\;$ No of electrons in $n = 3$ is 8.

Example - 33

Which shell would be the first to have a $g$ sub shell?

Solution :

Since the angular momentum quantum number, $l$ , must be less than the principal quantum number, $n$ , and since the $g$ sub shell designation represents a $l$ value of 4, the minimum n value possible is 5. In alphabetical notation (KLMNO), the fifth shell is designated the O shell.

Example - 34

In what ways do the spatial distributions of the orbitals in each pair differ from each other?

(a) $1s$ and $2s$

(b) $2s$ and $2p$

(c) $2 {p_x}$ and $2{p_y}$

Solution :

(a) The $2s$ orbital is large, and it has a spherical node, which the 1s orbital does not have.

(b)The $2p$ orbital is directed in space, aligned along one of the $x, y,$ or $z$ axes. It has planar node which includes the other two axes.

(c) Each orbital is aligned along the axis corresponding to its subscript.

Example - 35

The nucleus of an atom is located at $x = y = z = 0$ . (a) If the probability of finding an s-orbital electron in a tiny volume around $x = a, y = z = 0$ , is $1.0{\rm{ }} \times {\rm{ }}{10^{ - 5}}$ , what is the probability of finding the electron in the same sized volume around $x = z = 0, y = a$ ? (b) What would be the probability at the second site if the electron were in a ${p_z}$ orbital? Explain.

Solution :

(a) $1.0{\rm{ }} \times {\rm{ }}{10^{ - 5}}$ . Since the distance from the nucleus is the same, and the $s$ orbital is spherically symmetrical, the probability in each of the two volumes is the same.

(b) Since the ${p_z}$ orbital has a node at $z = 0$ , the probability of finding the electron in this volume is 0.

Example - 36

Suppose a particle has four quantum numbers such that the permitted values are those given below. How many particles could be fitted into the $n = 1$ shell? into the $n = 2$ shell? into the $n = 3$ shell?

$n$ : $1,2,3....$

$\;l$ : $(n - 1),{\rm{ }}(n - 3),{\rm{ }}(n - 5),.....$ but no negative number

$j\;$ : $\left( {l + \frac{1}{2}} \right)\;\;{\rm{or}}\;\;\left( {l - \frac{1}{2}} \right)$ or if the latter is not negative

$m$ : $- j$ in integral steps to $+ j$

Solution :

(a) For the $n = 1$ shell

$n = 1 \;\;\;\;\; \Rightarrow \;\;\;\;\; l = 0 \;\;\;\;\; \Rightarrow \;\;\;\;\; j = 1/2 \;\;\;\;\; \Rightarrow \;\;\;\;\;\;m = - 1/2\,\,{\rm{or }} + 1/2$

There are two sets possible and hence two particles can be filled in accordance with the Pauli’s exclusion principle.

(b) For the $n = 2$ shell

$n = 2 \;\;\;\;\; \Rightarrow \;\;\;\;\; l = 1 \;\;\;\;\; \Rightarrow \;\;\;\;\; j = 1/2\,\,\;\;\;\;\;{\rm{or}}\;\;\;\;\,\,\frac{3}{2} \;\;\;\;\;$

For $j{\rm{ }} = {\rm{ }}1/2,{\rm{ m = 1/2 }}\;\;\;{\rm{or}}\;\;{\rm{ }} - {\rm{1/2 }}$

For $j{\rm{ }} = {\rm{ }}3/2,{\rm{ m }} = {\rm{ }} - 3/2{\rm{ }}\;\;{\rm{or}}\;\;{\rm{ }} - 1/2{\rm{ }}\;\;{\rm{or}}\;\;{\rm{ }}1/2{\rm{ or }}1/2$

Thus there are six distinct values.

(c) For $n = 3$ shell.

To simplify the representation let us use a tree structure

There are twelve unique values and 12 particles can be filled.

Example - 37

The solution to the Schrödinger equation for an electron in the ground state of the hydrogen atom is ${\psi _{1s}} = \frac{1}{{\sqrt {\pi a_0^3} }}{e^{ - r/ao}}$

where $r$ is the distance from the nucleus and ${a_0}$ is $0.529{\rm{ }} \times {\rm{ }}{10^{ - 8{\rm{ }}}}{\rm{cm}}$ . The probability of finding an electron at any point in space is proportional to ${\left| \psi \right|^2}$ . Using calculus, show that the maximum probability of finding the electron in the 1s orbital of hydrogen occurs at $r{\rm{ }} = {\rm{ }}{a_0}$ .To solve this problem one needs to know how to find the maximum value of a function using calculus.

Solution :

In this question we are asked to show that the probability of finding an electron at $r{\rm{ }} = {\rm{ }}{a_0}$ is maximum. We know that the probability we define, has to be associated with a volume. Here the set of points for whom the distance from the origin is ${a_0}(r{\rm{ }} = {\rm{ }}{a_0})$ form a spherical surface. A surface has an area but no volume so there should not be any probability associated with it and the probability of finding the electron on it should be zero. This answer should be perfectly agreeable but this is not what the person who framed the question had in mind. Questions such as this one, which do not completely communicate what the person who framed the question desires are common in quantum mechanics. The reason for this is probably that most writers in their hurry neglect something fundamental like we saw in this problem. This is precisely why this problem has been included here. We can be linient to the author of the problem and miss interpret his question as, the proof that if we consider spherical shells of a thickness $dr$ then the shell with radius $r{\rm{ }} = {\rm{ }}{a_0}{\rm{ }}$ has the maximum probability of finding the electron in it. This is most probably what he had in mind. The probability of finding the electron in a shell of radius $r$ and thickness $dr$ is equal to the product of the probability density at that radial location and the volume $= {\left| {{\psi _{1s}}} \right|^2} \times 4\pi {r^2}dr$

We have to find the value of $r$ where this quantity has a maximum value. As $dr$ is the infinitesimal thickness of the shell we are considering for all shells, it does not vary with the radial location <ted>r</tex>.

Thus we only need to find the value of $r$ for which ${\left| \psi \right|_{1s}}^2 \times 4\pi {r^2}$ is maximum (recall that this is the radial probability density function). To find the maximum of function $f\left( r \right) = {\left( {\frac{1}{{\sqrt {\pi a_0^3} }}{e^{ - r/{a_0}}}} \right)^2} \times 4\pi {r^2}$

we follow the standard methods we learn in calculus.

Let us draw the graphs of ${r^2}$ and ${e^{ - 2r/{a_0}}}.$

The plot of the function $f\left( r \right) = \frac{{4{r^2}{e^{ - 2/{a_0}}}}}{{a_0^3}}$

This is the same as the plot of the radial probability density function of the 1s orbital in Fig - 23.

The function is zero at $r = 0$ and $r = \infty$ . The function is positive in between these values. The Rolle’s theorem of calculus tells us that a maximum or a minimum has to occur between two points which have the same value. Since the function is positive $(i.e{\rm{ }}f(r){\rm{ }} > {\rm{ }}0){\rm{ }}$ between $r = 0$ and $r = \infty$ only a maximum can occur. The location of this maximum is determined by differentiating $f(r)$ and determining the value of $r$ for which $\frac{{df\left( r \right)}}{{dr}} =0$ .

On differentiation :

$\frac{{df}}{{dr}} = \frac{4}{{a_0^3}}\frac{{d({r^2}{e^{ - 2r/ao}})}}{{dr}} = \frac{4}{{a_0^3}}\left( {2r({e^{ - 2r/a0}}) - \frac{{2{r^2}}}{{{a_0}}}({e^{ - 2r/a0}})} \right) = \frac{8}{{a_0^4}} \times r \times ({a_0} - r) \times {e^{ - 2r/a0}}$

This is equal to zero when $r = 0$ or $r = {a_0}$ or $r = \infty$

Thus the maximum occurs at $r = {a_0}$ .

Here ${a_0}{\rm{ }} = {\rm{ }}0.529{\rm{ }}{\AA}.$ This is the Bohr radius for 1s shell. The radial probability density function of $1s$ is maximum at the Bohr radius for $1s$ . The probability of finding the 1s electron in a shell of thickness $dr$ is maximum when the radius of that shell is equal to the Bohr radius for $1s$ .

Example - 38

How many unpaired electrons are there in the $N{i^{2 + }}{\rm{ }}$ ion?

Solution :

According to the rules written in section 3.3.3.2 we must first write the electronic configuration of $Ni$ . This is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^8}4{s^2}$ . Now we have to remove two electrons from the outer most sub shell, which is $4s$ . Thus the electronic configuration is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^8}$ . The box representation given below shows that there are two unpaired electrons.

Example - 39

Predict the magnetic moment for $C{o^{3 + }}$ .

solution :

By using the same logic as we did for the previous problem we get the electronic configuration of $C{o^{3 + }}$ as $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^6}$ . The box representation given below shows that there are four unpaired electrons. Thus the magnetic moment is $\mu = \sqrt {n\left( {n + 2} \right)} = \sqrt {4\left( {4 + 2} \right)} = 4.9B.M$

$3{d^6}$

Example - 40

Complete the following statements:

(a) Two electrons in the same_________ must have opposite spins.

(b) When $l = 3$ , ${m_l}$ may have values from __________ to __________.

(c) Orbitals with the same energy are said to be__________.

(d) The $2p$ orbitals of an atom have identical shapes but differ in their __________.

(e) A nodal surface is one at which the probability of finding an electron is __________.

Solution :

(a) orbital

(b) –3 to +3

(c) degenerate

(d) orientations in space

(e) zero.

Atomic Structure - Section III

From LocusWiki

The Quantum Mechanical Model of the Atom

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