2. Atomic Structure (Suggested Solutions)_2014

Atomic Structure (Suggested Solutions)_2014 8

Topic: Atomic Structure (Suggested Solutions) Section A: Multiple Choice Questions 1

D. The term nucleons is the sum of the number of protons and neutrons.

For an electrically neutral particle such as P, the number of protons is equal to the number of electrons. For a negatively charged particle such as Q-, the numerical value of the charge translated into the number of electrons gained. For a positively charged particle such as R2+, the numerical value of the charge translated into the number of electrons lost.

Particle Neutrons Protons Electrons

P 16.0 33 16 = 17 17 Q- 18.0 35 18 = 17 17 + 1 = 18 R2+ 17.0 34 17 = 17 17 2 = 15 S3- 16.0 31 16 = 15 15 + 3 = 18

2 D. Positively charged particles will be deflected toward negative terminal and negatively charged particles will be deflected towards positive terminal hence

option B and C are eliminated. Angle of deflection Mass

Charge

From the diagram, it is observed that particle X is deflected twice as much as particle Y. Hence the mass/electronic charge (m/e) ratio of Y must be twice of that of particle X.

Particle Mass Electronic Charge m/e ratio 4He+ 4.0 +1 4 / +1 = +4 2H+ 2.0 +1 2 / +1 = +2 1H+ 1.0 +1 1 / +1 = +1

The m/e ratios of the particles given in option A and D are calculated in the table above. It can be observed that the m/e ratio of 2H+ is twice of that of 1H+

while the m/e ratio of 4He+ is 4 times of that of 1H+ hence eliminating option A.

3 A. The larger the numerical value of the m/e ratio of the particle in an electric field, the lesser the deflection of the particle.

Particle Mass Electronic Charge m/e ratio 224

12Mg 24.0 +2 24 / +2 = +12

32713 Al

27.0 +3 27 / +3 = +9

2168 O

16.0 2 16 / 2 = 8

33115P

31.0 3 31 / 3 = 10.3

Please note that the sign of the m/e ratio indicates the deflection of the particle to which terminal i.e. the negative m/e ratio will be deflected towards the positive terminal and vice versa.


The m/e ratios of the particles are calculated in the table above. It can be

observed that the numeral value of the m/e ratio of 22412Mg is the largest and

hence it is the least deflected.

4 A. A Group III element will have the general valence electronic configuration of ns2np1. For the removal of the first three electrons, a general (steady) increase in ionisation energy is observed because the electrons are being removed from ion of increasing positive charge. As the electrons are removed, the outermost electrons are held more tightly by the constant nuclear charge; hence the ionization process becomes increasingly difficult. The removal of the 4th electron which is from the inner principal quantum shell disrupts the stable octet configuration. Hence it is energetically more difficult to remove which is indicated by a big jump in ionisation energy after the removal of the 3rd electron.

5 A. Valence electronic configuration of successive elements:

Group V: ns2np3

Group VI: ns2np4

Group VII: ns2np5

Group 0: ns2np6

Group I: (n+1)s1

Since elements from Group V to Group 0 are in the same period, the shielding effect remains almost constant. On the other hand, nuclear charge increases from Group V to Group 0 hence effective nuclear charge increases and it becomes increasingly difficult to remove the electron. Consequently the first ionisation energy generally increases from Group V to Group 0. However due to inter-electronic repulsion present in the doubly filled p orbital for the element in Group VI, a slight drop in ionisation energy is observed.

As for the first ionisation energy for the successive element in Group I, it will be much lower because

(i) the increased in shielding effect provided by the additional filled quantum principal shell outweighs the increased in nuclear charge.

OR (ii) the removal of the first electron is from a higher principal quantum

number orbital which is further away from the nucleus. Hence being less closely held, much lower energy is needed to remove the electron.

Option B is wrong: Let P, Q, R, S, T be the five elements from Group V. As they are listed in the order of increasing atomic mass, the arrangement of the elements in a group would be:

P

Q

R

S

T


Hence the first ionisation energy would decrease from P > Q > R > S > T as going down the group the increased in shielding effect provided by the additional filled quantum principal shell outweighs the increased in nuclear charge. Option C is wrong: A general (steady) increase in ionisation energy should be observed in the successive ionisation energies for a Group V element because electrons are being removed from ion of increasing positive charge. As the electrons are removed, the outermost electrons are held more tightly by the constant nuclear charge; hence the ionization process becomes increasingly difficult. Option D is wrong: Element with an atomic number of 5 will have the electronic configuration of 1s22s22p1 hence the removal of the second electron from 2s orbital which is closer to the nucleus would require greater ionisation energy than the removal of the fist electron from the 2p orbital.

6 A. The electronic configurations of the 4 options are as follows:

Na: 1s22s22p63s1

Mg: 1s22s22p63s2

Al: 1s22s22p63s23p1

Ar: 1s22s22p63s23p6

The removal of the first and second electrons for Mg, Al and Ar are all from the same principal quantum shell whereas for Na, the first electron is from the 3rd principal quantum shell and the second electron is from the 2nd principal quantum shell.

Comparatively, greater amount of energy would be required to remove the second electron from Na than the rest as the stable octet configuration of the inner principal quantum shell is disrupted. Hence the ratio of the second ionization energy to the first ionization energy would be the highest for Na.

7 B. From Figure 1, it is deduced that element X is from Group III due to a large difference in ionization energy of the 3rd and 4th electron, indicating that 4th electron is from the inner principal quantum shell.

Since Figure 2 shows the first ionisation energy of eight consecutive elements (including X) of Period 3 and element X is from Group III, it must be the 3rd point plotted on the graph of Figure 2.

8 B.

A common misconception is to write the electronic configuration of ion by filling the orbitals starting from the lowest energy level with the total number of electrons of the ions.

The correct method is 1st Step Write the electronic configuration of the electrically neutral atom 2nd Step Remove the electrons lost by atom to form ion from the orbitals with the highest energy (i.e. you remove electrons from 4s then 3d) Please be aware that when 4s and 3d orbitals are empty, 4s orbital is at a


lower energy level hence 4s orbitals are filled up before 3d orbitals. However once they are occupied, the 4s orbital is at a higher energy level hence electrons are generally removed from 4s orbital before 3d orbitals. Option A: V atom has 23 electrons: 1s22s22p63s23p6 4s23d3. Option B: Cu atom has 29 electrons: 1s22s22p63s23p6 4s13d10

Electronic configuration of Cu atom is not 1s22s22p63s23p6

4s23d9 because electronic configuration with fully filled 3d subshell is unusually stable due to symmetrical charge distribution around the metal centre.

Option C: Mn atom has 25 electrons: 1s22s22p63s23p6 4s23d5

Mn2+ ion has 25 2 = 23 electrons: 1s22s22p63s23p63d5

Option D: Cr atom has 24 electrons: 1s22s22p63s23p6 4s13d5

Electronic configuration of Cr atom is not 1s22s22p63s23p6

4s23d4 because electronic configuration with half-filled 3d subshell is unusually stable due to symmetrical charge distribution around the metal centre. Cr+ ion has 24 1 = 23 electrons: 1s22s22p63s23p63d5

Only Cu atom has unpaired electron in the 4s orbital. 9 D.

Option A: Cr atom has 24 electrons: 1s22s22p63s23p6 4s23d4. Cr3+ ion has 24 3 = 21 electrons: 1s22s22p63s23p63d3

3 unpaired electrons. Option B: Ni atom has 28 electrons: 1s22s22p63s23p6 4s23d8. Ni2+ ion has 28 2 = 26 electrons: 1s22s22p63s23p63d8.

2 unpaired electrons. Option C: Ca atom has 20 electrons: 1s22s22p63s23p6 4s2. Ca2+ ion has 20 2 = 18 electrons: 1s22s22p63s23p6.

0 unpaired electrons.

Option D: Co atom has 27 electrons: 1s22s22p63s23p6 4s23d7. Co3+ ion has 27 3 = 24 electrons: 1s22s22p63s23p6 3d6.

4 unpaired electrons. 10 D.

The oxidation of Ni in [NiO2]2- is +2.

Let the oxidation of Ni in [NiO2]2- be x.

x + 2(2) = 2 x = +2

Ni atom has 28 electrons: 1s22s22p63s23p6 4s23d8. Ni2+ ion has 28 2 = 26 electrons: 1s22s22p63s23p63d8.

11 C.

Option A: C- ion has 7 electrons: 1s22s22p3. On losing an electron, C atom has 7-1= 6 electrons: 1s22s22p2. 2 unpaired electrons in p orbitals. Option B: F atom has 9 electrons: 1s22s22p5. On losing an electron, F+ ion has 9-1= 8 electrons: 1s22s22p4. 2 unpaired electrons in p orbitals. Option C: N- ion has 8 electrons: 1s22s22p4. On losing an electron, N atom has 8-1= 7 electrons: 1s22s22p3. 3 unpaired electrons in p orbitals. Option D: O+ atom has 7 electrons: 1s22s22p3. On losing an electron, O2+ ion has 7-1= 6 electrons: 1s22s22p2. 2 unpaired electrons in p orbitals.


12 C. S atom has 16 electrons: 1s22s22p63s23p4. On gaining two electrons, S2- ion has 16+2= 18 electrons: 1s22s22p63s23p6.

Particle Neutrons Protons Electrons 31S2 31 16 = 15 16 16+2 = 18 35S2 35 16 = 19 16 16+2 = 18

The larger the numerical value of the m/e ratio of the particle in an electric field, the lesser the deflection of the particle.

Particle Mass Electronic Charge m/e ratio 31S2 31.0 -2 31 / -2 = -15.5 35S2 35.0 -2 35 / -2 = -17.5

Under the same magnetic field, the 35S2 ion will be deflected less than the 31S2 since the numerical value of the m/e ratio is larger.

13 A. Cu atom has 29 electrons: 1s22s22p63s23p63d104s1. Cu+ ion has 28 electrons: 1s22s22p63s23p63d10.

14

C. Option A: C+ ion has 5 electrons: 1s22s22p1; N- ion has 8 electrons: 1s22s22p4. C+ ion has 1 unpaired electron & N- has 2 unpaired electrons. Option B: O- ion has 9 electrons: 1s22s22p5; S2+ ion has 14 electrons: 1s22s22p63s23p2. O- ion has 1 unpaired electron & S2+ ion has 2 unpaired electrons. Option C: Cr atom has 24 electrons: 1s22s22p63s23p63d54s1; Fe3+ ion has 26-3 = 23 electrons: 1s22s22p63s23p63d5 Cr atom has 6 unpaired electrons & Fe3+ ion has 5 unpaired electrons. Option D: Mn4+ ion has 25-4=21 electrons: 1s22s22p63s23p63d3; Co atom has 27 electrons: 1s22s22p63s23p6 4s23d7. Mn4+ ion has 3 unpaired electrons & Co atom has 3 unpaired electrons.

15

C.

[ H = 1

1 H; D = 2

1 H; T = 3H; C =

12

6 C; O =16

8 O ]

Particle Neutrons Protons Electrons

CO32- 30 30 32

H2DO+ 9 11 10

OH- 8 9 10

TCO3- 32 31 32

16 C.

Option 1 O- ion has 8 + 1 = 9 electrons: 1s22s22p5. Option 2 N atom has 7 electrons: 1s22s22p3. Option 3 P atom has 15 electrons: 1s22s22p63s23p3.


17 A. Option 1 p orbitals of different quantum number all have the same shape.

p orbitals of different quantum number only differ in terms of size. Option 2 With the same principal quantum shell, a p orbital reside further away from the nucleus than a s orbital, hence it will have higher energy than a s orbital. Option 3 All orbitals contain a maximum number of 2 electrons.

18 D.

P atom has 15 electrons: 1s22s22p63s23p3. Option 1 The outer most subshell is a p orbital which has a dumb-bell shape hence the description is incorrect. Option 2 The outermost subshell is the 3p orbital with principal quantum number n = 3. Option 3 The outermost subshell is half filled: 3p3.

19 B. Isoelectronic refers to having same number of electrons. Option 1 Gaseous particle Y and gaseous particle Z are isoelectronic i.e. have the same number of electrons. Since Y has n protons and Z has (n-1) protons, the proton/electron (p/e) ratio is higher in Y than Z. Consequently Y would have a smaller radius than Z. Option 2 Since the proton/electron (p/e) ratio is higher in Y than Z, the electron would be more tightly held in Y than Z hence Y would have higher first ionisation energy than Z. Option 3 Since the proton/electron (p/e) ratio is higher in Y than Z, the ability of Y to attract electron to itself would be higher than Z. Hence Y should be more electronegative than Z.

20 D. Isotopes refer to atoms of the same element which contain the same number of protons and electrons but different number of neutrons.

Option 1 16O, 17O and 18O are isotopes so they will have the same number of electrons which translate into the same electronic configuration. Option 2


Since 16O, 17O and 18O are isotopes, the proton/electron (p/e) ratio is the same for all the isotopes hence they will have the same first ionisation energy. Option 3 16O, 17O and 18O are isotopes so they will have the same number of protons.

21 B. Option 1 N3- ion has 7 + 3 =10 electrons: 1s22s22p6. Mg atom has 12 electrons with electronic configuration of electrons: 1s22s22p63s2 hence Mg2+ ion has 12 2 =10 electrons: 1s22s22p6. F- ion has 9 + 1 = 10 electrons: 1s22s22p6. From the electronic configurations, all 3 ions do not contain any unpaired electrons in their valence shells. Option 2

Ion Mass Electronic Charge m/e ratio

N3- 14.0 3 14 / 3 = 4.7 Mg2+ 24.3 +2 24.3 / +2 = +12.2

F- 19.0 1 1 9/ 1 = 19 Please note that the sign of the m/e ratio indicate the deflection of the particle to which terminal i.e. the negative m/e ratio will be deflected towards the positive terminal and vice versa. The m/e ratios of the particles are calculated in the table above. It can be observed that the numeral value of the m/e ratio of F- ion is the largest and hence it is the least deflected. Option 3 All 3 ions are isoelectronic but nuclear charge increases from N3- to F- to Mg2+. As p/e ratio increases from N3- to F- to Mg2+, the ionic radius would decreases in the order of N3-, F- , Mg2+.

22 C.

Option 1 Nuclear charge does increases down the group in a Periodic table but it does not explain the decrease in first I.E. down a group in a Periodic Table. Option 2 Going down the group, the number of filled inner principal quantum shells increases hence the valence electron becomes progressively further from the nucleus. Consequently the first I.E. decrease down a group in a Periodic Table. Option 3 Going down the group, the number of filled inner principal quantum shells increases hence the valence electron becomes progressively further from the nucleus, which results decrease in first ionisation energy down a group in a Periodic Table.

23 A.

Option 1 E belongs to Group V because there is a large difference in ionization energy of the 5th & 6th electron, indicates that 6th electron is from the inner quantum shell. Option 2 E having 5 valence electrons can form a chloride with the following dot and cross diagram.


1 lone pair and 3 bonding pairs, hence the shape is trigonal pyramidal. Option 3 A Group V element has the general electronic configuration of ns2np3 so it will have half filled p-orbital.

24 A.

Option 1 Rb atom has 37 electrons. Rb+ ion has 37 1 = 36 electrons. Option 2 Sr atom has 38 electrons. Sr2+ ion has 38 2 = 36 electrons. On losing 2 electrons, Sr has one less electron shell. The nucleus attracts the valence electrons more strongly towards itself and hence, Sr2+ ion is much smaller than X2- ion even though it has same number of electrons. X2- ion is negatively charged. The repulsion between identical charges causes the ionic size to be larger than the Sr atom. Option 3 X- ion has higher p/e ratio, therefore the valence electrons experience stronger nuclear attraction than X2- ion.

Section B: Structured Questions 1 (a) (i) Isotopes are atoms of the same element thus having the same

number of protons but with different number of neutrons.

Comments: - Atoms with the same atomic / proton number but with different

nucleon / mass number are a description of isotopes rather than the definition of isotopes.

(ii)

Isotope protons neutrons electrons 14N 7 7 7 15N 7 8 7

(b) (i) The same number of electrons.

(ii) 1s2 2s2 2p6.

(iii) Nuclear charge of the fluoride ion is greater than in ion of nitrogen while the screening effect is the same for both ions. Hence, more energy is required to overcome the stronger attraction between the outermost electron and the nucleus of fluoride ion. Comments: - Nuclear charge is dependent on number of protons in nucleus. A common error during explanation is students tend to use the term nitrogen and ion of nitrogen or fluorine and fluoride quite freely. It

Cl Cl Cl

E


is important to note that nitrogen would represent N atom while ion of nitrogen would represent N3- ion which are two very different particles. Same goes for fluorine: F atom and fluoride: F- ion.

2 (a) Q belongs to Group VII. There is a large difference in ionization energy

of the 7th and 8th electron, indicating that 8th electron is from the inner quantum shell. Comments: - The group number of the element should be written in Roman numerals

rather than Arabic numerals. - The +/- sign must be placed in front.

A common misconception is to reason that the 8th ionisation energy is high hence element Q is from Group VII. It is important to realise that without the comparison with the 7th ionisation energy, it does not show that the eighth electron requires much higher energy than the previous electron. As a result it does not indicate that the eighth electron is from an inner quantum shell.

(b) (i) P: -2 Q: -1 S: +1 Comments: - Since P, Q, R and S are a sequence of consecutive elements and

Q belongs to Group VII, P must be from Group VI (6 valence electrons), R from Group 0 (8 valence electrons) and S from Group I (1 valence electron). Hence P, Q, R are in the same period while S is from the next period as illustrated below.

- Consequently to be isoelectronic with R, P must gain two electrons

hence having a charge of -2 while Q must gain one electron hence having a charge of -1 and S must lose one electron hence having a charge of +1.

(ii) P

2- > Q- > S+ Comments: - A common incorrect answer given by students is P > Q > S. It

should be noted that the symbols P, Q and S represent atoms and not ions.

- Assuming P has n number of protons, Q will have (n+1) number of protons, R will have (n+2) number of protons and S will have (n+3) number of protons. Since the ions of P, Q and S are isoelectronic, the proton/electron (p/e) ratio will increase from P2- to Q- to S+. The higher the p/e ratio, the more tightly held are the electrons, hence ionic radius decreases from P2- to Q- to S+.

3 (a) 1s2 2s2 2p5.

(b) Comments:

Can also accept either px or py orbital - No marks awarded if all three orbitals drawn.

P Q R S

x

y

z


5

4 Comments: Please take note that your graph should illustrate the following: - A general (steady) increase in ionisation energy due to electrons being

removed from ion of increasing positive charge. As the electrons are removed, the outermost electrons are held more tightly by the constant nuclear charge, hence the ionization process becomes more difficult.

- A slight jump in ionisation energy after the removal of the 1st electron. This is because the removal of the 1st electron is energetically easier due to the inter-electronic repulsion from the doubly filled 3p orbital.

- A slight jump in ionisation energy after the removal of the 4th electron. The removal of the 5th electron which is from the 3s orbital is more difficult to remove because it is of a lower energy as the 3s orbital is closer to the nucleus than the 3p orbital.

- A big jump in ionisation energy after the removal of the 6th electron. The removal of the 7th electron which is from the inner principal quantum shell disrupts the stable octet configuration. Hence it is energetically more difficult to remove.

0 1 2 3 4 5 6 7 No. of electrons

Ionisation energy/ kJmol1


Comments: - The electronic configuration of chlorine is 1s22s22p63s23p5. Please take note that

your graph should illustrate the following: - A general (steady) increase in ionisation energy due to electrons being removed

from ion of increasing positive charge. As the electrons are removed, the outermost electrons are held more tightly by the constant nuclear charge, hence the ionization process becomes increasingly difficult.

- A slight jump in ionisation energy after the removal of the 5th electron. The removal of the 6th electron which is from the 3s orbital is more difficult to remove because it is of a lower energy as the 3s orbital is closer to the nucleus than the 3p orbital.

- A big jump in ionisation energy after the removal of the 7th electron. The removal of the 8th electron which is from the inner principal quantum shell disrupts the stable octet configuration. Hence it is energetically more difficult to remove.

6 (a) Total number of electrons removed is 24.

Number of electrons corresponds to number of protons in an atom; hence atomic number of element X is 24. Therefore, X is Cr. Comments: - The phrase successive ionisation energies indicates that all the

electrons of the atom are removed.

(b) 1s22s22p63s23p64s13d5.

(c) The successive I.E. of X generally increases. Nuclear charge remains unchanged / no. of protons remains the same when electrons are removed. Stronger electrostatic forces of attraction between nucleus and valence electrons / resulting positive ions hold the valence electrons more tightly. OR Electrons are being removed from ion of increasing positive charge. As the electrons are removed, the remaining electrons are held more tightly by the constant nuclear charge. Hence more energy is required to remove the remaining electrons Comments: - As the question asked for general trend, therefore it is not required to

explain the slight jumps in the successive I.E. of X.

7 (a) 1s22s22p63s23p4. (b)

Comments: - As the question is only concern with shape and size of the orbital, the

axes need not be drawn. - It must be obvious that the s orbital increases in size as the principal

52

24


quantum number increases. - As only two p orbitals are required to be drawn to illustrate shape and

size, the question wants to see the comparison in terms of size between 2p and 3p orbital. A common error for student is to draw the px, py and pz orbital of the same quantum shell.

8 (a) Electronic configuration of F is1s22s22p5.

1s 2s

2px 2py 2pz Comments: - It is important that all axes must be labelled. This is especially so for

the drawing of the 2p orbitals otherwise it would be impossible to differentiate the three p orbitals.

- It must be obvious that the 2s orbital is greater in size than the 1s orbital.

- Valence shell orbitals include 2s and 2p orbitals only.

(b) (i) Element X is in Group I. There is a large difference in ionisation energy of the 1st and 2nd electron which indicates that the 2nd electron is from an inner quantum shell. Comments: - The group number of the element should be written in Roman

numerals rather than Arabic numerals. - A common misconception is to reason that the 2nd ionisation

energy is high hence element X is from Group I. It is important to realise that without the comparison with the 1st ionisation energy, it does not show that the second electron requires much higher energy than the previous electron. As a result it does not indicate that the second electron is from an inner quantum shell.

x x

y y

z z


(ii)

Comments: - The axis must be labelled. - Clear distinction in energy level between the different principal

quantum shells. - All orbitals must be clearly labelled. - All the 2p orbitals must be of the same energy. - All the 3p orbitals must be of the same energy.

9 (a) 1s22s22p63s23p63d6. Comments: Fe atom has 26 electrons while Fe2+ ion has 24 electrons. - A common misconception is to write the electronic configuration of Fe2+

ion by filling the orbitals starting from the lowest energy level with 24 electrons hence giving the answer as 1s22s22p63s23p64s23d4.

- The correct method is

1st Step Write the electronic configuration of Fe atom i.e. 1s22s22p63s23p6

4s23d6. 2nd Step Remove the 2 electrons lost by Fe atom to form Fe2+ ion from the orbitals with the highest energy which is the 4s orbital which gives the answer 1s22s22p63s23p63d6. Please be aware that when 4s and 3d orbitals are empty, 4s orbital is at a lower energy level hence 4s orbitals are filled up before 3d orbitals. However once they are occupied, the 4s orbital is at a higher energy level hence electrons are generally removed from 4s orbital before 3d orbitals.

(b)


Comments: - The axis must be labelled. - Clear distinction in energy level between the different principal quantum

shells. - All orbitals must be clearly labelled. - All the 2p orbitals must be of the same energy. - All the 3p orbitals must be of the same energy. - All the 3d orbitals must be of the same energy. - Except for the arrangement of the 6 electrons occupying the 3d orbitals

given in the answer above, no other arrangement is acceptable.

10 (a)

Comments: - As 208Pb2+ is positively charged, it should be deflected towards the

negative terminal. - Beam and angle must be clearly labelled.

(b) 232

16S will be deflected to a greater extent because it has a higher

electron charge / mass (e/m) ratio.

Hence angle of deflection of 232

16S 528

208

232

2

Comments:

- As 232

16S is negatively charged, it should be deflected towards the

positive terminal. - Each beam and angle must be clearly labelled.


11 (a) The technetium atom, Tc9943 contains 43 protons, 43 electrons and 56 neutrons.

The relative charge, relative mass and location of each sub-atomic particle is as follows:

Sub-atomic Particle Relative Charge Relative Mass Location

Proton +1 1 Nucleus

Neutron 0 1 Nucleus

Electron -1 1/1840 Orbital

Comments: - It is not acceptable to say that for electron, the relative mass of the charge

is 0 or its location is in electron shell.

(b) (i)

Hence angle of deflection of Tc9943 11.14

1

992

55

Comments: - The angle of deflection is dependent on mass / electronic charge

(m/e) ratio of the ion.

- Since the m/e ratio of Tc9943 is higher than

255

25Mn , Tc9943 should

logically be deflected lesser than 4. This should be a good reference when students check their answer.

- As Tc9943 is positively charged, it should be deflected towards the

negative terminal.

(ii)


Comments: - A common misconception is to write the electronic configuration of

Mn2+ ion by filling the orbitals starting from the lowest energy level with 23 electrons hence giving the answer as 1s22s22p63s23p64s23d3.

- The correct method is 1st Step Write the electronic configuration of Mn atom i.e. 1s22s22p63s23p6

4s23d5. 2nd Step Remove the 2 electrons lost by Mn atom to form Mn2+ ion from the orbitals with the highest energy which is the 4s orbital which gives the answer 1s22s22p63s23p63d5.

- For the drawing of the energy diagram, the following must be illustrated: 1. The axis must be labelled. 2. Clear distinction of difference in energy level between principal

quantum shells. 3. All orbitals must be clearly labelled. 4. All the 2p orbitals must be of the same energy. 5. All the 3p orbitals must be of the same energy. 6. All the 3d orbitals must be of the same energy. 7. Except for the arrangement of the 5 electrons occupying the 3d

orbitals given in the answer above, no other arrangement is acceptable.

(c) (i) Z belongs to Group VI. There is a sharp increase in ionisation

energy between the 6th and 7th IE. This shows that 7th electron is being removed from the next inner quantum shell.

Comments: - The group number of the element should be written in Roman

numerals rather than Arabic numerals. - A common misconception is to reason that the 7th ionisation energy

is high hence Z is from Group VI. It is important to realise that without the comparison with the 6th ionisation energy, it does not show that the seventh electron requires much higher energy than the previous electron. As a result it does not indicate that the seventh electron is from an inner quantum shell.

(ii) X is Silicon and its electronic configuration is 1s22s22p63s23p2.

Comments:

- X belongs to Group IV as 1. From the diagram, there is a large jump in ionisation energy

between the 4th and 5th electron which indicate that the 5th electron is from the inner quantum shell.

2. Only Group IV element exists as a macromolecule and is a

semi-conductor. - As it is mentioned that X is from period 3 of the periodic table, X

must be silicon. -


(iii) 1st I.E of Y is lower than nitrogen. Y has one more electron shells. There is weaker nuclear attraction on the valence electrons in Y as they are now further away from the nucleus.

Comments: - Y must be phosphorus as it belongs to Group V in period 3 because

there is a large jump in ionisation energy between the 5th and 6th electron which indicate that the 6th electron is from the inner principal quantum shell.

- (iv) Nitrogen is in Period 2 and does not have low-lying energetically

accessible d orbitals and can accommodate only 8 electrons. Comments: - Nitrogen cannot expand beyond its octet structure.

12 (a) Group VI.

Comments: - A has the highest third IE which means that 3rd electron is removed from

the inner quantum shell. Therefore A has 2 valence electrons; is in Group II.

(b) Sulfur.

1s2 2s2 2p6 3s2 3p4 Comments: - E is in Period 3 and in Group VI. Therefore the element must be sulfur.

(c) 2 >> 1

1 2 Comments: - The angle of deflection is dependent on mass / electronic charge (m/e)

ratio of the ion. - E- has a higher mass than A+. Both are singly charged. - Since the m/e ratio of E- is higher than A+, E- should be deflected lesser

than A+. - As A+ is positively charged, it should be deflected towards the negative

terminal. - As E- is negatively charged, it should be deflected towards the positive

terminal. 13 (a) (i) Isotopes are atoms of the same elements and have the same

number of protons but different number of neutrons.

E -

A+

+

_

Source


(ii) Isotopic mass is the mass of a particular isotope of an element. Comments: - Isotopic mass is dependent on the number of isotopes and the

percentage abundance of isotopes of an element.

(b) (i) R;QP; 158

3147

146

.

Comments: - If the particle has more electrons than protons, the particle becomes

negatively charged. If the particle has more protons than electrons, the particle becomes positively charged.

(ii) Unusual isotopes: P and R

Comments: - P is actually carbon whose usual isotope is carbon-12 and not

carbon-14 and R is actually oxygen whose usual isotope is oxygen-16 and not oxygen-15.

(c) (i) The polarity of plate A is - since positively charged H1 and He4

nuclides are attracted towards it. (ii) The deflection is dependent on the charge/mass (z/m) ratio. H1 has

z/m ratio of 1/1=1 while He4 has a z/m ratio of 2/4 = .

Since Mass

Charge deflection of Angle , H1 is deflected twice as much

as He4 .

(iii) The deflection is dependent on the charge/mass (z/m) ratio.

The z/m ratio of H2 =1/2 = z/m ratio of He4 .

Angle of deflection of H2 = Angle of deflection of He4 = 2o.

The z/m ratio of He3 =2/3 = 2/3 x z/m ratio of H1 .

Angle of deflection of He3 = 2/3 x Angle of deflection of H1 = (3

8)o

2. Atomic Structure (Suggested Solutions)_2014

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Transcript of 2. Atomic Structure (Suggested Solutions)_2014