Download Chapter 4: ATOMIC STRUCTURE 1(a) Define or state the (i) Orbital

Chapter 4: ATOMIC STRUCTURE
1(a)
Define or state the
(i)
Orbital
Refers to the region or volume in space surrounding the nucleus where
the probability to find the electron is high. An orbital is described by
combination of integer and a letter, corresponding to the values of three
quantum numbers for the orbitals which are principle quantum numbers
denoted as n, the azimuthal quantum numbers denotes as l, the magnetic
quantum number ml
(ii)
Principle quantum number, n
Is indicated by integers 1,2…. This quantum numbers related directly to
the size and energy of the orbital
(iii)
Azimuthal quantum number, l
Is indicated by the letters s, p, d,f,… correspond to the value of 0, 1, 2, 3..
this quantum number can defines the shape of orbital.
(iv)
The magnetic quantum number, ml
Relates to the orientation of the orbital in the space
(v)
An electron shell
Is the set of all orbitals with the same value of n, eg: 4s 4p 4d 4f
(vi)
A sub-shell
Is the set of one or more orbitals with the same value of n and l values,
eg: 4s and 4p are sub-shell of the n = 4. There is one orbital in an s subshell and three in a p sub-shell
(vii)
Aufbau’ rule
States that the electrons must fill orbitals of lowest energy available
before orbitals with higher energy are filled
(viii)
Pauli’s exclusion principle
States that for each orbital can accommodate a maximum of two electrons
with opposite spin or no two electrons in an atom can have the same set
of four quantum numbers n, l, ml and ms
(ix)
Hund’s rule
States that in a set of orbitals of the same energy (degenerate orbitals),
electrons will occupy the orbitals singularly first (same spin) before pairing
occurs.
(x)
Electron configuration
The way in which the electrons are distributed among the various orbitals
of an atom
(b)
2
The electronic configuration of a carbon atom is 1s2 2s2 2p2 . Write a complete set of
quantum numbers for;
[2]/4a
(i)
The highest energy electron of carbon
(2,1,-1,+1/2) or (2,1,0,+1/2)
(ii)
The lowest energy electron of carbon
(1,0,0,+1/2) or (1,0,0,-1/2)
(iii)
Draw all the orbitals occupied by the valence electrons of carbon atom
[3]/4b
The following combinations are not allowed. Write the new values of l to create an
allowable of angular momentum quantum numbers (l) combination:
[4]/4a
(a)
(b)
(c)
(d)
3
n=
n=
n=
n=
2;
4;
7;
4;
l=
l=
l=
l=
0;
4;
2;
1;
ml=
ml=
ml=
ml=
-1
+1
+3
-2
l=1
l=1,2 ,3
l=3,4,5,6
l=2,3
Identify the types of orbital given below:
Orbital shapes
[5]/4c
Type
S
Px, Pz, Py
dx2–y2, dxz,
dz2, dyx, dxy
4
(a)
Write the electronic configuration of chromium according to the Aufbau
Principle and the actual configuration observed.
[2]/4d
According to Aufbau Principle: 1s2 2s2 2p6 3s2 3p6 4s2 3d4
Actual: 1s2 2s2 2p6 3s2 3p6 4s1 3d5
(b)
Give a reason for the irregularity.
[1]
Half-filled orbital (4s1 3d5) is more stable than partially-filled d
orbitals.
5
(a)
Write the condensed orbital diagram of Cobalt
↿⇂
[Ar]
↿⇂
↿⇂
↿
4s
(b)
6
↿
[1]/4d
↿
3d
State the name of the rule that has to be applied in orbital 3d to determine
the answer in 5(a)
[1]
Hund’s rule
Write the orbital diagrams for Na+, O and S. State the number of unpaired electrons
in each atoms/ions.
[6]/4d
Na+
O
S
↿⇂
↿⇂
↓
1s
2s
↿⇂
↿⇂
↿⇂
↓
2p
↓
↿⇂
↿⇂
↓↿⇂
2
↿⇂↓ ↿
↓1s
2s
↓
↿⇂
↿⇂
1s
2s
↓
2 unpaired electrons↓
No unpaired electron
2 unpaired electrons
↿
2p
↿⇂
↓
↿⇂
2p↓
↿⇂
↿⇂
↿⇂
↿
↓
↓
3s
↓
3p
↓
↿
7
Complete the following table by write the condensed electronic configurations for
each of the following atoms and identify the number of inner, outer and valence
electrons.
[10]/4e
Element
B
F
V
Cu
Cr
Te
8
Condensed electron
configuration
[He] 2s2 2p1
[He] 2s2 2p5
[He] 4s2 3p3
[Ar] 4s1 3d10
[He] 4s1 3d5
[Kr] 4d10 5s2 4p4
No of valence
electrons
3
7
5
11
6
6
4
d
16
p
1
s
State the general electronic configuration for valence electron for lithium, carbon,
titanium, neon
[4]/4g
Lithium =ns1
Carbon= ns2np2
10
No of outer
electrons
3
7
2
1
1
2
Complete the following table with identify the group and block of elements that
corresponds to each of the following electronic configurations:
[5]/4f
Electronic configuration
Group
block
2
10
4
16
p
(a)[Ar] 4s 3d 4p
(b)[Xe] 6s2 4f14 5d2
(c)[Ne] 3s2 3p4
(d)[Ne]3s1
9
No of inner
electrons
2
2
18
18
18
46
titanium= ns2(n-1)d2
neon= ns2np6
Given:
P3- , S2- , Cl(a)
Describe the similarity between 3 species above
[1]/4h
Ions with the same number of electrons thus have the same
electronic configuration.
(b)
Arrange the ions in the order of increasing size.
[1]/4i
Cl- <S2-<P3-
(c )
Explain your answer in (b)
[2]/4i
Generally, when Zeff increased, attraction between valence electron
and nucleus increased, thus, size of ions decreased.
As consequence, P3-,is the largest species due to weaker attraction
between nucleus and valence electron. Meanwhile Cl- largest nuclear
charge,thus smallest species due to stronger attraction between
nucleus and valence electron
Zeff = Z – S
where Z = number of proton in the nucleus
S = inner-shell electrons
For Cl-,
11
Predict, with the reasons, how the radius of the following ions compares with the
radius of neon atom.
(i)
the fluorine ion, F¯
(ii)
the magnesium ion, Mg2+
(i)
12
F¯ and Ne are isoelectronic. However, Ne has a greater number of
proton compare to F¯. Therefore, the attraction for electrons in Ne
(Zeff) in Ne is stronger and consequently the atomic radius of Ne is
smaller.
(ii)
Mg2+ and Ne are isoelectronic. However, Mg2+ has 2 protons more
than Ne. Hence, the attraction for electron in Mg2+ is stronger.
Moreover, being positively charge, the electrons are more tightly
bound. As a result, Mg2+ is smaller than Ne.
Calculation of Zeff
Zeff for F¯= 9 – 2 = 7
Zeff for Mg2+ = 12 – 10 = 2
Zeff for Ne = 11 – 2 = 9
(a)
Arrange the following atoms in order of decreasing atomic radius
Nitrogen,Phosphorus,Silicon
(b)
Si>P>N
[1]/4i
Explain your answer in (a)
[2]
N is above P in group 15 and has a lower n. Thus it has a smaller
atomic radius than P. P is smaller than Si because of P has greater
effective nuclear charge, Zeff than Si.
13
State two factors that influence the IE.
[2]/4j
i.
Principal quantum number, n (number of valence shell)
ii.
Effective nuclear charge, Zeff
14
Explain the difference in first ionization energy between magnesium and calcium
[2]/4k
Electronic configuration;
Mg: [Ne]3s1
Ca: [Ar]4s1
Mg has larger 1st ionization energy because it has smaller atomic radius
than Ca. Valence electron in Ca is located in period 4. Valence electron in Mg
feels stronger nucleus attraction than in the Ca atom. Thus, it is harder to
remove an electron from Mg atom than in Ca atom.
15
Explain the difference in first ionization energy between phosphorous and sulfur
[2]/4k
Electronic configuration;
P: [Ne] 3s2 3p3 (Half-filled)
S: [Ne] 3s2 3p4 (partially-filled)
P has half filled 3p sub-shells, which gives greater stability compared to S
which has paired electrons in 3p orbital which causes e-e repulsion. P needs
more energy to remove an electron thus P has a higher 1st IE+
16
The first four successive ionization energies of Period 3 element, A is shown below:
4m
(a)
IE
1
2
3
4
5
(kJmol-1)
786
1580
3230
4360
16000
Identify the group number of A.Explain
[2]
4A (14). Because there are a big jump between IE4 and IE5, ,means
four electron removed is from outermost shell and indicating that the
element has four valence electron.
17
(b)
Name the element A
carbon
(c)
The electrons are lost when A atom becomes A3+ ions. Write down the
electron configuration of Al 3+
[1]
1s2 2s22p2
[1]
An element X has the following successive ionization energies.
IE1
1450
IE2
2162
IE3
2962
IE4
5532
4l
IE5
10539
(a)
Write the general valence electronic configuration for X.
ns2 np2
[1]
(b)
State the group number belongs to X
14 or 4A
(c)
Write the full electronic configuration for X if X is located in Period 4, [1]
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2
(d)
Give the symbol for X? [1]
[1]
Ge
18
Given below are the hypothetical values of successive ionization energies of an
element R:
IE
(kJ/mol)
19
2880
4520
9450
53000
64200
66400
(a)
Define the ionization energy
The minimum energy required to removed 1 mol of electron from an
atom in 1 mol of gaseous atom at standard condition
(b)
Explain why the ionization energies increased with the number of electron
removed.
The effective nuclear charge increases when successive electrons are
removed
(c)
Determine the group in which of R belongs to. Explain.
Group 3A. The first three successive ionization energies shows the
steady increased but very extreme increase is observed at the fourth
ionization energy. This indicates that the 4th electron requires an
extremely high amount of ionization energy as it is situated in the
inner shell. Thus, element R has 3 valence electron, it can be
concluded that R is a member of Group 3A.
Figure 1 shows the first ionization energy against the elements in periods 2 in the
periodic table. Explain why the first ionization energy of Be is higher than B.
Fig: First ionization energy, kJ/mol against element
There is general increase in the first ionization energy with increasing
proton umber of the elements. Going across the period from left to the
right, the atomic size of the elements are decreases and the effective
nuclear charge increases. As the result, the electrons are more tightly
bonded to the nucleus and are more difficult to remove. However, the first
ionization energy of Be is higher than B due to the first electron
to
be
removed from a fully filled s orbital. The s2 configuration has additional
stability, making the electron more difficult to remove compared to B which
only have partially filled p orbital. As the result the first ionization energy
for Be is higher than expected.
20
(a)
Write a balanced chemical equation to represent the first and second electron
affinity process for bromine.
[2]/4n
EA1 : Br (g) + e¯  Br¯ (g)
EA2 : Br¯ (g) + e¯  Br2¯ (g)
(c)
Second electron affinity always has a positive value (an endothermic
process). Explain.
[2]/4o
When an electron is added to a negative ion (anion), a strong
repulsion is felt. Thus, heat is absorbed to overcome the electrostatic
repulsion between the existing electrons. The energy absorbed >
energy released when electron is added thus 2nd EA is always
positive.