Page 37 - PRE-U STPM CHEMISTRY TERM 1

Chemistry Term 1 STPM
D is in Group 1 of the Periodic Table. There is a very The lines in the Lyman series are in the ultra-
nd
st
large increase between the 1 and 2 I.E. violet region, while the lines in the Balmer series
are in the visible region.
6 (a) 3d (ii) Using the equation:
Energy 4s ∆E = hf
23
–34
3
3p 987.3 × 10 = (6.63 × 10 )(6.02 × 10 )(f)
15 –1
3s f = 2.47 × 10 s
Or,
Using the relationship:
2p
c
λ = —
2s
f
3.00 × 10
= —–———— 8
1s 2.47 × 10 15
–7
(b) (i) 1s 2s 2p = 1.21 × 10 m
= 121 nm
(b) (i) The electronic configuration is:
1s 2s 2p 3s 3p 3d 4s 1
5
6
6
2
2
2
(ii)
4s
3d
(iii)
3p Energy
7 (a) The third electron will have the same spin as one
of the other two electrons and will experience 3s
repulsion. 2p
(b) A set of 3d orbitals that are fully half-filled has extra
stability compared to a partially filled 3d orbitals. 2s
8 (a) Hund’s rule states that in a set of degenerate orbitals 1s
electrons will occupy the orbitals singly first before
pairing occurs. (ii) The electronic arrangement violates the
Pauli’s exclusion principle states that an orbital can Aufbau’s principle which states that electrons
accommodate a maximum of two electrons only with will fill available orbitals with lower energy first
opposite spins. before orbitals with higher energies are filled. In
(b) The electronic configuration of N is 1s 2s 3p . an empty atom, the 4s orbital is of lower energy
2
5
2
2–
than the 3d. Thus, by right, two electrons must
 
 
2p be filled in 4s before the 3d orbitals are filled to

 
2
2
2
4
6
6
2s give it a configuration of 1s 2s 2p 3s 3p 3d
6
4s instead of 1s 2s 2p 3s 3p 3d 4s . This
2
2
2
1
6
5
2
 
1s abnormality arises because a half-filled 3d sub-
5
9 Each orbital can accept two electrons only. shell (3d ) is energetically more stable than a
4
2
The seven s electrons will have configuration of 1s , 2s , 3s 2 partially filled 3d sub-shell (3d ).
2
and 4s . 1
6
The twelve p electrons will have configuration of 2p and Chapter 3 Chemical Bonding
3p .
6
The ten d electrons will have configuration of 3d . Quick Check 3.1
10
2
2
2
6
10
Electronic configuration of X is 1s 2s 2p 3s 3p 3d 4s . 1
6
1 (a) (f)
10 (a) (i) n = ∞
O Cl –
x x • x x
• • •
– x x – x x
O • S • O Cl • I • Cl
1st Last • • •
x x x • •
n = 2 O Cl
Balmer series
1st Last
n = 1
Lyman series
360
12 Answers.indd 360 3/26/18 4:06 PM