Author: Luke
Last saved by: Luke Edwards-Stuart

Paddington Academy

TOPIC 1 ASSESSED HOMEWORK MARK SCHEME
 
1.	      (i)          = 10.8
OR ratio 10:11 = 1:4 OR 20:80 etc
Allow idea that there are 5 × 0.2 divisions between 10 and 11.
1
abundance of ^10B is 20(%)
OR
 = 10.8
10x + 1100 − 11x = 1080
∴ x = 1100 − 1080 = 20%
Correct answer scores M1 and M2.
1
(ii)     Same number of electrons (in outer shell or orbital)
Ignore electrons determine chemical properties.
Same electronic configuration / arrangement
Ignore protons unless wrong.
1
[3]

2.	(a)    (Total number of) protons and neutrons (in nucleus of atom)
(number of) nucleons
1
(b)     Zn
Do not allow Zn^−1 or Zn^+1 or ZN
Ignore numbers
1
(c)         m / z
Allow mass / charge
1
(relative) abundance / (relative) intensity
QoL
Allow M1 + M2 in any order
1

(d)    (i)       
M1 = topline
1
M2 = ÷ 4
1
= 207.3
Only
207.3 = 3 marks
1
(ii)     Lead / Pb
Not PB
1
(iii)    Same number of electrons (in outer shell) / same electronic configuration
Ignore electrons determine chemical properties
Ignore reference to p and n if correct
Penalise if incorrect
1
[9]

   3.       (a)     Average/mean mass of (1) atom(s) (of an element) 
1/12 mass of one atom of ^12C
1
If moles and atoms mixes Max = 1
1
OR
(Average) mass of one mole of atoms 
1/12 mass of one mole of ^12C
OR
(Weighted) average mass of all the isotopes
1/12 mass of one atom of ^12C
OR
Average mass of an atom/isotope compared to C-12 on a scale in
which an atom of C-12 has a mass of 12
This expression = 2 marks

(b)     d block
Allow 3d/D
Other numbers lose M1
Ignore transition metals
1
[Ar] 3d^24s^2
1
Can be written in full
Allow subscripts
3d^2 and 4s^2 can be in either order
27
1
(c)     
(= 1550)
1
(or ∑ their abundances)
If one graph reading error lose M1 and allow consequential M2 and M3.
If 2 GR errors penalise M1 and M2 but allow consequential M3
If not 17 or ∑ their abundances lose M2 and M3
1
= 91.2
91.2 = 3 marks provided working shown.
1
Zr/Zirconium
M4 -allow nearest consequential element from M3
accept Zr in any circumstance
1
(d)     High voltage supply
1
Removes electron(s) (to form ions)
1
Z^+ = 90 has shortest TOF
If not 90 lose M3 and M4
If charge is wrong on 90 isotope lose M3 only
Accept any symbol in place of Z
1
since lowest mass/lowest m/z
Allow lightest
1

(e)     (ions hit detector and) cause current/(ions) accept electrons/cause
electron flow
QWC
1
bigger current = more of that isotope/current proportional to abundance
Implication that current depends on the number of ions
1
[15]

4.        (a)    37
These answers only.
Allow answers in words.
1
48
Ignore any sum(s) shown to work out the answers.
1
(b)          Dissolved in volatile solvent/passed through hollow needle
1
Subjected to high voltage
1
(c)     (i)     s / block s / group s
Only
1
(ii)     1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^1
Allow 3d^10 before 4s^2
Allow in any order.
1
(d)     (85 × 2.5) + 87 ×1            3.5
M1 is for top line
1
1
= 85.6
Only
1
OR
(58 × 5) + 87 ×2            7
M1^85Rb 71.4% and ^87Rb 28.6%
M2 divide by 100
1
1
85.6
M3 = 85.6
1

(e)     Detector
Mark independently
Allow detection (plate).
1
Current / digital pulses / electrical signal related to abundance
Not electrical charge.
1
 [11]

5.	(a)     N^3- / N^–3
1
      (b)     F^–/ fluoride
Ignore fluorine/F
Penalise Fl
1
[2]

6.              H ^– = 1s^2 or 1s_2
1
[1] 

7.          (a)     Na(g) → Na^+(g) + e^–
OR Na(g) + e^– → Na^+(g) + 2e^–
(‑) on electron not essential
equation (1)
state symbols (1)
Ignore state symbols on electrons
2
(b)     Trend : Increases (1)
          Explanation : Increased nuclear charge or proton number (1)
Stronger attraction (between nucleus and (outer) e^–) (1)
Trend wrong
Allow M2 only if M3 correct (con)
3
(c)     How values deviate from trend: (both values) too low (1)
Explanation for Al: e^– removed from (3) p (1)
                             e^– or orbital is higher in energy or better
                               shielded than (3)s
                             or p electron is shielded by 3s electrons (1)
Allow e^– is further away
Mark independently
Explanation for S: e^– removed from (3)p electron pair (1)
                   repulsion between paired e^– (reduces energy required) (1)
Mark separately
If deviation wrong allow M2 and M4
If M3 and / or M5 right (con)
If used ‘d’ rather than ‘p’ orbital - lose M2 + M4 but may get M3, M5 (explanation marks)
5
[10]

 

8.          (a)     Heat / enthalpy / energy for removal of one electron (1)

from a gaseous atom (1)
can score in an equation
must have first mark to score the second
2
(b)      (i)      2 (1)
(ii)     Two elements (or Na / Mg) before the drop (in energy) to Al (1)
(iii)     ionisation energy of Al < that for Mg (1)
(iv)    fall in energy from P to S (1)
or discontinuity in trend
From Al to P there are 3 additional electrons (1)
or three elements
For second mark idea of block of 3 elements
5
[7] 

9.	(a)     (i)      Higher than P
1
(ii)     1s^2 2s^2 2p^6 3s^1
Allow any order
1

(iii)    Al^+(g) + e ^(−)   Al^2+(g) + 2e^(−)
OR
Al^+(g)   Al^2+(g) + e^(−)
OR
Al^+(g) − e^(−)   Al^2+(g)
1
(iv)    Electron in Si (removed from) (3)p orbital / electron (removed) 
from higher energy orbital or sub-shell / electron in silicon is more shielded
Accept converse arguments relating to Al
Penalise incorrect p-orbital
1
(b)     Sodium / Na
Allow Na^+
1
Electron (removed) from the 2^nd shell / 2p (orbital)
M2 is dependent on M1
Allow electron from shell nearer the nucleus (so more attraction)
1
(c)     Silicon / Si
Not SI
1
(d)     Heat or energy needed to overcome the attraction between the (negative) electron and the (positive) nucleus or protons
Not breaking bonds
QoL
Or words to that effect eg electron promoted to higher energy level (infinity) so energy must be supplied
1
[8]
10.          (a)     2s^2 2p^6;
If ignored the 1s^2 given and written 1s^22s^22p^6 mark as correct
Allow capitals and subscripts
1
(b)     (i)      Na^+(g) → Na^2+ (g) + e^(–);
One mark for equation and one mark for state symbols
         Na^+(g) + e^(–) → Na^2+ (g) + 2e^(–);
M2 dependent on M1
Allow Na^+(g) – e^(–) → Na(g)
Allow X^+(g) → X^2+ (g) + e = 1 mark
2
(ii)     Na^(2+) requires loss of e^– from a 2(p) orbital or 2^nd energy level or
2^nd shell and Mg^(2+) requires loss of e^– from a 3(s) orbital or 3^rd
energy level or 3^rd shell / Na^(2+) loses e from a lower (energy)
orbital/ or vice versa;
Not from 3p
1
         Less shielding (in Na);
Or vice versa for Mg
1
         e^(–) closer to nucleus/ more attraction (of electron to nucleus) (in Na);
M3 needs to be comparative
1
(iii)     Aluminium /Al;
1
(c)     Decreases;
If not decreases CE = 0
If blank, mark on
1
          Increasing nuclear charge/ increasing number of protons;
1
          Electrons in same shell or level/ same shielding/ similar shielding;
1
 [10]

11.	D
[1]
12.	D
[1]
13.		D
[1]
14.		A
[1]