• 9701 March 2016 Paper 42 Worked Solutions | Chemistry

Question 1
Complete the diagrams to show the energies of the electrons in a carbon atom, a C+ ion and a C– ion.



Question 3
In an isolated transition metal atom the five d orbitals have the same energy. When a transition metal ion forms a tetrahedral complex the d orbitals are split into two groups of different energies.



Question 5
Some reactions of chromium ions are shown below.



Question 6
Two elements, V and W, are in adjacent groups in the Periodic Table.






Solutions provided by Kashish Varshney, India

When sound travels through air, the air particles vibrate. A graph of displacement against time for a single air particle is shown.

Question 24

When sound travels through air, the air particles vibrate. A graph of displacement against time for a single air particle is shown.

Which graph best shows how the kinetic energy of the air particle varies with time?

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q23

Solution:

Answer: D.

Kinetic energy = ½ mv²

A graph of displacement against time for a single air particle is shown. 

The gradient of the displacement-time graph gives the velocity of the air particle at that point in time. This is done by calculating the gradient of the tangent at that point.

The gradient (and hence, velocity) is found to be zero at the maximum displacement (the tangent is horizontal) and maximum when the displacement is zero (the tangent is steepest).

Note that the velocity is NOT zero when the displacement is zero. At the points where displacement = 0, we need to draw a tangent to obtain the velocity. It can clearly be observed then that the line is steepest at these points – that is, the velocity is greatest when displacement = 0.

Thus, at time = 0, T and 2T the velocity is zero and hence kinetic energy is zero. [A and C incorrect]

But, between time = 0 and T or between time = T and 2T the displacement is zero (in case) twice. So, the velocity (and kinetic energy) reaches its maximum value 2 times in each of the 2 intervals. [B is incorrect]

A potentiometer is used as shown to compare the e.m.f.s of two cells. The balance points for cells X and Y are 0.70 m and 0.90 m respectively.

Question 41

A potentiometer is used as shown to compare the e.m.f.s of two cells.

The balance points for cells X and Y are 0.70 m and 0.90 m respectively.

If the e.m.f. of cell X is 1.1 V, what is the e.m.f. of cell Y?

A 0.69 V          B 0.86 V          C 0.99 V          D 1.4 V

Reference: Past Exam Paper – November 2007 Paper 1 Q34

Solution:

Answer: D.

At the balance length, the p.d. across the wire is equal to the e.m.f. of the cell.

p.d. ∝ L

For cell X: 1.1 V ∝ 0.70 m     (1)

For cell Y: emf ∝ 0.90 m        (2)

Divide (2) by (1),

emf / 1.1 = 0.90 / 0.70

emf = (0.90/0.70) × 1.1 = 1.4 V

Monochromatic light of wavelength λ is incident on two narrow slits S1 and S2, a small distance apart.

Question 34

Monochromatic light of wavelength λ is incident on two narrow slits S1 and S2, a small distance apart. A series of bright and dark fringes are observed on a screen a long distance away from the slits.

The n th dark fringe from the central bright fringe is observed at point P on the screen.

Which equation is correct for all positive values of n?

A S2P – S1P = nλ / 2

B S2P – S1P = nλ

C S2P – S1P = (n – ½)λ

D S2P – S1P = (n + ½)λ

Reference: Past Exam Paper – June 2017 Paper 13 Q28

Solution:

Answer: C. 

Dark fringes are formed when destructive interference occurs.

For destructive interference, the path difference should be 0.5λ, 1.5λ, 2.5λ, 3.5λ … [B is incorrect]

Consider option A.

Path difference = nλ /2

When n = 1, path difference = ½ λ

When n = 2, path difference = 2 λ / 2 = λ      [A is incorrect]

Consider option D.

When n = 1, path difference = (1 + ½) λ = 3λ / 2

This equation suggest that the first dark fringe occurs at 3λ / 2. [D is incorrect]

Consider option C.

When n = 1, path difference = (1 – ½) λ = λ / 2

This equation suggest that the path difference for the first dark fringe is equal to λ / 2 (= 0.5λ), which is correct.