Which molecules have an overall dipole moment? 1 carbon monoxide, CO 2 phosphine, PH3 3 carbon dioxide, CO2

Question 31 [Chemical Bonding > Covalent bonding and co-ordinate (dative covalent) bonding including shapes of simple molecules]

Which molecules have an overall dipole moment?

1 carbon monoxide, CO

2 phosphine, PH3

3 carbon dioxide, CO2

Reference: Past Exam Paper – Chemistry (9701) March 2016 Paper 12 Q31

Solution:

  Answer: B

   Firstly, to begin elimination, remove CO₂. The molecule is linear AND symmetrical. Therefore the dipoles of the otherwise polar C=O cancel out.

Coming to Carbon Monoxide, CO, it contains only one polar carbon-oxygen bond and therefore is a polar molecule.

PH₃ was the one molecule that confused the majority of the candidates. DO NOT make the mistake of assuming that all the three bonds are equally spaced, at 120°. If that were the case then the molecule would have been symmetrical and ultimately non polar. However the first protocol in dealing with an unknown compound is, draw it. You will notice that there is still a lone pair of electrons on the central P atom; this would distort the symmetrical shape and thus, PH₃ has an overall dipole.

After black and white photographic film has been developed, unreacted silver bromide is removed by reaction with sodium thiosulfate.

Question 19 [Group 7 > reactions of halide ions]

After black and white photographic film has been developed, unreacted silver bromide is removed by reaction with sodium thiosulfate.

AgBr + 2Na2S2O3 → 4Na⁺ + Br ^– + [Ag(S2O3)2]^3–

What is the function of the thiosulfate ion?

A to make the silver ions soluble

B to oxidise the silver ions

C to reduce the bromine

D to reduce the silver ions

Reference: Past Exam Paper – Chemistry (9701) March 2016 Paper 12 Q19

Solution:

     Answer:A
 

     Firstly, if confused work out oxidation changes for both silver and bromine to see if oxidation or reduction has taken place. Neither oxidation or reduction takes place and so automatically the only feasible answer is A.

   Secondly, AgBr is a precipitate, which you should be able to recall from qualitative tests and other chapters from inorganic chemistry. We see that Ag is then incorporated into this complex on the products side of the equation. The formula is enclosed in square brackets and we also observe a charge of -3. Automatically we see that this complex is in ionic form and therefore silver has gone from an insoluble solid to a soluble ion.

A rod PQ is attached at P to a vertical wall, as shown in Fig. 3.1. The length of the rod is 1.60 m. The weight W of the rod acts 0.64 m from P.

Question 42

A rod PQ is attached at P to a vertical wall, as shown in Fig. 3.1.

The length of the rod is 1.60 m. The weight W of the rod acts 0.64 m from P. The rod is kept horizontal and in equilibrium by a wire attached to Q and to the wall at R. The wire provides a force F on the rod of 44 N at 30° to the horizontal.

(a) Determine

(i) the vertical component of F,          [1]

(ii) the horizontal component of F.     [1]

(b) By taking moments about P, determine the weight W of the rod.           [2]

(c) Explain why the wall must exert a force on the rod at P.             [1]

(d) On Fig. 3.1, draw an arrow to represent the force acting on the rod at P. Label your arrow with the letter S.             [1]

Reference: Past Exam Paper – June 2015 Paper 22 Q3

Solution:

(a)

(i) Vertical component = 44 sin 30° = 22 N

(ii) Horizontal component = 44 cos 30° = 38(.1) N

(b)

{The vertical component of F acts at a distance of (0.64 + 0.96 =) 1.60m from the pivot.

Clockwise moment = Anti-clockwise moment}

W × 0.64 = 22 × 1.60

W = 55 N

(c) F has a horizontal component (not balanced by W)

{The horizontal component of F acts on the wall and this is not balanced by W. From Newton’s 3^rd law, there should be an equal and opposite force exerted by the wall on the rod.}

OR F has 38 N acting horizontally

OR 38 N acts on wall

OR vertical component of F does not balance W

OR F and W do not make a closed triangle of forces

(d) Line from P in the direction towards the point on wire vertically above W and direction up

{This force, along with F and W should form a system that is in equilibrium. For equilibrium, the resultant force and the resultant moment should be zero. If all the 3 forces pass through the same point, the resultant moment would be zero.

Moment = Force × perpendicular distance from line of action of force to pivot.

In this case, the pivot is that point where all the 3 forces pass. As they pass on the point, the ‘distance …’ is zero, and thus, the moment at that point is zero.}