Question

question 1.9
which of the following compounds contains the lead(iii) ion?
pbo
pb₂o
pb₂s
b and c only

Explanation:

Step1: Determine the charge of oxygen and sulfur

Oxygen (\(O\)) has a charge of \(- 2\) and sulfur (\(S\)) has a charge of \(- 2\) in these compounds.

Step2: Calculate the charge of lead (\(Pb\)) in \(PbO\)

Let the charge of \(Pb\) be \(x\). Using the formula for charge balance in a compound (\(x + (-2)=0\)), we get \(x = + 2\). So \(Pb\) has a \(+2\) charge (lead (II)) in \(PbO\).

Step3: Calculate the charge of lead (\(Pb\)) in \(Pb_{2}O\)

Let the charge of \(Pb\) be \(y\). Using charge balance (\(2y+(-2) = 0\)), we solve \(2y=2\), so \(y = + 1\). This is not lead (II).

Step4: Calculate the charge of lead (\(Pb\)) in \(Pb_{2}S\)

Let the charge of \(Pb\) be \(z\). Using charge balance (\(2z+(-2)=0\)), we solve \(2z = 2\), so \(z=+1\). This is not lead (II).

Since none of the options \(PbO\), \(Pb_{2}O\), \(Pb_{2}S\) actually contain the lead (II) ion (assuming there might be a mis - understanding in the problem setup, but based on standard charge - balancing calculations):

Answer:

None of the options are correct as per the charge - balancing calculations for lead (II) ion. If we assume a typo and consider common lead (II) compounds:

• For \(PbO\): Let \(Pb\) charge be \(x\), \(x+(-2) = 0\), \(x = + 2\) (lead (II) oxide). But if the problem had a correct set of options (maybe \(PbCl_{2}\) etc. is missing), but with the given options, there is an error. If we force - fit (assuming wrong sub - scripts in problem):

If we assume \(Pb_{2}O\) was \(PbO\) (a common lead (II) compound) then \(PbO\) contains lead (II) ion. But as per the given options in the problem (with the formulas as \(PbO\), \(Pb_{2}O\), \(Pb_{2}S\)) there is no correct answer. If we assume the problem intended \(PbO\) (lead (II) oxide) as the answer (ignoring the \(Pb_{2}O\) and \(Pb_{2}S\) incorrectness in terms of lead (II) based on standard formulas), then the answer is \(PbO\).