Question

question 4 (0.5 points)
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the correct names of $\ce{pbcl_{4}}$ and $\ce{nibr_{2}}$ are:

1. iron(iv) chloride and nitrogen bromide
2. lead(iv)chloride and nickel(ii)bromide
3. lead chloride (iv) and nitrogen bromide
4. iron chloride and nickel bromide (ii)

Explanation:

• For \(PbCl_4\):
• \(Pb\) is the chemical symbol for lead.
• In \(PbCl_4\), chlorine (\(Cl\)) has an oxidation state of \(- 1\). Let the oxidation state of \(Pb\) be \(x\). Using the formula \(x+4\times(-1) = 0\) (since the compound is neutral), we get \(x = + 4\). So it is lead(IV) chloride.
• For \(NiBr_2\):
• \(Ni\) is the chemical symbol for nickel.
• Bromine (\(Br\)) has an oxidation state of \(-1\). Let the oxidation state of \(Ni\) be \(y\). Using the formula \(y + 2\times(-1)=0\), we get \(y=+2\). So it is nickel(II) bromide.

Answer:

1. Lead(IV)chloride and Nickel(II)Bromide