The chemical formula for lead(II) nitrite is: Pb(NO2) 2 How many oxygen atoms are in each formula unit of lead (II) nitrite?
The correct answer and explanation is :
The chemical formula for lead(II) nitrite is Pb(NO₂)₂. To determine how many oxygen atoms are in each formula unit of lead(II) nitrite, we need to break down the formula and analyze its composition.
- Formula Breakdown:
- Pb represents lead, and its oxidation state is +2 in lead(II) nitrite.
- (NO₂)₂ represents two nitrite ions. Each nitrite ion (NO₂⁻) consists of one nitrogen (N) atom bonded to two oxygen (O) atoms.
- Oxygen Atoms in Each Nitrite Ion:
- Each NO₂ ion contains 2 oxygen atoms. Since there are two nitrite ions in the formula (NO₂)₂, the total number of oxygen atoms is: [
2 \, \text{(oxygen atoms per NO₂)} \times 2 \, \text{(NO₂ ions)} = 4 \, \text{oxygen atoms}
]
Thus, in each formula unit of lead(II) nitrite, there are 4 oxygen atoms.
Explanation:
Lead(II) nitrite, Pb(NO₂)₂, consists of one lead ion (Pb²⁺) and two nitrite ions (NO₂⁻). The nitrite ion (NO₂⁻) is a polyatomic ion that has one nitrogen atom (N) and two oxygen atoms (O) bonded to it. Each nitrite ion thus contains two oxygen atoms. Since there are two nitrite ions in the formula, the total number of oxygen atoms in the formula unit is calculated by multiplying the number of oxygen atoms in one nitrite ion by the number of nitrite ions.
Therefore, the total number of oxygen atoms in each formula unit of lead(II) nitrite (Pb(NO₂)₂) is 4. This is the number of oxygen atoms present in each molecular or empirical unit of the compound.