Question

the correct formula for zirconium(iv) carbonate is________ and the oxidation state of zirconium in the compound is_
zr₄co₃: +2
zr(co₃)₂: +4
zr(co₃)₄: +2
zr₂(co₃)₄: +4
zr₄co₃: +4

Explanation:

Step1: Analyze Zirconium(IV) Charge

Zirconium(IV) means Zr has a +4 oxidation state (given by the (IV) in the name). So we can eliminate options with +2 for Zr's oxidation state (first, third options).

Step2: Analyze Carbonate Ion Charge

Carbonate ion is $\ce{CO_3^{2-}}$. Let's check the formula formation. For a compound, total positive charge = total negative charge.

• For $\ce{Zr(CO_3)_2}$: Zr is +4, each $\ce{CO_3^{2-}}$ is -2. Two carbonates: $2\times(-2) = -4$. Zr is +4. So $+4 + (-4) = 0$, which is neutral.
• For $\ce{Zr_2(CO_3)_4}$: Zr is +4 (two Zr: $2\times4 = 8$), four carbonates: $4\times(-2) = -8$. $8 + (-8) = 0$, but the formula can be simplified (divide by 2: $\ce{Zr(CO_3)_2}$), so the simpler formula is preferred.
• The first and fifth options have incorrect formulas (e.g., $\ce{Zr_4CO_3}$ doesn't balance charges). The third option has Zr(CO₃)₄ with Zr +2 (wrong, since it's Zirconium(IV)) and charge imbalance (Zr +2, four carbonates: $4\times(-2) = -8$; $2 - 8 = -6

eq 0$).

Step3: Confirm the Correct Option

The second option $\ce{Zr(CO_3)_2}$ has Zr with +4 (matches Zirconium(IV)) and balanced charges (Zr +4, two $\ce{CO_3^{2-}}$: $2\times(-2) = -4$; $4 - 4 = 0$).

Answer:

B. $\ce{Zr(CO_3)_2}$; +4