what is the formula for dicyanobis (ethylenediamine) zirconium (IV) nitrate?
Thanks!!!
(CN)2(en)2Zr(IV)NO3 which I wrote just following your words. Then I would rearrange this to
[Zr(en)2(CN)2](NO3)2
Check me out on this.
What is the formula for dicyanobis(ethylenediamine)zirconium(IV) nitrate?
a. Zr[(CN)2(NO3)2(en)2]
b. [Zr(CN)2(NO3)2(en)2]
c. (NO3)2[Zr(CN)2(en)2]
d. [Zr(CN)2(en)2](NO3)2
e. [Zr(NO3)2(en)2](CN)2
Sorry, I think the multiple choice part didn't show up.
It's D isn't it? That's the equivalent of my response.
To determine the formula for dicyanobis(ethylenediamine)zirconium(IV) nitrate, let's break it down step by step.
1. Let's start with the central metal ion, which is zirconium (IV) (Zr4+). This tells us that zirconium has a charge of +4.
2. The ligand "dicyanobis(ethylenediamine)" indicates that the ligand has two cyanide (CN-) groups and two ethylenediamine (en) groups. The ethylenediamine ligand (en) consists of two nitrogen atoms bonded to two separate ethylene groups.
3. The coordination number of zirconium in the complex is 6, which means it can form six bonds with the surrounding ligands.
4. The nitrate anion (NO3-) is the counterion of the complex. It carries a charge of -1.
Putting it all together, the formula for dicyanobis(ethylenediamine)zirconium(IV) nitrate is [Zr(en)2(CN)2](NO3)2.