Question

Identify the formula of Bis(ethylenediamine)dithiocyanatoplatinum(IV)ion from following.

• A. \([\text{Pt(en)}_2(\text{NCS})]^{2+}\)
• B. \(\text{Pt}[(\text{en})_2(\text{SCN})_2]^{4+}\)
• C. \([\text{Pt(en)}_2(\text{SCN})_2]^{2+}\)
• D. \(\text{Pt}[(\text{en})_3(\text{SCN})_2]^{4+}\)

Hint:

Add the charges: Pt(IV) plus neutral en plus two SCN- gives +2.

Answer 1

The Correct Option is C

Step 1: Break the name into parts:
Ligands: ethylenediamine (en), neutral, number 2 (bis). Thiocyanato ($\text{SCN}^-$), number 2 (di). Metal: Pt in +4 state.

Step 2: Add up the charges:
$+4$ from platinum, 0 from two en, and $-2$ from two thiocyanates gives $+2$.

Step 3: Write it:
Everything goes inside one square bracket with the charge outside: $[\text{Pt(en)}_2(\text{SCN})_2]^{2+}$. Only option (C) has two en, two SCN and a 2+ charge.

Final Answer:
Option (C) gives the correct formula. \[ \boxed{[\text{Pt(en)}_2(\text{SCN})_2]^{2+}} \]