Question:medium

Write the formulas of: (a) Mercury(I) tetrathiocyanato-S-cobaltate(III)
(b) Dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate

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Mercury(I) always appears as $Hg_2^{2+}$ (diatomic ion). Ethane-1,2-diamine = en, a bidentate ligand.
Updated On: Jul 23, 2026
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Solution and Explanation

Step 1: Decoding part (a).
Mercury(I) cation is $Hg_2^{2+}$. The complex anion is tetrathiocyanato-S-cobaltate(III): four $SCN^-$ ligands bonded through sulfur, cobalt in $+3$ state. Anion formula: $[Co(SCN)_4]^-$ (charge $= +3 - 4 = -1$).
Step 2: Formula for (a).
One $Hg_2^{2+}$ balances two $[Co(SCN)_4]^-$... wait, charge balance: $Hg_2^{2+}$ ($+2$) needs two $[Co(SCN)_4]^-$ ($2 \times -1 = -2$). Formula: $Hg_2[Co(SCN)_4]$.
Step 3: Decoding part (b).
Dichloridobis(ethane-1,2-diamine)platinum(IV): two $Cl^-$ and two bidentate $en$ ligands around Pt(IV). Cation charge $= +4 - 2 = +2$. Two nitrate ($NO_3^-$) counter ions needed.
Step 4: Formula for (b) and final answers.
(a) $Hg_2[Co(SCN)_4]$; (b) $[Pt(en)_2Cl_2](NO_3)_2$.
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