Question

The Complex hexamine platinum (IV) chloride will give ___ number of ions on ionization

• A. 3
• B. 5
• C. 2
• D. 4

Answer 1

The Correct Option is B

The complex hexamine platinum (IV) chloride, $\text{[Pt(NH}_3\text{)}_6\text{]Cl}_4$, consists of a central platinum (IV) ion coordinated with six ammonia ligands (NH₃) and four chloride ions (Cl^-) as counterions.
When this complex dissociates in solution, it will release its constituent ions. The complex itself contains one platinum ion, six ammonia ligands, and four chloride ions.
Upon ionization, the hexamine platinum (IV) chloride complex will generate two types of ions:
$\text{[Pt(NH}_3\text{)}_6]^{4+}$ ion: This is the cationic part of the complex, consisting of the platinum (IV) ion coordinated with six ammonia ligands. It carries a charge of +4.
Four Cl^- ions: These are the anionic counterions in the complex. Each chloride ion carries a charge of -1.
Therefore, the hexamine platinum (IV) chloride complex will give a total of five ions upon ionization: one $\text{[Pt(NH}_3\text{)}_6]^{4+}$ cation and four Cl^- anions.
Hence, the correct answer is (B) 5.

Answer 2

Correct answer: 5

The complex is hexamine platinum (IV) chloride, which has the formula: 

$$[Pt(NH_3)_6]Cl_4$$

Upon ionization in water, it dissociates as:

$$[Pt(NH_3)_6]Cl_4 \rightarrow [Pt(NH_3)_6]^{4+} + 4Cl^-$$

This gives 1 complex cation and 4 chloride anions, totaling:

1 + 4 = 5 ions

Answer 3

The complex hexamine platinum(IV) chloride has the formula $[Pt(NH_3)_6]Cl_4$.

When this complex ionizes in solution, it dissociates into one complex cation and four chloride anions:

$[Pt(NH_3)_6]Cl_4 \rightarrow [Pt(NH_3)_6]^{4+} + 4Cl^-$

Therefore, the total number of ions produced is 1 (complex cation) + 4 (chloride anions) = 5.

Thus, the complex gives 5 ions on ionization.

Answer:

5