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Chapter 21: Problem 83

Give formulas for the following. a. hexakis(pyridine)cobalt(III) chloride b. pentaammineiodochromium(III) iodide c. tris(ethylenediamine)nickel(II) bromide d. potassium tetracyanonickelate(II) e. tetraamminedichloroplatinum(IV) tetrachloroplatinate(II)

Short Answer

Expert verified
The formulas for the given coordination compounds are: a. \[ [Co(C5H5N)_6]Cl_3 \] b. \[ [Cr(NH_3)_5I]I_2 \] c. \[ [Ni(en)_3]Br_2 \] d. \[ K_2[Ni(CN)_4] \] e. \[ [Pt(NH_3)_4Cl_2][PtCl_4] \]

Step by step solution

01

Identify the central metal ion and its oxidation state

The central metal ion is cobalt(III), which means cobalt with an oxidation state of +3 (Co^3+).
02

Identify the ligands and their charges

The ligand is hexakis(pyridine), which means six pyridine (C5H5N) ligands. Pyridine is a neutral ligand.
03

Write the formula

As there are six pyridine ligands, the formula for the complex ion will be [Co(C5H5N)6]^3+. The counter ion is chloride, with a -1 charge (Cl^-). To balance the charges, we need three Cl^-. Therefore, the final formula is: \[ [Co(C5H5N)_6]Cl_3 \] #b. pentaammineiodochromium(III) iodide#
04

Identify the central metal ion and its oxidation state

The central metal ion is chromium(III), which means chromium with an oxidation state of +3 (Cr^3+).
05

Identify the ligands and their charges

The ligands are pentaammine (5 NH3) and iodo (I^-). Ammine (NH3) is a neutral ligand, while iodide has a -1 charge.
06

Write the formula

Since the complex ion has an overall charge of +2 (Cr^3+ - I^- = +2), the formula will be [Cr(NH3)5I]^2+. The counter ion is iodide (I^-). To balance the charges, we need two I^-. Therefore, the final formula is: \[ [Cr(NH_3)_5I]I_2 \] #c. tris(ethylenediamine)nickel(II) bromide#
07

Identify the central metal ion and its oxidation state

The central metal ion is nickel(II), which means nickel with an oxidation state of +2 (Ni^2+).
08

Identify the ligands and their charges

The ligand is tris(ethylenediamine), which means three ethylenediamine (en) ligands. Ethylenediamine is a neutral ligand.
09

Write the formula

As there are three ethylenediamine ligands, the formula for the complex ion will be [Ni(en)3]^2+. The counter ion is bromide, with a -1 charge (Br^-). To balance the charges, we need two Br^-. Therefore, the final formula is: \[ [Ni(en)_3]Br_2 \] #d. potassium tetracyanonickelate(II)#
10

Identify the central metal ion and its oxidation state

The central metal ion is nickel(II), which means nickel with an oxidation state of +2 (Ni^2+).
11

Identify the ligands and their charges

The ligand is tetracyano, which means four cyanide (CN^-) ligands. Cyanide has a -1 charge.
12

Write the formula

Since there are four cyanide ligands, the formula for the complex ion will be [Ni(CN)4]^2-. The counter ion is potassium (K^+). To balance the charges, we need two K^+. Therefore, the final formula is: \[ K_2[Ni(CN)_4] \] #e. tetraamminedichloroplatinum(IV) tetrachloroplatinate(II)#
13

Identify the central metal ions and their oxidation states

The central metal ions are platinum(IV), with an oxidation state of +4 (Pt^4+) and platinum(II), with an oxidation state of +2 (Pt^2+).
14

Identify the ligands and their charges

For the first complex ion, the ligands are tetraammine (4 NH3) and two chlorides (2 Cl^-). Ammine is a neutral ligand, while chloride has a -1 charge. For the second complex ion, the ligands are four chlorides (4 Cl^-), which has a -1 charge.
15

Write the formulas

The formula for the first complex ion will be [Pt(NH3)4Cl2]^2+. The formula for the second complex ion is [PtCl4]^2-. Since these two ions have complementary charges, they balance each other out, and the final formula is: \[ [Pt(NH_3)_4Cl_2][PtCl_4] \]

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Most popular questions from this chapter

The compound $\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6} \mathrm{Cl}_{2}$ is green, whereas \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{2}\) is violet. Predict the predominant color of light absorbed by each compound. Which compound absorbs light with the shorter wavelength? Predict in which compound \(\Delta\) is greater and whether \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{NH}_{3}\) is the stronger field ligand. Do your conclusions agree with the spectrochemical series?

Carbon monoxide is toxic because it binds more strongly to iron in hemoglobin \((\mathrm{Hb})\) than does \(\mathrm{O}_{2} .\) Consider the following reactions and approximate standard free energy changes: $$\begin{aligned} \mathrm{Hb}+\mathrm{O}_{2} \longrightarrow \mathrm{HbO}_{2} & \Delta G^{\circ}=-70 \mathrm{kJ} \\ \mathrm{Hb}+\mathrm{CO} \longrightarrow \mathrm{HbCO} & \Delta G^{\circ}=-80 \mathrm{kJ} \end{aligned}$$ Using these data, estimate the equilibrium constant value at $25^{\circ} \mathrm{C}$ for the following reaction: $$\mathrm{HbO}_{2}(a q)+\mathrm{CO}(g) \rightleftharpoons \mathrm{HbCO}(a q)+\mathrm{O}_{2}(g)$$

The \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\) ion is diamagnetic, but \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) is para-magnetic. Explain.

When aqueous KI is added gradually to mercury(II) nitrate, an orange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. $\left(\text {Hint} : \mathrm{Hg}^{2+} \text { reacts with } \mathrm{I}^{-} \text { to form } \mathrm{HgI}_{4}^{2-} .\right)\( Would you expect \)\mathrm{Hg} \mathrm{I}_{4}^{2-}$ to form colored solutions? Explain.

When 6\(M\) ammonia is added gradually to aqueous copper(II) nitrate, a white precipitate forms. The precipitate dissolves as more 6\(M\) ammonia is added. Write balanced equations to explain these observations. [Hint: \(\mathrm{Cu}^{2+}\) reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+} . ]\)
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