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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

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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

Which of the following crystal field diagram(s) is(are) correct for the complex given? a. \(\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) (tetrahedral) b. \(\operatorname{Mn}(\mathrm{CN})_{6}^{3-}\) (strong field) c. \(\mathrm{Ni}(\mathrm{CN})_{4}^{2-}\) (square planar, diamagnetic)

Consider the complex ions \(\mathrm{Co}\left(\mathrm{NH}_{3}\right) 6^{3+}, \mathrm{Co}(\mathrm{CN})_{6}^{3-},\) and \(\mathrm{CoF}_{6}^{3-} .\) The wavelengths of absorbed electromagnetic radiation for these compounds (in no specific order) are \(770 \mathrm{nm},\) \(440 \mathrm{nm},\) and 290 \(\mathrm{nm} .\) Match the complex ion to the wave- length of absorbed electromagnetic radiation.

The complex ion NiCl \(_{4}^{2-}\) has two unpaired electrons, whereas \(\mathrm{Ni}(\mathrm{CN})_{4}^{2-}\) is diamagnetic. Propose structures for these two complex ions.

Until the discoveries of Alfred Werner, it was thought that carbon had to be present in a compound for it to be optically active. Werner prepared the following compound containing \(\mathrm{OH}^{-}\) ions as bridging groups and separated the optical isomers. a. Draw structures of the two optically active isomers of this compound. b. What are the oxidation states of the cobalt ions? c. How many unpaired electrons are present if the complex is the low-spin case?

When an aqueous solution of KCN is added to a solution containing \(\mathrm{Ni}^{2+}\) ions, a precipitate forms, which redissolves on addition of more \(\mathrm{KCN}\) solution. Write reactions describing what happens in this solution. [Hint: \(\mathrm{CN}^{-}\) is a Bronsted-Lowry base \(\left(K_{\mathrm{b}} \approx 10^{-5}\right)\) and a Lewis base.]

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