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just olya [345]
4 years ago
14

Problem PageQuestion Suppose a rhodium atom in the oxidation state formed a complex with one iodide anion and five ammonia molec

ules. Write the chemical formula of this complex.

Chemistry
2 answers:
Angelina_Jolie [31]4 years ago
5 0

Answer:

It forms an octahedral complex

Rh[NH3]5I

Explanation:

Or see structure attached below

Kamila [148]4 years ago
3 0

Answer:

[ Rhl(NH₃)₅]²⁺

Explanation:

Symbol of Rhodium atom = Rh

charge of Rhodium in complex = +3

charge on Iodine in complex = -1

Charge on NH₃= 0 (neutral)

Therefore, the total charge on the complex is given as follows;

Charge on complex= +3-1 + 0

                                        = +2

Therefore, the required formula of the complex = [ Rhl(NH₃)₅]²⁺

[Note: Inside the co-ordination complex,metal is written first and followed by negative and neutral ligands respectively and charge on complex is written on the right top of the complex. therefore, the chemical formular of the given complex is given as follows; [ Rhl(NH₃)₅]²⁺

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Determine the freezing point of a solution which contains 0.31 mol of sucrose in 175 g of water. kf = 1.86ï°c/m
barxatty [35]
Freezing point depression is a colligative property.

The formula that rules it is:

ΔT f = Kf * m

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And kf = 1.86 °C/m

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8 0
4 years ago
Which of elements would you expect to form the following with copper? Substitutional solid solution with complete solubility Sub
Kipish [7]

Answer:

Element Atomic Radius (nm) Crystal Structure Electronegativity Valence

Cu 0.1278 FCC 1.9 +2

C 0.071

H 0.046

O 0.060

Ag 0.1445 FCC 1.9 +1

Al 0.1431 FCC 1.5 +3

Co 0.1253 HCP 1.8 +2

Cr 0.1249 BCC 1.6 +3

Fe 0.1241 BCC 1.8 +2

Ni 0.1246 FCC 1.8 +2

Pd 0.1376 FCC 2.2 +2

Pt 0.1387 FCC 2.2 +2

Zn 0.1332 HCP 1.6 +2

The answers to the question are

(a) Copper will form substitutional solid solution with complete solubility with

Ni due to their very similar size, equal valency and similar electronegativities as well as Pd, and Pt but less likely with aluminium due to size and valency difference

(b) Copper will form substitutional solid solution of incomplete solubility with Ag, Al, Co, Cr, Fe, Zn

(c) Copper will form interstitial solid solution with

C, H, O due to the large difference between the size of copper atomic radius and the smaller atomic radii of C, H and O

Explanation:

Substitutional solid solution rules

the following are the Hume-Rothery rules for substitutional solid solutions:

1. Less than 15%difference between the atomic radius of the solute and solvent:

2. Similarity in the crystal structures of solute and solvent;

3. The valency of the solvent and solute must be similar before they can be said to be completely soluble in each other.

A lower valency metal is more likely to dissolve in one higher valency.

4. The electronegativities of the solute and solvent should be similar a wide variation in electronegativity will lead to the formation of intermetallic compounds rather than a solid solutions.

3 0
3 years ago
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