All categories

2 years ago

The compound Rh(CO)(H)(PH3)2 forms cis and trans isomers. Use this information

1 answer:

5 0

The geometry of a complex is determined by the number of ligands that surround the central atom/ion. There are four ligands in the complex. The coordination number of the complex is 4. The oxidation number of Rh is zero. The preferred geometrical shape of this complex is square planar.

Complexes are formed when Lewis bases called ligands become attached to a central metal atom or ion. The coordination number of the complex refers to the number of ligands that surround the central metal atom/ion.

In this case, four ligands surround the central Rhodium atom in a zero oxidation state.

Since we were told that the compound exhibits cis/trans isomerism then it can not have a tetrahedral geometry. It must have a square planar geometry as tetrahedral complexes do not exhibit cis/trans isomerism.

The image of the cis- and trans- geometrical isomers of the compound is attached to this answer.

Learn more: brainly.com/question/9616145

You might be interested in

Which characteristic is given by the angular momentum quantum number

Marianna [84]

Answer: orbital shape.

Justification:

1) There are four quantum numbers to describe the electrons. These are:

i) Principal quantum number (n)

ii) Azimuthal quantum number (ℓ), also called angular momentum quantum number.

iii) Magnetic quantum number (m)

iv) Spin quantum number (s)

2) The principal quantum number tells the main energy level. It can be 1, 2, 3, 4, 5, 6, 7. It is related with the orbial size. 1 is a small orbital, 7 is a big orbital.

2) The Azimuthal quantum number (ℓ) or angular momentum quantum number may be a number between 0 and n - 1.

It tells the kind of orbital, which is its shape

The correspondence is:

0 = s orbital,

1 = p orbital,

2 = d orbital,

3 = f orbital.

3) Magnetic quantum number (m) tells the orientation. It can be from - ℓ to + ℓ

For example when ℓ = 1, the orbital is p, and the magnetic quantum number may be -1, 0, or +1, which correspond to px, py, pz: the orientation of the p orbital in the space.

4) Spin quantum number (s) can be either +1/2 or -1/2.

7 0

2 years ago

Read 2 more answers

Suppose that some FeSCN2+ is added to the above solution to shift the equilibrium. When equilibrium is re-established, the follo

kiruha [24]

Explanation:

The given reaction equation will be as follows.

$[FeSCN^{2+}] \rightleftharpoons [Fe^{3+}] + [SCN^{-}]$

Let is assume that at equilibrium the concentrations of given species are as follows.

$[Fe^{3+}] = 8.17 \times 10^{-3}$ M

$[SCN^{-}] = 8.60 \times 10^{-3}$ M

$[FeSCN^{2+}] = 6.25 \times 10^{-2}$ M

Now, first calculate the value of $K_{eq}$ as follows.

$K_{eq} = \frac{[Fe^{3+}][SCN^{-}]}{[FeSCN^{2+}]}$

= $\frac{8.17 \times 10^{-3} \times 8.60 \times 10^{-3}}{6.25 \times 10^{-2}}$

= $11.24 \times 10^{-4}$

Now, according to the concentration values at the re-established equilibrium the value for $[FeSCN^{2+}]$ will be calculated as follows.

$K_{eq} = \frac{[Fe^{3+}][SCN^{-}]}{[FeSCN^{2+}]}$

$11.24 \times 10^{-4} = \frac{8.12 \times 10^{-3} \times 7.84 \times 10^{-3}}{[FeSCN^{2+}]}$

$[FeSCN^{2+}] = 5.66 \times 10^{-2}$ M

Thus, we can conclude that the concentration of $[FeSCN^{2+}]$ in the new equilibrium mixture is $5.66 \times 10^{-2}$ M.

7 0

3 years ago

How much energy must be removed from a 94.4 g sample of benzene (molar mass= 78.11 g/mol) at 322.0 K to solidify the sample and

Kay [80]

Answer : The energy removed must be, 29.4 kJ

Explanation :

The process involved in this problem are :

$(1):C_6H_6(l)(322K)\rightarrow C_6H_6(l)(279K)\\\\(2):C_6H_6(l)(279K)\rightarrow C_6H_6(s)(279K)\\\\(3):C_6H_6(s)(279K)\rightarrow C_6H_6(s)(205K)$

The expression used will be:

$Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]$

where,

$Q$ = heat released for the reaction = ?

m = mass of benzene = 94.4 g

$c_{p,s}$ = specific heat of solid benzene = $1.51J/g^oC=1.51J/g.K$

$c_{p,l}$ = specific heat of liquid benzene = $1.73J/g^oC=1.73J/g.K$

$\Delta H_{fusion}$ = enthalpy change for fusion = $-9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g$

Now put all the given values in the above expression, we get:

$Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]$

$Q=-29427.312J=-29.4kJ$

Negative sign indicates that the heat is removed from the system.

Therefore, the energy removed must be, 29.4 kJ

3 0

3 years ago

What is a model that is used to predict possible genotypes and phenotype of offspring?

frutty [35]

This is called the pedigree chart.

7 0

2 years ago

A system gains 652 kJ of heat, resulting in a change in internal energy of the system equal to +241 kJ. How much work is done?

levacccp [35]

Answer:

-411 kj

Explanation:

We solve by using this formula

∆U = ∆Q + ∆W

This formula is the first law of thermodynamics

Change in internal energy U = +241

Heat gained by system Q = 652

Putting the value into the equation

+241 = 652 + W

Workdone = 241 - 652

Workdone = -411 kj

Since work done is negative it means that work was done by the system

3 0

2 years ago