ORGANIC CHEM +SG +SAPLING >IP<
ORGANIC CHEM +SG +SAPLING >IP<
6th Edition
ISBN: 9781319171179
Author: LOUDON
Publisher: MAC HIGHER
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Chapter 18, Problem 18.9P
Interpretation Introduction

(a)

Interpretation:

The dn for the metal in [W(CO)5]2 complex is to be stated.

Concept introduction:

The number of unshared valence electrons on the metal is the number n in the dn notation. For example, if the metal in a complex has 6 unshared valence electrons, the complex is said to have d6 configuration.

Expert Solution
Check Mark

Answer to Problem 18.9P

The dn notation for the [W(CO)5]2 complex is d8.

Explanation of Solution

The formula for calculation of n in the dn notation is given below.

n = (ValenceelectronsinneutralMOxidationstateofM)

The formula for the calculation of oxidation state is given below

XC=XM+XL×n

• n is the number of ligands.

XC is the charge on the complex.

XM is the charge on the metal.

XL is the charge on the ligand.

Therefore, the oxidation state of W in [W(CO)5]2 is calculated below.

XC=XM+(XCO×nCO)XM0(0)=2XM=2

Now, valence electronic configuration of W is 5d46s2, so, valence electrons in tungsten are 6.

L-type ligands have no effect on the dn notation. Since carbonyl ligand is a L- type ligand it would not have any effect in the dn.

Value of n for the metal complex is calculated below.

n=(ValenceelectronsinneutralMOxidationstateofM)=6(2)=8

Therefore, the given complex is a d8 complex

Conclusion

The value of n in the [W(CO)5]2 complex is 8.

Interpretation Introduction

(b)

Interpretation:

The dn for the metal in Pd(PPh3)4 complex is to be stated.

Concept introduction:

The number of unshared valence electrons on the metal is the number n in the dn notation. For example, if the metal in a complex has 6 unshared valence electron the complex is said to have d6 configuration.

Expert Solution
Check Mark

Answer to Problem 18.9P

The dn notation for the Pd(PPh3)4 complex is d10.

Explanation of Solution

The formula for calculation of n in the dn notation is given below.

n = (ValenceelectronsinneutralMOxidationstateofM)

The oxidation state of the metal is calculated by the formula given below.

OxidationstateofM=(Netchargepresent over the complexSum of charges of X-type ligand in dissosciated state)

Since phosphine is a neutral ligand and there is no overall charge on the complex.

The formula for the calculation of oxidation state is given below

XC=XM+XL×n

• n is the number of ligands

XC is the charge on the complex

XM is the charge on the metal

XL is the charge on the ligand

So, the oxidation state of Pd in the complex is calculated as shown below.

XM+XPPh3×nPPh3=XCXM+0(4)=0XM=0

Now, valence electronic configuration of Pd is 4d10 so, valence electrons =10

Value of n for the metal complex is calculated below.

n=(ValenceelectroninneutralMOxidationstateofM)=100=10

Therefore, the given complex is a d10 complex.

Conclusion

The value of n in the Pd(PPh3)4 complex is 10.

Interpretation Introduction

(c)

Interpretation:

The dn for the metal in [Rh(PPh)3(H)2Cl] is to be stated.

Concept introduction:

The number of unshared valence electrons on the metal is the number n in the dn notation. For example, if the metal in a complex has 6 unshared valence electron the complex is said to have d6 configuration.

Expert Solution
Check Mark

Answer to Problem 18.9P

The dn notation for the complex is d6.

Explanation of Solution

The given complex is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 18, Problem 18.9P

Figure 1

The formula for calculation of n in the dn notation is given below.

n = (ValenceelectronsinneutralMOxidationstateofM)

The oxidation state of metal is calculated by the formula given below.

OxidationstateofM=(Netchargepresent over the complexSum of charges of X-type ligand in dissosciated state)

Since, phosphine is a neutral ligand.and there is no overall charge on the complex. The charge assosciated with chlorine and hydride ligands is 1.

The formula for the calculation of oxidation state is given below.

XC=XM+XL×n

• n is the number of ligands.

XC is the charge on the complex.

XM is the charge on the metal.

XL is the charge on the ligand.

So, the oxidation state of Rh in the complex is calculated below.

XM+(XCl×nCl)+(XH×nH)=XCXM+1(1)+2(1)=0XM12=0XM=+3

The valence electronic configuration of Rh is 4d85s1 so, number of valence electrons is 9.

Value of n for the metal complex is calculated below.

n=(ValenceelectroninneutralMOxidationstateofM)=93=6

Therefore, the given complex is a d6 complex.

Conclusion

The value of n in the [Rh(PPh)3(H)2Cl] complex is 6.

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Chapter 18 Solutions

ORGANIC CHEM +SG +SAPLING >IP<

Ch. 18 - Prob. 18.11PCh. 18 - Prob. 18.12PCh. 18 - Prob. 18.13PCh. 18 - Prob. 18.14PCh. 18 - Prob. 18.15PCh. 18 - Prob. 18.16PCh. 18 - Prob. 18.17PCh. 18 - Prob. 18.18PCh. 18 - Prob. 18.19PCh. 18 - Prob. 18.20PCh. 18 - Prob. 18.21PCh. 18 - Prob. 18.22PCh. 18 - Prob. 18.23PCh. 18 - Prob. 18.24PCh. 18 - Prob. 18.25PCh. 18 - Prob. 18.26PCh. 18 - Prob. 18.27PCh. 18 - Prob. 18.28PCh. 18 - Prob. 18.29PCh. 18 - Prob. 18.30PCh. 18 - Prob. 18.31PCh. 18 - Prob. 18.32PCh. 18 - Prob. 18.33PCh. 18 - Prob. 18.34PCh. 18 - Prob. 18.35PCh. 18 - Prob. 18.36PCh. 18 - Prob. 18.37PCh. 18 - Prob. 18.38PCh. 18 - Prob. 18.39PCh. 18 - Prob. 18.40PCh. 18 - Prob. 18.41PCh. 18 - Prob. 18.42PCh. 18 - Prob. 18.43PCh. 18 - Prob. 18.44PCh. 18 - Prob. 18.45PCh. 18 - Prob. 18.46APCh. 18 - Prob. 18.47APCh. 18 - Prob. 18.48APCh. 18 - Prob. 18.49APCh. 18 - Prob. 18.50APCh. 18 - Prob. 18.51APCh. 18 - Prob. 18.52APCh. 18 - Prob. 18.53APCh. 18 - Prob. 18.54APCh. 18 - Prob. 18.55APCh. 18 - Prob. 18.56APCh. 18 - Prob. 18.57APCh. 18 - Prob. 18.58APCh. 18 - Prob. 18.59APCh. 18 - Prob. 18.60APCh. 18 - Prob. 18.61APCh. 18 - Prob. 18.62APCh. 18 - Prob. 18.63APCh. 18 - Prob. 18.64APCh. 18 - Prob. 18.65APCh. 18 - Prob. 18.66APCh. 18 - Prob. 18.67APCh. 18 - Prob. 18.68APCh. 18 - Prob. 18.69APCh. 18 - Prob. 18.70APCh. 18 - Prob. 18.71APCh. 18 - Prob. 18.72APCh. 18 - Prob. 18.73APCh. 18 - Prob. 18.74APCh. 18 - Prob. 18.75APCh. 18 - Prob. 18.76APCh. 18 - Prob. 18.77APCh. 18 - Prob. 18.78APCh. 18 - Prob. 18.79APCh. 18 - Prob. 18.80APCh. 18 - Prob. 18.81APCh. 18 - Prob. 18.82APCh. 18 - Prob. 18.83APCh. 18 - Prob. 18.84APCh. 18 - Prob. 18.85APCh. 18 - Prob. 18.86APCh. 18 - Prob. 18.87APCh. 18 - Prob. 18.88APCh. 18 - Prob. 18.89APCh. 18 - Prob. 18.90APCh. 18 - Prob. 18.91APCh. 18 - Prob. 18.92AP
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