Concept explainers
(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.
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 = (Valence electrons in neutral M−Oxidation state of M)
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)XM−0(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=(Valence electrons in neutral M−Oxidation state of M)=6−(−2)=8
Therefore, the given complex is a d8 complex
The value of n in the [W(CO)5]2− complex is 8.
(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.
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 = (Valence electrons in neutral M−Oxidation state of M)
The oxidation state of the metal is calculated by the formula given below.
Oxidation state of M=(Net charge present over the complex−Sum 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=(Valence electron in neutral M−Oxidation state of M )=10−0=10
Therefore, the given complex is a d10 complex.
The value of n in the Pd(PPh3)4 complex is 10.
(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.
Answer to Problem 18.9P
The dn notation for the complex is d6.
Explanation of Solution
The given complex is shown below.
Figure 1
The formula for calculation of n in the dn notation is given below.
n = (Valence electrons in neutral M−Oxidation state of M)
The oxidation state of metal is calculated by the formula given below.
Oxidation state of M=(Net charge present over the complex−Sum 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)=0XM−1−2=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=(Valence electron in neutral M−Oxidation state of M )=9−3=6
Therefore, the given complex is a d6 complex.
The value of n in the [Rh(PPh)3(H)2Cl] complex is 6.
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