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(show all your works and please do all questions from part a to part e. Sorry fo

ID: 1016346 • Letter: #

Question

(show all your works and please do all questions from part a to part e. Sorry for long questions, but I give thanks to anyone who help me answer all questions below)

a. what is the oxidation state of the central metal ion for [Cr(NH3)5Cl]Cl2

b. what is the complete electron configuration for this central metal ion

c. clearly identify(type and periodic row)which orbitals(empty or filled)in part b are valence orbitals for this compound

d. given that [Cr((NH3)5Cl]Cl2 is a six-coordinated complex, how many empty orbitals must be available on the central chromium metal(Lewis acid)to accept the lone pairs from the ammine and chloro Lewis base ligands

e. what is the hybridization of the orbitals listed in part d

Explanation / Answer

a. what is the oxidation state of the central metal ion for [Cr(NH3)5Cl]Cl2

Assume the oxidation state of central metal is X.

The oxidation state of complex = 0

The oxidation state of NH3= 0

The oxidation state of Cl = -1

[Cr(NH3)5Cl]Cl2

X + (5*0) + (-1*3)= 0

X=+3

Thus the oxidation state of Cr = +3

b. what is the complete electron configuration for this central metal ion    

the electron configuration of Cr(24):

[Ar] 3d5 4s1

Cr3+:

[Ar] 3d 3 4s0

c. clearly identify(type and periodic row)which orbitals(empty or filled)in part b are valence orbitals for this compound

for this central atom Cr3+, empty and filled valence orbital are:

3d, 4s, 4p; 4d

3 d is partially filled while 4s,4p are empty

d. given that [Cr((NH3)5Cl]Cl2 is a six-coordinated complex, how many empty orbitals must be available on the central chromium metal(Lewis acid)to accept the lone pairs from the ammine and chloro Lewis base ligands

there are 6 empty orbitals msut be available on the central chromium metal(Lewis acid)to accept the lone pairs from the ammine and chloro Lewis base ligands, these are two -3d ; one 4s and three -4p

e. what is the hybridization of the orbitals listed in part d

the hybridization is d2sp3