Number of complexes from the following with even number of unpaired electrons is __________ .
[Given atomic numbers: V = 23, Cr = 24, Fe = 26, Ni = 28, Cu = 29] [2024]
2
4
1
5
(A)
Complex | Electronic configuration of central metal ion without crystal field splitting |
Electronic configuration of central metal ion after crystal field splitting |
Number of unpaired d electrons in complex |
2 | |||
4 | |||
5 | |||
3 | |||
1 |
The correct sequence of ligands in the order of decreasing field strength is : [2024]
(C)
Ligands can be arranged in a series in the order of increasing field strength as given below:
If an iron (III) complex with the formula has no electron in its orbital, then the value of x + y is [2024]
4
3
5
6
(D)
Iron in +3 oxidation state forms complexes with C.N. = 6.
So
Ammonia is a neutral ligand and cyanide is a unit negatively charged ligand.
Since sum of oxidation number = charge
Thus
The complex is
The number of unpaired -electrons in is ________ . [2024]
1
0
4
2
(B)
Oxidation number of Co = +3
With , ligands like acts as strong field ligands. With strong field ligands, complexes split as
So it has no unpaired electrons.
The correct order of ligands arranged in increasing field strength. [2024]
(C)
In general, ligands can be arranged in a series in the order of increasing field strength as given below:
Which one of the following complexes will exhibit the least paramagnetic behaviour ?
[Atomic number, Cr = 24, Mn = 25, Fe = 26, Co = 27] [2024]
(A)
Complex | Electronic configuration of central metal ion without splitting |
Electronic configuration after splitting |
Number of unpaired electrons |
3 | |||
5 | |||
4 | |||
4 |
Paramagnetic nature number of unpaired electrons.
The number of complexes from the following with no electrons in the orbital is _______ .
[2024]
1
4
2
3
(D)
Compound | Electronic configuration of central atom without tetrahedral splitting |
Electronic configuration of central atom after tetrahedral splitting |
Three complexes donot have electron in orbitals
Match List I with List II.
List I | List II | ||
Tetrahedral Complex | Electronic Configuration | ||
A. | I. | ||
B. | II. | ||
C. | III. | ||
D. | IV. |
Choose the correct answer from the options given below: [2024]
(A) - (III), (B) - (I), (C) - (IV), (D) - (II)
(A) - (III), (B) - (IV), (C) - (II), (D) - (I)
(A) - (IV), (B) - (III), (C) - (I), (D) - (II)
(A) - (I), (B) - (III), (C) - (IV), (D) - (II)
(A)
Complex | Electronic configuration of central metal without tetrahedral splitting |
Electronic |
Number of complexes with even number of electrons in orbitals is - [2024]
3
5
2
1
(A)
Complex | Electronic configuration of central metal ion without octahedral splitting |
Electronic configuration of central metal ion after octahedral splitting |
Number of electrons in |
4 | |||
5 | |||
6 | |||
6 | |||
3 |
Note: In complexes with C.N = 6, undergoes hybridization (except in complexes with C.N = 6, where undergoes hybridization).
The Spin only magnetic moment value of square planar complex is ________ B.M. (Nearest integer) (Given atomic number for Pt = 78) [2024]
(0)
In , is in +2 oxidation state.
Metal ions with or configuration with coordination number 4 are always hybridized and diamagnetic, irrespective of ligand. This is because for these or is high on account of poor shielding of and/or electrons, hence all ligands are sufficiently attracted towards the metal ion resulting large crystal field splitting which causes in pairing up of electrons.