Consider the following ligands NH2^-, acac, OH^-, Gly, O2^-, Phen, DMG, NO2^-, CO3^2-, Cl^-, CH3COO^-, en, SO4^2-. Then calculate value of "P + Q - R - S"
Here
P : Total number of ligands which

Question

Consider the following ligands NH2^-, acac, OH^-, Gly, O2^-, Phen, DMG, NO2^-, CO3^2-, Cl^-, CH3COO^-, en, SO4^2-. Then calculate value of "P + Q - R - S"
Here
P : Total number of ligands which act as bridging as well as monodentate only.
Q : Total number of flexidentate ligands
R : Total number of bidentate ligands only
S : Total number of unsymmetrical bidentate ligands

Prashant Kotian · Accepted Answer

Correct answer: **ANSWER: 1**<BR>SOLUTION: P = 4 (NH2^-, O2^-, OH^-, CL^-). Q = 4 (SO4^2-, NO2^-, CO3^2-, CH3COO^-). R = 5 (EN, ACAC, PHEN, DMG, GLY). S = 2 (DMG, GLY). VALUE = 4 + 4 - 5 - 2 = 1.. Step 1: Determine the oxidation state and d-electron configuration of the metal in each complex: Co^2+ in (a) is 3d^7; Ni^2+ in (b) and (d) is 3d^8; Co^3+ in (c) is 3d^6. Step 2: In [Co(NH3)6]^3+, the strong-field NH3 ligands force all six d-electrons of Co^3+ to pair up in the lower t_2g orbitals, resulting in a low-spin t_2g^6 e_g^0 configuration. Step 3: Since there are zero unpaired electrons, [Co(NH3)6]^3+ is diamagnetic, while the other configurations retain unpaired electrons and are paramagnetic.

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