What is the decreasing order of field strength of given ligands?
(A) \( S^{2-} \)
(B) Ethylenediamine
(C) \( NCS^- \)
(D) \( CN^- \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (D), (B), (C), (A)
Ligands differ in how strongly they split the d orbitals of the metal ion. Arranged in order of increasing field strength, they form the spectrochemical series, which is found by experiment from the light absorbed by complexes.
Picking out the four given ligands: \( \ce{CN-} \) is near the strong end, ethylenediamine (en) just below it, \( \ce{NCS-} \) (bonded through N) in the middle, and \( \ce{S^2-} \) near the weak end. In decreasing order of field strength:
\( \displaystyle \ce{CN-} \gt \text{en} \gt \ce{NCS-} \gt \ce{S^2-} \)
Shortcut: \( \ce{CN-} \) is the strongest and \( \ce{S^2-} \) the weakest of the four, so the order starts with (D) and ends with (A): Options (3) and (4). Since en is stronger than \( \ce{NCS-} \), Option (3) remains.
NCERT (2026-27 edition):
Crystal Field Theory – …
