Target Exam

CUET

Subject

Chemistry

Chapter

Inorganic: Coordination Compounds

Question:

Match the complexes given in List-I with their properties given in List-II

List-I Complex

List-II Property

(A) $[FeF_6]^{3-}$

(I) $d^6$, high spin

(B) $[CoF_6]^{3-}$

(II) $d^5$, high spin

(C) $[Mn(CN)_6]^{3-}$

(III) $d^6$, low spin

(D) $[Co(CN)_6]^{3-}$

(IV) $d^4$, low spin

Choose the correct answer from the options given below:

Options:

(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

(A)-(IV), (B)-(I), (C)-(II), (D)-(III)

(A)-(II), (B)-(I), (C)-(IV), (D)-(III)

(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

Correct Answer:

(A)-(II), (B)-(I), (C)-(IV), (D)-(III)

Explanation:

The correct answer is Option (3) → (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

List-I Complex

List-II Property

(A) $[FeF_6]^{3-}$

(II) $d^5$, high spin

(B) $[CoF_6]^{3-}$

(I) $d^6$, high spin

(C) $[Mn(CN)_6]^{3-}$

(IV) $d^4$, low spin

(D) $[Co(CN)_6]^{3-}$

(III) $d^6$, low spin

Determine oxidation states

(A) [FeF₆]³⁻

Fe + 6(–1) = –3
Fe – 6 = –3 → Fe³⁺ (d⁵)
F⁻ = weak ligand → high-spin
d⁵, high spin(II)

(B) [CoF₆]³⁻

Co + 6(–1) = –3
Co – 6 = –3 → Co³⁺ (d⁶)
F⁻ = weak ligand → high-spin
d⁶, high spin(I)

(C) [Mn(CN)₆]³⁻

Mn + 6(–1) = –3
Mn – 6 = –3 → Mn³⁺ (d⁴)
CN⁻ = strong ligand → low-spin
d⁴, low spin(IV)

(D) [Co(CN)₆]³⁻

Co + 6(–1) = –3
Co – 6 = –3 → Co³⁺ (d⁶)
CN⁻ = strong ligand → low-spin
d⁶, low spin(III)