Q7 / 75 · JEE Main 2026 (24 January Shift 1) · Chemistry
Given below are two statements:
Statement I: Hybridisation, shape and spin only magnetic moment of $\mathrm{K}_{3}\left[\mathrm{Co}\left(\mathrm{CO}_{3}\right)_{3}\right]$ is $\mathrm{sp}^{3} \mathrm{~d}^{2}$, octahedral and 4.9 BM respectively.
Statement II: Geometry, hybridisation and spin only magnetic moment values $(\mathrm{BM})$ of the ions $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-},\left[\mathrm{MnBr}_{4}\right]^{2-}$ and $\left[\mathrm{CoF}_{6}\right]^{3-}$ respectively are square planar, tetrahedral, octahedral; $\mathrm{dsp}^{2}, \mathrm{sp}^{3}, \mathrm{sp}^{3} \mathrm{~d}^{2}$ and $0,5.9,4.9$.
In the light of the above statements, choose the correct answer from the options given below
Statement I: Hybridisation, shape and spin only magnetic moment of $\mathrm{K}_{3}\left[\mathrm{Co}\left(\mathrm{CO}_{3}\right)_{3}\right]$ is $\mathrm{sp}^{3} \mathrm{~d}^{2}$, octahedral and 4.9 BM respectively.
Statement II: Geometry, hybridisation and spin only magnetic moment values $(\mathrm{BM})$ of the ions $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-},\left[\mathrm{MnBr}_{4}\right]^{2-}$ and $\left[\mathrm{CoF}_{6}\right]^{3-}$ respectively are square planar, tetrahedral, octahedral; $\mathrm{dsp}^{2}, \mathrm{sp}^{3}, \mathrm{sp}^{3} \mathrm{~d}^{2}$ and $0,5.9,4.9$.
In the light of the above statements, choose the correct answer from the options given below
- AStatement I is true but Statement II is false
- BBoth Statement I and Statement II are false
- Correct answerCBoth Statement I and Statement II are true
- DStatement I is false but Statement II is true
Solution
In Statement I, for $K_3[Co(CO_3)_3]$, Cobalt is in $+3$ oxidation state ($Co^{3+}: 3d^6$). Carbonate ($CO_3^{2-}$) is a weak field ligand. Thus, no pairing occurs, resulting in 4 unpaired electrons. Hybridisation is $sp^3d^2$, shape is octahedral, and $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9$ BM. Statement I is true.
In Statement II:
For $[Ni(CN)_4]^{2-}$, $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand, causing pairing. Hybridisation is $dsp^2$, geometry is square planar, and $\mu = 0$ BM.
For $[MnBr_4]^{2-}$, $Mn^{2+}$ is $3d^5$. $Br^-$ is a weak field ligand. Hybridisation is $sp^3$, geometry is tetrahedral, and $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9$ BM.
For $[CoF_6]^{3-}$, $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand. Hybridisation is $sp^3d^2$, geometry is octahedral, and $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9$ BM.
Statement II is true.
Both statements are true.
In Statement II:
For $[Ni(CN)_4]^{2-}$, $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand, causing pairing. Hybridisation is $dsp^2$, geometry is square planar, and $\mu = 0$ BM.
For $[MnBr_4]^{2-}$, $Mn^{2+}$ is $3d^5$. $Br^-$ is a weak field ligand. Hybridisation is $sp^3$, geometry is tetrahedral, and $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9$ BM.
For $[CoF_6]^{3-}$, $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand. Hybridisation is $sp^3d^2$, geometry is octahedral, and $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9$ BM.
Statement II is true.
Both statements are true.
Hence, the correct answer is option: 'C'.