274093
What will be the theoretical value of magnetic moment $(\mu)$, when $\mathrm{CN}^{-}$ligands join $\mathrm{Fe}^{3+}$ ion to yield complex.
1 $2.83 \mathrm{BM}$
2 $3.87 \mathrm{BM}$
3 $5.92 \mathrm{BM}$
4 $1.73 \mathrm{BM}$
Explanation:
(D) : When $\mathrm{CN}^{-}$ligands join $\mathrm{Fe}^{3+}$ ion to yield the complex ion $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$. Electronic configuration of $\mathrm{Fe}^{3+}$ is $[\mathrm{Ar}] 3 \mathrm{~d}^{5}$. As we know $\mathrm{CN}^{-}$is the strong field ligand. Number of unpaired electron $(\mathrm{n})=1$ Magnetic moment $\left(\mu_{s}\right)=\sqrt{n(n+2)} B M$ $\mu_{\mathrm{s}}=\sqrt{1(1+2)} \mathrm{BM}$ $\mu_{\mathrm{s}}=1.73 \mathrm{BM} .$
(D) : Paramagnetic species contains unpaired electron. (a) $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-} \Rightarrow \mathrm{Fe}^{2+}: \mathrm{d}^{6}$ $\mathrm{CN}$ act as strong field ligand cause pairing of metal $\mathrm{d}$ electron. Number of unpaired electron $=0$ (b) $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right], \mathrm{CO}$ act as strong field ligand cause pairing of metal d electron. Number of unpaired electron $=0$ (c) $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]^{2-}$ Number of unpaired electron $=0$ (d) $\left[\mathrm{CoF}_{6}\right]^{3-}$ Cobalt is present in +3 oxidation state, and $\mathrm{F}^{-}$act as weak field ligand so no pairing of metal d electron. Number of unpaired electron $=4$ Hence, $\left[\mathrm{CoF}_{6}\right]^{3-}$ is act as paramagnetic.
AIIMS-2014
COORDINATION COMPOUNDS
274095
A complex $\left[\mathrm{CoL}_{6}\right]^{n+}$ where $\mathrm{L}$ is neutral ligand has a magnetic moment $\mu=4.5 \mathrm{~B}$. M. Hence,
1 Co must be in +2 oxidation state.
2 L must be af strong ligand.
3 The complex must be highly destabilised.
4 Co must be in +3 oxidation state.
Explanation:
(D) : $\left[\mathrm{CoL}_{6}\right]^{\mathrm{n}+}$ has a magnetic moment $\mu=4.5 \mathrm{BM}$. It means cobalt having 4 unpaired electron. This is possible when metal is present in +3 oxidation state and ligand (L) must be weak field ligand. $\mathrm{Co}^{3+} \Rightarrow \mathrm{d}^{6}$ (weak field ligand) Number of unpaired electron $=4$ $\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{4(4+2)}=4.5 \text { BM. }$
AIIMS-2010
COORDINATION COMPOUNDS
274096
A compound of a metal ion $M^{x+}(Z=24)$ has a spin only magnetic moment of $\sqrt{15}$ Bohr Magnetons. The number of unpaired electrons in the compound are
1 2
2 4
3 5
4 3
Explanation:
(D) : Spin only magnetic moment $\sqrt{\mathrm{n}(\mathrm{n}+2)}$ B.M. For given metal ion, $\mu=\sqrt{15}$ B.M. $\sqrt{15}=\sqrt{\mathrm{n}(\mathrm{n}+2)}$ $\mathrm{n}^{2}+2 \mathrm{n}=15$ $\mathrm{n}=3$ Hence, number of unpaired electron $=3$
274093
What will be the theoretical value of magnetic moment $(\mu)$, when $\mathrm{CN}^{-}$ligands join $\mathrm{Fe}^{3+}$ ion to yield complex.
1 $2.83 \mathrm{BM}$
2 $3.87 \mathrm{BM}$
3 $5.92 \mathrm{BM}$
4 $1.73 \mathrm{BM}$
Explanation:
(D) : When $\mathrm{CN}^{-}$ligands join $\mathrm{Fe}^{3+}$ ion to yield the complex ion $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$. Electronic configuration of $\mathrm{Fe}^{3+}$ is $[\mathrm{Ar}] 3 \mathrm{~d}^{5}$. As we know $\mathrm{CN}^{-}$is the strong field ligand. Number of unpaired electron $(\mathrm{n})=1$ Magnetic moment $\left(\mu_{s}\right)=\sqrt{n(n+2)} B M$ $\mu_{\mathrm{s}}=\sqrt{1(1+2)} \mathrm{BM}$ $\mu_{\mathrm{s}}=1.73 \mathrm{BM} .$
(D) : Paramagnetic species contains unpaired electron. (a) $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-} \Rightarrow \mathrm{Fe}^{2+}: \mathrm{d}^{6}$ $\mathrm{CN}$ act as strong field ligand cause pairing of metal $\mathrm{d}$ electron. Number of unpaired electron $=0$ (b) $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right], \mathrm{CO}$ act as strong field ligand cause pairing of metal d electron. Number of unpaired electron $=0$ (c) $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]^{2-}$ Number of unpaired electron $=0$ (d) $\left[\mathrm{CoF}_{6}\right]^{3-}$ Cobalt is present in +3 oxidation state, and $\mathrm{F}^{-}$act as weak field ligand so no pairing of metal d electron. Number of unpaired electron $=4$ Hence, $\left[\mathrm{CoF}_{6}\right]^{3-}$ is act as paramagnetic.
AIIMS-2014
COORDINATION COMPOUNDS
274095
A complex $\left[\mathrm{CoL}_{6}\right]^{n+}$ where $\mathrm{L}$ is neutral ligand has a magnetic moment $\mu=4.5 \mathrm{~B}$. M. Hence,
1 Co must be in +2 oxidation state.
2 L must be af strong ligand.
3 The complex must be highly destabilised.
4 Co must be in +3 oxidation state.
Explanation:
(D) : $\left[\mathrm{CoL}_{6}\right]^{\mathrm{n}+}$ has a magnetic moment $\mu=4.5 \mathrm{BM}$. It means cobalt having 4 unpaired electron. This is possible when metal is present in +3 oxidation state and ligand (L) must be weak field ligand. $\mathrm{Co}^{3+} \Rightarrow \mathrm{d}^{6}$ (weak field ligand) Number of unpaired electron $=4$ $\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{4(4+2)}=4.5 \text { BM. }$
AIIMS-2010
COORDINATION COMPOUNDS
274096
A compound of a metal ion $M^{x+}(Z=24)$ has a spin only magnetic moment of $\sqrt{15}$ Bohr Magnetons. The number of unpaired electrons in the compound are
1 2
2 4
3 5
4 3
Explanation:
(D) : Spin only magnetic moment $\sqrt{\mathrm{n}(\mathrm{n}+2)}$ B.M. For given metal ion, $\mu=\sqrt{15}$ B.M. $\sqrt{15}=\sqrt{\mathrm{n}(\mathrm{n}+2)}$ $\mathrm{n}^{2}+2 \mathrm{n}=15$ $\mathrm{n}=3$ Hence, number of unpaired electron $=3$
274093
What will be the theoretical value of magnetic moment $(\mu)$, when $\mathrm{CN}^{-}$ligands join $\mathrm{Fe}^{3+}$ ion to yield complex.
1 $2.83 \mathrm{BM}$
2 $3.87 \mathrm{BM}$
3 $5.92 \mathrm{BM}$
4 $1.73 \mathrm{BM}$
Explanation:
(D) : When $\mathrm{CN}^{-}$ligands join $\mathrm{Fe}^{3+}$ ion to yield the complex ion $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$. Electronic configuration of $\mathrm{Fe}^{3+}$ is $[\mathrm{Ar}] 3 \mathrm{~d}^{5}$. As we know $\mathrm{CN}^{-}$is the strong field ligand. Number of unpaired electron $(\mathrm{n})=1$ Magnetic moment $\left(\mu_{s}\right)=\sqrt{n(n+2)} B M$ $\mu_{\mathrm{s}}=\sqrt{1(1+2)} \mathrm{BM}$ $\mu_{\mathrm{s}}=1.73 \mathrm{BM} .$
(D) : Paramagnetic species contains unpaired electron. (a) $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-} \Rightarrow \mathrm{Fe}^{2+}: \mathrm{d}^{6}$ $\mathrm{CN}$ act as strong field ligand cause pairing of metal $\mathrm{d}$ electron. Number of unpaired electron $=0$ (b) $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right], \mathrm{CO}$ act as strong field ligand cause pairing of metal d electron. Number of unpaired electron $=0$ (c) $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]^{2-}$ Number of unpaired electron $=0$ (d) $\left[\mathrm{CoF}_{6}\right]^{3-}$ Cobalt is present in +3 oxidation state, and $\mathrm{F}^{-}$act as weak field ligand so no pairing of metal d electron. Number of unpaired electron $=4$ Hence, $\left[\mathrm{CoF}_{6}\right]^{3-}$ is act as paramagnetic.
AIIMS-2014
COORDINATION COMPOUNDS
274095
A complex $\left[\mathrm{CoL}_{6}\right]^{n+}$ where $\mathrm{L}$ is neutral ligand has a magnetic moment $\mu=4.5 \mathrm{~B}$. M. Hence,
1 Co must be in +2 oxidation state.
2 L must be af strong ligand.
3 The complex must be highly destabilised.
4 Co must be in +3 oxidation state.
Explanation:
(D) : $\left[\mathrm{CoL}_{6}\right]^{\mathrm{n}+}$ has a magnetic moment $\mu=4.5 \mathrm{BM}$. It means cobalt having 4 unpaired electron. This is possible when metal is present in +3 oxidation state and ligand (L) must be weak field ligand. $\mathrm{Co}^{3+} \Rightarrow \mathrm{d}^{6}$ (weak field ligand) Number of unpaired electron $=4$ $\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{4(4+2)}=4.5 \text { BM. }$
AIIMS-2010
COORDINATION COMPOUNDS
274096
A compound of a metal ion $M^{x+}(Z=24)$ has a spin only magnetic moment of $\sqrt{15}$ Bohr Magnetons. The number of unpaired electrons in the compound are
1 2
2 4
3 5
4 3
Explanation:
(D) : Spin only magnetic moment $\sqrt{\mathrm{n}(\mathrm{n}+2)}$ B.M. For given metal ion, $\mu=\sqrt{15}$ B.M. $\sqrt{15}=\sqrt{\mathrm{n}(\mathrm{n}+2)}$ $\mathrm{n}^{2}+2 \mathrm{n}=15$ $\mathrm{n}=3$ Hence, number of unpaired electron $=3$
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COORDINATION COMPOUNDS
274093
What will be the theoretical value of magnetic moment $(\mu)$, when $\mathrm{CN}^{-}$ligands join $\mathrm{Fe}^{3+}$ ion to yield complex.
1 $2.83 \mathrm{BM}$
2 $3.87 \mathrm{BM}$
3 $5.92 \mathrm{BM}$
4 $1.73 \mathrm{BM}$
Explanation:
(D) : When $\mathrm{CN}^{-}$ligands join $\mathrm{Fe}^{3+}$ ion to yield the complex ion $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$. Electronic configuration of $\mathrm{Fe}^{3+}$ is $[\mathrm{Ar}] 3 \mathrm{~d}^{5}$. As we know $\mathrm{CN}^{-}$is the strong field ligand. Number of unpaired electron $(\mathrm{n})=1$ Magnetic moment $\left(\mu_{s}\right)=\sqrt{n(n+2)} B M$ $\mu_{\mathrm{s}}=\sqrt{1(1+2)} \mathrm{BM}$ $\mu_{\mathrm{s}}=1.73 \mathrm{BM} .$
(D) : Paramagnetic species contains unpaired electron. (a) $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-} \Rightarrow \mathrm{Fe}^{2+}: \mathrm{d}^{6}$ $\mathrm{CN}$ act as strong field ligand cause pairing of metal $\mathrm{d}$ electron. Number of unpaired electron $=0$ (b) $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right], \mathrm{CO}$ act as strong field ligand cause pairing of metal d electron. Number of unpaired electron $=0$ (c) $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]^{2-}$ Number of unpaired electron $=0$ (d) $\left[\mathrm{CoF}_{6}\right]^{3-}$ Cobalt is present in +3 oxidation state, and $\mathrm{F}^{-}$act as weak field ligand so no pairing of metal d electron. Number of unpaired electron $=4$ Hence, $\left[\mathrm{CoF}_{6}\right]^{3-}$ is act as paramagnetic.
AIIMS-2014
COORDINATION COMPOUNDS
274095
A complex $\left[\mathrm{CoL}_{6}\right]^{n+}$ where $\mathrm{L}$ is neutral ligand has a magnetic moment $\mu=4.5 \mathrm{~B}$. M. Hence,
1 Co must be in +2 oxidation state.
2 L must be af strong ligand.
3 The complex must be highly destabilised.
4 Co must be in +3 oxidation state.
Explanation:
(D) : $\left[\mathrm{CoL}_{6}\right]^{\mathrm{n}+}$ has a magnetic moment $\mu=4.5 \mathrm{BM}$. It means cobalt having 4 unpaired electron. This is possible when metal is present in +3 oxidation state and ligand (L) must be weak field ligand. $\mathrm{Co}^{3+} \Rightarrow \mathrm{d}^{6}$ (weak field ligand) Number of unpaired electron $=4$ $\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{4(4+2)}=4.5 \text { BM. }$
AIIMS-2010
COORDINATION COMPOUNDS
274096
A compound of a metal ion $M^{x+}(Z=24)$ has a spin only magnetic moment of $\sqrt{15}$ Bohr Magnetons. The number of unpaired electrons in the compound are
1 2
2 4
3 5
4 3
Explanation:
(D) : Spin only magnetic moment $\sqrt{\mathrm{n}(\mathrm{n}+2)}$ B.M. For given metal ion, $\mu=\sqrt{15}$ B.M. $\sqrt{15}=\sqrt{\mathrm{n}(\mathrm{n}+2)}$ $\mathrm{n}^{2}+2 \mathrm{n}=15$ $\mathrm{n}=3$ Hence, number of unpaired electron $=3$
