(B) : a) $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} \Rightarrow$ Low spin complex pairing occur Number of unpaired electron $=1$ b) $\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3-} \Rightarrow$ Number of unpaired electron $=3$ c) $\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-} \Rightarrow$ Low spin complex, pairing occurs Number of unpaired electron $=0$ d) $\left[\mathrm{Sc}(\mathrm{CN})_{6}\right]^{3-}$ $\mathrm{Sc}^{3+}: \mathrm{d}^{0}$ Number of unpaired electron $=0$ Hence, $\quad\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3-}$ contains maximum unpaired electrons due to this it has maximum paramagnetic in nature.
AMU-2012
COORDINATION COMPOUNDS
274098
What is the magnetic moment of $\mathrm{Fe}^{3+}$ ion in $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ ?
1 1.73 B.M.
2 5.9 B.M.
3 Diamagnetic
4 None of these
Explanation:
(A) : $\mathrm{Fe}^{3+}$ is a $\mathrm{d}^{5}$ system and presence of strong field ligand $\mathrm{CN}^{-}$, causes pairing up of all electrons. Thus no. of unpaired electrons $n=1$ $\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}$ $\mu=\sqrt{3}=1.73$ B.M.
AMU - 2010
COORDINATION COMPOUNDS
274107
Among the following, which one is paramagnetic and has tetrahedral geometry?
(B) : $\mathrm{CN}^{-}$and $\mathrm{CO}$ act as strong field ligand and do pairing of electrons. $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ and $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ are low spin complex and have no unpaired electron. Therefore, both are diamagnetic in nature. In $\left[\mathrm{CoCl}_{2}(\mathrm{en})_{2}\right]^{+}$, cobalt is present in +3 oxidation state and with $\mathrm{Co}^{3+}$ ethylene diamine (en) act as strong field ligand. $\left(\Delta_{\mathrm{o}}>\mathrm{p}\right)$ No. of unpaired electron is zero so it behaves as diamagnetic. $\left[\mathrm{NiCl}_{4}\right]^{2-} \mathrm{Cl}^{-}$act as weak field ligand and pairing of electron are not possible. This complex $\left[\mathrm{NiCl}_{4}\right]^{2-}$ has two unpaired electron so it behaves as paramagnetic and its hybridization is $\mathrm{sp}^{3}$ therefore $\left[\mathrm{NiCl}_{4}\right]^{2-}$ is tetrahedral complex.
J and K CET-(2011)
COORDINATION COMPOUNDS
274108
Which one of the following complex ions has the highest magnetic moment?
(A) : When Complex having maximum number of unpaired electron then its magnetic moment is also maximum. (a) $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+} \Rightarrow \mathrm{Cr}^{3+}: \mathrm{d}^{3}$ No. of unpaired electron $=3$ Magnetic moment $=\sqrt{3(3+2)}=3.8$ B.M. (b) $\quad\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}, \mathrm{CN}$ act as strong field ligand $\left(\Delta_{\mathrm{o}}>\mathrm{p}\right)$ then pairing of electrons occur. $\mathrm{Fe}^{3+}: \mathrm{d}^{5}$ No. of unpaired electron $=1$ Magnetic moment $=1.73$ B.M. (c) $\quad\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{2-} \Rightarrow \mathrm{Fe}^{2+}: \mathrm{d}^{6}$ (d) Magnetic moment $=0$ B.M. $\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}: \mathrm{Zn}^{2+}=\mathrm{d}^{10}$ No. of unpaired electron $=0$ Magnetic moment $=0$ B.M. Hence, $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ complex has highest magnetic moment.
(B) : a) $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} \Rightarrow$ Low spin complex pairing occur Number of unpaired electron $=1$ b) $\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3-} \Rightarrow$ Number of unpaired electron $=3$ c) $\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-} \Rightarrow$ Low spin complex, pairing occurs Number of unpaired electron $=0$ d) $\left[\mathrm{Sc}(\mathrm{CN})_{6}\right]^{3-}$ $\mathrm{Sc}^{3+}: \mathrm{d}^{0}$ Number of unpaired electron $=0$ Hence, $\quad\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3-}$ contains maximum unpaired electrons due to this it has maximum paramagnetic in nature.
AMU-2012
COORDINATION COMPOUNDS
274098
What is the magnetic moment of $\mathrm{Fe}^{3+}$ ion in $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ ?
1 1.73 B.M.
2 5.9 B.M.
3 Diamagnetic
4 None of these
Explanation:
(A) : $\mathrm{Fe}^{3+}$ is a $\mathrm{d}^{5}$ system and presence of strong field ligand $\mathrm{CN}^{-}$, causes pairing up of all electrons. Thus no. of unpaired electrons $n=1$ $\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}$ $\mu=\sqrt{3}=1.73$ B.M.
AMU - 2010
COORDINATION COMPOUNDS
274107
Among the following, which one is paramagnetic and has tetrahedral geometry?
(B) : $\mathrm{CN}^{-}$and $\mathrm{CO}$ act as strong field ligand and do pairing of electrons. $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ and $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ are low spin complex and have no unpaired electron. Therefore, both are diamagnetic in nature. In $\left[\mathrm{CoCl}_{2}(\mathrm{en})_{2}\right]^{+}$, cobalt is present in +3 oxidation state and with $\mathrm{Co}^{3+}$ ethylene diamine (en) act as strong field ligand. $\left(\Delta_{\mathrm{o}}>\mathrm{p}\right)$ No. of unpaired electron is zero so it behaves as diamagnetic. $\left[\mathrm{NiCl}_{4}\right]^{2-} \mathrm{Cl}^{-}$act as weak field ligand and pairing of electron are not possible. This complex $\left[\mathrm{NiCl}_{4}\right]^{2-}$ has two unpaired electron so it behaves as paramagnetic and its hybridization is $\mathrm{sp}^{3}$ therefore $\left[\mathrm{NiCl}_{4}\right]^{2-}$ is tetrahedral complex.
J and K CET-(2011)
COORDINATION COMPOUNDS
274108
Which one of the following complex ions has the highest magnetic moment?
(A) : When Complex having maximum number of unpaired electron then its magnetic moment is also maximum. (a) $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+} \Rightarrow \mathrm{Cr}^{3+}: \mathrm{d}^{3}$ No. of unpaired electron $=3$ Magnetic moment $=\sqrt{3(3+2)}=3.8$ B.M. (b) $\quad\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}, \mathrm{CN}$ act as strong field ligand $\left(\Delta_{\mathrm{o}}>\mathrm{p}\right)$ then pairing of electrons occur. $\mathrm{Fe}^{3+}: \mathrm{d}^{5}$ No. of unpaired electron $=1$ Magnetic moment $=1.73$ B.M. (c) $\quad\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{2-} \Rightarrow \mathrm{Fe}^{2+}: \mathrm{d}^{6}$ (d) Magnetic moment $=0$ B.M. $\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}: \mathrm{Zn}^{2+}=\mathrm{d}^{10}$ No. of unpaired electron $=0$ Magnetic moment $=0$ B.M. Hence, $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ complex has highest magnetic moment.
(B) : a) $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} \Rightarrow$ Low spin complex pairing occur Number of unpaired electron $=1$ b) $\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3-} \Rightarrow$ Number of unpaired electron $=3$ c) $\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-} \Rightarrow$ Low spin complex, pairing occurs Number of unpaired electron $=0$ d) $\left[\mathrm{Sc}(\mathrm{CN})_{6}\right]^{3-}$ $\mathrm{Sc}^{3+}: \mathrm{d}^{0}$ Number of unpaired electron $=0$ Hence, $\quad\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3-}$ contains maximum unpaired electrons due to this it has maximum paramagnetic in nature.
AMU-2012
COORDINATION COMPOUNDS
274098
What is the magnetic moment of $\mathrm{Fe}^{3+}$ ion in $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ ?
1 1.73 B.M.
2 5.9 B.M.
3 Diamagnetic
4 None of these
Explanation:
(A) : $\mathrm{Fe}^{3+}$ is a $\mathrm{d}^{5}$ system and presence of strong field ligand $\mathrm{CN}^{-}$, causes pairing up of all electrons. Thus no. of unpaired electrons $n=1$ $\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}$ $\mu=\sqrt{3}=1.73$ B.M.
AMU - 2010
COORDINATION COMPOUNDS
274107
Among the following, which one is paramagnetic and has tetrahedral geometry?
(B) : $\mathrm{CN}^{-}$and $\mathrm{CO}$ act as strong field ligand and do pairing of electrons. $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ and $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ are low spin complex and have no unpaired electron. Therefore, both are diamagnetic in nature. In $\left[\mathrm{CoCl}_{2}(\mathrm{en})_{2}\right]^{+}$, cobalt is present in +3 oxidation state and with $\mathrm{Co}^{3+}$ ethylene diamine (en) act as strong field ligand. $\left(\Delta_{\mathrm{o}}>\mathrm{p}\right)$ No. of unpaired electron is zero so it behaves as diamagnetic. $\left[\mathrm{NiCl}_{4}\right]^{2-} \mathrm{Cl}^{-}$act as weak field ligand and pairing of electron are not possible. This complex $\left[\mathrm{NiCl}_{4}\right]^{2-}$ has two unpaired electron so it behaves as paramagnetic and its hybridization is $\mathrm{sp}^{3}$ therefore $\left[\mathrm{NiCl}_{4}\right]^{2-}$ is tetrahedral complex.
J and K CET-(2011)
COORDINATION COMPOUNDS
274108
Which one of the following complex ions has the highest magnetic moment?
(A) : When Complex having maximum number of unpaired electron then its magnetic moment is also maximum. (a) $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+} \Rightarrow \mathrm{Cr}^{3+}: \mathrm{d}^{3}$ No. of unpaired electron $=3$ Magnetic moment $=\sqrt{3(3+2)}=3.8$ B.M. (b) $\quad\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}, \mathrm{CN}$ act as strong field ligand $\left(\Delta_{\mathrm{o}}>\mathrm{p}\right)$ then pairing of electrons occur. $\mathrm{Fe}^{3+}: \mathrm{d}^{5}$ No. of unpaired electron $=1$ Magnetic moment $=1.73$ B.M. (c) $\quad\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{2-} \Rightarrow \mathrm{Fe}^{2+}: \mathrm{d}^{6}$ (d) Magnetic moment $=0$ B.M. $\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}: \mathrm{Zn}^{2+}=\mathrm{d}^{10}$ No. of unpaired electron $=0$ Magnetic moment $=0$ B.M. Hence, $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ complex has highest magnetic moment.
(B) : a) $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} \Rightarrow$ Low spin complex pairing occur Number of unpaired electron $=1$ b) $\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3-} \Rightarrow$ Number of unpaired electron $=3$ c) $\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-} \Rightarrow$ Low spin complex, pairing occurs Number of unpaired electron $=0$ d) $\left[\mathrm{Sc}(\mathrm{CN})_{6}\right]^{3-}$ $\mathrm{Sc}^{3+}: \mathrm{d}^{0}$ Number of unpaired electron $=0$ Hence, $\quad\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3-}$ contains maximum unpaired electrons due to this it has maximum paramagnetic in nature.
AMU-2012
COORDINATION COMPOUNDS
274098
What is the magnetic moment of $\mathrm{Fe}^{3+}$ ion in $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ ?
1 1.73 B.M.
2 5.9 B.M.
3 Diamagnetic
4 None of these
Explanation:
(A) : $\mathrm{Fe}^{3+}$ is a $\mathrm{d}^{5}$ system and presence of strong field ligand $\mathrm{CN}^{-}$, causes pairing up of all electrons. Thus no. of unpaired electrons $n=1$ $\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}$ $\mu=\sqrt{3}=1.73$ B.M.
AMU - 2010
COORDINATION COMPOUNDS
274107
Among the following, which one is paramagnetic and has tetrahedral geometry?
(B) : $\mathrm{CN}^{-}$and $\mathrm{CO}$ act as strong field ligand and do pairing of electrons. $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ and $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ are low spin complex and have no unpaired electron. Therefore, both are diamagnetic in nature. In $\left[\mathrm{CoCl}_{2}(\mathrm{en})_{2}\right]^{+}$, cobalt is present in +3 oxidation state and with $\mathrm{Co}^{3+}$ ethylene diamine (en) act as strong field ligand. $\left(\Delta_{\mathrm{o}}>\mathrm{p}\right)$ No. of unpaired electron is zero so it behaves as diamagnetic. $\left[\mathrm{NiCl}_{4}\right]^{2-} \mathrm{Cl}^{-}$act as weak field ligand and pairing of electron are not possible. This complex $\left[\mathrm{NiCl}_{4}\right]^{2-}$ has two unpaired electron so it behaves as paramagnetic and its hybridization is $\mathrm{sp}^{3}$ therefore $\left[\mathrm{NiCl}_{4}\right]^{2-}$ is tetrahedral complex.
J and K CET-(2011)
COORDINATION COMPOUNDS
274108
Which one of the following complex ions has the highest magnetic moment?
(A) : When Complex having maximum number of unpaired electron then its magnetic moment is also maximum. (a) $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+} \Rightarrow \mathrm{Cr}^{3+}: \mathrm{d}^{3}$ No. of unpaired electron $=3$ Magnetic moment $=\sqrt{3(3+2)}=3.8$ B.M. (b) $\quad\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}, \mathrm{CN}$ act as strong field ligand $\left(\Delta_{\mathrm{o}}>\mathrm{p}\right)$ then pairing of electrons occur. $\mathrm{Fe}^{3+}: \mathrm{d}^{5}$ No. of unpaired electron $=1$ Magnetic moment $=1.73$ B.M. (c) $\quad\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{2-} \Rightarrow \mathrm{Fe}^{2+}: \mathrm{d}^{6}$ (d) Magnetic moment $=0$ B.M. $\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}: \mathrm{Zn}^{2+}=\mathrm{d}^{10}$ No. of unpaired electron $=0$ Magnetic moment $=0$ B.M. Hence, $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ complex has highest magnetic moment.
