(A) : Diamagnatic substance having no unpaired electrons. (a) $\left[\mathrm{Co}(\mathrm{ox})_{3}\right]^{3-}, \Rightarrow \mathrm{Co}^{3+}=\mathrm{d}^{6}$ In $\left[\mathrm{Co}(\mathrm{ox})_{3}\right]^{3-}, \mathrm{Co}^{3+}$ act as strong field ligand with oxalate. $\mathrm{Co}^{3+}$ having no unpaired electron and act as diamagnetic. (b) $\quad\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}, \Rightarrow \mathrm{Fe}^{2+}=\mathrm{d}^{6}$ $\mathrm{Fe}^{2+}$ with $\mathrm{CN}^{-}$act as strong field ligand so no unpaired electron is present. Hence it is also act as diamagnetic complex.
AIIMS-26 May
COORDINATION COMPOUNDS
274056
Which of the following is not expected to show paramagnetism?
(B) : (a) $\mathrm{H}_{2} \mathrm{O}$ act as weak field ligand. So pairing is not possible. This complex having unpaired electron it behave like paramagnetic. (b) In $\left[\mathrm{Ni}(\mathrm{Co})_{4}\right]$, Co act as strong field ligand $\Delta_{\mathrm{o}}>\mathrm{p}$. $\mathrm{Ni}: 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$ $3 \mathrm{~d}^{10}$ No. unpaired electron so $\left[\mathrm{Ni}(\mathrm{Co})_{4}\right]$ act as diamagnetic. (c) $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}, \mathrm{NH}_{3}$ act as weak field ligand so pairing of electron is not possible. $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$ have 2 unpaired electron so behave as paramagnetic. (d) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}, \mathrm{NH}_{3}$ act as weak field ligand $\Delta_{\mathrm{o}}<\mathrm{p}$ $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ has 3 unpaired electron so it behave as paramagnetic.
JIPMER-2017
COORDINATION COMPOUNDS
274058
The paramagnetic species is
1 $\mathrm{SiO}_{2}$
2 $\mathrm{TiO}_{2}$
3 $\mathrm{BaO}_{2}$
4 $\mathrm{KO}_{2}$
Explanation:
(D) : The species which has unpaired electron, is paramagnetic in nature. $\mathrm{KO}_{2}$ : Let, $\mathrm{x}$ be the oxidation state of $\mathrm{K}$. $\mathrm{x}+2(-2)=0$ $\mathrm{x}=+4$ $\mathrm{~K}^{4+}=1 \mathrm{~s}^{2}, 2 \mathrm{~s}^{2}, 2 \mathrm{p}^{6}, 3 \mathrm{~s}^{2}, 3 \mathrm{p}^{3}$ Unpaired electron $(\mathrm{n})=3$, Paramagnetic
UPTU/UPSEE-2017
COORDINATION COMPOUNDS
274059
What will be the spin only magnetic moment of high spin and low spin $d^{5}$ electronic system in an octahedral complex?
(B) : High spin $\mathrm{d}^{5}$ octahedral complex No. of unpaired electron $=5$ Spin only magnetic moment $=\sqrt{5(5+2)}=5.92$ B.M. Low spin $\mathrm{d}^{5}$ octahedral complex No. of unpaired electron $=1$ Spin only Magnetic moment $=\sqrt{1(1+2)}=1.73$ B.M.
(A) : Diamagnatic substance having no unpaired electrons. (a) $\left[\mathrm{Co}(\mathrm{ox})_{3}\right]^{3-}, \Rightarrow \mathrm{Co}^{3+}=\mathrm{d}^{6}$ In $\left[\mathrm{Co}(\mathrm{ox})_{3}\right]^{3-}, \mathrm{Co}^{3+}$ act as strong field ligand with oxalate. $\mathrm{Co}^{3+}$ having no unpaired electron and act as diamagnetic. (b) $\quad\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}, \Rightarrow \mathrm{Fe}^{2+}=\mathrm{d}^{6}$ $\mathrm{Fe}^{2+}$ with $\mathrm{CN}^{-}$act as strong field ligand so no unpaired electron is present. Hence it is also act as diamagnetic complex.
AIIMS-26 May
COORDINATION COMPOUNDS
274056
Which of the following is not expected to show paramagnetism?
(B) : (a) $\mathrm{H}_{2} \mathrm{O}$ act as weak field ligand. So pairing is not possible. This complex having unpaired electron it behave like paramagnetic. (b) In $\left[\mathrm{Ni}(\mathrm{Co})_{4}\right]$, Co act as strong field ligand $\Delta_{\mathrm{o}}>\mathrm{p}$. $\mathrm{Ni}: 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$ $3 \mathrm{~d}^{10}$ No. unpaired electron so $\left[\mathrm{Ni}(\mathrm{Co})_{4}\right]$ act as diamagnetic. (c) $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}, \mathrm{NH}_{3}$ act as weak field ligand so pairing of electron is not possible. $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$ have 2 unpaired electron so behave as paramagnetic. (d) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}, \mathrm{NH}_{3}$ act as weak field ligand $\Delta_{\mathrm{o}}<\mathrm{p}$ $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ has 3 unpaired electron so it behave as paramagnetic.
JIPMER-2017
COORDINATION COMPOUNDS
274058
The paramagnetic species is
1 $\mathrm{SiO}_{2}$
2 $\mathrm{TiO}_{2}$
3 $\mathrm{BaO}_{2}$
4 $\mathrm{KO}_{2}$
Explanation:
(D) : The species which has unpaired electron, is paramagnetic in nature. $\mathrm{KO}_{2}$ : Let, $\mathrm{x}$ be the oxidation state of $\mathrm{K}$. $\mathrm{x}+2(-2)=0$ $\mathrm{x}=+4$ $\mathrm{~K}^{4+}=1 \mathrm{~s}^{2}, 2 \mathrm{~s}^{2}, 2 \mathrm{p}^{6}, 3 \mathrm{~s}^{2}, 3 \mathrm{p}^{3}$ Unpaired electron $(\mathrm{n})=3$, Paramagnetic
UPTU/UPSEE-2017
COORDINATION COMPOUNDS
274059
What will be the spin only magnetic moment of high spin and low spin $d^{5}$ electronic system in an octahedral complex?
(B) : High spin $\mathrm{d}^{5}$ octahedral complex No. of unpaired electron $=5$ Spin only magnetic moment $=\sqrt{5(5+2)}=5.92$ B.M. Low spin $\mathrm{d}^{5}$ octahedral complex No. of unpaired electron $=1$ Spin only Magnetic moment $=\sqrt{1(1+2)}=1.73$ B.M.
(A) : Diamagnatic substance having no unpaired electrons. (a) $\left[\mathrm{Co}(\mathrm{ox})_{3}\right]^{3-}, \Rightarrow \mathrm{Co}^{3+}=\mathrm{d}^{6}$ In $\left[\mathrm{Co}(\mathrm{ox})_{3}\right]^{3-}, \mathrm{Co}^{3+}$ act as strong field ligand with oxalate. $\mathrm{Co}^{3+}$ having no unpaired electron and act as diamagnetic. (b) $\quad\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}, \Rightarrow \mathrm{Fe}^{2+}=\mathrm{d}^{6}$ $\mathrm{Fe}^{2+}$ with $\mathrm{CN}^{-}$act as strong field ligand so no unpaired electron is present. Hence it is also act as diamagnetic complex.
AIIMS-26 May
COORDINATION COMPOUNDS
274056
Which of the following is not expected to show paramagnetism?
(B) : (a) $\mathrm{H}_{2} \mathrm{O}$ act as weak field ligand. So pairing is not possible. This complex having unpaired electron it behave like paramagnetic. (b) In $\left[\mathrm{Ni}(\mathrm{Co})_{4}\right]$, Co act as strong field ligand $\Delta_{\mathrm{o}}>\mathrm{p}$. $\mathrm{Ni}: 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$ $3 \mathrm{~d}^{10}$ No. unpaired electron so $\left[\mathrm{Ni}(\mathrm{Co})_{4}\right]$ act as diamagnetic. (c) $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}, \mathrm{NH}_{3}$ act as weak field ligand so pairing of electron is not possible. $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$ have 2 unpaired electron so behave as paramagnetic. (d) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}, \mathrm{NH}_{3}$ act as weak field ligand $\Delta_{\mathrm{o}}<\mathrm{p}$ $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ has 3 unpaired electron so it behave as paramagnetic.
JIPMER-2017
COORDINATION COMPOUNDS
274058
The paramagnetic species is
1 $\mathrm{SiO}_{2}$
2 $\mathrm{TiO}_{2}$
3 $\mathrm{BaO}_{2}$
4 $\mathrm{KO}_{2}$
Explanation:
(D) : The species which has unpaired electron, is paramagnetic in nature. $\mathrm{KO}_{2}$ : Let, $\mathrm{x}$ be the oxidation state of $\mathrm{K}$. $\mathrm{x}+2(-2)=0$ $\mathrm{x}=+4$ $\mathrm{~K}^{4+}=1 \mathrm{~s}^{2}, 2 \mathrm{~s}^{2}, 2 \mathrm{p}^{6}, 3 \mathrm{~s}^{2}, 3 \mathrm{p}^{3}$ Unpaired electron $(\mathrm{n})=3$, Paramagnetic
UPTU/UPSEE-2017
COORDINATION COMPOUNDS
274059
What will be the spin only magnetic moment of high spin and low spin $d^{5}$ electronic system in an octahedral complex?
(B) : High spin $\mathrm{d}^{5}$ octahedral complex No. of unpaired electron $=5$ Spin only magnetic moment $=\sqrt{5(5+2)}=5.92$ B.M. Low spin $\mathrm{d}^{5}$ octahedral complex No. of unpaired electron $=1$ Spin only Magnetic moment $=\sqrt{1(1+2)}=1.73$ B.M.
(A) : Diamagnatic substance having no unpaired electrons. (a) $\left[\mathrm{Co}(\mathrm{ox})_{3}\right]^{3-}, \Rightarrow \mathrm{Co}^{3+}=\mathrm{d}^{6}$ In $\left[\mathrm{Co}(\mathrm{ox})_{3}\right]^{3-}, \mathrm{Co}^{3+}$ act as strong field ligand with oxalate. $\mathrm{Co}^{3+}$ having no unpaired electron and act as diamagnetic. (b) $\quad\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}, \Rightarrow \mathrm{Fe}^{2+}=\mathrm{d}^{6}$ $\mathrm{Fe}^{2+}$ with $\mathrm{CN}^{-}$act as strong field ligand so no unpaired electron is present. Hence it is also act as diamagnetic complex.
AIIMS-26 May
COORDINATION COMPOUNDS
274056
Which of the following is not expected to show paramagnetism?
(B) : (a) $\mathrm{H}_{2} \mathrm{O}$ act as weak field ligand. So pairing is not possible. This complex having unpaired electron it behave like paramagnetic. (b) In $\left[\mathrm{Ni}(\mathrm{Co})_{4}\right]$, Co act as strong field ligand $\Delta_{\mathrm{o}}>\mathrm{p}$. $\mathrm{Ni}: 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$ $3 \mathrm{~d}^{10}$ No. unpaired electron so $\left[\mathrm{Ni}(\mathrm{Co})_{4}\right]$ act as diamagnetic. (c) $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}, \mathrm{NH}_{3}$ act as weak field ligand so pairing of electron is not possible. $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$ have 2 unpaired electron so behave as paramagnetic. (d) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}, \mathrm{NH}_{3}$ act as weak field ligand $\Delta_{\mathrm{o}}<\mathrm{p}$ $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ has 3 unpaired electron so it behave as paramagnetic.
JIPMER-2017
COORDINATION COMPOUNDS
274058
The paramagnetic species is
1 $\mathrm{SiO}_{2}$
2 $\mathrm{TiO}_{2}$
3 $\mathrm{BaO}_{2}$
4 $\mathrm{KO}_{2}$
Explanation:
(D) : The species which has unpaired electron, is paramagnetic in nature. $\mathrm{KO}_{2}$ : Let, $\mathrm{x}$ be the oxidation state of $\mathrm{K}$. $\mathrm{x}+2(-2)=0$ $\mathrm{x}=+4$ $\mathrm{~K}^{4+}=1 \mathrm{~s}^{2}, 2 \mathrm{~s}^{2}, 2 \mathrm{p}^{6}, 3 \mathrm{~s}^{2}, 3 \mathrm{p}^{3}$ Unpaired electron $(\mathrm{n})=3$, Paramagnetic
UPTU/UPSEE-2017
COORDINATION COMPOUNDS
274059
What will be the spin only magnetic moment of high spin and low spin $d^{5}$ electronic system in an octahedral complex?
(B) : High spin $\mathrm{d}^{5}$ octahedral complex No. of unpaired electron $=5$ Spin only magnetic moment $=\sqrt{5(5+2)}=5.92$ B.M. Low spin $\mathrm{d}^{5}$ octahedral complex No. of unpaired electron $=1$ Spin only Magnetic moment $=\sqrt{1(1+2)}=1.73$ B.M.
